3.957 \(\int e^{c+d x} \coth ^2(a+b x) \, dx\)

Optimal. Leaf size=94 \[ -\frac {4 e^{c+d x} \, _2F_1\left (1,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )}{d}+\frac {4 e^{c+d x} \, _2F_1\left (2,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )}{d}+\frac {e^{c+d x}}{d} \]

[Out]

exp(d*x+c)/d-4*exp(d*x+c)*hypergeom([1, 1/2*d/b],[1+1/2*d/b],exp(2*b*x+2*a))/d+4*exp(d*x+c)*hypergeom([2, 1/2*
d/b],[1+1/2*d/b],exp(2*b*x+2*a))/d

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Rubi [A]  time = 0.11, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5485, 2194, 2251} \[ -\frac {4 e^{c+d x} \, _2F_1\left (1,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )}{d}+\frac {4 e^{c+d x} \, _2F_1\left (2,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )}{d}+\frac {e^{c+d x}}{d} \]

Antiderivative was successfully verified.

[In]

Int[E^(c + d*x)*Coth[a + b*x]^2,x]

[Out]

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rule 5485

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

Rubi steps

\begin {align*} \int e^{c+d x} \coth ^2(a+b x) \, dx &=\int \left (e^{c+d x}+\frac {4 e^{c+d x}}{\left (-1+e^{2 (a+b x)}\right )^2}+\frac {4 e^{c+d x}}{-1+e^{2 (a+b x)}}\right ) \, dx\\ &=4 \int \frac {e^{c+d x}}{\left (-1+e^{2 (a+b x)}\right )^2} \, dx+4 \int \frac {e^{c+d x}}{-1+e^{2 (a+b x)}} \, dx+\int e^{c+d x} \, dx\\ &=\frac {e^{c+d x}}{d}-\frac {4 e^{c+d x} \, _2F_1\left (1,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d}+\frac {4 e^{c+d x} \, _2F_1\left (2,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 145, normalized size = 1.54 \[ -\frac {2 d e^{2 a+c} \left (\frac {e^{d x} \, _2F_1\left (1,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )}{d}-\frac {e^{x (2 b+d)} \, _2F_1\left (1,\frac {d}{2 b}+1;\frac {d}{2 b}+2;e^{2 (a+b x)}\right )}{2 b+d}\right )}{\left (e^{2 a}-1\right ) b}+\frac {\text {csch}(a) \sinh (b x) e^{c+d x} \text {csch}(a+b x)}{b}+\frac {e^{c+d x}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c + d*x)*Coth[a + b*x]^2,x]

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cosh \left (b x + a\right )^{2} \operatorname {csch}\left (b x + a\right )^{2} e^{\left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.68, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{d x +c} \left (\cosh ^{2}\left (b x +a \right )\right ) \mathrm {csch}\left (b x +a \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a)^2,x)

[Out]

int(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 16 \, b d \int -\frac {e^{\left (d x + c\right )}}{8 \, b^{2} - 6 \, b d + d^{2} - {\left (8 \, b^{2} - 6 \, b d + d^{2}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 3 \, {\left (8 \, b^{2} - 6 \, b d + d^{2}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 3 \, {\left (8 \, b^{2} - 6 \, b d + d^{2}\right )} e^{\left (2 \, b x + 2 \, a\right )}}\,{d x} + \frac {{\left (8 \, b^{2} e^{c} + 10 \, b d e^{c} + d^{2} e^{c} + {\left (8 \, b^{2} e^{c} - 6 \, b d e^{c} + d^{2} e^{c}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, {\left (8 \, b^{2} e^{c} + 2 \, b d e^{c} - d^{2} e^{c}\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )} e^{\left (d x\right )}}{8 \, b^{2} d - 6 \, b d^{2} + d^{3} + {\left (8 \, b^{2} d - 6 \, b d^{2} + d^{3}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, {\left (8 \, b^{2} d - 6 \, b d^{2} + d^{3}\right )} e^{\left (2 \, b x + 2 \, a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*b*d*integrate(-e^(d*x + c)/(8*b^2 - 6*b*d + d^2 - (8*b^2 - 6*b*d + d^2)*e^(6*b*x + 6*a) + 3*(8*b^2 - 6*b*d
+ d^2)*e^(4*b*x + 4*a) - 3*(8*b^2 - 6*b*d + d^2)*e^(2*b*x + 2*a)), x) + (8*b^2*e^c + 10*b*d*e^c + d^2*e^c + (8
*b^2*e^c - 6*b*d*e^c + d^2*e^c)*e^(4*b*x + 4*a) - 2*(8*b^2*e^c + 2*b*d*e^c - d^2*e^c)*e^(2*b*x + 2*a))*e^(d*x)
/(8*b^2*d - 6*b*d^2 + d^3 + (8*b^2*d - 6*b*d^2 + d^3)*e^(4*b*x + 4*a) - 2*(8*b^2*d - 6*b*d^2 + d^3)*e^(2*b*x +
2*a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)^2*exp(c + d*x))/sinh(a + b*x)^2,x)

[Out]

int((cosh(a + b*x)^2*exp(c + d*x))/sinh(a + b*x)^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)**2*csch(b*x+a)**2,x)

[Out]

Timed out

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