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

Optimal result

Integrand size = 18, antiderivative size = 199 \[ \int F^{c (a+b x)} \sin ^3(d+e x) \, dx=-\frac {6 e^3 F^{c (a+b x)} \cos (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}+\frac {6 b c e^2 F^{c (a+b x)} \log (F) \sin (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}-\frac {3 e F^{c (a+b x)} \cos (d+e x) \sin ^2(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^3(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)} \] Output:

-6*e^3*F^(c*(b*x+a))*cos(e*x+d)/(9*e^4+10*b^2*c^2*e^2*ln(F)^2+b^4*c^4*ln(F 
)^4)+6*b*c*e^2*F^(c*(b*x+a))*ln(F)*sin(e*x+d)/(9*e^4+10*b^2*c^2*e^2*ln(F)^ 
2+b^4*c^4*ln(F)^4)-3*e*F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d)^2/(9*e^2+b^2*c^ 
2*ln(F)^2)+b*c*F^(c*(b*x+a))*ln(F)*sin(e*x+d)^3/(9*e^2+b^2*c^2*ln(F)^2)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.77 \[ \int F^{c (a+b x)} \sin ^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (-3 e \cos (d+e x) \left (9 e^2+b^2 c^2 \log ^2(F)\right )+3 \cos (3 (d+e x)) \left (e^3+b^2 c^2 e \log ^2(F)\right )-2 b c \log (F) \left (-13 e^2-b^2 c^2 \log ^2(F)+\cos (2 (d+e x)) \left (e^2+b^2 c^2 \log ^2(F)\right )\right ) \sin (d+e x)\right )}{4 \left (9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)\right )} \] Input:

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

Output:

(F^(c*(a + b*x))*(-3*e*Cos[d + e*x]*(9*e^2 + b^2*c^2*Log[F]^2) + 3*Cos[3*( 
d + e*x)]*(e^3 + b^2*c^2*e*Log[F]^2) - 2*b*c*Log[F]*(-13*e^2 - b^2*c^2*Log 
[F]^2 + Cos[2*(d + e*x)]*(e^2 + b^2*c^2*Log[F]^2))*Sin[d + e*x]))/(4*(9*e^ 
4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4934, 4932}

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.

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

\(\Big \downarrow \) 4934

\(\displaystyle \frac {6 e^2 \int F^{c (a+b x)} \sin (d+e x)dx}{b^2 c^2 \log ^2(F)+9 e^2}+\frac {b c \log (F) \sin ^3(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}-\frac {3 e \sin ^2(d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}\)

\(\Big \downarrow \) 4932

\(\displaystyle \frac {b c \log (F) \sin ^3(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}-\frac {3 e \sin ^2(d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}+\frac {6 e^2 \left (\frac {b c \log (F) \sin (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}-\frac {e \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}\right )}{b^2 c^2 \log ^2(F)+9 e^2}\)

Input:

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

Output:

(-3*e*F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x]^2)/(9*e^2 + b^2*c^2*Log[F] 
^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x]^3)/(9*e^2 + b^2*c^2*Log[F]^ 
2) + (6*e^2*(-((e*F^(c*(a + b*x))*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)) 
+ (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)))/(9* 
e^2 + b^2*c^2*Log[F]^2)
 

Defintions of rubi rules used

rule 4932

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> 
 Simp[b*c*Log[F]*F^(c*(a + b*x))*(Sin[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x 
] - Simp[e*F^(c*(a + b*x))*(Cos[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] /; F 
reeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]
 

rule 4934

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbo 
l] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Sin[d + e*x]^n/(e^2*n^2 + b^2*c^2*Lo 
g[F]^2)), x] + (-Simp[e*n*F^(c*(a + b*x))*Cos[d + e*x]*(Sin[d + e*x]^(n - 1 
)/(e^2*n^2 + b^2*c^2*Log[F]^2)), x] + Simp[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c 
^2*Log[F]^2)   Int[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x]) /; FreeQ[{ 
F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[n, 1]
 

