\(\int F^{c (a+b x)} (f \csc (d+\frac {i b c x \log (F)}{-2+n}))^n \, dx\) [117]

Optimal result

Integrand size = 31, antiderivative size = 141 \[ \int F^{c (a+b x)} \left (f \csc \left (d+\frac {i b c x \log (F)}{-2+n}\right )\right )^n \, dx=\frac {f^2 F^{c (a+b x)} (2-n) \left (f \csc \left (d-\frac {i b c x \log (F)}{2-n}\right )\right )^{-2+n}}{b c (1-n) \log (F)}+\frac {i f F^{c (a+b x)} (2-n) \cos \left (d-\frac {i b c x \log (F)}{2-n}\right ) \left (f \csc \left (d-\frac {i b c x \log (F)}{2-n}\right )\right )^{-1+n}}{b c (1-n) \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.57 \[ \int F^{c (a+b x)} \left (f \csc \left (d+\frac {i b c x \log (F)}{-2+n}\right )\right )^n \, dx=\frac {e^{-2 i d} F^{c (a+b x)} \left (e^{2 i d}-F^{\frac {2 b c x}{-2+n}}\right ) (-2+n) \left (f \csc \left (d+\frac {i b c x \log (F)}{-2+n}\right )\right )^n}{2 b c (-1+n) \log (F)} \] Input:

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

Output:

(F^(c*(a + b*x))*(E^((2*I)*d) - F^((2*b*c*x)/(-2 + n)))*(-2 + n)*(f*Csc[d 
+ (I*b*c*x*Log[F])/(-2 + n)])^n)/(2*b*c*E^((2*I)*d)*(-1 + n)*Log[F])
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {7271, 4947}

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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 4947

\(\displaystyle \csc ^{-n}\left (d-\frac {i b c x \log (F)}{2-n}\right ) \left (f \csc \left (d-\frac {i b c x \log (F)}{2-n}\right )\right )^n \left (\frac {(2-n) F^{c (a+b x)} \csc ^{n-2}\left (d-\frac {i b c x \log (F)}{2-n}\right )}{b c (1-n) \log (F)}-\frac {i (2-n) F^{c (a+b x)} \cos \left (d-\frac {i b c x \log (F)}{2-n}\right ) \csc ^{n-1}\left (d-\frac {i b c x \log (F)}{2-n}\right )}{b c (1-n) \log (F)}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 4947

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

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [F]

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

Input:

int(F^(c*(b*x+a))*(f*csc(d+I*b*c*x*ln(F)/(-2+n)))^n,x)
 

Output:

int(F^(c*(b*x+a))*(f*csc(d+I*b*c*x*ln(F)/(-2+n)))^n,x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.94 \[ \int F^{c (a+b x)} \left (f \csc \left (d+\frac {i b c x \log (F)}{-2+n}\right )\right )^n \, dx=\frac {{\left ({\left (n - 2\right )} e^{\left (-\frac {2 \, {\left (b c x \log \left (F\right ) - i \, d n + 2 i \, d\right )}}{n - 2}\right )} - n + 2\right )} F^{b c x + a c} \left (\frac {2 i \, f e^{\left (-\frac {b c x \log \left (F\right ) - i \, d n + 2 i \, d}{n - 2}\right )}}{e^{\left (-\frac {2 \, {\left (b c x \log \left (F\right ) - i \, d n + 2 i \, d\right )}}{n - 2}\right )} - 1}\right )^{n} e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) - i \, d n + 2 i \, d\right )}}{n - 2}\right )}}{2 \, {\left (b c n - b c\right )} \log \left (F\right )} \] Input:

integrate(F^(c*(b*x+a))*(f*csc(d+I*b*c*x*log(F)/(-2+n)))^n,x, algorithm="f 
ricas")
 

Output:

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

Sympy [F]

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

integrate(F**(c*(b*x+a))*(f*csc(d+I*b*c*x*ln(F)/(-2+n)))**n,x)
 

Output:

Integral(F**(c*(a + b*x))*(f*csc(I*b*c*x*log(F)/(n - 2) + d))**n, x)
 

Maxima [F]

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

integrate(F^(c*(b*x+a))*(f*csc(d+I*b*c*x*log(F)/(-2+n)))^n,x, algorithm="m 
axima")
 

Output:

integrate((f*csc(I*b*c*x*log(F)/(n - 2) + d))^n*F^((b*x + a)*c), x)
 

Giac [F]

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

integrate(F^(c*(b*x+a))*(f*csc(d+I*b*c*x*log(F)/(-2+n)))^n,x, algorithm="g 
iac")
 

Output:

integrate((f*csc(I*b*c*x*log(F)/(n - 2) + d))^n*F^((b*x + a)*c), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(F^(c*(b*x+a))*(f*csc(d+I*b*c*x*log(F)/(-2+n)))^n,x)
 

Output:

f**(a*c + n)*int(f**(b*c*x)*csc((log(f)*b*c*i*x + d*n - 2*d)/(n - 2))**n,x 
)