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

Optimal. Leaf size=125 \[ \frac {2 e^{-2 a-x (2 b-d)+c} \, _2F_1\left (1,\frac {1}{2} \left (\frac {d}{b}-2\right );\frac {d}{2 b};e^{2 (a+b x)}\right )}{2 b-d}-\frac {7 e^{-2 a-x (2 b-d)+c}}{4 (2 b-d)}+\frac {e^{2 a+x (2 b+d)+c}}{4 (2 b+d)}+\frac {e^{c+d x}}{d} \]

[Out]

-7/4*exp(-2*a+c-(2*b-d)*x)/(2*b-d)+exp(d*x+c)/d+1/4*exp(2*a+c+(2*b+d)*x)/(2*b+d)+2*exp(-2*a+c-(2*b-d)*x)*hyper
geom([1, -1+1/2*d/b],[1/2*d/b],exp(2*b*x+2*a))/(2*b-d)

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Rubi [A]  time = 0.25, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5511, 2194, 2227, 2251} \[ \frac {2 e^{-2 a-x (2 b-d)+c} \, _2F_1\left (1,\frac {1}{2} \left (\frac {d}{b}-2\right );\frac {d}{2 b};e^{2 (a+b x)}\right )}{2 b-d}-\frac {7 e^{-2 a-x (2 b-d)+c}}{4 (2 b-d)}+\frac {e^{2 a+x (2 b+d)+c}}{4 (2 b+d)}+\frac {e^{c+d x}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*E^(-2*a + c - (2*b - d)*x))/(4*(2*b - d)) + E^(c + d*x)/d + E^(2*a + c + (2*b + d)*x)/(4*(2*b + d)) + (2*E
^(-2*a + c - (2*b - d)*x)*Hypergeometric2F1[1, (-2 + d/b)/2, d/(2*b), E^(2*(a + b*x))])/(2*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 ^2(a+b x) \coth (a+b x) \, dx &=\int \left (\frac {7}{4} e^{-2 a+c-(2 b-d) x}+e^{-2 a+c-(2 b-d) x+2 (a+b x)}+\frac {1}{4} e^{-2 a+c-(2 b-d) x+4 (a+b x)}+\frac {2 e^{-2 a+c-(2 b-d) x}}{-1+e^{2 (a+b x)}}\right ) \, dx\\ &=\frac {1}{4} \int e^{-2 a+c-(2 b-d) x+4 (a+b x)} \, dx+\frac {7}{4} \int e^{-2 a+c-(2 b-d) x} \, dx+2 \int \frac {e^{-2 a+c-(2 b-d) x}}{-1+e^{2 (a+b x)}} \, dx+\int e^{-2 a+c-(2 b-d) x+2 (a+b x)} \, dx\\ &=-\frac {7 e^{-2 a+c-(2 b-d) x}}{4 (2 b-d)}+\frac {2 e^{-2 a+c-(2 b-d) x} \, _2F_1\left (1,\frac {1}{2} \left (-2+\frac {d}{b}\right );\frac {d}{2 b};e^{2 (a+b x)}\right )}{2 b-d}+\frac {1}{4} \int e^{2 a+c+(2 b+d) x} \, dx+\int e^{c+d x} \, dx\\ &=-\frac {7 e^{-2 a+c-(2 b-d) x}}{4 (2 b-d)}+\frac {e^{c+d x}}{d}+\frac {e^{2 a+c+(2 b+d) x}}{4 (2 b+d)}+\frac {2 e^{-2 a+c-(2 b-d) x} \, _2F_1\left (1,\frac {1}{2} \left (-2+\frac {d}{b}\right );\frac {d}{2 b};e^{2 (a+b x)}\right )}{2 b-d}\\ \end {align*}

