\(\int F^{c (a+b x)} \cos ^2(d+e x) \sin ^2(d+e x) \, dx\) [95]

Optimal result

Integrand size = 26, antiderivative size = 83 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^2(d+e x) \, dx=\frac {F^{c (a+b x)}}{8 b c \log (F)}-\frac {F^{c (a+b x)} (b c \cos (4 d+4 e x) \log (F)+4 e \sin (4 d+4 e x))}{8 \left (16 e^2+b^2 c^2 \log ^2(F)\right )} \] Output:

1/8*F^(c*(b*x+a))/b/c/ln(F)-F^(c*(b*x+a))*(b*c*cos(4*e*x+4*d)*ln(F)+4*e*si 
n(4*e*x+4*d))/(128*e^2+8*b^2*c^2*ln(F)^2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.07 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^2(d+e x) \, dx=\frac {F^{c (a+b x)} \left (16 e^2+b^2 c^2 \log ^2(F)-b^2 c^2 \cos (4 (d+e x)) \log ^2(F)-4 b c e \log (F) \sin (4 (d+e x))\right )}{4 \left (32 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)\right )} \] Input:

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

Output:

(F^(c*(a + b*x))*(16*e^2 + b^2*c^2*Log[F]^2 - b^2*c^2*Cos[4*(d + e*x)]*Log 
[F]^2 - 4*b*c*e*Log[F]*Sin[4*(d + e*x)]))/(4*(32*b*c*e^2*Log[F] + 2*b^3*c^ 
3*Log[F]^3))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {4972, 2009}

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.

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

\(\Big \downarrow \) 4972

\(\displaystyle \int \left (\frac {1}{8} F^{c (a+b x)}-\frac {1}{8} \cos (4 d+4 e x) F^{c (a+b x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {e \sin (4 d+4 e x) F^{c (a+b x)}}{2 \left (b^2 c^2 \log ^2(F)+16 e^2\right )}-\frac {b c \log (F) \cos (4 d+4 e x) F^{c (a+b x)}}{8 \left (b^2 c^2 \log ^2(F)+16 e^2\right )}+\frac {F^{c (a+b x)}}{8 b c \log (F)}\)

Input:

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

Output:

F^(c*(a + b*x))/(8*b*c*Log[F]) - (b*c*F^(c*(a + b*x))*Cos[4*d + 4*e*x]*Log 
[F])/(8*(16*e^2 + b^2*c^2*Log[F]^2)) - (e*F^(c*(a + b*x))*Sin[4*d + 4*e*x] 
)/(2*(16*e^2 + b^2*c^2*Log[F]^2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4972

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

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(-\frac {F^{c \left (b x +a \right )} \left (\frac {b^{2} c^{2} \ln \left (F \right )^{2} \cos \left (4 e x +4 d \right )}{4}-\frac {b^{2} c^{2} \ln \left (F \right )^{2}}{4}+\sin \left (4 e x +4 d \right ) b c e \ln \left (F \right )-4 e^{2}\right )}{2 \ln \left (F \right ) b c \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(95\)
risch	\(\frac {F^{c \left (b x +a \right )}}{8 b c \ln \left (F \right )}-\frac {F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (4 e x +4 d \right )}{8 \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {e \,F^{c \left (b x +a \right )} \sin \left (4 e x +4 d \right )}{2 \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(106\)
orering	\(\frac {\left (3 b^{2} c^{2} \ln \left (F \right )^{2}+16 e^{2}\right ) F^{c \left (b x +a \right )} \cos \left (e x +d \right )^{2} \sin \left (e x +d \right )^{2}}{\ln \left (F \right ) b c \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {3 \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (e x +d \right )^{2} \sin \left (e x +d \right )^{2}-2 F^{c \left (b x +a \right )} \cos \left (e x +d \right ) \sin \left (e x +d \right )^{3} e +2 F^{c \left (b x +a \right )} \cos \left (e x +d \right )^{3} \sin \left (e x +d \right ) e \right )}{16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \cos \left (e x +d \right )^{2} \sin \left (e x +d \right )^{2}-4 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (e x +d \right ) \sin \left (e x +d \right )^{3} e +4 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (e x +d \right )^{3} \sin \left (e x +d \right ) e +2 F^{c \left (b x +a \right )} e^{2} \sin \left (e x +d \right )^{4}-12 F^{c \left (b x +a \right )} \cos \left (e x +d \right )^{2} \sin \left (e x +d \right )^{2} e^{2}+2 F^{c \left (b x +a \right )} \cos \left (e x +d \right )^{4} e^{2}}{\ln \left (F \right ) b c \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(380\)
norman	\(\frac {-\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {28 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {28 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5}}{16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{7}}{16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right ) b c \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {2 e^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{8}}{\ln \left (F \right ) b c \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 \left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{\ln \left (F \right ) b c \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 \left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{6}}{\ln \left (F \right ) b c \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {4 \left (2 b^{2} c^{2} \ln \left (F \right )^{2}-3 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{\ln \left (F \right ) b c \left (16 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{4}}\)	\(495\)

