3.4.21 \(\int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [321]

Optimal. Leaf size=418 \[ -\frac {3 i f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}+\frac {3 i f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {6 f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}+\frac {6 f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}-\frac {6 i f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}+\frac {6 i f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))} \]

[Out]

-(f*x+e)^3/b/d/(a+b*sin(d*x+c))-3*I*f*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/d^2/(a^2-b^2)^(
1/2)+3*I*f*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/d^2/(a^2-b^2)^(1/2)-6*f^2*(f*x+e)*polylog(
2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/d^3/(a^2-b^2)^(1/2)+6*f^2*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+
(a^2-b^2)^(1/2)))/b/d^3/(a^2-b^2)^(1/2)-6*I*f^3*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/d^4/(a^2-b
^2)^(1/2)+6*I*f^3*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/d^4/(a^2-b^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.57, antiderivative size = 418, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {4507, 3404, 2296, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {6 i f^3 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^4 \sqrt {a^2-b^2}}+\frac {6 i f^3 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^4 \sqrt {a^2-b^2}}-\frac {6 f^2 (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3 \sqrt {a^2-b^2}}+\frac {6 f^2 (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^3 \sqrt {a^2-b^2}}-\frac {3 i f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {3 i f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2 \sqrt {a^2-b^2}}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Cos[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

((-3*I)*f*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) + ((3*I)*f
*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) - (6*f^2*(e + f*x)*
PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) + (6*f^2*(e + f*x)*PolyLog[2,
(I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) - ((6*I)*f^3*PolyLog[3, (I*b*E^(I*(c +
d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^4) + ((6*I)*f^3*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt
[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^4) - (e + f*x)^3/(b*d*(a + b*Sin[c + d*x]))

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 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 4507

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

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {(3 f) \int \frac {(e+f x)^2}{a+b \sin (c+d x)} \, dx}{b d}\\ &=-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {(6 f) \int \frac {e^{i (c+d x)} (e+f x)^2}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b d}\\ &=-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}-\frac {(6 i f) \int \frac {e^{i (c+d x)} (e+f x)^2}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\sqrt {a^2-b^2} d}+\frac {(6 i f) \int \frac {e^{i (c+d x)} (e+f x)^2}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\sqrt {a^2-b^2} d}\\ &=-\frac {3 i f (e+f x)^2 \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^2}+\frac {3 i f (e+f x)^2 \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^2}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {\left (6 i f^2\right ) \int (e+f x) \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^2}-\frac {\left (6 i f^2\right ) \int (e+f x) \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^2}\\ &=-\frac {3 i f (e+f x)^2 \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^2}+\frac {3 i f (e+f x)^2 \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^2}-\frac {6 f^2 (e+f x) \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^3}+\frac {6 f^2 (e+f x) \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^3}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {\left (6 f^3\right ) \int \text {Li}_2\left (\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^3}-\frac {\left (6 f^3\right ) \int \text {Li}_2\left (\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^3}\\ &=-\frac {3 i f (e+f x)^2 \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^2}+\frac {3 i f (e+f x)^2 \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^2}-\frac {6 f^2 (e+f x) \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^3}+\frac {6 f^2 (e+f x) \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^3}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}-\frac {\left (6 i f^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \sqrt {a^2-b^2} d^4}+\frac {\left (6 i f^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \sqrt {a^2-b^2} d^4}\\ &=-\frac {3 i f (e+f x)^2 \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^2}+\frac {3 i f (e+f x)^2 \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^2}-\frac {6 f^2 (e+f x) \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^3}+\frac {6 f^2 (e+f x) \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^3}-\frac {6 i f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}+\frac {6 i f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.05, size = 703, normalized size = 1.68 \begin {gather*} \frac {3 i f \left (2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right ) (\cos (c)+i \sin (c))-2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right ) (\cos (c)+i \sin (c))-i \left (-2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right ) (\cos (c)+i \sin (c))+2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right ) (\cos (c)+i \sin (c))+d^2 \left (\sqrt {a^2-b^2} f x (2 e+f x) \left (-\log \left (1+\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right )+\log \left (1-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right )\right ) (\cos (c)+i \sin (c))+2 e^2 \tan ^{-1}\left (\frac {b \cos (c+d x)+i (a+b \sin (c+d x))}{\sqrt {a^2-b^2}}\right ) \sqrt {\left (-a^2+b^2\right ) (\cos (2 c)+i \sin (2 c))}\right )\right )\right )}{b \sqrt {a^2-b^2} d^4 \sqrt {\left (-a^2+b^2\right ) (\cos (2 c)+i \sin (2 c))}}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Cos[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

((3*I)*f*(2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, -((b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + S
qrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) - 2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*Po
lyLog[2, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a
*Sin[c])]*(Cos[c] + I*Sin[c]) - I*(-2*Sqrt[a^2 - b^2]*f^2*PolyLog[3, -((b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))
/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) + 2*Sqrt[a^2 - b^2]*
f^2*PolyLog[3, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^
2] + a*Sin[c])]*(Cos[c] + I*Sin[c]) + d^2*(Sqrt[a^2 - b^2]*f*x*(2*e + f*x)*(-Log[1 + (b*(Cos[2*c + d*x] + I*Si
n[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c])] + Log[1 - (b*(Cos[2*c + d*x
] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])])*(Cos[c] + I*Sin
[c]) + 2*e^2*ArcTan[(b*Cos[c + d*x] + I*(a + b*Sin[c + d*x]))/Sqrt[a^2 - b^2]]*Sqrt[(-a^2 + b^2)*(Cos[2*c] + I
*Sin[2*c])]))))/(b*Sqrt[a^2 - b^2]*d^4*Sqrt[(-a^2 + b^2)*(Cos[2*c] + I*Sin[2*c])]) - (e + f*x)^3/(b*d*(a + b*S
in[c + d*x]))

