3.3.48 \(\int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [248]

Optimal. Leaf size=751 \[ \frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}+\frac {3 a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}-\frac {3 a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}+\frac {3 a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]

[Out]

3/2*a^2*f*ln(a+b*sin(d*x+c))/b/(a^2-b^2)^2/d^2-f*ln(a+b*sin(d*x+c))/b/(a^2-b^2)/d^2+3/2*I*a^3*(f*x+e)*ln(1-I*b
*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d-3/2*I*a*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^
(1/2)))/b/(a^2-b^2)^(3/2)/d-3/2*I*a^3*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d
+3/2*I*a*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d+3/2*a^3*f*polylog(2,I*b*exp(
I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2-3/2*a*f*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2))
)/b/(a^2-b^2)^(3/2)/d^2-3/2*a^3*f*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2+3/2*
a*f*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-1/2*a*(f*x+e)*cos(d*x+c)/(a^2-b^2)
/d/(a+b*sin(d*x+c))^2-1/2*a*f/b/(a^2-b^2)/d^2/(a+b*sin(d*x+c))-3/2*a^2*(f*x+e)*cos(d*x+c)/(a^2-b^2)^2/d/(a+b*s
in(d*x+c))+(f*x+e)*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 1.97, antiderivative size = 751, normalized size of antiderivative = 1.00, number of steps used = 48, number of rules used = 11, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {6874, 3406, 3405, 3404, 2296, 2221, 2317, 2438, 2747, 31, 32} \begin {gather*} -\frac {3 a f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {3 a f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {3 a^3 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{5/2}}-\frac {3 a^3 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d^2 \left (a^2-b^2\right )^{5/2}}-\frac {a f}{2 b d^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b d^2 \left (a^2-b^2\right )^2}-\frac {f \log (a+b \sin (c+d x))}{b d^2 \left (a^2-b^2\right )}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d \left (a^2-b^2\right )^{3/2}}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d \left (a^2-b^2\right )^{3/2}}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {a (e+f x) \cos (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}+\frac {(e+f x) \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d \left (a^2-b^2\right )^{5/2}}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d \left (a^2-b^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

(((3*I)/2)*a^3*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d) - (((3*
I)/2)*a*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) - (((3*I)/2)*a
^3*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d) + (((3*I)/2)*a*(e +
f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) + (3*a^2*f*Log[a + b*Sin[c
+ d*x]])/(2*b*(a^2 - b^2)^2*d^2) - (f*Log[a + b*Sin[c + d*x]])/(b*(a^2 - b^2)*d^2) + (3*a^3*f*PolyLog[2, (I*b
*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(5/2)*d^2) - (3*a*f*PolyLog[2, (I*b*E^(I*(c + d*x))
)/(a - Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(3/2)*d^2) - (3*a^3*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^
2 - b^2])])/(2*b*(a^2 - b^2)^(5/2)*d^2) + (3*a*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(2*b
*(a^2 - b^2)^(3/2)*d^2) - (a*(e + f*x)*Cos[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - (a*f)/(2*b*(a^
2 - b^2)*d^2*(a + b*Sin[c + d*x])) - (3*a^2*(e + f*x)*Cos[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])) +
((e + f*x)*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2296

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3404

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[(c + d*x)^m*(E
^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3405

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(c + d*x)^m*(Cos[
e + f*x]/(f*(a^2 - b^2)*(a + b*Sin[e + f*x]))), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[b*d*(m/(f*(a^2 - b^2))), Int[(c + d*x)^(m - 1)*(Cos[e + f*x]/(a + b*Sin[e + f*x])), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3406

