\(\int F^{c (a+b x)} \csc (d+e x) \, dx\) [5]
Optimal result
Integrand size = 16, antiderivative size = 81 \[
\int F^{c (a+b x)} \csc (d+e x) \, dx=-\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{e-i b c \log (F)}
\]
[Out]
-2*exp(I*(e*x+d))*F^(c*(b*x+a))*hypergeom([1, 1/2*(e-I*b*c*ln(F))/e],[3/2-1/2*I*b*c*ln(F)/e],exp(2*I*(e*x+d)))
/(e-I*b*c*ln(F))
Rubi [A] (verified)
Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00,
number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4538}
\[
\int F^{c (a+b x)} \csc (d+e x) \, dx=-\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{e-i b c \log (F)}
\]
[In]
Int[F^(c*(a + b*x))*Csc[d + e*x],x]
[Out]
(-2*E^(I*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[1, (e - I*b*c*Log[F])/(2*e), (3 - (I*b*c*Log[F])/e)/2, E
^((2*I)*(d + e*x))])/(e - I*b*c*Log[F])
Rule 4538
Int[Csc[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(-2*I)^n*E^(I*n*(d + e*
x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hypergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 1 + n/2 - I*b*c*(L
og[F]/(2*e)), E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
Rubi steps \begin{align*}
\text {integral}& = -\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{e-i b c \log (F)} \\
\end{align*}
Mathematica [A] (verified)
Time = 2.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.41
\[
\int F^{c (a+b x)} \csc (d+e x) \, dx=\frac {i F^{c (a+b x)} \left (\operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},-\cos (d+e x)-i \sin (d+e x)\right )-\operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},\cos (d+e x)+i \sin (d+e x)\right )\right )}{b c \log (F)}
\]
[In]
Integrate[F^(c*(a + b*x))*Csc[d + e*x],x]
[Out]
(I*F^(c*(a + b*x))*(Hypergeometric2F1[1, ((-I)*b*c*Log[F])/e, 1 - (I*b*c*Log[F])/e, -Cos[d + e*x] - I*Sin[d +
e*x]] - Hypergeometric2F1[1, ((-I)*b*c*Log[F])/e, 1 - (I*b*c*Log[F])/e, Cos[d + e*x] + I*Sin[d + e*x]]))/(b*c*
Log[F])
Maple [F]
\[\int F^{c \left (x b +a \right )} \csc \left (e x +d \right )d x\]
[In]
int(F^(c*(b*x+a))*csc(e*x+d),x)
[Out]
int(F^(c*(b*x+a))*csc(e*x+d),x)
Fricas [F]
\[
\int F^{c (a+b x)} \csc (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right ) \,d x }
\]
[In]
integrate(F^(c*(b*x+a))*csc(e*x+d),x, algorithm="fricas")
[Out]
integral(F^(b*c*x + a*c)*csc(e*x + d), x)
Sympy [F]
\[
\int F^{c (a+b x)} \csc (d+e x) \, dx=\int F^{c \left (a + b x\right )} \csc {\left (d + e x \right )}\, dx
\]
[In]
integrate(F**(c*(b*x+a))*csc(e*x+d),x)
[Out]
Integral(F**(c*(a + b*x))*csc(d + e*x), x)
Maxima [F]
\[
\int F^{c (a+b x)} \csc (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right ) \,d x }
\]
[In]
integrate(F^(c*(b*x+a))*csc(e*x+d),x, algorithm="maxima")
[Out]
2*(F^(b*c*x)*F^(a*c)*b*c*log(F)*sin(e*x + d) + F^(b*c*x)*F^(a*c)*e*cos(e*x + d) - (F^(b*c*x)*F^(a*c)*b*c*log(F
)*sin(e*x + d) + F^(b*c*x)*F^(a*c)*e*cos(e*x + d))*cos(2*e*x + 2*d) - 2*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*
e^3 + (F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^3)*cos(2*e*x + 2*d)^2 + (F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^
3)*sin(2*e*x + 2*d)^2 - 2*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^3)*cos(2*e*x + 2*d))*integrate((F^(b*c*x)*b*
c*cos(e*x + d)*log(F) - F^(b*c*x)*e*sin(e*x + d) + (F^(b*c*x)*b*c*cos(e*x + d)*log(F) - F^(b*c*x)*e*sin(e*x +
d))*cos(4*e*x + 4*d) - 2*(F^(b*c*x)*b*c*cos(e*x + d)*log(F) - F^(b*c*x)*e*sin(e*x + d))*cos(2*e*x + 2*d) + (F^
(b*c*x)*b*c*log(F)*sin(e*x + d) + F^(b*c*x)*e*cos(e*x + d))*sin(4*e*x + 4*d) - 2*(F^(b*c*x)*b*c*log(F)*sin(e*x
+ d) + F^(b*c*x)*e*cos(e*x + d))*sin(2*e*x + 2*d))/(b^2*c^2*log(F)^2 + (b^2*c^2*log(F)^2 + e^2)*cos(4*e*x + 4
*d)^2 + 4*(b^2*c^2*log(F)^2 + e^2)*cos(2*e*x + 2*d)^2 + (b^2*c^2*log(F)^2 + e^2)*sin(4*e*x + 4*d)^2 - 4*(b^2*c
^2*log(F)^2 + e^2)*sin(4*e*x + 4*d)*sin(2*e*x + 2*d) + 4*(b^2*c^2*log(F)^2 + e^2)*sin(2*e*x + 2*d)^2 + e^2 + 2
*(b^2*c^2*log(F)^2 + e^2 - 2*(b^2*c^2*log(F)^2 + e^2)*cos(2*e*x + 2*d))*cos(4*e*x + 4*d) - 4*(b^2*c^2*log(F)^2
+ e^2)*cos(2*e*x + 2*d)), x) + (F^(b*c*x)*F^(a*c)*b*c*cos(e*x + d)*log(F) - F^(b*c*x)*F^(a*c)*e*sin(e*x + d))
*sin(2*e*x + 2*d))/(b^2*c^2*log(F)^2 + (b^2*c^2*log(F)^2 + e^2)*cos(2*e*x + 2*d)^2 + (b^2*c^2*log(F)^2 + e^2)*
sin(2*e*x + 2*d)^2 + e^2 - 2*(b^2*c^2*log(F)^2 + e^2)*cos(2*e*x + 2*d))
Giac [F]
\[
\int F^{c (a+b x)} \csc (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right ) \,d x }
\]
[In]
integrate(F^(c*(b*x+a))*csc(e*x+d),x, algorithm="giac")
[Out]
integrate(F^((b*x + a)*c)*csc(e*x + d), x)
Mupad [F(-1)]
Timed out. \[
\int F^{c (a+b x)} \csc (d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{\sin \left (d+e\,x\right )} \,d x
\]
[In]
int(F^(c*(a + b*x))/sin(d + e*x),x)
[Out]
int(F^(c*(a + b*x))/sin(d + e*x), x)