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

Optimal. Leaf size=160 \[ \frac {6 e^{-a-x (b-d)+c} \, _2F_1\left (1,-\frac {b-d}{2 b};\frac {b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}-\frac {4 e^{-a-x (b-d)+c} \, _2F_1\left (2,-\frac {b-d}{2 b};\frac {b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}-\frac {5 e^{-a-x (b-d)+c}}{2 (b-d)}+\frac {e^{a+x (b+d)+c}}{2 (b+d)} \]

[Out]

-5/2*exp(-a+c-(b-d)*x)/(b-d)+1/2*exp(a+c+(b+d)*x)/(b+d)+6*exp(-a+c-(b-d)*x)*hypergeom([1, 1/2*(-b+d)/b],[1/2*(
b+d)/b],exp(2*b*x+2*a))/(b-d)-4*exp(-a+c-(b-d)*x)*hypergeom([2, 1/2*(-b+d)/b],[1/2*(b+d)/b],exp(2*b*x+2*a))/(b
-d)

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Rubi [A]  time = 0.31, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5511, 2194, 2227, 2251} \[ \frac {6 e^{-a-x (b-d)+c} \, _2F_1\left (1,-\frac {b-d}{2 b};\frac {b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}-\frac {4 e^{-a-x (b-d)+c} \, _2F_1\left (2,-\frac {b-d}{2 b};\frac {b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}-\frac {5 e^{-a-x (b-d)+c}}{2 (b-d)}+\frac {e^{a+x (b+d)+c}}{2 (b+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*E^(-a + c - (b - d)*x))/(2*(b - d)) + E^(a + c + (b + d)*x)/(2*(b + d)) + (6*E^(-a + c - (b - d)*x)*Hyperg
eometric2F1[1, -(b - d)/(2*b), (b + d)/(2*b), E^(2*(a + b*x))])/(b - d) - (4*E^(-a + c - (b - d)*x)*Hypergeome
tric2F1[2, -(b - d)/(2*b), (b + d)/(2*b), E^(2*(a + b*x))])/(b - 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 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, 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 5511

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol]
:> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
IGtQ[m, 0] && IGtQ[n, 0] && HyperbolicQ[G] && HyperbolicQ[H]

Rubi steps

\begin {align*} \int e^{c+d x} \cosh (a+b x) \coth ^2(a+b x) \, dx &=\int \left (\frac {5}{2} e^{-a+c-(b-d) x}+\frac {1}{2} e^{-a+c-(b-d) x+2 (a+b x)}+\frac {4 e^{-a+c-(b-d) x}}{\left (-1+e^{2 (a+b x)}\right )^2}+\frac {6 e^{-a+c-(b-d) x}}{-1+e^{2 (a+b x)}}\right ) \, dx\\ &=\frac {1}{2} \int e^{-a+c-(b-d) x+2 (a+b x)} \, dx+\frac {5}{2} \int e^{-a+c-(b-d) x} \, dx+4 \int \frac {e^{-a+c-(b-d) x}}{\left (-1+e^{2 (a+b x)}\right )^2} \, dx+6 \int \frac {e^{-a+c-(b-d) x}}{-1+e^{2 (a+b x)}} \, dx\\ &=-\frac {5 e^{-a+c-(b-d) x}}{2 (b-d)}+\frac {6 e^{-a+c-(b-d) x} \, _2F_1\left (1,-\frac {b-d}{2 b};\frac {b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}-\frac {4 e^{-a+c-(b-d) x} \, _2F_1\left (2,-\frac {b-d}{2 b};\frac {b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}+\frac {1}{2} \int e^{a+c+(b+d) x} \, dx\\ &=-\frac {5 e^{-a+c-(b-d) x}}{2 (b-d)}+\frac {e^{a+c+(b+d) x}}{2 (b+d)}+\frac {6 e^{-a+c-(b-d) x} \, _2F_1\left (1,-\frac {b-d}{2 b};\frac {b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}-\frac {4 e^{-a+c-(b-d) x} \, _2F_1\left (2,-\frac {b-d}{2 b};\frac {b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}\\ \end {align*}

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Mathematica [A]  time = 1.26, size = 145, normalized size = 0.91 \[ \frac {e^{c-\frac {a d}{b}} \text {csch}(a+b x) \left (e^{d \left (\frac {a}{b}+x\right )} \left (b^2 \cosh (2 (a+b x))-b d \sinh (2 (a+b x))-3 b^2+2 d^2\right )-4 d (b-d) e^{\frac {(b+d) (a+b x)}{b}} \sinh (a+b x) \, _2F_1\left (1,\frac {b+d}{2 b};\frac {3 b+d}{2 b};e^{2 (a+b x)}\right )\right )}{2 b (b-d) (b+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(c - (a*d)/b)*Csch[a + b*x]*(-4*(b - d)*d*E^(((b + d)*(a + b*x))/b)*Hypergeometric2F1[1, (b + d)/(2*b), (3*
b + d)/(2*b), E^(2*(a + b*x))]*Sinh[a + b*x] + E^(d*(a/b + x))*(-3*b^2 + 2*d^2 + b^2*Cosh[2*(a + b*x)] - b*d*S
inh[2*(a + b*x)])))/(2*b*(b - d)*(b + d))

