\(\int F^{c (a+b x)} \cos ^2(d+e x) \sin ^3(d+e x) \, dx\) [94]

Optimal result

Integrand size = 26, antiderivative size = 172 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^3(d+e x) \, dx=-\frac {F^{c (a+b x)} (e \cos (d+e x)-b c \log (F) \sin (d+e x))}{8 \left (e^2+b^2 c^2 \log ^2(F)\right )}-\frac {F^{c (a+b x)} (3 e \cos (3 d+3 e x)-b c \log (F) \sin (3 d+3 e x))}{16 \left (9 e^2+b^2 c^2 \log ^2(F)\right )}+\frac {F^{c (a+b x)} (5 e \cos (5 d+5 e x)-b c \log (F) \sin (5 d+5 e x))}{16 \left (25 e^2+b^2 c^2 \log ^2(F)\right )} \] Output:

-1/8*F^(c*(b*x+a))*(e*cos(e*x+d)-b*c*ln(F)*sin(e*x+d))/(e^2+b^2*c^2*ln(F)^ 
2)-F^(c*(b*x+a))*(3*e*cos(3*e*x+3*d)-b*c*ln(F)*sin(3*e*x+3*d))/(144*e^2+16 
*b^2*c^2*ln(F)^2)+F^(c*(b*x+a))*(5*e*cos(5*e*x+5*d)-b*c*ln(F)*sin(5*e*x+5* 
d))/(400*e^2+16*b^2*c^2*ln(F)^2)
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.84 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^3(d+e x) \, dx=\frac {1}{16} F^{c (a+b x)} \left (\frac {2 (-e \cos (d+e x)+b c \log (F) \sin (d+e x))}{e^2+b^2 c^2 \log ^2(F)}+\frac {-3 e \cos (3 (d+e x))+b c \log (F) \sin (3 (d+e x))}{9 e^2+b^2 c^2 \log ^2(F)}+\frac {5 e \cos (5 (d+e x))-b c \log (F) \sin (5 (d+e x))}{25 e^2+b^2 c^2 \log ^2(F)}\right ) \] Input:

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

Output:

(F^(c*(a + b*x))*((2*(-(e*Cos[d + e*x]) + b*c*Log[F]*Sin[d + e*x]))/(e^2 + 
 b^2*c^2*Log[F]^2) + (-3*e*Cos[3*(d + e*x)] + b*c*Log[F]*Sin[3*(d + e*x)]) 
/(9*e^2 + b^2*c^2*Log[F]^2) + (5*e*Cos[5*(d + e*x)] - b*c*Log[F]*Sin[5*(d 
+ e*x)])/(25*e^2 + b^2*c^2*Log[F]^2)))/16
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.47, 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.

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

\(\Big \downarrow \) 4972

\(\displaystyle \int \left (\frac {1}{8} \sin (d+e x) F^{c (a+b x)}+\frac {1}{16} \sin (3 d+3 e x) F^{c (a+b x)}-\frac {1}{16} \sin (5 d+5 e x) F^{c (a+b x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b c \log (F) \sin (d+e x) F^{c (a+b x)}}{8 \left (b^2 c^2 \log ^2(F)+e^2\right )}+\frac {b c \log (F) \sin (3 d+3 e x) F^{c (a+b x)}}{16 \left (b^2 c^2 \log ^2(F)+9 e^2\right )}-\frac {b c \log (F) \sin (5 d+5 e x) F^{c (a+b x)}}{16 \left (b^2 c^2 \log ^2(F)+25 e^2\right )}-\frac {e \cos (d+e x) F^{c (a+b x)}}{8 \left (b^2 c^2 \log ^2(F)+e^2\right )}-\frac {3 e \cos (3 d+3 e x) F^{c (a+b x)}}{16 \left (b^2 c^2 \log ^2(F)+9 e^2\right )}+\frac {5 e \cos (5 d+5 e x) F^{c (a+b x)}}{16 \left (b^2 c^2 \log ^2(F)+25 e^2\right )}\)

Input:

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

Output:

-1/8*(e*F^(c*(a + b*x))*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2) - (3*e*F^(c 
*(a + b*x))*Cos[3*d + 3*e*x])/(16*(9*e^2 + b^2*c^2*Log[F]^2)) + (5*e*F^(c* 
(a + b*x))*Cos[5*d + 5*e*x])/(16*(25*e^2 + b^2*c^2*Log[F]^2)) + (b*c*F^(c* 
(a + b*x))*Log[F]*Sin[d + e*x])/(8*(e^2 + b^2*c^2*Log[F]^2)) + (b*c*F^(c*( 
a + b*x))*Log[F]*Sin[3*d + 3*e*x])/(16*(9*e^2 + b^2*c^2*Log[F]^2)) - (b*c* 
F^(c*(a + b*x))*Log[F]*Sin[5*d + 5*e*x])/(16*(25*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 = 3.52 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.40

method	result	size
risch	\(-\frac {e \,F^{c \left (b x +a \right )} \cos \left (e x +d \right )}{8 \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {c b \ln \left (F \right ) F^{c \left (b x +a \right )} \sin \left (e x +d \right )}{8 e^{2}+8 b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {5 e \,F^{c \left (b x +a \right )} \cos \left (5 e x +5 d \right )}{16 \left (b^{2} c^{2} \ln \left (F \right )^{2}+25 e^{2}\right )}-\frac {c b \ln \left (F \right ) F^{c \left (b x +a \right )} \sin \left (5 e x +5 d \right )}{16 \left (b^{2} c^{2} \ln \left (F \right )^{2}+25 e^{2}\right )}-\frac {3 e \,F^{c \left (b x +a \right )} \cos \left (3 e x +3 d \right )}{16 \left (9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {c b \ln \left (F \right ) F^{c \left (b x +a \right )} \sin \left (3 e x +3 d \right )}{144 e^{2}+16 b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(241\)
parallelrisch	\(-\frac {3 \left (\left (\frac {\ln \left (F \right )^{5} b^{5} c^{5}}{3}+\frac {10 \ln \left (F \right )^{3} b^{3} c^{3} e^{2}}{3}+3 \ln \left (F \right ) b c \,e^{4}\right ) \sin \left (5 e x +5 d \right )+\left (-\frac {5 \ln \left (F \right )^{4} b^{4} c^{4} e}{3}-\frac {50 \ln \left (F \right )^{2} b^{2} c^{2} e^{3}}{3}-15 e^{5}\right ) \cos \left (5 e x +5 d \right )+\left (-\frac {b c \ln \left (F \right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) \sin \left (3 e x +3 d \right )}{3}+\left (b^{2} c^{2} \ln \left (F \right )^{2} e +e^{3}\right ) \cos \left (3 e x +3 d \right )-\frac {2 \left (9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) \left (b c \ln \left (F \right ) \sin \left (e x +d \right )-e \cos \left (e x +d \right )\right )}{3}\right ) \left (b^{2} c^{2} \ln \left (F \right )^{2}+25 e^{2}\right )\right ) F^{c \left (b x +a \right )}}{16 \ln \left (F \right )^{6} b^{6} c^{6}+560 \ln \left (F \right )^{4} b^{4} c^{4} e^{2}+4144 \ln \left (F \right )^{2} b^{2} c^{2} e^{4}+3600 e^{6}}\)	\(269\)
default	\(\frac {F^{a c} \left (\frac {-\frac {3 e \,{\mathrm e}^{b c x \ln \left (F \right )}}{9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {3 e \,{\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {3 e x}{2}+\frac {3 d}{2}\right )^{2}}{9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 b c \ln \left (F \right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {3 e x}{2}+\frac {3 d}{2}\right )}{9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {3 e x}{2}+\frac {3 d}{2}\right )^{2}}+\frac {-\frac {2 e \,{\mathrm e}^{b c x \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e \,{\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 b c \ln \left (F \right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}+\frac {\frac {5 e \,{\mathrm e}^{b c x \ln \left (F \right )}}{b^{2} c^{2} \ln \left (F \right )^{2}+25 e^{2}}-\frac {5 e \,{\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {5 e x}{2}+\frac {5 d}{2}\right )^{2}}{b^{2} c^{2} \ln \left (F \right )^{2}+25 e^{2}}-\frac {2 b c \ln \left (F \right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {5 e x}{2}+\frac {5 d}{2}\right )}{b^{2} c^{2} \ln \left (F \right )^{2}+25 e^{2}}}{1+\tan \left (\frac {5 e x}{2}+\frac {5 d}{2}\right )^{2}}\right )}{16}\)	\(384\)
orering	\(\text {Expression too large to display}\)	\(1889\)