Input:

int(F^(c*(b*x+a))*cos(e*x+d)^2*sin(e*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2*F^(c*(b*x+a))*(1/4*b^2*c^2*ln(F)^2*cos(4*e*x+4*d)-1/4*b^2*c^2*ln(F)^2 
+sin(4*e*x+4*d)*b*c*e*ln(F)-4*e^2)/ln(F)/b/c/(16*e^2+b^2*c^2*ln(F)^2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.36 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^2(d+e x) \, dx=-\frac {{\left ({\left (b^{2} c^{2} \cos \left (e x + d\right )^{4} - b^{2} c^{2} \cos \left (e x + d\right )^{2}\right )} \log \left (F\right )^{2} + 2 \, {\left (2 \, b c e \cos \left (e x + d\right )^{3} - b c e \cos \left (e x + d\right )\right )} \log \left (F\right ) \sin \left (e x + d\right ) - 2 \, e^{2}\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + 16 \, b c e^{2} \log \left (F\right )} \] Input:

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

Output:

-((b^2*c^2*cos(e*x + d)^4 - b^2*c^2*cos(e*x + d)^2)*log(F)^2 + 2*(2*b*c*e* 
cos(e*x + d)^3 - b*c*e*cos(e*x + d))*log(F)*sin(e*x + d) - 2*e^2)*F^(b*c*x 
 + a*c)/(b^3*c^3*log(F)^3 + 16*b*c*e^2*log(F))
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 24.04 (sec) , antiderivative size = 1380, normalized size of antiderivative = 16.63 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^2(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((x*sin(d)**2*cos(d)**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0 
)), (x*sin(d + e*x)**4/8 + x*sin(d + e*x)**2*cos(d + e*x)**2/4 + x*cos(d + 
 e*x)**4/8 + sin(d + e*x)**3*cos(d + e*x)/(8*e) - sin(d + e*x)*cos(d + e*x 
)**3/(8*e), Eq(F, 1)), (F**(a*c)*(x*sin(d + e*x)**4/8 + x*sin(d + e*x)**2* 
cos(d + e*x)**2/4 + x*cos(d + e*x)**4/8 + sin(d + e*x)**3*cos(d + e*x)/(8* 
e) - sin(d + e*x)*cos(d + e*x)**3/(8*e)), Eq(b, 0)), (x*sin(d + e*x)**4/8 
+ x*sin(d + e*x)**2*cos(d + e*x)**2/4 + x*cos(d + e*x)**4/8 + sin(d + e*x) 
**3*cos(d + e*x)/(8*e) - sin(d + e*x)*cos(d + e*x)**3/(8*e), Eq(c, 0)), (- 
F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/4 - d)**4/16 + I*F**(a*c + b*c*x)*x* 
sin(I*b*c*x*log(F)/4 - d)**3*cos(I*b*c*x*log(F)/4 - d)/4 + 3*F**(a*c + b*c 
*x)*x*sin(I*b*c*x*log(F)/4 - d)**2*cos(I*b*c*x*log(F)/4 - d)**2/8 - I*F**( 
a*c + b*c*x)*x*sin(I*b*c*x*log(F)/4 - d)*cos(I*b*c*x*log(F)/4 - d)**3/4 - 
F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F)/4 - d)**4/16 + F**(a*c + b*c*x)*sin( 
I*b*c*x*log(F)/4 - d)**4/(6*b*c*log(F)) - 5*I*F**(a*c + b*c*x)*sin(I*b*c*x 
*log(F)/4 - d)**3*cos(I*b*c*x*log(F)/4 - d)/(12*b*c*log(F)) + 5*I*F**(a*c 
+ b*c*x)*sin(I*b*c*x*log(F)/4 - d)*cos(I*b*c*x*log(F)/4 - d)**3/(12*b*c*lo 
g(F)) + F**(a*c + b*c*x)*cos(I*b*c*x*log(F)/4 - d)**4/(6*b*c*log(F)), Eq(e 
, -I*b*c*log(F)/4)), (-F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/4 + d)**4/16 
+ I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/4 + d)**3*cos(I*b*c*x*log(F)/4 + 
 d)/4 + 3*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/4 + d)**2*cos(I*b*c*x*l...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (79) = 158\).

Time = 0.07 (sec) , antiderivative size = 356, normalized size of antiderivative = 4.29 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^2(d+e x) \, dx=-\frac {{\left (F^{a c} b^{2} c^{2} \cos \left (4 \, d\right ) \log \left (F\right )^{2} + 4 \, F^{a c} b c e \log \left (F\right ) \sin \left (4 \, d\right )\right )} F^{b c x} \cos \left (4 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \cos \left (4 \, d\right ) \log \left (F\right )^{2} - 4 \, F^{a c} b c e \log \left (F\right ) \sin \left (4 \, d\right )\right )} F^{b c x} \cos \left (4 \, e x + 8 \, d\right ) - {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (4 \, d\right ) - 4 \, F^{a c} b c e \cos \left (4 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (4 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (4 \, d\right ) + 4 \, F^{a c} b c e \cos \left (4 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (4 \, e x + 8 \, d\right ) - 2 \, {\left (F^{a c} b^{2} c^{2} \cos \left (4 \, d\right )^{2} \log \left (F\right )^{2} + F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (4 \, d\right )^{2} + 16 \, {\left (F^{a c} \cos \left (4 \, d\right )^{2} + F^{a c} \sin \left (4 \, d\right )^{2}\right )} e^{2}\right )} F^{b c x}}{16 \, {\left (b^{3} c^{3} \cos \left (4 \, d\right )^{2} \log \left (F\right )^{3} + b^{3} c^{3} \log \left (F\right )^{3} \sin \left (4 \, d\right )^{2} + 16 \, {\left (b c \cos \left (4 \, d\right )^{2} \log \left (F\right ) + b c \log \left (F\right ) \sin \left (4 \, d\right )^{2}\right )} e^{2}\right )}} \] Input:

integrate(F^(c*(b*x+a))*cos(e*x+d)^2*sin(e*x+d)^2,x, algorithm="maxima")
 

Output:

-1/16*((F^(a*c)*b^2*c^2*cos(4*d)*log(F)^2 + 4*F^(a*c)*b*c*e*log(F)*sin(4*d 
))*F^(b*c*x)*cos(4*e*x) + (F^(a*c)*b^2*c^2*cos(4*d)*log(F)^2 - 4*F^(a*c)*b 
*c*e*log(F)*sin(4*d))*F^(b*c*x)*cos(4*e*x + 8*d) - (F^(a*c)*b^2*c^2*log(F) 
^2*sin(4*d) - 4*F^(a*c)*b*c*e*cos(4*d)*log(F))*F^(b*c*x)*sin(4*e*x) + (F^( 
a*c)*b^2*c^2*log(F)^2*sin(4*d) + 4*F^(a*c)*b*c*e*cos(4*d)*log(F))*F^(b*c*x 
)*sin(4*e*x + 8*d) - 2*(F^(a*c)*b^2*c^2*cos(4*d)^2*log(F)^2 + F^(a*c)*b^2* 
c^2*log(F)^2*sin(4*d)^2 + 16*(F^(a*c)*cos(4*d)^2 + F^(a*c)*sin(4*d)^2)*e^2 
)*F^(b*c*x))/(b^3*c^3*cos(4*d)^2*log(F)^3 + b^3*c^3*log(F)^3*sin(4*d)^2 + 
16*(b*c*cos(4*d)^2*log(F) + b*c*log(F)*sin(4*d)^2)*e^2)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 916, normalized size of antiderivative = 11.04 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^2(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*(2*b*c*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1 
/2*pi*a*c + 4*e*x + 4*d)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sg 
n(F) - pi*b*c + 8*e)^2) + (pi*b*c*sgn(F) - pi*b*c + 8*e)*sin(1/2*pi*b*c*x* 
sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c + 4*e*x + 4*d)/(4*b 
^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 8*e)^2))*e^(b*c*x*log(abs 
(F)) + a*c*log(abs(F))) - 1/8*(2*b*c*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c* 
x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c - 4*e*x - 4*d)*log(abs(F))/(4*b^2*c^2*l 
og(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 8*e)^2) + (pi*b*c*sgn(F) - pi*b*c 
 - 8*e)*sin(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*p 
i*a*c - 4*e*x - 4*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 
8*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 1/4*(2*b*c*cos(-1/2*pi* 
b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)*log(abs(F))/ 
(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2) - (pi*b*c*sgn(F) - 
pi*b*c)*sin(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2* 
pi*a*c)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2))*e^(b*c*x*l 
og(abs(F)) + a*c*log(abs(F))) + I*(-I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi* 
b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c + 4*I*e*x + 4*I*d)/(16*I*pi*b*c 
*sgn(F) - 16*I*pi*b*c + 32*b*c*log(abs(F)) + 128*I*e) + I*e^(-1/2*I*pi*b*c 
*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c - 4*I*e*x 
- 4*I*d)/(-16*I*pi*b*c*sgn(F) + 16*I*pi*b*c + 32*b*c*log(abs(F)) - 128*...
 

Mupad [B] (verification not implemented)

Time = 16.60 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^2(d+e x) \, dx=\frac {F^{c\,\left (a+b\,x\right )}\,\left (16\,e^2+b^2\,c^2\,{\ln \left (F\right )}^2-b^2\,c^2\,{\ln \left (F\right )}^2\,\cos \left (4\,d+4\,e\,x\right )-4\,b\,c\,e\,\ln \left (F\right )\,\sin \left (4\,d+4\,e\,x\right )\right )}{8\,b\,c\,\ln \left (F\right )\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+16\,e^2\right )} \] Input:

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

Output:

(F^(c*(a + b*x))*(16*e^2 + b^2*c^2*log(F)^2 - b^2*c^2*log(F)^2*cos(4*d + 4 
*e*x) - 4*b*c*e*log(F)*sin(4*d + 4*e*x)))/(8*b*c*log(F)*(16*e^2 + b^2*c^2* 
log(F)^2))
 

Reduce [F]

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

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

Output:

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