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*(c + d*x)^m
*Cos[e + f*x]*((a + b*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(a^2 - b^2))), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^
m*(a + b*Sin[e + f*x])^(n + 1), x], x] - Dist[b*((n + 2)/((n + 1)*(a^2 - b^2))), Int[(c + d*x)^m*Sin[e + f*x]*
(a + b*Sin[e + f*x])^(n + 1), x], x] + Dist[b*d*(m/(f*(n + 1)*(a^2 - b^2))), Int[(c + d*x)^(m - 1)*Cos[e + f*x
]*(a + b*Sin[e + f*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && ILtQ[n, -2] &&
IGtQ[m, 0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (-\frac {a (e+f x)}{b (a+b \sin (c+d x))^3}+\frac {e+f x}{b (a+b \sin (c+d x))^2}\right ) \, dx\\ &=\frac {\int \frac {e+f x}{(a+b \sin (c+d x))^2} \, dx}{b}-\frac {a \int \frac {e+f x}{(a+b \sin (c+d x))^3} \, dx}{b}\\ &=-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {a \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}+\frac {a \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}-\frac {a^2 \int \frac {e+f x}{(a+b \sin (c+d x))^2} \, dx}{b \left (a^2-b^2\right )}-\frac {f \int \frac {\cos (c+d x)}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right ) d}+\frac {(a f) \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right ) d}\\ &=-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a^2 (e+f x) \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {a^3 \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )^2}+\frac {a \int \left (-\frac {a (e+f x)}{b (a+b \sin (c+d x))^2}+\frac {e+f x}{b (a+b \sin (c+d x))}\right ) \, dx}{2 \left (a^2-b^2\right )}+\frac {(2 a) \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b \left (a^2-b^2\right )}-\frac {f \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (c+d x)\right )}{b \left (a^2-b^2\right ) d^2}+\frac {(a f) \text {Subst}\left (\int \frac {1}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{2 b \left (a^2-b^2\right ) d^2}+\frac {\left (a^2 f\right ) \int \frac {\cos (c+d x)}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {a^2 (e+f x) \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (2 a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b \left (a^2-b^2\right )^2}-\frac {(2 i a) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {(2 i a) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {a \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}-\frac {a^2 \int \frac {e+f x}{(a+b \sin (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}+\frac {\left (a^2 f\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (c+d x)\right )}{b \left (a^2-b^2\right )^2 d^2}\\ &=-\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {a^2 f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\left (2 i a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{5/2}}-\frac {\left (2 i a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{5/2}}-\frac {a^3 \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{2 b \left (a^2-b^2\right )^2}+\frac {a \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b \left (a^2-b^2\right )}+\frac {\left (a^2 f\right ) \int \frac {\cos (c+d x)}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2 d}+\frac {(i a f) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right )^{3/2} d}-\frac {(i a f) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right )^{3/2} d}\\ &=\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d}-\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d}+\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {a^2 f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {a^3 \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b \left (a^2-b^2\right )^2}-\frac {(i a) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {(i a) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {\left (a^2 f\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (c+d x)\right )}{2 b \left (a^2-b^2\right )^2 d^2}+\frac {(a f) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {(a f) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {\left (i a^3 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right )^{5/2} d}+\frac {\left (i a^3 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right )^{5/2} d}\\ &=\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}-\frac {a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\left (i a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{5/2}}-\frac {\left (i a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{5/2}}-\frac {\left (a^3 f\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \left (a^2-b^2\right )^{5/2} d^2}+\frac {\left (a^3 f\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \left (a^2-b^2\right )^{5/2} d^2}+\frac {(i a f) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {(i a f) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{2 b \left (a^2-b^2\right )^{3/2} d}\\ &=\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}+\frac {a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d^2}-\frac {a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d^2}+\frac {a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {(a f) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {(a f) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {\left (i a^3 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {\left (i a^3 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{2 b \left (a^2-b^2\right )^{5/2} d}\\ &=\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}+\frac {a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d^2}-\frac {3 a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d^2}+\frac {3 a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (a^3 f\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}+\frac {\left (a^3 f\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}\\ &=\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}+\frac {3 a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}-\frac {3 a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}+\frac {3 a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 11.33, size = 959, normalized size = 1.28 \begin {gather*} \frac {\frac {\left (-3 a b d (e+f x)+\left (a^2+2 b^2\right ) f \cos (c+d x)\right ) \left (-\frac {2 \left (a^2+2 b^2\right ) f \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {2 \left (a^2 f+2 b^2 f+a b (-3 d e+3 c f)\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {\left (a^2+2 b^2\right ) f \log \left (\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )}{b}+\frac {\left (a^2+2 b^2\right ) f \log \left (\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \sin (c+d x))\right )}{b}+\frac {3 i a b f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{-i a+b+\sqrt {-a^2+b^2}}\right )+\text {Li}_2\left (\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {-a^2+b^2}\right )}\right )\right )}{\sqrt {-a^2+b^2}}-\frac {3 i a b f \left (\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b+\sqrt {-a^2+b^2}}\right )+\text {Li}_2\left (\frac {a \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a-i \left (b+\sqrt {-a^2+b^2}\right )}\right )\right )}{\sqrt {-a^2+b^2}}-\frac {3 i a b f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {-b+\sqrt {-a^2+b^2}-a \tan \left (\frac {1}{2} (c+d x)\right )}{i a-b+\sqrt {-a^2+b^2}}\right )+\text {Li}_2\left (\frac {a \left (i+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{i a-b+\sqrt {-a^2+b^2}}\right )\right )}{\sqrt {-a^2+b^2}}+\frac {3 i a b f \left (\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b-\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b-\sqrt {-a^2+b^2}}\right )+\text {Li}_2\left (\frac {a+i a \tan \left (\frac {1}{2} (c+d x)\right )}{a+i \left (-b+\sqrt {-a^2+b^2}\right )}\right )\right )}{\sqrt {-a^2+b^2}}\right )}{\left (a^2-b^2\right )^2 \left (\left (a^2+2 b^2\right ) f \cos (c+d x)+3 a b \left (-d e+c f-i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )+i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}-\frac {a d (e+f x) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))^2}+\frac {a \left (-a^2+b^2\right ) f-b \left (a^2+2 b^2\right ) d (e+f x) \cos (c+d x)}{(a-b)^2 b (a+b)^2 (a+b \sin (c+d x))}}{2 d^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

