\(\int F^{c (a+b x)} (g \cos (d+\frac {i b c x \log (F)}{2+n})+f \sin (d+\frac {i b c x \log (F)}{2+n}))^n \, dx\) [34]

Optimal result

Integrand size = 51, antiderivative size = 163 \[ \int F^{c (a+b x)} \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )^n \, dx=\frac {F^{c (a+b x)} (2+n) \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )^{1+n} \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )-i \left (f \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )-g \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )\right )}{b c \left (f^2+g^2\right ) (1+n) \log (F)} \] Output:

F^(c*(b*x+a))*(2+n)*(g*cos(d+I*b*c*x*ln(F)/(2+n))+f*sin(d+I*b*c*x*ln(F)/(2 
+n)))^(1+n)*(g*cos(d+I*b*c*x*ln(F)/(2+n))+f*sin(d+I*b*c*x*ln(F)/(2+n))-I*( 
f*cos(d+I*b*c*x*ln(F)/(2+n))-g*sin(d+I*b*c*x*ln(F)/(2+n))))/b/c/(f^2+g^2)/ 
(1+n)/ln(F)
 

Mathematica [A] (warning: unable to verify)

Time = 6.09 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.01 \[ \int F^{c (a+b x)} \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )^n \, dx=\frac {F^{c \left (a+\frac {b n x}{2+n}\right )} (2+n) \left (F^{-\frac {b c x}{2+n}} (f+i g) (-i \cos (d)+\sin (d))+F^{\frac {b c x}{2+n}} (f-i g) (i \cos (d)+\sin (d))\right )^n \left (F^{\frac {2 b c x}{2+n}} (f-i g)-(f+i g) \cos (2 d)+(-i f+g) \sin (2 d)\right ) (\cosh (n \log (2))-\sinh (n \log (2)))}{2 b c (f-i g) (1+n) \log (F)} \] Input:

Integrate[F^(c*(a + b*x))*(g*Cos[d + (I*b*c*x*Log[F])/(2 + n)] + f*Sin[d + 
 (I*b*c*x*Log[F])/(2 + n)])^n,x]
 

Output:

(F^(c*(a + (b*n*x)/(2 + n)))*(2 + n)*(((f + I*g)*((-I)*Cos[d] + Sin[d]))/F 
^((b*c*x)/(2 + n)) + F^((b*c*x)/(2 + n))*(f - I*g)*(I*Cos[d] + Sin[d]))^n* 
(F^((2*b*c*x)/(2 + n))*(f - I*g) - (f + I*g)*Cos[2*d] + ((-I)*f + g)*Sin[2 
*d])*(Cosh[n*Log[2]] - Sinh[n*Log[2]]))/(2*b*c*(f - I*g)*(1 + n)*Log[F])
 

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.

\(\displaystyle \int F^{c (a+b x)} \left (f \sin \left (d+\frac {i b c x \log (F)}{n+2}\right )+g \cos \left (d+\frac {i b c x \log (F)}{n+2}\right )\right )^n \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int F^{a c+b c x} \left (f \sin \left (d+\frac {i b c x \log (F)}{n+2}\right )+g \cos \left (d+\frac {i b c x \log (F)}{n+2}\right )\right )^ndx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int F^{a c+b c x} \left (f \sin \left (d+\frac {i b c x \log (F)}{n+2}\right )+g \cos \left (d+\frac {i b c x \log (F)}{n+2}\right )\right )^ndx\)

Input:

Int[F^(c*(a + b*x))*(g*Cos[d + (I*b*c*x*Log[F])/(2 + n)] + f*Sin[d + (I*b* 
c*x*Log[F])/(2 + n)])^n,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]

\[\int F^{c \left (b x +a \right )} \left (g \cos \left (d +\frac {i b c x \ln \left (F \right )}{n +2}\right )+f \sin \left (d +\frac {i b c x \ln \left (F \right )}{n +2}\right )\right )^{n}d x\]

Input:

int(F^(c*(b*x+a))*(g*cos(d+I*b*c*x*ln(F)/(n+2))+f*sin(d+I*b*c*x*ln(F)/(n+2 
)))^n,x)
 