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cosh \left (b x + a\right )^{3} \operatorname {csch}\left (b x + a\right )^{2} e^{\left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(cosh(b*x + a)^3*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 \[ \int \cosh \left (b x + a\right )^{3} \operatorname {csch}\left (b x + a\right )^{2} e^{\left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)^3*csch(b*x + a)^2*e^(d*x + c), x)

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maple [F]  time = 0.80, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{d x +c} \left (\cosh ^{3}\left (b x +a \right )\right ) \mathrm {csch}\left (b x +a \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(d*x+c)*cosh(b*x+a)^3*csch(b*x+a)^2,x)

[Out]

int(exp(d*x+c)*cosh(b*x+a)^3*csch(b*x+a)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 16 \, b d \int \frac {e^{\left (d x + c\right )}}{{\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (7 \, b x + 7 \, a\right )} - 3 \, {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (5 \, b x + 5 \, a\right )} + 3 \, {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (3 \, b x + 3 \, a\right )} - {\left (15 \, b^{2} - 8 \, b d + d^{2}\right )} e^{\left (b x + a\right )}}\,{d x} - \frac {{\left (15 \, b^{3} e^{c} + 39 \, b^{2} d e^{c} + 25 \, b d^{2} e^{c} + d^{3} e^{c} - {\left (15 \, b^{3} e^{c} - 23 \, b^{2} d e^{c} + 9 \, b d^{2} e^{c} - d^{3} e^{c}\right )} e^{\left (6 \, b x + 6 \, a\right )} + {\left (105 \, b^{3} e^{c} - 11 \, b^{2} d e^{c} - 17 \, b d^{2} e^{c} + 3 \, d^{3} e^{c}\right )} e^{\left (4 \, b x + 4 \, a\right )} - {\left (105 \, b^{3} e^{c} + 59 \, b^{2} d e^{c} - b d^{2} e^{c} - 3 \, d^{3} e^{c}\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )} e^{\left (d x\right )}}{2 \, {\left ({\left (15 \, b^{4} - 8 \, b^{3} d - 14 \, b^{2} d^{2} + 8 \, b d^{3} - d^{4}\right )} e^{\left (5 \, b x + 5 \, a\right )} - 2 \, {\left (15 \, b^{4} - 8 \, b^{3} d - 14 \, b^{2} d^{2} + 8 \, b d^{3} - d^{4}\right )} e^{\left (3 \, b x + 3 \, a\right )} + {\left (15 \, b^{4} - 8 \, b^{3} d - 14 \, b^{2} d^{2} + 8 \, b d^{3} - d^{4}\right )} e^{\left (b x + a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

16*b*d*integrate(e^(d*x + c)/((15*b^2 - 8*b*d + d^2)*e^(7*b*x + 7*a) - 3*(15*b^2 - 8*b*d + d^2)*e^(5*b*x + 5*a
) + 3*(15*b^2 - 8*b*d + d^2)*e^(3*b*x + 3*a) - (15*b^2 - 8*b*d + d^2)*e^(b*x + a)), x) - 1/2*(15*b^3*e^c + 39*
b^2*d*e^c + 25*b*d^2*e^c + d^3*e^c - (15*b^3*e^c - 23*b^2*d*e^c + 9*b*d^2*e^c - d^3*e^c)*e^(6*b*x + 6*a) + (10
5*b^3*e^c - 11*b^2*d*e^c - 17*b*d^2*e^c + 3*d^3*e^c)*e^(4*b*x + 4*a) - (105*b^3*e^c + 59*b^2*d*e^c - b*d^2*e^c
- 3*d^3*e^c)*e^(2*b*x + 2*a))*e^(d*x)/((15*b^4 - 8*b^3*d - 14*b^2*d^2 + 8*b*d^3 - d^4)*e^(5*b*x + 5*a) - 2*(1
5*b^4 - 8*b^3*d - 14*b^2*d^2 + 8*b*d^3 - d^4)*e^(3*b*x + 3*a) + (15*b^4 - 8*b^3*d - 14*b^2*d^2 + 8*b*d^3 - d^4
)*e^(b*x + a))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {e}}^{c+d\,x}}{{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)^3*exp(c + d*x))/sinh(a + b*x)^2,x)

[Out]

int((cosh(a + b*x)^3*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)**3*csch(b*x+a)**2,x)

[Out]

Timed out

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