3.245 \(\int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=574 \[ \frac {a^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}-\frac {a^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}-\frac {f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {a f \log (a+b \sin (c+d x))}{b d^2 \left (a^2-b^2\right )}+\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{3/2}}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d \left (a^2-b^2\right )^{3/2}}-\frac {i (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}+\frac {i (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d \sqrt {a^2-b^2}}-\frac {a (e+f x) \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.62, antiderivative size = 574, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6742, 3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ \frac {a^2 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}-\frac {a^2 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}-\frac {f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {a f \log (a+b \sin (c+d x))}{b d^2 \left (a^2-b^2\right )}+\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{3/2}}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d \left (a^2-b^2\right )^{3/2}}-\frac {i (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}+\frac {i (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d \sqrt {a^2-b^2}}-\frac {a (e+f x) \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

(I*a^2*(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) - (I*(e + f*x)*
Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d) - (I*a^2*(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) + (I*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a
+ Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d) + (a*f*Log[a + b*Sin[c + d*x]])/(b*(a^2 - b^2)*d^2) + (a^2*f*PolyL
og[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^2) - (f*PolyLog[2, (I*b*E^(I*(c + d
*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) - (a^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2
- b^2])])/(b*(a^2 - b^2)^(3/2)*d^2) + (f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2
- b^2]*d^2) - (a*(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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2264

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 2279

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 2391

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 2668

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 3323

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 3324

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 6742

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

________________________________________________________________________________________

Mathematica [B]  time = 15.65, size = 2141, normalized size = 3.73 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

(-(a*d*e*Cos[c + d*x]) + a*c*f*Cos[c + d*x] - a*f*(c + d*x)*Cos[c + d*x])/((a - b)*(a + b)*d^2*(a + b*Sin[c +
d*x])) + (((2*a*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (2*(-(b*d*e) + a*f + b*c
*f)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (a*f*Log[Sec[(c + d*x)/2]^2])/b - (a*f
*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])])/b - (I*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] + (I*b*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*T
an[(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] + (I*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] - (I*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]))]))/Sqrt[-a^2 +
b^2])*(-((b*e)/((a^2 - b^2)*(a + b*Sin[c + d*x]))) + (b*c*f)/((a^2 - b^2)*d*(a + b*Sin[c + d*x])) - (b*f*(c +
d*x))/((a^2 - b^2)*d*(a + b*Sin[c + d*x])) + (a*f*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x]))))/(d*((a
*f*Tan[(c + d*x)/2])/b - (a*f*Cos[(c + d*x)/2]^2*(b*Cos[c + d*x]*Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/2]^2*(a +
b*Sin[c + d*x])*Tan[(c + d*x)/2]))/(b*(a + b*Sin[c + d*x])) + (a^2*f*Sec[(c + d*x)/2]^2)/((a^2 - b^2)*(1 + (b
+ a*Tan[(c + d*x)/2])^2/(a^2 - b^2))) - (a*(-(b*d*e) + a*f + b*c*f)*Sec[(c + d*x)/2]^2)/((a^2 - b^2)*(1 + (b +
a*Tan[(c + d*x)/2])^2/(a^2 - b^2))) + (I*b*f*(((-1/2*I)*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*
a - b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) - (Log[1 - (a*(I + Tan[(c + d*x)/2]))
/(I*a - b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(2*(I + Tan[(c + d*x)/2])) + (a*Log[1 - I*Tan[(c + d*x)/2]]
*Sec[(c + d*x)/2]^2)/(2*(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]))))/Sqrt[-a^2 + b^2] - (I*b*f*(((I/2)*Log[(
b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d
*x)/2]) - ((I/2)*a*Log[1 - (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(a
+ I*a*Tan[(c + d*x)/2]) + (a*Log[1 + I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(2*(b - Sqrt[-a^2 + b^2] + a*Tan[
(c + d*x)/2]))))/Sqrt[-a^2 + b^2] - (I*b*f*(((I/2)*Log[1 - (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2
+ b^2]))]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) - ((I/2)*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]
)/((-I)*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) + (a*Log[1 - I*Tan[(c + d*x)/2
]]*Sec[(c + d*x)/2]^2)/(2*(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]))))/Sqrt[-a^2 + b^2] + (I*b*f*(((-1/2*I)*
Log[1 - (a*(1 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d*x)/
2]) + ((I/2)*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)
/(1 + I*Tan[(c + d*x)/2]) + (a*Log[1 + I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(2*(b + Sqrt[-a^2 + b^2] + a*Ta
n[(c + d*x)/2]))))/Sqrt[-a^2 + b^2]))

________________________________________________________________________________________

fricas [B]  time = 0.70, size = 1514, normalized size = 2.64 \[ \text {result too large to display} \]

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))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

integrate((f*x + e)*sin(d*x + c)/(b*sin(d*x + c) + a)^2, x)

________________________________________________________________________________________

maple [A]  time = 1.45, size = 750, normalized size = 1.31 \[ \frac {2 i a \left (f x +e \right ) \left (b -i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{b \left (-a^{2}+b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {a 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 )}{b \,d^{2} \left (a^{2}-b^{2}\right )}-\frac {2 a f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{b \,d^{2} \left (a^{2}-b^{2}\right )}+\frac {2 i b c f \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 ) \sqrt {-a^{2}+b^{2}}}-\frac {2 i b 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 ) \sqrt {-a^{2}+b^{2}}}+\frac {i b 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 )}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}-\frac {i b 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 )}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {b 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}{d \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {b 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}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}-\frac {b 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}{d \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}-\frac {b 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}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}} \]

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))^2,x)

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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))^2,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 details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

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))^2,x)

[Out]

\text{Hanged}

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**2,x)

[Out]

Timed out

________________________________________________________________________________________