(((-3*a*b*d*(e + f*x) + (a^2 + 2*b^2)*f*Cos[c + d*x])*((-2*(a^2 + 2*b^2)*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqr
t[a^2 - b^2]])/Sqrt[a^2 - b^2] + (2*(a^2*f + 2*b^2*f + a*b*(-3*d*e + 3*c*f))*ArcTan[(b + a*Tan[(c + d*x)/2])/S
qrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - ((a^2 + 2*b^2)*f*Log[Sec[(c + d*x)/2]^2])/b + ((a^2 + 2*b^2)*f*Log[Sec[(c +
d*x)/2]^2*(a + b*Sin[c + d*x])])/b + ((3*I)*a*b*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*
Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])] + PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[
-a^2 + b^2]))]))/Sqrt[-a^2 + b^2] - ((3*I)*a*b*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*Ta
n[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])] + PolyLog[2, (a*(1 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqrt[-a^2
+ b^2]))]))/Sqrt[-a^2 + b^2] - ((3*I)*a*b*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(-b + Sqrt[-a^2 + b^2] - a*Tan[(c
+ d*x)/2])/(I*a - b + Sqrt[-a^2 + b^2])] + PolyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])
]))/Sqrt[-a^2 + b^2] + ((3*I)*a*b*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2
])/(I*a + b - Sqrt[-a^2 + b^2])] + PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]))/Sq
rt[-a^2 + b^2]))/((a^2 - b^2)^2*((a^2 + 2*b^2)*f*Cos[c + d*x] + 3*a*b*(-(d*e) + c*f - I*f*Log[1 - I*Tan[(c + d
*x)/2]] + I*f*Log[1 + I*Tan[(c + d*x)/2]]))) - (a*d*(e + f*x)*Cos[c + d*x])/((a - b)*(a + b)*(a + b*Sin[c + d*
x])^2) + (a*(-a^2 + b^2)*f - b*(a^2 + 2*b^2)*d*(e + f*x)*Cos[c + d*x])/((a - b)^2*b*(a + b)^2*(a + b*Sin[c + d
*x])))/(2*d^2)

