\(\int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [303]

Optimal result

Integrand size = 28, antiderivative size = 536 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {(e+f x)^2}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {2 \left (a^2-b^2\right ) f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 \left (a^2-b^2\right ) f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 a f^2 \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}-\frac {f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d} \] Output:

1/4*(f*x+e)^2/b/d+1/3*I*(a^2-b^2)*(f*x+e)^3/b^3/f+2*a*f*(f*x+e)*cos(d*x+c) 
/b^2/d^2-(a^2-b^2)*(f*x+e)^2*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)^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3 
/d+2*I*(a^2-b^2)*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2) 
))/b^3/d^2+2*I*(a^2-b^2)*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^ 
2)^(1/2)))/b^3/d^2-2*(a^2-b^2)*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^ 
2)^(1/2)))/b^3/d^3-2*(a^2-b^2)*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^ 
2)^(1/2)))/b^3/d^3-2*a*f^2*sin(d*x+c)/b^2/d^3+a*(f*x+e)^2*sin(d*x+c)/b^2/d 
-1/2*f*(f*x+e)*cos(d*x+c)*sin(d*x+c)/b/d^2+1/4*f^2*sin(d*x+c)^2/b/d^3-1/2* 
(f*x+e)^2*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. \(2283\) vs. \(2(536)=1072\).

Time = 5.69 (sec) , antiderivative size = 2283, normalized size of antiderivative = 4.26 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

((48*I)*a^2*d^3*e^2*E^((2*I)*c)*x - (48*I)*b^2*d^3*e^2*E^((2*I)*c)*x + (48 
*I)*a^2*d^3*e*E^((2*I)*c)*f*x^2 - (48*I)*b^2*d^3*e*E^((2*I)*c)*f*x^2 + (16 
*I)*a^2*d^3*E^((2*I)*c)*f^2*x^3 - (16*I)*b^2*d^3*E^((2*I)*c)*f^2*x^3 + (24 
*I)*a*b*d^2*e^2*E^(I*c)*Cos[d*x] - (24*I)*a*b*d^2*e^2*E^((3*I)*c)*Cos[d*x] 
 + 48*a*b*d*e*E^(I*c)*f*Cos[d*x] + 48*a*b*d*e*E^((3*I)*c)*f*Cos[d*x] - (48 
*I)*a*b*E^(I*c)*f^2*Cos[d*x] + (48*I)*a*b*E^((3*I)*c)*f^2*Cos[d*x] + (48*I 
)*a*b*d^2*e*E^(I*c)*f*x*Cos[d*x] - (48*I)*a*b*d^2*e*E^((3*I)*c)*f*x*Cos[d* 
x] + 48*a*b*d*E^(I*c)*f^2*x*Cos[d*x] + 48*a*b*d*E^((3*I)*c)*f^2*x*Cos[d*x] 
 + (24*I)*a*b*d^2*E^(I*c)*f^2*x^2*Cos[d*x] - (24*I)*a*b*d^2*E^((3*I)*c)*f^ 
2*x^2*Cos[d*x] + 6*b^2*d^2*e^2*Cos[2*d*x] + 6*b^2*d^2*e^2*E^((4*I)*c)*Cos[ 
2*d*x] - (6*I)*b^2*d*e*f*Cos[2*d*x] + (6*I)*b^2*d*e*E^((4*I)*c)*f*Cos[2*d* 
x] - 3*b^2*f^2*Cos[2*d*x] - 3*b^2*E^((4*I)*c)*f^2*Cos[2*d*x] + 12*b^2*d^2* 
e*f*x*Cos[2*d*x] + 12*b^2*d^2*e*E^((4*I)*c)*f*x*Cos[2*d*x] - (6*I)*b^2*d*f 
^2*x*Cos[2*d*x] + (6*I)*b^2*d*E^((4*I)*c)*f^2*x*Cos[2*d*x] + 6*b^2*d^2*f^2 
*x^2*Cos[2*d*x] + 6*b^2*d^2*E^((4*I)*c)*f^2*x^2*Cos[2*d*x] - 48*a^2*d^2*e^ 
2*E^((2*I)*c)*Log[b - (2*I)*a*E^(I*(c + d*x)) - b*E^((2*I)*(c + d*x))] + 4 
8*b^2*d^2*e^2*E^((2*I)*c)*Log[b - (2*I)*a*E^(I*(c + d*x)) - b*E^((2*I)*(c 
+ d*x))] - 96*a^2*d^2*e*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)])] + 96*b^2*d^2*e*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...
 

Rubi [A] (verified)

Time = 2.36 (sec) , antiderivative size = 483, normalized size of antiderivative = 0.90, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {5036, 3042, 3777, 25, 3042, 3777, 3042, 3117, 4904, 3042, 3791, 17, 5030, 2620, 3011, 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.

