### 3.895 $$\int \frac{F^{c (a+b x)}}{f+i f \sinh (d+e x)} \, dx$$

Optimal. Leaf size=85 $\frac{2 e^{\frac{1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{\frac{1}{2} (2 d+2 e x+i \pi )}\right )}{f (b c \log (F)+e)}$

[Out]

(2*E^((2*d + I*Pi + 2*e*x)/2)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^
((2*d + I*Pi + 2*e*x)/2)])/(f*(e + b*c*Log[F]))

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Rubi [A]  time = 0.0752858, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.08, Rules used = {5496, 5492} $\frac{2 e^{\frac{1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{\frac{1}{2} (2 d+2 e x+i \pi )}\right )}{f (b c \log (F)+e)}$

Antiderivative was successfully veriﬁed.

[In]

Int[F^(c*(a + b*x))/(f + I*f*Sinh[d + e*x]),x]

[Out]

(2*E^((2*d + I*Pi + 2*e*x)/2)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^
((2*d + I*Pi + 2*e*x)/2)])/(f*(e + b*c*Log[F]))

Rule 5496

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sinh[(d_.) + (e_.)*(x_)])^(n_.), x_Symbol] :> Dist[2^n*f^n
, Int[F^(c*(a + b*x))*Cosh[d/2 + (e*x)/2 - (f*Pi)/(4*g)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] &
& EqQ[f^2 + g^2, 0] && ILtQ[n, 0]

Rule 5492

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(n*(d + e*x))*F
^(c*(a + b*x))*Hypergeometric2F1[n, n/2 + (b*c*Log[F])/(2*e), 1 + n/2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))])
/(e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{F^{c (a+b x)}}{f+i f \sinh (d+e x)} \, dx &=\frac{\int F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \, dx}{2 f}\\ &=\frac{2 e^{\frac{1}{2} (2 d+i \pi +2 e x)} F^{c (a+b x)} \, _2F_1\left (2,1+\frac{b c \log (F)}{e};2+\frac{b c \log (F)}{e};-e^{\frac{1}{2} (2 d+i \pi +2 e x)}\right )}{f (e+b c \log (F))}\\ \end{align*}

Mathematica [A]  time = 3.36968, size = 104, normalized size = 1.22 $\frac{2 F^{c (a+b x)} \left (\, _2F_1\left (1,\frac{b c \log (F)}{e};\frac{b c \log (F)}{e}+1;-i e^{d+e x}\right )+\frac{\cosh \left (\frac{e x}{2}\right )-\sinh \left (\frac{e x}{2}\right )}{\left (1-i e^d\right ) \sinh \left (\frac{e x}{2}\right )+\left (-1-i e^d\right ) \cosh \left (\frac{e x}{2}\right )}\right )}{e f}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[F^(c*(a + b*x))/(f + I*f*Sinh[d + e*x]),x]

[Out]

(2*F^(c*(a + b*x))*(Hypergeometric2F1[1, (b*c*Log[F])/e, 1 + (b*c*Log[F])/e, (-I)*E^(d + e*x)] + (Cosh[(e*x)/2
] - Sinh[(e*x)/2])/((-1 - I*E^d)*Cosh[(e*x)/2] + (1 - I*E^d)*Sinh[(e*x)/2])))/(e*f)

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{c \left ( bx+a \right ) }}{f+if\sinh \left ( ex+d \right ) }}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d)),x)