Input:

int(F^(c*(b*x+a))*cos(e*x+d)^2*sin(e*x+d)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/8/(e^2+b^2*c^2*ln(F)^2)*e*F^(c*(b*x+a))*cos(e*x+d)+1/8/(e^2+b^2*c^2*ln( 
F)^2)*c*b*ln(F)*F^(c*(b*x+a))*sin(e*x+d)+5/16/(b^2*c^2*ln(F)^2+25*e^2)*e*F 
^(c*(b*x+a))*cos(5*e*x+5*d)-1/16/(b^2*c^2*ln(F)^2+25*e^2)*c*b*ln(F)*F^(c*( 
b*x+a))*sin(5*e*x+5*d)-3/16/(9*e^2+b^2*c^2*ln(F)^2)*e*F^(c*(b*x+a))*cos(3* 
e*x+3*d)+1/16/(9*e^2+b^2*c^2*ln(F)^2)*c*b*ln(F)*F^(c*(b*x+a))*sin(3*e*x+3* 
d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (164) = 328\).

Time = 0.21 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.02 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^3(d+e x) \, dx=\frac {{\left (45 \, e^{5} \cos \left (e x + d\right )^{5} - 75 \, e^{5} \cos \left (e x + d\right )^{3} + {\left (5 \, b^{4} c^{4} e \cos \left (e x + d\right )^{5} - 7 \, b^{4} c^{4} e \cos \left (e x + d\right )^{3} + 2 \, b^{4} c^{4} e \cos \left (e x + d\right )\right )} \log \left (F\right )^{4} + 2 \, {\left (25 \, b^{2} c^{2} e^{3} \cos \left (e x + d\right )^{5} - 41 \, b^{2} c^{2} e^{3} \cos \left (e x + d\right )^{3} + 13 \, b^{2} c^{2} e^{3} \cos \left (e x + d\right )\right )} \log \left (F\right )^{2} - {\left ({\left (b^{5} c^{5} \cos \left (e x + d\right )^{4} - b^{5} c^{5} \cos \left (e x + d\right )^{2}\right )} \log \left (F\right )^{5} + 2 \, {\left (5 \, b^{3} c^{3} e^{2} \cos \left (e x + d\right )^{4} - 7 \, b^{3} c^{3} e^{2} \cos \left (e x + d\right )^{2} - b^{3} c^{3} e^{2}\right )} \log \left (F\right )^{3} + {\left (9 \, b c e^{4} \cos \left (e x + d\right )^{4} - 13 \, b c e^{4} \cos \left (e x + d\right )^{2} - 26 \, b c e^{4}\right )} \log \left (F\right )\right )} \sin \left (e x + d\right )\right )} F^{b c x + a c}}{b^{6} c^{6} \log \left (F\right )^{6} + 35 \, b^{4} c^{4} e^{2} \log \left (F\right )^{4} + 259 \, b^{2} c^{2} e^{4} \log \left (F\right )^{2} + 225 \, e^{6}} \] Input:

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

Output:

(45*e^5*cos(e*x + d)^5 - 75*e^5*cos(e*x + d)^3 + (5*b^4*c^4*e*cos(e*x + d) 
^5 - 7*b^4*c^4*e*cos(e*x + d)^3 + 2*b^4*c^4*e*cos(e*x + d))*log(F)^4 + 2*( 
25*b^2*c^2*e^3*cos(e*x + d)^5 - 41*b^2*c^2*e^3*cos(e*x + d)^3 + 13*b^2*c^2 
*e^3*cos(e*x + d))*log(F)^2 - ((b^5*c^5*cos(e*x + d)^4 - b^5*c^5*cos(e*x + 
 d)^2)*log(F)^5 + 2*(5*b^3*c^3*e^2*cos(e*x + d)^4 - 7*b^3*c^3*e^2*cos(e*x 
+ d)^2 - b^3*c^3*e^2)*log(F)^3 + (9*b*c*e^4*cos(e*x + d)^4 - 13*b*c*e^4*co 
s(e*x + d)^2 - 26*b*c*e^4)*log(F))*sin(e*x + d))*F^(b*c*x + a*c)/(b^6*c^6* 
log(F)^6 + 35*b^4*c^4*e^2*log(F)^4 + 259*b^2*c^2*e^4*log(F)^2 + 225*e^6)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1800 vs. \(2 (164) = 328\).

Time = 0.14 (sec) , antiderivative size = 1800, normalized size of antiderivative = 10.47 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^3(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/32*((F^(a*c)*b^5*c^5*log(F)^5*sin(5*d) - 5*F^(a*c)*b^4*c^4*e*cos(5*d)*l 
og(F)^4 + 10*F^(a*c)*b^3*c^3*e^2*log(F)^3*sin(5*d) - 50*F^(a*c)*b^2*c^2*e^ 
3*cos(5*d)*log(F)^2 + 9*F^(a*c)*b*c*e^4*log(F)*sin(5*d) - 45*F^(a*c)*e^5*c 
os(5*d))*F^(b*c*x)*cos(5*e*x) - (F^(a*c)*b^5*c^5*log(F)^5*sin(5*d) + 5*F^( 
a*c)*b^4*c^4*e*cos(5*d)*log(F)^4 + 10*F^(a*c)*b^3*c^3*e^2*log(F)^3*sin(5*d 
) + 50*F^(a*c)*b^2*c^2*e^3*cos(5*d)*log(F)^2 + 9*F^(a*c)*b*c*e^4*log(F)*si 
n(5*d) + 45*F^(a*c)*e^5*cos(5*d))*F^(b*c*x)*cos(5*e*x + 10*d) + (F^(a*c)*b 
^5*c^5*log(F)^5*sin(5*d) + 3*F^(a*c)*b^4*c^4*e*cos(5*d)*log(F)^4 + 26*F^(a 
*c)*b^3*c^3*e^2*log(F)^3*sin(5*d) + 78*F^(a*c)*b^2*c^2*e^3*cos(5*d)*log(F) 
^2 + 25*F^(a*c)*b*c*e^4*log(F)*sin(5*d) + 75*F^(a*c)*e^5*cos(5*d))*F^(b*c* 
x)*cos(3*e*x + 8*d) - (F^(a*c)*b^5*c^5*log(F)^5*sin(5*d) - 3*F^(a*c)*b^4*c 
^4*e*cos(5*d)*log(F)^4 + 26*F^(a*c)*b^3*c^3*e^2*log(F)^3*sin(5*d) - 78*F^( 
a*c)*b^2*c^2*e^3*cos(5*d)*log(F)^2 + 25*F^(a*c)*b*c*e^4*log(F)*sin(5*d) - 
75*F^(a*c)*e^5*cos(5*d))*F^(b*c*x)*cos(3*e*x - 2*d) + 2*(F^(a*c)*b^5*c^5*l 
og(F)^5*sin(5*d) + F^(a*c)*b^4*c^4*e*cos(5*d)*log(F)^4 + 34*F^(a*c)*b^3*c^ 
3*e^2*log(F)^3*sin(5*d) + 34*F^(a*c)*b^2*c^2*e^3*cos(5*d)*log(F)^2 + 225*F 
^(a*c)*b*c*e^4*log(F)*sin(5*d) + 225*F^(a*c)*e^5*cos(5*d))*F^(b*c*x)*cos(e 
*x + 6*d) - 2*(F^(a*c)*b^5*c^5*log(F)^5*sin(5*d) - F^(a*c)*b^4*c^4*e*cos(5 
*d)*log(F)^4 + 34*F^(a*c)*b^3*c^3*e^2*log(F)^3*sin(5*d) - 34*F^(a*c)*b^2*c 
^2*e^3*cos(5*d)*log(F)^2 + 225*F^(a*c)*b*c*e^4*log(F)*sin(5*d) - 225*F^...
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 1915, normalized size of antiderivative = 11.13 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^3(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/16*(2*b*c*log(abs(F))*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 + 5*e*x + 5*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*s 
gn(F) - pi*b*c + 10*e)^2) - (pi*b*c*sgn(F) - pi*b*c + 10*e)*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 + 5*e*x + 5*d)/( 
4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 10*e)^2))*e^(b*c*x*log 
(abs(F)) + a*c*log(abs(F))) + 1/16*(2*b*c*log(abs(F))*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 + 3*e*x + 3*d)/(4*b^2* 
c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 6*e)^2) - (pi*b*c*sgn(F) - p 
i*b*c + 6*e)*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 + 3*e*x + 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b 
*c + 6*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 1/8*(2*b*c*log(abs 
(F))*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 + e*x + d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 2*e)^2) 
 - (pi*b*c*sgn(F) - pi*b*c + 2*e)*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 + e*x + d)/(4*b^2*c^2*log(abs(F))^2 + (pi* 
b*c*sgn(F) - pi*b*c + 2*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - 1 
/8*(2*b*c*log(abs(F))*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 - e*x - d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - 
 pi*b*c - 2*e)^2) - (pi*b*c*sgn(F) - pi*b*c - 2*e)*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 - e*x - d)/(4*b^2*c^2*...
 