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Maple [A]
time = 2.34, size = 1084, normalized size = 1.44

method	result	size
risch	\(\frac {i \left (4 i b \,a^{3} d f x \,{\mathrm e}^{i \left (d x +c \right )}-3 i b^{3} a d e \,{\mathrm e}^{3 i \left (d x +c \right )}+5 i b^{3} a d e \,{\mathrm e}^{i \left (d x +c \right )}-3 i b^{3} a d f x \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a^{4} d f x \,{\mathrm e}^{2 i \left (d x +c \right )}+5 b^{2} a^{2} d f x \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b^{4} d f x \,{\mathrm e}^{2 i \left (d x +c \right )}+5 i b^{3} a d f x \,{\mathrm e}^{i \left (d x +c \right )}+2 i a^{4} f \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i b^{2} f \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}+4 i b \,a^{3} d e \,{\mathrm e}^{i \left (d x +c \right )}+2 a^{4} d e \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,a^{3} f \,{\mathrm e}^{3 i \left (d x +c \right )}+5 b^{2} a^{2} d e \,{\mathrm e}^{2 i \left (d x +c \right )}-b^{3} a f \,{\mathrm e}^{3 i \left (d x +c \right )}+2 b^{4} d e \,{\mathrm e}^{2 i \left (d x +c \right )}-a^{2} b^{2} d f x -2 b^{4} d f x -b \,a^{3} f \,{\mathrm e}^{i \left (d x +c \right )}-a^{2} b^{2} d e +b^{3} a f \,{\mathrm e}^{i \left (d x +c \right )}-2 b^{4} d e \right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} \left (a^{2}-b^{2}\right )^{2} d^{2} b}-\frac {a^{2} f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{b \,d^{2} \left (-a^{2}+b^{2}\right )^{2}}+\frac {a^{2} f \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{2 b \,d^{2} \left (-a^{2}+b^{2}\right )^{2}}-\frac {2 b f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left (-a^{2}+b^{2}\right )^{2}}+\frac {b f \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{d^{2} \left (-a^{2}+b^{2}\right )^{2}}-\frac {3 b a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{2 d \left (-a^{2}+b^{2}\right )^{\frac {5}{2}}}-\frac {3 b a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{2 d^{2} \left (-a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {3 i b a f c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} \left (-a^{2}+b^{2}\right )^{\frac {5}{2}}}-\frac {3 i b a e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d \left (-a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {3 b a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{2 d \left (-a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {3 b a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{2 d^{2} \left (-a^{2}+b^{2}\right )^{\frac {5}{2}}}-\frac {3 i b a f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{2 d^{2} \left (-a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {3 i b a f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{2 d^{2} \left (-a^{2}+b^{2}\right )^{\frac {5}{2}}}\)	\(1084\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