[Out]

int(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -4 \, F^{a c} b c e \int \frac{F^{b c x}}{i \, b^{2} c^{2} f \log \left (F\right )^{2} - 3 i \, b c e f \log \left (F\right ) + 2 i \, e^{2} f +{\left (b^{2} c^{2} f e^{\left (3 \, d\right )} \log \left (F\right )^{2} - 3 \, b c e f e^{\left (3 \, d\right )} \log \left (F\right ) + 2 \, e^{2} f e^{\left (3 \, d\right )}\right )} e^{\left (3 \, e x\right )} +{\left (-3 i \, b^{2} c^{2} f e^{\left (2 \, d\right )} \log \left (F\right )^{2} + 9 i \, b c e f e^{\left (2 \, d\right )} \log \left (F\right ) - 6 i \, e^{2} f e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )} - 3 \,{\left (b^{2} c^{2} f e^{d} \log \left (F\right )^{2} - 3 \, b c e f e^{d} \log \left (F\right ) + 2 \, e^{2} f e^{d}\right )} e^{\left (e x\right )}}\,{d x} \log \left (F\right ) + \frac{{\left (4 i \, F^{a c} e + 2 \,{\left (F^{a c} b c e^{d} \log \left (F\right ) - 2 \, F^{a c} e e^{d}\right )} e^{\left (e x\right )}\right )} F^{b c x}}{-i \, b^{2} c^{2} f \log \left (F\right )^{2} + 3 i \, b c e f \log \left (F\right ) - 2 i \, e^{2} f +{\left (i \, b^{2} c^{2} f e^{\left (2 \, d\right )} \log \left (F\right )^{2} - 3 i \, b c e f e^{\left (2 \, d\right )} \log \left (F\right ) + 2 i \, e^{2} f e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )} + 2 \,{\left (b^{2} c^{2} f e^{d} \log \left (F\right )^{2} - 3 \, b c e f e^{d} \log \left (F\right ) + 2 \, e^{2} f e^{d}\right )} e^{\left (e x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d)),x, algorithm="maxima")

[Out]

-4*F^(a*c)*b*c*e*integrate(F^(b*c*x)/(I*b^2*c^2*f*log(F)^2 - 3*I*b*c*e*f*log(F) + 2*I*e^2*f + (b^2*c^2*f*e^(3*
d)*log(F)^2 - 3*b*c*e*f*e^(3*d)*log(F) + 2*e^2*f*e^(3*d))*e^(3*e*x) + (-3*I*b^2*c^2*f*e^(2*d)*log(F)^2 + 9*I*b
*c*e*f*e^(2*d)*log(F) - 6*I*e^2*f*e^(2*d))*e^(2*e*x) - 3*(b^2*c^2*f*e^d*log(F)^2 - 3*b*c*e*f*e^d*log(F) + 2*e^
2*f*e^d)*e^(e*x)), x)*log(F) + (4*I*F^(a*c)*e + 2*(F^(a*c)*b*c*e^d*log(F) - 2*F^(a*c)*e*e^d)*e^(e*x))*F^(b*c*x
)/(-I*b^2*c^2*f*log(F)^2 + 3*I*b*c*e*f*log(F) - 2*I*e^2*f + (I*b^2*c^2*f*e^(2*d)*log(F)^2 - 3*I*b*c*e*f*e^(2*d
)*log(F) + 2*I*e^2*f*e^(2*d))*e^(2*e*x) + 2*(b^2*c^2*f*e^d*log(F)^2 - 3*b*c*e*f*e^d*log(F) + 2*e^2*f*e^d)*e^(e
*x))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (e f e^{\left (e x + d\right )} - i \, e f\right )}{\rm integral}\left (-\frac{2 i \, F^{b c x + a c} b c \log \left (F\right )}{e f e^{\left (e x + d\right )} - i \, e f}, x\right ) + 2 i \, F^{b c x + a c}}{e f e^{\left (e x + d\right )} - i \, e f} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{F^{a c} F^{b c x}}{i \sinh{\left (d + e x \right )} + 1}\, dx}{f} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))/(f+I*f*sinh(e*x+d)),x)

[Out]

Integral(F**(a*c)*F**(b*c*x)/(I*sinh(d + e*x) + 1), x)/f

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{i \, f \sinh \left (e x + d\right ) + f}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d)),x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)/(I*f*sinh(e*x + d) + f), x)