3.326 \(\int F^{c (a+b x)} \text {csch}^2(d+e x) \, dx\)

Optimal. Leaf size=68 \[ \frac {4 e^{2 (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,\frac {b c \log (F)}{2 e}+1;\frac {b c \log (F)}{2 e}+2;e^{2 (d+e x)}\right )}{b c \log (F)+2 e} \]

[Out]

4*exp(2*e*x+2*d)*F^(c*(b*x+a))*hypergeom([2, 1+1/2*b*c*ln(F)/e],[2+1/2*b*c*ln(F)/e],exp(2*e*x+2*d))/(b*c*ln(F)
+2*e)

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {5493} \[ \frac {4 e^{2 (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,\frac {b c \log (F)}{2 e}+1;\frac {b c \log (F)}{2 e}+2;e^{2 (d+e x)}\right )}{b c \log (F)+2 e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Csch[d + e*x]^2,x]

[Out]

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

Rule 5493

Int[Csch[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), 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 F^{c (a+b x)} \text {csch}^2(d+e x) \, dx &=\frac {4 e^{2 (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1+\frac {b c \log (F)}{2 e};2+\frac {b c \log (F)}{2 e};e^{2 (d+e x)}\right )}{2 e+b c \log (F)}\\ \end {align*}

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Mathematica [A]  time = 2.82, size = 87, normalized size = 1.28 \[ -\frac {2 F^{c (a+b x)} \left (\left (e^{2 d}-1\right ) \, _2F_1\left (1,\frac {b c \log (F)}{2 e};\frac {b c \log (F)}{2 e}+1;e^{2 (d+e x)}\right )+\sinh (d) \text {csch}(d+e x) (\cosh (e x)-\sinh (e x))\right )}{\left (e^{2 d}-1\right ) e} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))*Csch[d + e*x]^2,x]

[Out]

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

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (F^{b c x + a c} \operatorname {csch}\left (e x + d\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*csch(e*x+d)^2,x, algorithm="fricas")

[Out]

integral(F^(b*c*x + a*c)*csch(e*x + d)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*csch(e*x+d)^2,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)*csch(e*x + d)^2, x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int F^{c \left (b x +a \right )} \mathrm {csch}\left (e x +d \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*csch(e*x+d)^2,x)

[Out]

int(F^(c*(b*x+a))*csch(e*x+d)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 16 \, F^{a c} b c e \int -\frac {F^{b c x}}{b^{2} c^{2} \log \relax (F)^{2} - 6 \, b c e \log \relax (F) + 8 \, e^{2} - {\left (b^{2} c^{2} e^{\left (6 \, d\right )} \log \relax (F)^{2} - 6 \, b c e e^{\left (6 \, d\right )} \log \relax (F) + 8 \, e^{2} e^{\left (6 \, d\right )}\right )} e^{\left (6 \, e x\right )} + 3 \, {\left (b^{2} c^{2} e^{\left (4 \, d\right )} \log \relax (F)^{2} - 6 \, b c e e^{\left (4 \, d\right )} \log \relax (F) + 8 \, e^{2} e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} - 3 \, {\left (b^{2} c^{2} e^{\left (2 \, d\right )} \log \relax (F)^{2} - 6 \, b c e e^{\left (2 \, d\right )} \log \relax (F) + 8 \, e^{2} e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}}\,{d x} \log \relax (F) + \frac {4 \, {\left (4 \, F^{a c} e + {\left (F^{a c} b c e^{\left (2 \, d\right )} \log \relax (F) - 4 \, F^{a c} e e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}\right )} F^{b c x}}{b^{2} c^{2} \log \relax (F)^{2} - 6 \, b c e \log \relax (F) + 8 \, e^{2} + {\left (b^{2} c^{2} e^{\left (4 \, d\right )} \log \relax (F)^{2} - 6 \, b c e e^{\left (4 \, d\right )} \log \relax (F) + 8 \, e^{2} e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} - 2 \, {\left (b^{2} c^{2} e^{\left (2 \, d\right )} \log \relax (F)^{2} - 6 \, b c e e^{\left (2 \, d\right )} \log \relax (F) + 8 \, e^{2} e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*csch(e*x+d)^2,x, algorithm="maxima")

[Out]

16*F^(a*c)*b*c*e*integrate(-F^(b*c*x)/(b^2*c^2*log(F)^2 - 6*b*c*e*log(F) + 8*e^2 - (b^2*c^2*e^(6*d)*log(F)^2 -
6*b*c*e*e^(6*d)*log(F) + 8*e^2*e^(6*d))*e^(6*e*x) + 3*(b^2*c^2*e^(4*d)*log(F)^2 - 6*b*c*e*e^(4*d)*log(F) + 8*
e^2*e^(4*d))*e^(4*e*x) - 3*(b^2*c^2*e^(2*d)*log(F)^2 - 6*b*c*e*e^(2*d)*log(F) + 8*e^2*e^(2*d))*e^(2*e*x)), x)*
log(F) + 4*(4*F^(a*c)*e + (F^(a*c)*b*c*e^(2*d)*log(F) - 4*F^(a*c)*e*e^(2*d))*e^(2*e*x))*F^(b*c*x)/(b^2*c^2*log
(F)^2 - 6*b*c*e*log(F) + 8*e^2 + (b^2*c^2*e^(4*d)*log(F)^2 - 6*b*c*e*e^(4*d)*log(F) + 8*e^2*e^(4*d))*e^(4*e*x)
- 2*(b^2*c^2*e^(2*d)*log(F)^2 - 6*b*c*e*e^(2*d)*log(F) + 8*e^2*e^(2*d))*e^(2*e*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{c \left (a + b x\right )} \operatorname {csch}^{2}{\left (d + e x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*csch(e*x+d)**2,x)

[Out]

Integral(F**(c*(a + b*x))*csch(d + e*x)**2, x)

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