Output:

int(F^(c*(b*x+a))*(g*cos(d+I*b*c*x*ln(F)/(n+2))+f*sin(d+I*b*c*x*ln(F)/(n+2 
)))^n,x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.93 \[ \int F^{c (a+b x)} \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )^n \, dx=-\frac {{\left ({\left (i \, f + g\right )} n + {\left ({\left (-i \, f + g\right )} n - 2 i \, f + 2 \, g\right )} e^{\left (-\frac {2 \, {\left (b c x \log \left (F\right ) - i \, d n - 2 i \, d\right )}}{n + 2}\right )} + 2 i \, f + 2 \, g\right )} \left (\frac {1}{2} \, {\left ({\left (-i \, f + g\right )} e^{\left (-\frac {2 \, {\left (b c x \log \left (F\right ) - i \, d n - 2 i \, d\right )}}{n + 2}\right )} + i \, f + g\right )} e^{\left (\frac {b c x \log \left (F\right ) - i \, d n - 2 i \, d}{n + 2}\right )}\right )^{n} F^{b c x + a c}}{2 \, {\left (-i \, b c f - b c g + {\left (-i \, b c f - b c g\right )} n\right )} \log \left (F\right )} \] Input:

integrate(F^(c*(b*x+a))*(g*cos(d+I*b*c*x*log(F)/(2+n))+f*sin(d+I*b*c*x*log 
(F)/(2+n)))^n,x, algorithm="fricas")
 

Output:

-1/2*((I*f + g)*n + ((-I*f + g)*n - 2*I*f + 2*g)*e^(-2*(b*c*x*log(F) - I*d 
*n - 2*I*d)/(n + 2)) + 2*I*f + 2*g)*(1/2*((-I*f + g)*e^(-2*(b*c*x*log(F) - 
 I*d*n - 2*I*d)/(n + 2)) + I*f + g)*e^((b*c*x*log(F) - I*d*n - 2*I*d)/(n + 
 2)))^n*F^(b*c*x + a*c)/((-I*b*c*f - b*c*g + (-I*b*c*f - b*c*g)*n)*log(F))
 

Sympy [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )^n \, dx=\text {Timed out} \] Input:

integrate(F**(c*(b*x+a))*(g*cos(d+I*b*c*x*ln(F)/(2+n))+f*sin(d+I*b*c*x*ln( 
F)/(2+n)))**n,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(F^(c*(b*x+a))*(g*cos(d+I*b*c*x*log(F)/(2+n))+f*sin(d+I*b*c*x*log 
(F)/(2+n)))^n,x, algorithm="maxima")
 

Output:

integrate((g*cos(I*b*c*x*log(F)/(n + 2) + d) + f*sin(I*b*c*x*log(F)/(n + 2 
) + d))^n*F^((b*x + a)*c), x)
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1509 vs. \(2 (147) = 294\).

Time = 55.96 (sec) , antiderivative size = 1509, normalized size of antiderivative = 9.26 \[ \int F^{c (a+b x)} \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )^n \, dx=\text {Too large to display} \] Input:

integrate(F^(c*(b*x+a))*(g*cos(d+I*b*c*x*log(F)/(2+n))+f*sin(d+I*b*c*x*log 
(F)/(2+n)))^n,x, algorithm="giac")
 

Output:

1/2*(F^(a*c)*f*n*e^((2*b*c*x*log(F) + I*d*n^2 - n^2*log(2) + n^2*log(I*f*e 
^(2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) + g*e^(2*(b*c*x*log(F) - I*d*n 
 - 2*I*d)/(n + 2)) - I*f + g) + 2*I*d*n - 2*n*log(2) + 2*n*log(I*f*e^(2*(b 
*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) + g*e^(2*(b*c*x*log(F) - I*d*n - 2*I 
*d)/(n + 2)) - I*f + g))/(n + 2) + 2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2 
)) - I*F^(a*c)*g*n*e^((2*b*c*x*log(F) + I*d*n^2 - n^2*log(2) + n^2*log(I*f 
*e^(2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) + g*e^(2*(b*c*x*log(F) - I*d 
*n - 2*I*d)/(n + 2)) - I*f + g) + 2*I*d*n - 2*n*log(2) + 2*n*log(I*f*e^(2* 
(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) + g*e^(2*(b*c*x*log(F) - I*d*n - 2 
*I*d)/(n + 2)) - I*f + g))/(n + 2) + 2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 
 2)) - F^(a*c)*f*n*e^((2*b*c*x*log(F) + I*d*n^2 - n^2*log(2) + n^2*log(I*f 
*e^(2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) + g*e^(2*(b*c*x*log(F) - I*d 
*n - 2*I*d)/(n + 2)) - I*f + g) + 2*I*d*n - 2*n*log(2) + 2*n*log(I*f*e^(2* 
(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) + g*e^(2*(b*c*x*log(F) - I*d*n - 2 
*I*d)/(n + 2)) - I*f + g))/(n + 2)) - I*F^(a*c)*g*n*e^((2*b*c*x*log(F) + I 
*d*n^2 - n^2*log(2) + n^2*log(I*f*e^(2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 
 2)) + g*e^(2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) - I*f + g) + 2*I*d*n 
 - 2*n*log(2) + 2*n*log(I*f*e^(2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) + 
 g*e^(2*(b*c*x*log(F) - I*d*n - 2*I*d)/(n + 2)) - I*f + g))/(n + 2)) + 2*F 
^(a*c)*f*e^((2*b*c*x*log(F) + I*d*n^2 - n^2*log(2) + n^2*log(I*f*e^(2*(...
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )^n \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (g\,\cos \left (d+\frac {b\,c\,x\,\ln \left (F\right )\,1{}\mathrm {i}}{n+2}\right )+f\,\sin \left (d+\frac {b\,c\,x\,\ln \left (F\right )\,1{}\mathrm {i}}{n+2}\right )\right )}^n \,d x \] Input:

int(F^(c*(a + b*x))*(g*cos(d + (b*c*x*log(F)*1i)/(n + 2)) + f*sin(d + (b*c 
*x*log(F)*1i)/(n + 2)))^n,x)
 

Output:

int(F^(c*(a + b*x))*(g*cos(d + (b*c*x*log(F)*1i)/(n + 2)) + f*sin(d + (b*c 
*x*log(F)*1i)/(n + 2)))^n, x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \left (g \cos \left (d+\frac {i b c x \log (F)}{2+n}\right )+f \sin \left (d+\frac {i b c x \log (F)}{2+n}\right )\right )^n \, dx=\text {too large to display} \] Input:

int(F^(c*(b*x+a))*(g*cos(d+I*b*c*x*log(F)/(2+n))+f*sin(d+I*b*c*x*log(F)/(2 
+n)))^n,x)
 

Output:

(f**(a*c)*(f**(b*c*x)*(cos((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*g + sin(( 
log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f)**n*f**2*i*n**2 + 2*f**(b*c*x)*(cos 
((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*g + sin((log(f)*b*c*i*x + d*n + 2*d 
)/(n + 2))*f)**n*f**2*i*n + f**(b*c*x)*(cos((log(f)*b*c*i*x + d*n + 2*d)/( 
n + 2))*g + sin((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f)**n*f*g*n**2 + 4*f 
**(b*c*x)*(cos((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*g + sin((log(f)*b*c*i 
*x + d*n + 2*d)/(n + 2))*f)**n*f*g*n + 4*f**(b*c*x)*(cos((log(f)*b*c*i*x + 
 d*n + 2*d)/(n + 2))*g + sin((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f)**n*f 
*g + int((f**(b*c*x)*(cos((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*g + sin((l 
og(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f)**n*cos((log(f)*b*c*i*x + d*n + 2*d) 
/(n + 2)))/(cos((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f**2*g*i*n**2 + 2*co 
s((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f**2*g*i*n + 2*cos((log(f)*b*c*i*x 
 + d*n + 2*d)/(n + 2))*f*g**2*n**2 + 4*cos((log(f)*b*c*i*x + d*n + 2*d)/(n 
 + 2))*f*g**2*n + 4*cos((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f*g**2 - cos 
((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*g**3*i*n**2 - 2*cos((log(f)*b*c*i*x 
 + d*n + 2*d)/(n + 2))*g**3*i*n + sin((log(f)*b*c*i*x + d*n + 2*d)/(n + 2) 
)*f**3*i*n**2 + 2*sin((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f**3*i*n + 2*s 
in((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f**2*g*n**2 + 4*sin((log(f)*b*c*i 
*x + d*n + 2*d)/(n + 2))*f**2*g*n + 4*sin((log(f)*b*c*i*x + d*n + 2*d)/(n 
+ 2))*f**2*g - sin((log(f)*b*c*i*x + d*n + 2*d)/(n + 2))*f*g**2*i*n**2 ...