\(\int \frac {(e+f x)^2 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [299]

Optimal result

Integrand size = 28, antiderivative size = 460 \[ \int \frac {(e+f x)^2 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a (e+f x)^3}{3 b^2 f}-\frac {2 f^2 \cos (c+d x)}{b d^3}+\frac {(e+f x)^2 \cos (c+d x)}{b d}+\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d}-\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {2 \sqrt {a^2-b^2} f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {2 \sqrt {a^2-b^2} f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2}+\frac {2 i \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3}-\frac {2 i \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^3}-\frac {2 f (e+f x) \sin (c+d x)}{b d^2} \] Output:

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

Mathematica [A] (verified)

Time = 3.70 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.17 \[ \int \frac {(e+f x)^2 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac {3 i \left (-a^2+b^2\right ) \left (-2 \sqrt {a^2-b^2} d f (e+f x) \operatorname {PolyLog}\left (2,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+2 \sqrt {a^2-b^2} d f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-i \left (d^2 \left (2 \sqrt {-a^2+b^2} e^2 \arctan \left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+\sqrt {a^2-b^2} f x (2 e+f x) \left (\log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-\log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )\right )\right )+2 \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-2 \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )\right )\right )}{\sqrt {-\left (a^2-b^2\right )^2} d^3}+\frac {3 b \cos (d x) \left (\left (-2 f^2+d^2 (e+f x)^2\right ) \cos (c)-2 d f (e+f x) \sin (c)\right )}{d^3}-\frac {3 b \left (2 d f (e+f x) \cos (c)+\left (-2 f^2+d^2 (e+f x)^2\right ) \sin (c)\right ) \sin (d x)}{d^3}}{3 b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.95 (sec) , antiderivative size = 435, normalized size of antiderivative = 0.95, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5036, 17, 3042, 3777, 3042, 3777, 25, 3042, 3118, 3804, 2694, 27, 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.

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

\(\Big \downarrow \) 5036

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

\(\Big \downarrow \) 17

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3118

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

(a*(e + f*x)^3)/(3*b^2*f) - (2*(a^2 - b^2)*(((-1/2*I)*b*(((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)*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)))/Sqrt[a^2 - 
 b^2] + ((I/2)*b*(((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)*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)))/Sqrt[a^2 - b^2]))/b^2 - (-(((e + f*x)^2*Cos[c + d 
*x])/d) + (2*f*((f*Cos[c + d*x])/d^2 + ((e + f*x)*Sin[c + d*x])/d))/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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 2694

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]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[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 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 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 3804

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[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 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]

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

Input:

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

Output:

int((f*x+e)^2*cos(d*x+c)^2/(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. 1636 vs. \(2 (400) = 800\).

Time = 0.27 (sec) , antiderivative size = 1636, normalized size of antiderivative = 3.56 \[ \int \frac {(e+f x)^2 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/6*(2*a*d^3*f^2*x^3 + 6*a*d^3*e*f*x^2 + 6*a*d^3*e^2*x - 6*b*f^2*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*f^2*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*f^2*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*f^2*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*(I*b*d*f^2*x + I*b*d*e*f) 
*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos( 
d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 6*(-I*b* 
d*f^2*x - I*b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*si 
n(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*b*d*f^2*x - I*b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I* 
a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt 
(-(a^2 - b^2)/b^2) - b)/b + 1) - 6*(I*b*d*f^2*x + I*b*d*e*f)*sqrt(-(a^2 - 
b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I* 
b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 3*(b*d^2*e^2 - 2*b*c* 
d*e*f + b*c^2*f^2)*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*(b*d^2*e^2 - 2*b*c*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((f*x+e)^2*cos(d*x+c)^2/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

( - 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a 
**3*d**2*e**2 + 3*cos(c + d*x)*a**3*b*d**2*e**2 - 12*cos(c + d*x)*a**2*b** 
2*d*e*f - 12*cos(c + d*x)*a**2*b**2*d*f**2*x + 6*cos(c + d*x)*a*b**3*d**2* 
e*f*x + 3*cos(c + d*x)*a*b**3*d**2*f**2*x**2 - 6*cos(c + d*x)*a*b**3*f**2 
+ 12*cos(c + d*x)*b**4*d*e*f + 12*cos(c + d*x)*b**4*d*f**2*x + 12*int(x**2 
/(tan((c + d*x)/2)**4*a + 2*tan((c + d*x)/2)**3*b + 2*tan((c + d*x)/2)**2* 
a + 2*tan((c + d*x)/2)*b + a),x)*a**3*b**2*d**3*f**2 - 12*int(x**2/(tan((c 
 + d*x)/2)**4*a + 2*tan((c + d*x)/2)**3*b + 2*tan((c + d*x)/2)**2*a + 2*ta 
n((c + d*x)/2)*b + a),x)*a*b**4*d**3*f**2 + 24*int((tan((c + d*x)/2)*x**2) 
/(tan((c + d*x)/2)**4*a + 2*tan((c + d*x)/2)**3*b + 2*tan((c + d*x)/2)**2* 
a + 2*tan((c + d*x)/2)*b + a),x)*a**2*b**3*d**3*f**2 - 24*int((tan((c + d* 
x)/2)*x**2)/(tan((c + d*x)/2)**4*a + 2*tan((c + d*x)/2)**3*b + 2*tan((c + 
d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a),x)*b**5*d**3*f**2 + 48*int((tan(( 
c + d*x)/2)*x)/(tan((c + d*x)/2)**4*a + 2*tan((c + d*x)/2)**3*b + 2*tan((c 
 + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a),x)*a**2*b**3*d**3*e*f - 48*int 
((tan((c + d*x)/2)*x)/(tan((c + d*x)/2)**4*a + 2*tan((c + d*x)/2)**3*b + 2 
*tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a),x)*b**5*d**3*e*f + 24*i 
nt(x/(tan((c + d*x)/2)**4*a + 2*tan((c + d*x)/2)**3*b + 2*tan((c + d*x)/2) 
**2*a + 2*tan((c + d*x)/2)*b + a),x)*a**3*b**2*d**3*e*f - 24*int(x/(tan((c 
 + d*x)/2)**4*a + 2*tan((c + d*x)/2)**3*b + 2*tan((c + d*x)/2)**2*a + 2...