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.72

method	result	size
parallelrisch	\(\frac {3 \left (\left (b^{2} c^{2} \ln \left (F \right )^{2} e +e^{3}\right ) \cos \left (3 e x +3 d \right )-\frac {b c \ln \left (F \right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) \sin \left (3 e x +3 d \right )}{3}+\left (9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) \left (b c \ln \left (F \right ) \sin \left (e x +d \right )-e \cos \left (e x +d \right )\right )\right ) F^{c \left (b x +a \right )}}{4 b^{4} c^{4} \ln \left (F \right )^{4}+40 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+36 e^{4}}\)	\(143\)
risch	\(-\frac {3 e \,F^{c \left (b x +a \right )} \cos \left (e x +d \right )}{4 \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {3 b c \,F^{c \left (b x +a \right )} \ln \left (F \right ) \sin \left (e x +d \right )}{4 \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {3 e \,F^{c \left (b x +a \right )} \cos \left (3 e x +3 d \right )}{4 \left (9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {c b \ln \left (F \right ) F^{c \left (b x +a \right )} \sin \left (3 e x +3 d \right )}{4 \left (9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(158\)
norman	\(\frac {-\frac {6 e^{3} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {6 e^{3} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{6}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {6 e \left (2 b^{2} c^{2} \ln \left (F \right )^{2}+3 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {6 e \left (2 b^{2} c^{2} \ln \left (F \right )^{2}+3 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{4}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {12 b c \ln \left (F \right ) e^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {12 b c \ln \left (F \right ) e^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{5}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {8 \ln \left (F \right ) b c \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{3}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}}{\left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}\right )^{3}}\)	\(483\)
orering	\(\frac {4 \ln \left (F \right ) b c \left (b^{2} c^{2} \ln \left (F \right )^{2}+5 e^{2}\right ) F^{c \left (b x +a \right )} \sin \left (e x +d \right )^{3}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {2 \left (3 b^{2} c^{2} \ln \left (F \right )^{2}+5 e^{2}\right ) \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \sin \left (e x +d \right )^{3}+3 F^{c \left (b x +a \right )} \sin \left (e x +d \right )^{2} e \cos \left (e x +d \right )\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {4 b c \ln \left (F \right ) \left (F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \sin \left (e x +d \right )^{3}+6 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \sin \left (e x +d \right )^{2} e \cos \left (e x +d \right )+6 F^{c \left (b x +a \right )} \sin \left (e x +d \right ) e^{2} \cos \left (e x +d \right )^{2}-3 F^{c \left (b x +a \right )} \sin \left (e x +d \right )^{3} e^{2}\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {F^{c \left (b x +a \right )} b^{3} c^{3} \ln \left (F \right )^{3} \sin \left (e x +d \right )^{3}+9 F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \sin \left (e x +d \right )^{2} e \cos \left (e x +d \right )+18 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \sin \left (e x +d \right ) e^{2} \cos \left (e x +d \right )^{2}-9 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \sin \left (e x +d \right )^{3} e^{2}+6 F^{c \left (b x +a \right )} e^{3} \cos \left (e x +d \right )^{3}-21 F^{c \left (b x +a \right )} \sin \left (e x +d \right )^{2} e^{3} \cos \left (e x +d \right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}\)	\(537\)
default	\(\frac {F^{a c} \left (\frac {-\frac {4 e \,{\mathrm e}^{b c x \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 e \,{\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {8 b c \ln \left (F \right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}+\frac {\frac {4 e \left (b^{2} c^{2} \ln \left (F \right )^{2}+3 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {4 e \left (11 b^{2} c^{2} \ln \left (F \right )^{2}+9 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {4 e \left (11 b^{2} c^{2} \ln \left (F \right )^{2}+9 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{4}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {4 e \left (b^{2} c^{2} \ln \left (F \right )^{2}+3 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{6}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {8 b c \ln \left (F \right ) \left (b^{2} c^{2} \ln \left (F \right )^{2}+3 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {8 b c \ln \left (F \right ) \left (b^{2} c^{2} \ln \left (F \right )^{2}+3 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{5}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {16 \ln \left (F \right ) b c \left (b^{2} c^{2} \ln \left (F \right )^{2}-e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{3}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}}{\left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}\right )^{3}}\right )}{4}\)	\(649\)

Input:

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

Output:

3*((b^2*c^2*ln(F)^2*e+e^3)*cos(3*e*x+3*d)-1/3*b*c*ln(F)*(e^2+b^2*c^2*ln(F) 
^2)*sin(3*e*x+3*d)+(9*e^2+b^2*c^2*ln(F)^2)*(b*c*ln(F)*sin(e*x+d)-e*cos(e*x 
+d)))*F^(c*(b*x+a))/(4*b^4*c^4*ln(F)^4+40*b^2*c^2*e^2*ln(F)^2+36*e^4)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.86 \[ \int F^{c (a+b x)} \sin ^3(d+e x) \, dx=\frac {{\left (3 \, e^{3} \cos \left (e x + d\right )^{3} - 9 \, e^{3} \cos \left (e x + d\right ) + 3 \, {\left (b^{2} c^{2} e \cos \left (e x + d\right )^{3} - b^{2} c^{2} e \cos \left (e x + d\right )\right )} \log \left (F\right )^{2} - {\left ({\left (b^{3} c^{3} \cos \left (e x + d\right )^{2} - b^{3} c^{3}\right )} \log \left (F\right )^{3} + {\left (b c e^{2} \cos \left (e x + d\right )^{2} - 7 \, b c e^{2}\right )} \log \left (F\right )\right )} \sin \left (e x + d\right )\right )} F^{b c x + a c}}{b^{4} c^{4} \log \left (F\right )^{4} + 10 \, b^{2} c^{2} e^{2} \log \left (F\right )^{2} + 9 \, e^{4}} \] Input:

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

Output:

(3*e^3*cos(e*x + d)^3 - 9*e^3*cos(e*x + d) + 3*(b^2*c^2*e*cos(e*x + d)^3 - 
 b^2*c^2*e*cos(e*x + d))*log(F)^2 - ((b^3*c^3*cos(e*x + d)^2 - b^3*c^3)*lo 
g(F)^3 + (b*c*e^2*cos(e*x + d)^2 - 7*b*c*e^2)*log(F))*sin(e*x + d))*F^(b*c 
*x + a*c)/(b^4*c^4*log(F)^4 + 10*b^2*c^2*e^2*log(F)^2 + 9*e^4)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.83 (sec) , antiderivative size = 1579, normalized size of antiderivative = 7.93 \[ \int F^{c (a+b x)} \sin ^3(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((x*sin(d)**3, Eq(F, 1) & Eq(e, 0)), (F**(a*c)*x*sin(d)**3, Eq(b, 
 0) & Eq(e, 0)), (x*sin(d)**3, Eq(c, 0) & Eq(e, 0)), (-3*F**(a*c + b*c*x)* 
x*sin(I*b*c*x*log(F) - d)**3/8 + 3*I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F) 
 - d)**2*cos(I*b*c*x*log(F) - d)/8 - 3*F**(a*c + b*c*x)*x*sin(I*b*c*x*log( 
F) - d)*cos(I*b*c*x*log(F) - d)**2/8 + 3*I*F**(a*c + b*c*x)*x*cos(I*b*c*x* 
log(F) - d)**3/8 + F**(a*c + b*c*x)*sin(I*b*c*x*log(F) - d)**3/(8*b*c*log( 
F)) - 3*I*F**(a*c + b*c*x)*sin(I*b*c*x*log(F) - d)**2*cos(I*b*c*x*log(F) - 
 d)/(4*b*c*log(F)) - 3*I*F**(a*c + b*c*x)*cos(I*b*c*x*log(F) - d)**3/(8*b* 
c*log(F)), Eq(e, -I*b*c*log(F))), (-F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/ 
3 - d)**3/8 + 3*I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/3 - d)**2*cos(I*b* 
c*x*log(F)/3 - d)/8 + 3*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/3 - d)*cos(I 
*b*c*x*log(F)/3 - d)**2/8 - I*F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F)/3 - d) 
**3/8 - 9*F**(a*c + b*c*x)*sin(I*b*c*x*log(F)/3 - d)**3/(8*b*c*log(F)) + 3 
*I*F**(a*c + b*c*x)*sin(I*b*c*x*log(F)/3 - d)**2*cos(I*b*c*x*log(F)/3 - d) 
/(4*b*c*log(F)) + I*F**(a*c + b*c*x)*cos(I*b*c*x*log(F)/3 - d)**3/(8*b*c*l 
og(F)), Eq(e, -I*b*c*log(F)/3)), (F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/3 
+ d)**3/8 - 3*I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/3 + d)**2*cos(I*b*c* 
x*log(F)/3 + d)/8 - 3*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/3 + d)*cos(I*b 
*c*x*log(F)/3 + d)**2/8 + I*F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F)/3 + d)** 
3/8 + 9*F**(a*c + b*c*x)*sin(I*b*c*x*log(F)/3 + d)**3/(8*b*c*log(F)) - ...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 813 vs. \(2 (199) = 398\).

