\(\int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [302]

Optimal result

Integrand size = 28, antiderivative size = 737 \[ \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {3 f^3 x}{8 b d^3}+\frac {(e+f x)^3}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac {3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {3 i \left (a^2-b^2\right ) f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {3 i \left (a^2-b^2\right ) f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {6 \left (a^2-b^2\right ) f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {6 \left (a^2-b^2\right ) f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {6 i \left (a^2-b^2\right ) f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^4}-\frac {6 i \left (a^2-b^2\right ) f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^4}-\frac {6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac {3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac {3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d} \] Output:

-3/8*f^3*x/b/d^3+1/4*(f*x+e)^3/b/d-6*I*(a^2-b^2)*f^3*polylog(4,I*b*exp(I*( 
d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d^4-6*a*f^3*cos(d*x+c)/b^2/d^4+3*a*f*(f*x 
+e)^2*cos(d*x+c)/b^2/d^2-(a^2-b^2)*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a-(a 
^2-b^2)^(1/2)))/b^3/d-(a^2-b^2)*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2- 
b^2)^(1/2)))/b^3/d-6*I*(a^2-b^2)*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a-(a^2- 
b^2)^(1/2)))/b^3/d^4+3*I*(a^2-b^2)*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c) 
)/(a+(a^2-b^2)^(1/2)))/b^3/d^2-6*(a^2-b^2)*f^2*(f*x+e)*polylog(3,I*b*exp(I 
*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d^3-6*(a^2-b^2)*f^2*(f*x+e)*polylog(3,I 
*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d^3+1/4*I*(a^2-b^2)*(f*x+e)^4/b 
^3/f+3*I*(a^2-b^2)*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^( 
1/2)))/b^3/d^2-6*a*f^2*(f*x+e)*sin(d*x+c)/b^2/d^3+a*(f*x+e)^3*sin(d*x+c)/b 
^2/d+3/8*f^3*cos(d*x+c)*sin(d*x+c)/b/d^4-3/4*f*(f*x+e)^2*cos(d*x+c)*sin(d* 
x+c)/b/d^2+3/4*f^2*(f*x+e)*sin(d*x+c)^2/b/d^3-1/2*(f*x+e)^3*sin(d*x+c)^2/b 
/d
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2452\) vs. \(2(737)=1474\).

Time = 10.42 (sec) , antiderivative size = 2452, normalized size of antiderivative = 3.33 \[ \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

(-32*(a^2 - b^2)*e^3*x*Cot[c] - 48*(a^2 - b^2)*e^2*f*x^2*Cot[c] - 32*(a^2 
- b^2)*e*f^2*x^3*Cot[c] - 8*(a^2 - b^2)*f^3*x^4*Cot[c] + (16*(a^2 - b^2)*( 
(4*I)*d^4*e^3*E^((2*I)*c)*x + (6*I)*d^4*e^2*E^((2*I)*c)*f*x^2 + (4*I)*d^4* 
e*E^((2*I)*c)*f^2*x^3 + I*d^4*E^((2*I)*c)*f^3*x^4 + (2*I)*d^3*e^3*ArcTan[( 
2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] - (2*I)*d^3*e^3*E^((2 
*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] + d^3* 
e^3*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] - d^ 
3*e^3*E^((2*I)*c)*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + 
d*x)))^2] + 6*d^3*e^2*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqr 
t[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e^2*E^((2*I)*c)*f*x*Log[1 + (b*E^(I* 
(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e*f^ 
2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2 
*I)*c)])] - 6*d^3*e*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a 
*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 2*d^3*f^3*x^3*Log[1 + (b*E^( 
I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 2*d^3*E^ 
((2*I)*c)*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 
+ b^2)*E^((2*I)*c)])] + 6*d^3*e^2*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E 
^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e^2*E^((2*I)*c)*f*x*Log[ 
1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] 
+ 6*d^3*e*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a...
 

Rubi [A] (verified)

Time = 3.65 (sec) , antiderivative size = 667, normalized size of antiderivative = 0.91, number of steps used = 25, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {5036, 3042, 3777, 25, 3042, 3777, 3042, 3777, 25, 3042, 3118, 4904, 3042, 3792, 17, 3042, 3115, 24, 5030, 2620, 3011, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

\(\Big \downarrow \) 5036

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \int (e+f x)^3 \cos (c+d x)dx}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \int (e+f x)^3 \sin \left (c+d x+\frac {\pi }{2}\right )dx}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 3777

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {3 f \int -(e+f x)^2 \sin (c+d x)dx}{d}+\frac {(e+f x)^3 \sin (c+d x)}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \int (e+f x)^2 \sin (c+d x)dx}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \int (e+f x)^2 \sin (c+d x)dx}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 3777

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \int (e+f x) \cos (c+d x)dx}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \int (e+f x) \sin \left (c+d x+\frac {\pi }{2}\right )dx}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 3777

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \int -\sin (c+d x)dx}{d}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\)

\(\Big \downarrow \) 3118

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\)

\(\Big \downarrow \) 4904

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \int (e+f x)^2 \sin ^2(c+d x)dx}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \int (e+f x)^2 \sin (c+d x)^2dx}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\)