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

\(\Big \downarrow \) 5036

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3117

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

\(\Big \downarrow \) 4904

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3791

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

\(\Big \downarrow \) 17

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

\(\Big \downarrow \) 5030

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

-(((a^2 - b^2)*(((-1/3*I)*(e + f*x)^3)/(b*f) + ((e + f*x)^2*Log[1 - (I*b*E 
^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) + ((e + f*x)^2*Log[1 - (I*b* 
E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d) - (2*f*((I*(e + f*x)*PolyLo 
g[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d - (f*PolyLog[3, (I*b* 
E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d^2))/(b*d) - (2*f*((I*(e + f*x)* 
PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/d - (f*PolyLog[3, 
 (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/d^2))/(b*d)))/b^2) + (a*((( 
e + f*x)^2*Sin[c + d*x])/d - (2*f*(-(((e + f*x)*Cos[c + d*x])/d) + (f*Sin[ 
c + d*x])/d^2))/d))/b^2 - (((e + f*x)^2*Sin[c + d*x]^2)/(2*d) - (f*((e + f 
*x)^2/(4*f) - ((e + f*x)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (f*Sin[c + d*x 
]^2)/(4*d^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 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 3117

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 3791

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

Maple [F]

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

Input:

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

Output:

int((f*x+e)^2*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. 1779 vs. \(2 (488) = 976\).

Time = 0.25 (sec) , antiderivative size = 1779, normalized size of antiderivative = 3.32 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(b^2*d^2*f^2*x^2 + 2*b^2*d^2*e*f*x + 4*(a^2 - b^2)*f^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))*sqr 
t(-(a^2 - b^2)/b^2))/b) + 4*(a^2 - b^2)*f^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) + 4*(a^2 - b^2)*f^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) + 4*(a^ 
2 - b^2)*f^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) - (2*b^2*d^2*f^2*x^2 + 
 4*b^2*d^2*e*f*x + 2*b^2*d^2*e^2 - b^2*f^2)*cos(d*x + c)^2 - 8*(a*b*d*f^2* 
x + a*b*d*e*f)*cos(d*x + c) + 4*(-I*(a^2 - b^2)*d*f^2*x - I*(a^2 - b^2)*d* 
e*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) + 4*(-I*(a^2 - b^2)*d*f^2*x - 
 I*(a^2 - b^2)*d*e*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) + 4*(I*(a^2 
- b^2)*d*f^2*x + I*(a^2 - b^2)*d*e*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) + 4*(I*(a^2 - b^2)*d*f^2*x + I*(a^2 - b^2)*d*e*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) + 2*((a^2 - b^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + ( 
a^2 - b^2)*c^2*f^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqr...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((f*x+e)^2*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]

\[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \cos \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {too large to display} \] Input:

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

Output:

( - 36*cos(c + d*x)*sin(c + d*x)*a**4*b**3*d**2*e*f*x - 18*cos(c + d*x)*si 
n(c + d*x)*a**4*b**3*d**2*f**2*x**2 + 9*cos(c + d*x)*sin(c + d*x)*a**4*b** 
3*f**2 - 18*cos(c + d*x)*sin(c + d*x)*a**3*b**4*d*e*f - 18*cos(c + d*x)*si 
n(c + d*x)*a**3*b**4*d*f**2*x + 84*cos(c + d*x)*sin(c + d*x)*a**2*b**5*d** 
2*e*f*x + 42*cos(c + d*x)*sin(c + d*x)*a**2*b**5*d**2*f**2*x**2 - 21*cos(c 
 + d*x)*sin(c + d*x)*a**2*b**5*f**2 + 12*cos(c + d*x)*sin(c + d*x)*a*b**6* 
d*e*f + 12*cos(c + d*x)*sin(c + d*x)*a*b**6*d*f**2*x - 48*cos(c + d*x)*sin 
(c + d*x)*b**7*d**2*e*f*x - 24*cos(c + d*x)*sin(c + d*x)*b**7*d**2*f**2*x* 
*2 + 12*cos(c + d*x)*sin(c + d*x)*b**7*f**2 - 72*cos(c + d*x)*a**4*b**3*d* 
e*f - 72*cos(c + d*x)*a**4*b**3*d*f**2*x + 48*cos(c + d*x)*a**3*b**4*d**2* 
e*f*x + 24*cos(c + d*x)*a**3*b**4*d**2*f**2*x**2 - 48*cos(c + d*x)*a**3*b* 
*4*f**2 + 288*cos(c + d*x)*a**2*b**5*d*e*f + 288*cos(c + d*x)*a**2*b**5*d* 
f**2*x - 48*cos(c + d*x)*a*b**6*d**2*e*f*x - 24*cos(c + d*x)*a*b**6*d**2*f 
**2*x**2 + 48*cos(c + d*x)*a*b**6*f**2 - 192*cos(c + d*x)*b**7*d*e*f - 192 
*cos(c + d*x)*b**7*d*f**2*x + 96*int(x**2/(tan((c + d*x)/2)**6*a + 2*tan(( 
c + d*x)/2)**5*b + 3*tan((c + d*x)/2)**4*a + 4*tan((c + d*x)/2)**3*b + 3*t 
an((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a),x)*a**5*b**3*d**3*f**2 - 
288*int(x**2/(tan((c + d*x)/2)**6*a + 2*tan((c + d*x)/2)**5*b + 3*tan((c + 
 d*x)/2)**4*a + 4*tan((c + d*x)/2)**3*b + 3*tan((c + d*x)/2)**2*a + 2*tan( 
(c + d*x)/2)*b + a),x)*a**3*b**5*d**3*f**2 + 192*int(x**2/(tan((c + d*x...