Time = 0.13 (sec) , antiderivative size = 813, normalized size of antiderivative = 4.09 \[ \int F^{c (a+b x)} \sin ^3(d+e x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/8*((F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) - 3*F^(a*c)*b^2*c^2*e*cos(3*d)*lo 
g(F)^2 + F^(a*c)*b*c*e^2*log(F)*sin(3*d) - 3*F^(a*c)*e^3*cos(3*d))*F^(b*c* 
x)*cos(3*e*x) - (F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) + 3*F^(a*c)*b^2*c^2*e*c 
os(3*d)*log(F)^2 + F^(a*c)*b*c*e^2*log(F)*sin(3*d) + 3*F^(a*c)*e^3*cos(3*d 
))*F^(b*c*x)*cos(3*e*x + 6*d) + 3*(F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) + F^( 
a*c)*b^2*c^2*e*cos(3*d)*log(F)^2 + 9*F^(a*c)*b*c*e^2*log(F)*sin(3*d) + 9*F 
^(a*c)*e^3*cos(3*d))*F^(b*c*x)*cos(e*x + 4*d) - 3*(F^(a*c)*b^3*c^3*log(F)^ 
3*sin(3*d) - F^(a*c)*b^2*c^2*e*cos(3*d)*log(F)^2 + 9*F^(a*c)*b*c*e^2*log(F 
)*sin(3*d) - 9*F^(a*c)*e^3*cos(3*d))*F^(b*c*x)*cos(e*x - 2*d) + (F^(a*c)*b 
^3*c^3*cos(3*d)*log(F)^3 + 3*F^(a*c)*b^2*c^2*e*log(F)^2*sin(3*d) + F^(a*c) 
*b*c*e^2*cos(3*d)*log(F) + 3*F^(a*c)*e^3*sin(3*d))*F^(b*c*x)*sin(3*e*x) + 
(F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 - 3*F^(a*c)*b^2*c^2*e*log(F)^2*sin(3*d) 
 + F^(a*c)*b*c*e^2*cos(3*d)*log(F) - 3*F^(a*c)*e^3*sin(3*d))*F^(b*c*x)*sin 
(3*e*x + 6*d) - 3*(F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 - F^(a*c)*b^2*c^2*e*l 
og(F)^2*sin(3*d) + 9*F^(a*c)*b*c*e^2*cos(3*d)*log(F) - 9*F^(a*c)*e^3*sin(3 
*d))*F^(b*c*x)*sin(e*x + 4*d) - 3*(F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 + F^( 
a*c)*b^2*c^2*e*log(F)^2*sin(3*d) + 9*F^(a*c)*b*c*e^2*cos(3*d)*log(F) + 9*F 
^(a*c)*e^3*sin(3*d))*F^(b*c*x)*sin(e*x - 2*d))/(b^4*c^4*cos(3*d)^2*log(F)^ 
4 + b^4*c^4*log(F)^4*sin(3*d)^2 + 9*(cos(3*d)^2 + sin(3*d)^2)*e^4 + 10*(b^ 
2*c^2*cos(3*d)^2*log(F)^2 + b^2*c^2*log(F)^2*sin(3*d)^2)*e^2)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 1275, normalized size of antiderivative = 6.41 \[ \int F^{c (a+b x)} \sin ^3(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(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*sg 
n(F) - pi*b*c + 6*e)^2) - (pi*b*c*sgn(F) - pi*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))) + 3/4*(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(a 
bs(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))) - 3/4*(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*sg 
n(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*s 
gn(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))) + 1/4*(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) - pi*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(a...
 

Mupad [B] (verification not implemented)

Time = 20.05 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.95 \[ \int F^{c (a+b x)} \sin ^3(d+e x) \, dx=-\frac {3\,F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (e\,x\right )-\sin \left (e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )}{8\,\left (e+b\,c\,\ln \left (F\right )\,1{}\mathrm {i}\right )}+\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (3\,e\,x\right )+\sin \left (3\,e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,\left (b\,c\,\ln \left (F\right )+e\,3{}\mathrm {i}\right )}+\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (3\,e\,x\right )-\sin \left (3\,e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )}{8\,\left (3\,e+b\,c\,\ln \left (F\right )\,1{}\mathrm {i}\right )}-\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (e\,x\right )+\sin \left (e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8\,\left (b\,c\,\ln \left (F\right )+e\,1{}\mathrm {i}\right )} \] Input:

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

Output:

(F^(c*(a + b*x))*(cos(3*e*x) + sin(3*e*x)*1i)*(cos(3*d) + sin(3*d)*1i)*1i) 
/(8*(e*3i + b*c*log(F))) - (3*F^(c*(a + b*x))*(cos(e*x) - sin(e*x)*1i)*(co 
s(d) - sin(d)*1i))/(8*(e + b*c*log(F)*1i)) + (F^(c*(a + b*x))*(cos(3*e*x) 
- sin(3*e*x)*1i)*(cos(3*d) - sin(3*d)*1i))/(8*(3*e + b*c*log(F)*1i)) - (F^ 
(c*(a + b*x))*(cos(e*x) + sin(e*x)*1i)*(cos(d) + sin(d)*1i)*3i)/(8*(e*1i + 
 b*c*log(F)))
 

Reduce [F]

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

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

Output:

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