Mupad [B] (verification not implemented)

Time = 17.04 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.69 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \sin ^3(d+e x) \, dx=-\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (e\,x\right )-\sin \left (e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )}{16\,\left (e+b\,c\,\ln \left (F\right )\,1{}\mathrm {i}\right )}-\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (3\,e\,x\right )+\sin \left (3\,e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (b\,c\,\ln \left (F\right )+e\,3{}\mathrm {i}\right )}-\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (3\,e\,x\right )-\sin \left (3\,e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )}{32\,\left (3\,e+b\,c\,\ln \left (F\right )\,1{}\mathrm {i}\right )}+\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (5\,e\,x\right )+\sin \left (5\,e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,d\right )+\sin \left (5\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (b\,c\,\ln \left (F\right )+e\,5{}\mathrm {i}\right )}+\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (5\,e\,x\right )-\sin \left (5\,e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,d\right )-\sin \left (5\,d\right )\,1{}\mathrm {i}\right )}{32\,\left (5\,e+b\,c\,\ln \left (F\right )\,1{}\mathrm {i}\right )}-\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (e\,x\right )+\sin \left (e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,\left (b\,c\,\ln \left (F\right )+e\,1{}\mathrm {i}\right )} \] Input:

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

Output:

(F^(c*(a + b*x))*(cos(5*e*x) + sin(5*e*x)*1i)*(cos(5*d) + sin(5*d)*1i)*1i) 
/(32*(e*5i + b*c*log(F))) - (F^(c*(a + b*x))*(cos(3*e*x) + sin(3*e*x)*1i)* 
(cos(3*d) + sin(3*d)*1i)*1i)/(32*(e*3i + b*c*log(F))) - (F^(c*(a + b*x))*( 
cos(3*e*x) - sin(3*e*x)*1i)*(cos(3*d) - sin(3*d)*1i))/(32*(3*e + b*c*log(F 
)*1i)) - (F^(c*(a + b*x))*(cos(e*x) - sin(e*x)*1i)*(cos(d) - sin(d)*1i))/( 
16*(e + b*c*log(F)*1i)) + (F^(c*(a + b*x))*(cos(5*e*x) - sin(5*e*x)*1i)*(c 
os(5*d) - sin(5*d)*1i))/(32*(5*e + b*c*log(F)*1i)) - (F^(c*(a + b*x))*(cos 
(e*x) + sin(e*x)*1i)*(cos(d) + sin(d)*1i)*1i)/(16*(e*1i + b*c*log(F)))
 

Reduce [F]

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

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

Output:

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