3.9.96 \(\int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx\) [896]

Optimal. Leaf size=196 \[ \frac {2 e^{\frac {1}{2} (2 d+i \pi +2 e x)} F^{c (a+b x)} \, _2F_1\left (2,1+\frac {b c \log (F)}{e};2+\frac {b c \log (F)}{e};-e^{\frac {1}{2} (2 d+i \pi +2 e x)}\right ) (e-b c \log (F))}{3 e^2 f^2}+\frac {b c F^{c (a+b x)} \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right ) \tanh \left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e f^2} \]

[Out]

2/3*exp(d+1/2*I*Pi+e*x)*F^(c*(b*x+a))*hypergeom([2, 1+b*c*ln(F)/e],[2+b*c*ln(F)/e],-exp(d+1/2*I*Pi+e*x))*(e-b*
c*ln(F))/e^2/f^2+1/6*b*c*F^(c*(b*x+a))*ln(F)*sech(1/2*d+1/4*I*Pi+1/2*e*x)^2/e^2/f^2+1/6*F^(c*(b*x+a))*sech(1/2
*d+1/4*I*Pi+1/2*e*x)^2*tanh(1/2*d+1/4*I*Pi+1/2*e*x)/e/f^2

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Rubi [A]
time = 0.08, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5604, 5598, 5600} \begin {gather*} \frac {2 e^{\frac {1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} (e-b c \log (F)) \, _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 )}{3 e^2 f^2}+\frac {b c \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac {\tanh \left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) F^{c (a+b x)}}{6 e f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(f + I*f*Sinh[d + e*x])^2,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)]*(e - b*c*Log[F]))/(3*e^2*f^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d/2 + (I/4)*Pi + (e
*x)/2]^2)/(6*e^2*f^2) + (F^(c*(a + b*x))*Sech[d/2 + (I/4)*Pi + (e*x)/2]^2*Tanh[d/2 + (I/4)*Pi + (e*x)/2])/(6*e
*f^2)

Rule 5598

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

Rule 5600

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))/(e*n + b*c*Log[F]))*Hypergeometric2F1[n, n/2 + b*c*(Log[F]/(2*e)), 1 + n/2 + b*c*(Log[F]/(2*e))
, -E^(2*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rule 5604

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 - f*(Pi/(4*g)) + e*(x/2)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] &
& EqQ[f^2 + g^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx &=\frac {\int F^{c (a+b x)} \text {sech}^4\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right ) \, dx}{4 f^2}\\ &=\frac {b c F^{c (a+b x)} \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right ) \tanh \left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e f^2}+\frac {\left (1-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right ) \, dx}{6 f^2}\\ &=\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 ) (e-b c \log (F))}{3 e^2 f^2}+\frac {b c F^{c (a+b x)} \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right ) \tanh \left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e f^2}\\ \end {align*}

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Mathematica [A]
time = 1.66, size = 255, normalized size = 1.30 \begin {gather*} \frac {F^{c (a+b x)} \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right ) \left (e (i e+b c \log (F)) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )+(1-i) \left (-1+(1+i) \, _2F_1\left (1,\frac {b c \log (F)}{e};1+\frac {b c \log (F)}{e};-i (\cosh (d+e x)+\sinh (d+e x))\right )\right ) \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )^3+2 e^2 \sinh \left (\frac {1}{2} (d+e x)\right )+2 \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )^2 \sinh \left (\frac {1}{2} (d+e x)\right )\right )}{3 e^3 (f+i f \sinh (d+e x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))/(f + I*f*Sinh[d + e*x])^2,x]

[Out]

(F^(c*(a + b*x))*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])*(e*(I*e + b*c*Log[F])*(Cosh[(d + e*x)/2] + I*Sinh[(
d + e*x)/2]) + (1 - I)*(-1 + (1 + I)*Hypergeometric2F1[1, (b*c*Log[F])/e, 1 + (b*c*Log[F])/e, (-I)*(Cosh[d + e
*x] + Sinh[d + e*x])])*(e^2 - b^2*c^2*Log[F]^2)*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])^3 + 2*e^2*Sinh[(d +
e*x)/2] + 2*(e^2 - b^2*c^2*Log[F]^2)*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])^2*Sinh[(d + e*x)/2]))/(3*e^3*(f
+ I*f*Sinh[d + e*x])^2)

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Maple [F]
time = 2.89, size = 0, normalized size = 0.00 \[\int \frac {F^{c \left (b x +a \right )}}{\left (f +i f \sinh \left (e x +d \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x)

[Out]

int(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x, algorithm="maxima")

[Out]

