3.883 \(\int e^{c+d x} \text {csch}^3(a+b x) \, dx\)

Optimal. Leaf size=100 \[ \frac {(b-d) e^{a+b x+c+d x} \, _2F_1\left (1,\frac {b+d}{2 b};\frac {1}{2} \left (\frac {d}{b}+3\right );e^{2 (a+b x)}\right )}{b^2}-\frac {d e^{c+d x} \text {csch}(a+b x)}{2 b^2}-\frac {e^{c+d x} \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

[Out]

-1/2*d*exp(d*x+c)*csch(b*x+a)/b^2-1/2*exp(d*x+c)*coth(b*x+a)*csch(b*x+a)/b+(b-d)*exp(b*x+d*x+a+c)*hypergeom([1
, 1/2*(b+d)/b],[3/2+1/2*d/b],exp(2*b*x+2*a))/b^2

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Rubi [A]  time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5491, 5493} \[ \frac {(b-d) e^{a+b x+c+d x} \, _2F_1\left (1,\frac {b+d}{2 b};\frac {1}{2} \left (\frac {d}{b}+3\right );e^{2 (a+b x)}\right )}{b^2}-\frac {d e^{c+d x} \text {csch}(a+b x)}{2 b^2}-\frac {e^{c+d x} \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[E^(c + d*x)*Csch[a + b*x]^3,x]

[Out]

-(d*E^(c + d*x)*Csch[a + b*x])/(2*b^2) - (E^(c + d*x)*Coth[a + b*x]*Csch[a + b*x])/(2*b) + ((b - d)*E^(a + c +
b*x + d*x)*Hypergeometric2F1[1, (b + d)/(2*b), (3 + d/b)/2, E^(2*(a + b*x))])/b^2

Rule 5491

Int[Csch[(d_.) + (e_.)*(x_)]^(n_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a +
b*x))*Csch[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))*Csch[d + e*x]^(n - 2), x], x] - Simp[(F^(c*(a + b*x))*Csch[d + e*x]^(n - 1)*Co
sh[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] && G
tQ[n, 1] && NeQ[n, 2]

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 e^{c+d x} \text {csch}^3(a+b x) \, dx &=-\frac {d e^{c+d x} \text {csch}(a+b x)}{2 b^2}-\frac {e^{c+d x} \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {1}{2} \left (1-\frac {d^2}{b^2}\right ) \int e^{c+d x} \text {csch}(a+b x) \, dx\\ &=-\frac {d e^{c+d x} \text {csch}(a+b x)}{2 b^2}-\frac {e^{c+d x} \coth (a+b x) \text {csch}(a+b x)}{2 b}+\frac {(b-d) e^{a+c+b x+d x} \, _2F_1\left (1,\frac {b+d}{2 b};\frac {1}{2} \left (3+\frac {d}{b}\right );e^{2 (a+b x)}\right )}{b^2}\\ \end {align*}

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Mathematica [A]  time = 2.49, size = 94, normalized size = 0.94 \[ \frac {e^c \left (\frac {2 \text {csch}(a) (b-d) e^{x (b+d)} \, _2F_1\left (1,\frac {b+d}{2 b};\frac {3 b+d}{2 b};e^{2 b x} (\cosh (a)+\sinh (a))^2\right )}{\coth (a)-1}-e^{d x} \text {csch}(a+b x) (b \coth (a+b x)+d)\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c + d*x)*Csch[a + b*x]^3,x]

[Out]

(E^c*(-(E^(d*x)*(d + b*Coth[a + b*x])*Csch[a + b*x]) + (2*(b - d)*E^((b + d)*x)*Csch[a]*Hypergeometric2F1[1, (
b + d)/(2*b), (3*b + d)/(2*b), E^(2*b*x)*(Cosh[a] + Sinh[a])^2])/(-1 + Coth[a])))/(2*b^2)

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {csch}\left (b x + a\right )^{3} e^{\left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

integral(csch(b*x + a)^3*e^(d*x + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}\left (b x + a\right )^{3} e^{\left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*csch(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(csch(b*x + a)^3*e^(d*x + c), x)

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{d x +c} \mathrm {csch}\left (b x +a \right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(d*x+c)*csch(b*x+a)^3,x)

[Out]

int(exp(d*x+c)*csch(b*x+a)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 48 \, {\left (b^{2} e^{c} + b d e^{c}\right )} \int \frac {e^{\left (b x + d x + a\right )}}{15 \, b^{2} - 8 \, b d + d^{2} + {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (8 \, b x + 8 \, a\right )} - 4 \, {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 6 \, {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 4 \, {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (2 \, b x + 2 \, a\right )}}\,{d x} + \frac {8 \, {\left ({\left (5 \, b e^{c} - d e^{c}\right )} e^{\left (3 \, b x + 3 \, a\right )} - 6 \, b e^{\left (b x + a + c\right )}\right )} e^{\left (d x\right )}}{15 \, b^{2} - 8 \, b d + d^{2} - {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 3 \, {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 3 \, {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (2 \, b x + 2 \, a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

48*(b^2*e^c + b*d*e^c)*integrate(e^(b*x + d*x + a)/(15*b^2 - 8*b*d + d^2 + (15*b^2 - 8*b*d + d^2)*e^(8*b*x + 8
*a) - 4*(15*b^2 - 8*b*d + d^2)*e^(6*b*x + 6*a) + 6*(15*b^2 - 8*b*d + d^2)*e^(4*b*x + 4*a) - 4*(15*b^2 - 8*b*d
+ d^2)*e^(2*b*x + 2*a)), x) + 8*((5*b*e^c - d*e^c)*e^(3*b*x + 3*a) - 6*b*e^(b*x + a + c))*e^(d*x)/(15*b^2 - 8*
b*d + d^2 - (15*b^2 - 8*b*d + d^2)*e^(6*b*x + 6*a) + 3*(15*b^2 - 8*b*d + d^2)*e^(4*b*x + 4*a) - 3*(15*b^2 - 8*
b*d + d^2)*e^(2*b*x + 2*a))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{c+d\,x}}{{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c + d*x)/sinh(a + b*x)^3,x)

[Out]

int(exp(c + d*x)/sinh(a + b*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{c} \int e^{d x} \operatorname {csch}^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*csch(b*x+a)**3,x)

[Out]

exp(c)*Integral(exp(d*x)*csch(a + b*x)**3, x)

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