I*(4*I*b*a^3*d*f*x*exp(I*(d*x+c))-3*I*b^3*a*d*e*exp(3*I*(d*x+c))+5*I*b^3*a*d*e*exp(I*(d*x+c))-3*I*b^3*a*d*f*x*
exp(3*I*(d*x+c))+2*a^4*d*f*x*exp(2*I*(d*x+c))+5*b^2*a^2*d*f*x*exp(2*I*(d*x+c))+2*b^4*d*f*x*exp(2*I*(d*x+c))+5*
I*b^3*a*d*f*x*exp(I*(d*x+c))+2*I*a^4*f*exp(2*I*(d*x+c))-2*I*b^2*f*exp(2*I*(d*x+c))*a^2+4*I*b*a^3*d*e*exp(I*(d*
x+c))+2*a^4*d*e*exp(2*I*(d*x+c))+b*a^3*f*exp(3*I*(d*x+c))+5*b^2*a^2*d*e*exp(2*I*(d*x+c))-b^3*a*f*exp(3*I*(d*x+
c))+2*b^4*d*e*exp(2*I*(d*x+c))-a^2*b^2*d*f*x-2*b^4*d*f*x-b*a^3*f*exp(I*(d*x+c))-a^2*b^2*d*e+b^3*a*f*exp(I*(d*x
+c))-2*b^4*d*e)/(-I*b*exp(2*I*(d*x+c))+I*b+2*a*exp(I*(d*x+c)))^2/(a^2-b^2)^2/d^2/b-1/b/d^2/(-a^2+b^2)^2*a^2*f*
ln(exp(I*(d*x+c)))+1/2/b/d^2/(-a^2+b^2)^2*a^2*f*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-2*b/d^2/(-a^2+
b^2)^2*f*ln(exp(I*(d*x+c)))+b/d^2/(-a^2+b^2)^2*f*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-3/2*b/d/(-a^2
+b^2)^(5/2)*a*f*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x-3/2*b/d^2/(-a^2+b^2)^(5/2
)*a*f*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c+3*I*b/d^2/(-a^2+b^2)^(5/2)*a*f*c*ar
ctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))-3*I*b/d/(-a^2+b^2)^(5/2)*a*e*arctan(1/2*(2*I*b*exp(I*(d*
x+c))-2*a)/(-a^2+b^2)^(1/2))+3/2*b/d/(-a^2+b^2)^(5/2)*a*f*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^
2+b^2)^(1/2)))*x+3/2*b/d^2/(-a^2+b^2)^(5/2)*a*f*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/
2)))*c-3/2*I*b/d^2/(-a^2+b^2)^(5/2)*a*f*dilog((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))+
3/2*I*b/d^2/(-a^2+b^2)^(5/2)*a*f*dilog((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2443 vs. \(2 (658) = 1316\).
time = 0.67, size = 2443, normalized size = 3.25 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(3*(I*a*b^5*f*cos(d*x + c)^2 - 2*I*a^2*b^4*f*sin(d*x + c) - I*(a^3*b^3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)
*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b
+ 1) + 3*(-I*a*b^5*f*cos(d*x + c)^2 + 2*I*a^2*b^4*f*sin(d*x + c) + I*(a^3*b^3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^
2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/
b + 1) + 3*(-I*a*b^5*f*cos(d*x + c)^2 + 2*I*a^2*b^4*f*sin(d*x + c) + I*(a^3*b^3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/
b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) -
b)/b + 1) + 3*(I*a*b^5*f*cos(d*x + c)^2 - 2*I*a^2*b^4*f*sin(d*x + c) - I*(a^3*b^3 + a*b^5)*f)*sqrt(-(a^2 - b^2
)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)
- b)/b + 1) + 3*((a^3*b^3 + a*b^5)*d*f*x + (a^3*b^3 + a*b^5)*c*f - (a*b^5*d*f*x + a*b^5*c*f)*cos(d*x + c)^2 +
2*(a^2*b^4*d*f*x + a^2*b^4*c*f)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) +
(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - 3*((a^3*b^3 + a*b^5)*d*f*x + (a^3*b^3 +