\(\Big \downarrow \) 3792

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {1}{2} \int (e+f x)^2dx+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\)

\(\Big \downarrow \) 17

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (-\frac {f^2 \int \sin (c+d x)^2dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (-\frac {f^2 \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}\)

\(\Big \downarrow \) 5030

\(\displaystyle -\frac {\left (a^2-b^2\right ) \left (\int \frac {e^{i (c+d x)} (e+f x)^3}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx+\int \frac {e^{i (c+d x)} (e+f x)^3}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx-\frac {i (e+f x)^4}{4 b f}\right )}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {\left (a^2-b^2\right ) \left (-\frac {3 f \int (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}-\frac {3 f \int (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^4}{4 b f}\right )}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {\left (a^2-b^2\right ) \left (-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^4}{4 b f}\right )}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}\)

\(\Big \downarrow \) 7163

\(\displaystyle -\frac {\left (a^2-b^2\right ) \left (-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^4}{4 b f}\right )}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {\left (a^2-b^2\right ) \left (-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^4}{4 b f}\right )}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {\left (a^2-b^2\right ) \left (-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^4}{4 b f}\right )}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}\)

Input:

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

Output:

-(((a^2 - b^2)*(((-1/4*I)*(e + f*x)^4)/(b*f) + ((e + f*x)^3*Log[1 - (I*b*E 
^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) + ((e + f*x)^3*Log[1 - (I*b* 
E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d) - (3*f*((I*(e + f*x)^2*Poly 
Log[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d - ((2*I)*f*(((-I)*( 
e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d + (f*P 
olyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d^2))/d))/(b*d) - 
(3*f*((I*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2] 
)])/d - ((2*I)*f*(((-I)*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sq 
rt[a^2 - b^2])])/d + (f*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b 
^2])])/d^2))/d))/(b*d)))/b^2) + (a*(((e + f*x)^3*Sin[c + d*x])/d - (3*f*(- 
(((e + f*x)^2*Cos[c + d*x])/d) + (2*f*((f*Cos[c + d*x])/d^2 + ((e + f*x)*S 
in[c + d*x])/d))/d))/d))/b^2 - (((e + f*x)^3*Sin[c + d*x]^2)/(2*d) - (3*f* 
((e + f*x)^3/(6*f) - ((e + f*x)^2*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (f*(e 
 + f*x)*Sin[c + d*x]^2)/(2*d^2) - (f^2*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/ 
(2*d)))/(2*d^2)))/(2*d))/b
 

Defintions of rubi rules used

rule 17

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

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2620

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] - Si 
mp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ 
[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3792

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo 
l] :> Simp[d*m*(c + d*x)^(m - 1)*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Sim 
p[b*(c + d*x)^m*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^ 
2*((n - 1)/n)   Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[d^2 
*m*((m - 1)/(f^2*n^2))   Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x]) 
/; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]
 

rule 4904

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

rule 5030

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

rule 5036

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

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

rule 7163

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

Maple [F]

\[\int \frac {\left (f x +e \right )^{3} \cos \left (d x +c \right )^{3}}{a +b \sin \left (d x +c \right )}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2684 vs. \(2 (673) = 1346\).

Time = 0.30 (sec) , antiderivative size = 2684, normalized size of antiderivative = 3.64 \[ \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*(2*b^2*d^3*f^3*x^3 + 6*b^2*d^3*e*f^2*x^2 - 24*I*(a^2 - b^2)*f^3*polyl 
og(4, -(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) - 24*I*(a^2 - b^2)*f^3*polylog(4, -(I*a*c 
os(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) + 24*I*(a^2 - b^2)*f^3*polylog(4, -(-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) + 24*I*(a^2 - b^2)*f^3*polylog(4, -(-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) - 2* 
(2*b^2*d^3*f^3*x^3 + 6*b^2*d^3*e*f^2*x^2 + 2*b^2*d^3*e^3 - 3*b^2*d*e*f^2 + 
 3*(2*b^2*d^3*e^2*f - b^2*d*f^3)*x)*cos(d*x + c)^2 + 3*(2*b^2*d^3*e^2*f - 
b^2*d*f^3)*x - 24*(a*b*d^2*f^3*x^2 + 2*a*b*d^2*e*f^2*x + a*b*d^2*e^2*f - 2 
*a*b*f^3)*cos(d*x + c) + 12*(-I*(a^2 - b^2)*d^2*f^3*x^2 - 2*I*(a^2 - b^2)* 
d^2*e*f^2*x - I*(a^2 - b^2)*d^2*e^2*f)*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) + 12*(-I*(a^2 - b^2)*d^2*f^3*x^2 - 2*I*(a^2 - b^2)*d^2*e*f^2*x - I*(a 
^2 - b^2)*d^2*e^2*f)*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) + 12*(I*(a^2 
- b^2)*d^2*f^3*x^2 + 2*I*(a^2 - b^2)*d^2*e*f^2*x + I*(a^2 - b^2)*d^2*e^2*f 
)*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) + 12*(I*(a^2 - b^2)*d^2*f^3*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \cos \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\left (f x +e \right )^{3} \cos \left (d x +c \right )^{3}}{\sin \left (d x +c \right ) b +a}d x \] Input:

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

Output:

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