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Mathematica [A]  time = 1.10, size = 172, normalized size = 1.38 \[ -\frac {e^{c-\frac {a d}{b}} \left (2 \left (4 b^2-d^2\right ) e^{d \left (\frac {a}{b}+x\right )} \, _2F_1\left (1,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )+2 d (2 b-d) e^{\left (\frac {d}{b}+2\right ) (a+b x)} \, _2F_1\left (1,\frac {d}{2 b}+1;\frac {d}{2 b}+2;e^{2 (a+b x)}\right )+d e^{d \left (\frac {a}{b}+x\right )} (d \sinh (2 (a+b x))-2 b \cosh (2 (a+b x)))\right )}{8 b^2 d-2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((E^(c - (a*d)/b)*(2*(4*b^2 - d^2)*E^(d*(a/b + x))*Hypergeometric2F1[1, d/(2*b), 1 + d/(2*b), E^(2*(a + b*x))
] + 2*(2*b - d)*d*E^((2 + d/b)*(a + b*x))*Hypergeometric2F1[1, 1 + d/(2*b), 2 + d/(2*b), E^(2*(a + b*x))] + d*
E^(d*(a/b + x))*(-2*b*Cosh[2*(a + b*x)] + d*Sinh[2*(a + b*x)])))/(8*b^2*d - 2*d^3))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cosh \left (b x + a\right )^{3} \operatorname {csch}\left (b x + a\right ) e^{\left (d x + c\right )}, x\right ) \]

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),x, algorithm="fricas")

[Out]

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

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

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),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.83, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{d x +c} \left (\cosh ^{3}\left (b x +a \right )\right ) \mathrm {csch}\left (b x +a \right )\, dx \]

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),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -4 \, b \int \frac {e^{\left (d x + c\right )}}{{\left (4 \, b - d\right )} e^{\left (6 \, b x + 6 \, a\right )} - 2 \, {\left (4 \, b - d\right )} e^{\left (4 \, b x + 4 \, a\right )} + {\left (4 \, b - d\right )} e^{\left (2 \, b x + 2 \, a\right )}}\,{d x} + \frac {{\left (24 \, b^{2} d e^{c} + 14 \, b d^{2} e^{c} + d^{3} e^{c} + {\left (8 \, b^{2} d e^{c} - 6 \, b d^{2} e^{c} + d^{3} e^{c}\right )} e^{\left (6 \, b x + 6 \, a\right )} + {\left (64 \, b^{3} e^{c} - 24 \, b^{2} d e^{c} - 10 \, b d^{2} e^{c} + 3 \, d^{3} e^{c}\right )} e^{\left (4 \, b x + 4 \, a\right )} - {\left (64 \, b^{3} e^{c} + 40 \, b^{2} d e^{c} - 2 \, b d^{2} e^{c} - 3 \, d^{3} e^{c}\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )} e^{\left (d x\right )}}{4 \, {\left ({\left (16 \, b^{3} d - 4 \, b^{2} d^{2} - 4 \, b d^{3} + d^{4}\right )} e^{\left (4 \, b x + 4 \, a\right )} - {\left (16 \, b^{3} d - 4 \, b^{2} d^{2} - 4 \, b d^{3} + d^{4}\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )}} \]

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),x, algorithm="maxima")

[Out]

-4*b*integrate(e^(d*x + c)/((4*b - d)*e^(6*b*x + 6*a) - 2*(4*b - d)*e^(4*b*x + 4*a) + (4*b - d)*e^(2*b*x + 2*a
)), x) + 1/4*(24*b^2*d*e^c + 14*b*d^2*e^c + d^3*e^c + (8*b^2*d*e^c - 6*b*d^2*e^c + d^3*e^c)*e^(6*b*x + 6*a) +
(64*b^3*e^c - 24*b^2*d*e^c - 10*b*d^2*e^c + 3*d^3*e^c)*e^(4*b*x + 4*a) - (64*b^3*e^c + 40*b^2*d*e^c - 2*b*d^2*
e^c - 3*d^3*e^c)*e^(2*b*x + 2*a))*e^(d*x)/((16*b^3*d - 4*b^2*d^2 - 4*b*d^3 + d^4)*e^(4*b*x + 4*a) - (16*b^3*d
- 4*b^2*d^2 - 4*b*d^3 + d^4)*e^(2*b*x + 2*a))

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

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),x)

[Out]

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

[Out]

Timed out

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