________________________________________________________________________________________

Maple [F]
time = 0.98, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{3} \cos \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*cos(d*x+c)/(a+b*sin(d*x+c))^2,x)

[Out]

int((f*x+e)^3*cos(d*x+c)/(a+b*sin(d*x+c))^2,x)

________________________________________________________________________________________

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)^3*cos(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 de

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2299 vs. \(2 (365) = 730\).
time = 0.56, size = 2299, normalized size = 5.50 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cos(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(2*(a^2 - b^2)*d^3*f^3*x^3 + 6*(a^2 - b^2)*d^3*f^2*x^2*e + 6*(a^2 - b^2)*d^3*f*x*e^2 + 2*(a^2 - b^2)*d^3*
e^3 - 6*(I*a*b*d*f^3*x + I*a*b*d*f^2*e + (I*b^2*d*f^3*x + I*b^2*d*f^2*e)*sin(d*x + c))*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) - 6*(-I*a*b*d*f^3*x - I*a*b*d*f^2*e + (-I*b^2*d*f^3*x - I*b^2*d*f^2*e)*sin(d*x + c))*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) - 6*(-I*a*b*d*f^3*x - I*a*b*d*f^2*e + (-I*b^2*d*f^3*x - I*b^2*d*f^2*e)*sin(d*x + c))*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) - 6*(I*a*b*d*f^3*x + I*a*b*d*f^2*e + (I*b^2*d*f^3*x + I*b^2*d*f^2*e)*sin(d*x + c))*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*b*c^2*f^3 - 2*a*b*c*d*f^2*e + a*b*d^2*f*e^2 + (b^2*c^2*f^3 - 2*b^2*c*d*f^2*e + b^2*d^2*f*e^2)
*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) - 3*(a*b*c^2*f^3 - 2*a*b*c*d*f^2*e + a*b*d^2*f*e^2 + (b^2*c^2*f^3 - 2*b^2*c*d*f^2*e + b^2*d^2*f*e^2)*s
in(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) + 3*(a*b*c^2*f^3 - 2*a*b*c*d*f^2*e + a*b*d^2*f*e^2 + (b^2*c^2*f^3 - 2*b^2*c*d*f^2*e + b^2*d^2*f*e^2)*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) + 3*(a*b*c^2*f^3 - 2*a*b*c*d*f^2*e + a*b*d^2*f*e^2 + (b^2*c^2*f^3 - 2*b^2*c*d*f^2*e + b^2*d^2*f*e^2)*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) + 3*(a*b*d^2*f^3*x^2 - a*b*c^2*f^3 + 2*(a*b*d^2*f^2*x + a*b*c*d*f^2)*e + (b^2*d^2*f^3*x^2 - b^2*c^2*f^3 +
2*(b^2*d^2*f^2*x + b^2*c*d*f^2)*e)*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*b*d^2*f^3*x^2 - a*b*c^2*f^3 + 2*
(a*b*d^2*f^2*x + a*b*c*d*f^2)*e + (b^2*d^2*f^3*x^2 - b^2*c^2*f^3 + 2*(b^2*d^2*f^2*x + b^2*c*d*f^2)*e)*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))*sqr
t(-(a^2 - b^2)/b^2) - b)/b) + 3*(a*b*d^2*f^3*x^2 - a*b*c^2*f^3 + 2*(a*b*d^2*f^2*x + a*b*c*d*f^2)*e + (b^2*d^2*
f^3*x^2 - b^2*c^2*f^3 + 2*(b^2*d^2*f^2*x + b^2*c*d*f^2)*e)*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*b*d^2*f
^3*x^2 - a*b*c^2*f^3 + 2*(a*b*d^2*f^2*x + a*b*c*d*f^2)*e + (b^2*d^2*f^3*x^2 - b^2*c^2*f^3 + 2*(b^2*d^2*f^2*x +
b^2*c*d*f^2)*e)*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) - 6*(b^2*f^3*sin(d*x + c) + a*b*f^3)*sqrt(-(a^2 - b^2)/b
^2)*polylog(3, -(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) + 6*(b^2*f^3*sin(d*x + c) + a*b*f^3)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(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) - 6*(b^2*f^3*sin(d*x + c) + a*b*f^3)*sqrt(
-(a^2 - b^2)/b^2)*polylog(3, -(-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) + 6*(b^2*f^3*sin(d*x + c) + a*b*f^3)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-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))/((a^2*b^2 - b^4)*d^4*sin(
d*x + c) + (a^3*b - a*b^3)*d^4)

________________________________________________________________________________________

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)**3*cos(d*x+c)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

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)^3*cos(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((f*x + e)^3*cos(d*x + c)/(b*sin(d*x + c) + a)^2, x)

________________________________________________________________________________________

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((cos(c + d*x)*(e + f*x)^3)/(a + b*sin(c + d*x))^2,x)

[Out]

\text{Hanged}

________________________________________________________________________________________