\(\int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx\) [40]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=-\frac {e^{-i d} F^{a c} (f x)^m \Gamma (1+m,x (i e-b c \log (F))) (x (i e-b c \log (F)))^{-m}}{2 (e+i b c \log (F))}-\frac {e^{i d} F^{a c} (f x)^m \Gamma (1+m,-x (i e+b c \log (F))) (-x (i e+b c \log (F)))^{-m}}{2 (e-i b c \log (F))} \] Output:

-1/2*F^(a*c)*(f*x)^m*GAMMA(1+m,x*(I*e-b*c*ln(F)))/exp(I*d)/((x*(I*e-b*c*ln 
(F)))^m)/(e+I*b*c*ln(F))-1/2*exp(I*d)*F^(a*c)*(f*x)^m*GAMMA(1+m,-x*(I*e+b* 
c*ln(F)))/(e-I*b*c*ln(F))/((-x*(I*e+b*c*ln(F)))^m)
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03 \[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\frac {1}{2} F^{a c} (f x)^m (x (-i e-b c \log (F)))^{-m} \left (-i x \Gamma (1+m,i e x-b c x \log (F)) (i x (e+i b c \log (F)))^{-1-m} (-i e x-b c x \log (F))^m (\cos (d)-i \sin (d))-\frac {\Gamma (1+m,-i e x-b c x \log (F)) (\cos (d)+i \sin (d))}{e-i b c \log (F)}\right ) \] Input:

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

Output:

(F^(a*c)*(f*x)^m*((-I)*x*Gamma[1 + m, I*e*x - b*c*x*Log[F]]*(I*x*(e + I*b* 
c*Log[F]))^(-1 - m)*((-I)*e*x - b*c*x*Log[F])^m*(Cos[d] - I*Sin[d]) - (Gam 
ma[1 + m, (-I)*e*x - b*c*x*Log[F]]*(Cos[d] + I*Sin[d]))/(e - I*b*c*Log[F]) 
))/(2*(x*((-I)*e - b*c*Log[F]))^m)
 

Rubi [F]

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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.94 \[ \int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx=\frac {{\left (i \, b c \log \left (F\right ) - e\right )} e^{\left (a c \log \left (F\right ) - m \log \left (-\frac {b c \log \left (F\right ) - i \, e}{f}\right ) - i \, d\right )} \Gamma \left (m + 1, -b c x \log \left (F\right ) + i \, e x\right ) + {\left (-i \, b c \log \left (F\right ) - e\right )} e^{\left (a c \log \left (F\right ) - m \log \left (-\frac {b c \log \left (F\right ) + i \, e}{f}\right ) + i \, d\right )} \Gamma \left (m + 1, -b c x \log \left (F\right ) - i \, e x\right )}{2 \, {\left (b^{2} c^{2} \log \left (F\right )^{2} + e^{2}\right )}} \] Input:

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

Output:

1/2*((I*b*c*log(F) - e)*e^(a*c*log(F) - m*log(-(b*c*log(F) - I*e)/f) - I*d 
)*gamma(m + 1, -b*c*x*log(F) + I*e*x) + (-I*b*c*log(F) - e)*e^(a*c*log(F) 
- m*log(-(b*c*log(F) + I*e)/f) + I*d)*gamma(m + 1, -b*c*x*log(F) - I*e*x)) 
/(b^2*c^2*log(F)^2 + e^2)
 

Sympy [F]

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

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

Output:

Integral(F**(c*(a + b*x))*(f*x)**m*sin(d + e*x), x)
 

Maxima [F]

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

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

Output:

integrate((f*x)^m*F^((b*x + a)*c)*sin(e*x + d), x)
 

Giac [F]

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

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

Output:

integrate((f*x)^m*F^((b*x + a)*c)*sin(e*x + d), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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