-16*(I*F^(a*c)*b^2*c^2*e*log(F)^2 + I*F^(a*c)*b*c*e^2*log(F))*integrate(F^(b*c*x)/(I*b^3*c^3*f^2*log(F)^3 - 9*
I*b^2*c^2*f^2*e*log(F)^2 + 26*I*b*c*f^2*e^2*log(F) - 24*I*f^2*e^3 - (b^3*c^3*f^2*e^(5*d)*log(F)^3 - 9*b^2*c^2*
f^2*e^(5*d + 1)*log(F)^2 + 26*b*c*f^2*e^(5*d + 2)*log(F) - 24*f^2*e^(5*d + 3))*e^(5*x*e) - 5*(-I*b^3*c^3*f^2*e
^(4*d)*log(F)^3 + 9*I*b^2*c^2*f^2*e^(4*d + 1)*log(F)^2 - 26*I*b*c*f^2*e^(4*d + 2)*log(F) + 24*I*f^2*e^(4*d + 3
))*e^(4*x*e) + 10*(b^3*c^3*f^2*e^(3*d)*log(F)^3 - 9*b^2*c^2*f^2*e^(3*d + 1)*log(F)^2 + 26*b*c*f^2*e^(3*d + 2)*
log(F) - 24*f^2*e^(3*d + 3))*e^(3*x*e) - 10*(I*b^3*c^3*f^2*e^(2*d)*log(F)^3 - 9*I*b^2*c^2*f^2*e^(2*d + 1)*log(
F)^2 + 26*I*b*c*f^2*e^(2*d + 2)*log(F) - 24*I*f^2*e^(2*d + 3))*e^(2*x*e) - 5*(b^3*c^3*f^2*e^d*log(F)^3 - 9*b^2
*c^2*f^2*e^(d + 1)*log(F)^2 + 26*b*c*f^2*e^(d + 2)*log(F) - 24*f^2*e^(d + 3))*e^(x*e)), x) + 4*(4*F^(a*c)*b*c*
e*log(F) + 4*F^(a*c)*e^2 - (F^(a*c)*b^2*c^2*e^(2*d)*log(F)^2 - 7*F^(a*c)*b*c*e^(2*d + 1)*log(F) + 12*F^(a*c)*e
^(2*d + 2))*e^(2*x*e) + 4*(-I*F^(a*c)*b*c*e^(d + 1)*log(F) + 4*I*F^(a*c)*e^(d + 2))*e^(x*e))*F^(b*c*x)/(b^3*c^
3*f^2*log(F)^3 - 9*b^2*c^2*f^2*e*log(F)^2 + 26*b*c*f^2*e^2*log(F) - 24*f^2*e^3 + (b^3*c^3*f^2*e^(4*d)*log(F)^3
- 9*b^2*c^2*f^2*e^(4*d + 1)*log(F)^2 + 26*b*c*f^2*e^(4*d + 2)*log(F) - 24*f^2*e^(4*d + 3))*e^(4*x*e) + 4*(-I*
b^3*c^3*f^2*e^(3*d)*log(F)^3 + 9*I*b^2*c^2*f^2*e^(3*d + 1)*log(F)^2 - 26*I*b*c*f^2*e^(3*d + 2)*log(F) + 24*I*f
^2*e^(3*d + 3))*e^(3*x*e) - 6*(b^3*c^3*f^2*e^(2*d)*log(F)^3 - 9*b^2*c^2*f^2*e^(2*d + 1)*log(F)^2 + 26*b*c*f^2*
e^(2*d + 2)*log(F) - 24*f^2*e^(2*d + 3))*e^(2*x*e) + 4*(I*b^3*c^3*f^2*e^d*log(F)^3 - 9*I*b^2*c^2*f^2*e^(d + 1)
*log(F)^2 + 26*I*b*c*f^2*e^(d + 2)*log(F) - 24*I*f^2*e^(d + 3))*e^(x*e))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x, algorithm="fricas")

[Out]

-(2*((I*b^2*c^2*e^(2*x*e + 2*d) + 2*b^2*c^2*e^(x*e + d) - I*b^2*c^2)*log(F)^2 + (I*b*c*e^(2*x*e + 2*d + 1) + b
*c*e^(x*e + d + 1))*log(F) + I*e^2 - 3*e^(x*e + d + 2))*F^(b*c*x + a*c) + 3*(-I*f^2*e^3 - f^2*e^(3*x*e + 3*d +
3) + 3*I*f^2*e^(2*x*e + 2*d + 3) + 3*f^2*e^(x*e + d + 3))*integral(-2*(-I*b^3*c^3*log(F)^3 + I*b*c*e^2*log(F)
)*F^(b*c*x + a*c)/(-3*I*f^2*e^3 + 3*f^2*e^(x*e + d + 3)), x))/(3*I*f^2*e^3 + 3*f^2*e^(3*x*e + 3*d + 3) - 9*I*f
^2*e^(2*x*e + 2*d + 3) - 9*f^2*e^(x*e + d + 3))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {F^{a c} F^{b c x}}{\sinh ^{2}{\left (d + e x \right )} - 2 i \sinh {\left (d + e x \right )} - 1}\, dx}{f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))/(f+I*f*sinh(e*x+d))**2,x)

[Out]

-Integral(F**(a*c)*F**(b*c*x)/(sinh(d + e*x)**2 - 2*I*sinh(d + e*x) - 1), x)/f**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)/(I*f*sinh(e*x + d) + f)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (f+f\,\mathrm {sinh}\left (d+e\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/(f + f*sinh(d + e*x)*1i)^2,x)

[Out]

int(F^(c*(a + b*x))/(f + f*sinh(d + e*x)*1i)^2, x)

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