3.1.28 \(\int F^{c (a+b x)} (f x)^m \sin (d+e x) \, dx\) [28]

3.1.28.1 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*(b*c*ln 
(F)+I*e))/(e-I*b*c*ln(F))/((-x*(b*c*ln(F)+I*e))^m)
 

3.1.28.2 Mathematica [A] (verified)

Time = 0.59 (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)
 

3.1.28.3 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
 

3.1.28.3.1 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]
 

3.1.28.4 Maple [F]

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

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)
 

3.1.28.5 Fricas [A] (verification not implemented)

Time = 0.10 (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)
 

3.1.28.6 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)
 

3.1.28.7 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)
 

3.1.28.8 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)
 

3.1.28.9 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)