a*b^5)*c*f - (a*b^5*d*f*x + a*b^5*c*f)*cos(d*x + c)^2 + 2*(a^2*b^4*d*f*x + a^2*b^4*c*f)*sin(d*x + c))*sqrt(-(a
^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)
/b^2) - b)/b) + 3*((a^3*b^3 + a*b^5)*d*f*x + (a^3*b^3 + a*b^5)*c*f - (a*b^5*d*f*x + a*b^5*c*f)*cos(d*x + c)^2
+ 2*(a^2*b^4*d*f*x + a^2*b^4*c*f)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c
) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - 3*((a^3*b^3 + a*b^5)*d*f*x + (a^3*b^3
+ a*b^5)*c*f - (a*b^5*d*f*x + a*b^5*c*f)*cos(d*x + c)^2 + 2*(a^2*b^4*d*f*x + a^2*b^4*c*f)*sin(d*x + c))*sqrt(
-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 -
b^2)/b^2) - b)/b) - 2*(a^6 - 2*a^4*b^2 + a^2*b^4)*f - 2*((2*a^5*b - a^3*b^3 - a*b^5)*d*f*x + (2*a^5*b - a^3*b^
3 - a*b^5)*d*e)*cos(d*x + c) - ((a^4*b^2 + a^2*b^4 - 2*b^6)*f*cos(d*x + c)^2 - 2*(a^5*b + a^3*b^3 - 2*a*b^5)*f
*sin(d*x + c) - (a^6 + 2*a^4*b^2 - a^2*b^4 - 2*b^6)*f - 3*((a^3*b^3 + a*b^5)*c*f - (a*b^5*c*f - a*b^5*d*e)*cos
(d*x + c)^2 - (a^3*b^3 + a*b^5)*d*e + 2*(a^2*b^4*c*f - a^2*b^4*d*e)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))*log(
2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - ((a^4*b^2 + a^2*b^4 - 2*b^6)*f*c
os(d*x + c)^2 - 2*(a^5*b + a^3*b^3 - 2*a*b^5)*f*sin(d*x + c) - (a^6 + 2*a^4*b^2 - a^2*b^4 - 2*b^6)*f - 3*((a^3
*b^3 + a*b^5)*c*f - (a*b^5*c*f - a*b^5*d*e)*cos(d*x + c)^2 - (a^3*b^3 + a*b^5)*d*e + 2*(a^2*b^4*c*f - a^2*b^4*
d*e)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b
^2) - 2*I*a) - ((a^4*b^2 + a^2*b^4 - 2*b^6)*f*cos(d*x + c)^2 - 2*(a^5*b + a^3*b^3 - 2*a*b^5)*f*sin(d*x + c) -
(a^6 + 2*a^4*b^2 - a^2*b^4 - 2*b^6)*f + 3*((a^3*b^3 + a*b^5)*c*f - (a*b^5*c*f - a*b^5*d*e)*cos(d*x + c)^2 - (a
^3*b^3 + a*b^5)*d*e + 2*(a^2*b^4*c*f - a^2*b^4*d*e)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*cos(d*x + c
) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - ((a^4*b^2 + a^2*b^4 - 2*b^6)*f*cos(d*x + c)^2 -
2*(a^5*b + a^3*b^3 - 2*a*b^5)*f*sin(d*x + c) - (a^6 + 2*a^4*b^2 - a^2*b^4 - 2*b^6)*f + 3*((a^3*b^3 + a*b^5)*c
*f - (a*b^5*c*f - a*b^5*d*e)*cos(d*x + c)^2 - (a^3*b^3 + a*b^5)*d*e + 2*(a^2*b^4*c*f - a^2*b^4*d*e)*sin(d*x +
c))*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) -
2*((a^5*b - 2*a^3*b^3 + a*b^5)*f + ((a^4*b^2 + a^2*b^4 - 2*b^6)*d*f*x + (a^4*b^2 + a^2*b^4 - 2*b^6)*d*e)*cos(
d*x + c))*sin(d*x + c))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d^2*cos(d*x + c)^2 - 2*(a^7*b^2 - 3*a^5*b^4 +
3*a^3*b^6 - a*b^8)*d^2*sin(d*x + c) - (a^8*b - 2*a^6*b^3 + 2*a^2*b^7 - b^9)*d^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((f*x + e)*sin(d*x + c)/(b*sin(d*x + c) + a)^3, x)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(e + f*x))/(a + b*sin(c + d*x))^3,x)

[Out]

\text{Hanged}

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