3.212 \(\int e^{a+b x} \coth ^3(a+b x) \, dx\)

Optimal. Leaf size=81 \[ \frac {e^{a+b x}}{b}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

exp(b*x+a)/b-2*exp(b*x+a)/b/(1-exp(2*b*x+2*a))^2+3*exp(b*x+a)/b/(1-exp(2*b*x+2*a))-3*arctanh(exp(b*x+a))/b

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Rubi [A]  time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2282, 390, 1158, 12, 288, 207} \[ \frac {e^{a+b x}}{b}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Coth[a + b*x]^3,x]

[Out]

E^(a + b*x)/b - (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))^2) + (3*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (3*A
rcTanh[E^(a + b*x)])/b

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1158

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + c*
x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, -Simp[(R*x*(d + e*x
^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*
(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {align*} \int e^{a+b x} \coth ^3(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (1+\frac {2 \left (1+3 x^4\right )}{\left (-1+x^2\right )^3}\right ) \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}+\frac {2 \operatorname {Subst}\left (\int \frac {1+3 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {12 x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {6 \operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}

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Mathematica [C]  time = 2.37, size = 286, normalized size = 3.53 \[ -\frac {e^{-5 (a+b x)} \left (256 e^{8 (a+b x)} \left (e^{2 (a+b x)}+1\right )^3 \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {11}{2};e^{2 (a+b x)}\right )+384 e^{8 (a+b x)} \left (5 e^{2 (a+b x)}+7\right ) \left (e^{2 (a+b x)}+1\right )^2 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};e^{2 (a+b x)}\right )-21 \left (507305 e^{2 (a+b x)}+173916 e^{4 (a+b x)}-154296 e^{6 (a+b x)}-73885 e^{8 (a+b x)}+4887 e^{10 (a+b x)}+252105\right )-\frac {315 \left (-28218 e^{2 (a+b x)}+1173 e^{4 (a+b x)}+17748 e^{6 (a+b x)}+4299 e^{8 (a+b x)}-1434 e^{10 (a+b x)}+7 e^{12 (a+b x)}-16807\right ) \tanh ^{-1}\left (\sqrt {e^{2 (a+b x)}}\right )}{\sqrt {e^{2 (a+b x)}}}\right )}{60480 b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(a + b*x)*Coth[a + b*x]^3,x]

[Out]

-1/60480*(-21*(252105 + 507305*E^(2*(a + b*x)) + 173916*E^(4*(a + b*x)) - 154296*E^(6*(a + b*x)) - 73885*E^(8*
(a + b*x)) + 4887*E^(10*(a + b*x))) - (315*(-16807 - 28218*E^(2*(a + b*x)) + 1173*E^(4*(a + b*x)) + 17748*E^(6
*(a + b*x)) + 4299*E^(8*(a + b*x)) - 1434*E^(10*(a + b*x)) + 7*E^(12*(a + b*x)))*ArcTanh[Sqrt[E^(2*(a + b*x))]
])/Sqrt[E^(2*(a + b*x))] + 384*E^(8*(a + b*x))*(1 + E^(2*(a + b*x)))^2*(7 + 5*E^(2*(a + b*x)))*HypergeometricP
FQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, E^(2*(a + b*x))] + 256*E^(8*(a + b*x))*(1 + E^(2*(a + b*x)))^3*Hypergeo
metricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 11/2}, E^(2*(a + b*x))])/(b*E^(5*(a + b*x)))

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fricas [B]  time = 0.63, size = 459, normalized size = 5.67 \[ \frac {2 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \, \sinh \left (b x + a\right )^{5} + 10 \, {\left (2 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{3} - 10 \, \cosh \left (b x + a\right )^{3} + 10 \, {\left (2 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 15 \, \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right ) + 4 \, \cosh \left (b x + a\right )}{2 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*cosh(b*x + a)^5 + 10*cosh(b*x + a)*sinh(b*x + a)^4 + 2*sinh(b*x + a)^5 + 10*(2*cosh(b*x + a)^2 - 1)*sin
h(b*x + a)^3 - 10*cosh(b*x + a)^3 + 10*(2*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^2 - 3*(cosh(b*x + a
)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b
*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 3*
(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)
^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x +
a) - 1) + 2*(5*cosh(b*x + a)^4 - 15*cosh(b*x + a)^2 + 2)*sinh(b*x + a) + 4*cosh(b*x + a))/(b*cosh(b*x + a)^4
+ 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)*si
nh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a) + b)

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giac [A]  time = 0.13, size = 72, normalized size = 0.89 \[ -\frac {\frac {2 \, {\left (3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 2 \, e^{\left (b x + a\right )} + 3 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 3 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(3*e^(3*b*x + 3*a) - e^(b*x + a))/(e^(2*b*x + 2*a) - 1)^2 - 2*e^(b*x + a) + 3*log(e^(b*x + a) + 1) - 3
*log(abs(e^(b*x + a) - 1)))/b

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maple [A]  time = 0.27, size = 89, normalized size = 1.10 \[ \frac {\frac {\cosh ^{2}\left (b x +a \right )}{\sinh \left (b x +a \right )}-\frac {2}{\sinh \left (b x +a \right )}+\frac {\cosh ^{3}\left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}-\frac {3 \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {3 \,\mathrm {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}-3 \arctanh \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*coth(b*x+a)^3,x)

[Out]

1/b*(cosh(b*x+a)^2/sinh(b*x+a)-2/sinh(b*x+a)+cosh(b*x+a)^3/sinh(b*x+a)^2-3/sinh(b*x+a)^2*cosh(b*x+a)+3/2*csch(
b*x+a)*coth(b*x+a)-3*arctanh(exp(b*x+a)))

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maxima [A]  time = 0.31, size = 88, normalized size = 1.09 \[ \frac {e^{\left (b x + a\right )}}{b} - \frac {3 \, \log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac {3 \, \log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac {3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(b*x + a)/b - 3/2*log(e^(b*x + a) + 1)/b + 3/2*log(e^(b*x + a) - 1)/b - (3*e^(3*b*x + 3*a) - e^(b*x + a))/(b
*(e^(4*b*x + 4*a) - 2*e^(2*b*x + 2*a) + 1))

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mupad [B]  time = 1.11, size = 97, normalized size = 1.20 \[ \frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {3\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + b*x)^3*exp(a + b*x),x)

[Out]

exp(a + b*x)/b - (3*atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b))/(-b^2)^(1/2) - (2*exp(a + b*x))/(b*(exp(4*a + 4*b*
x) - 2*exp(2*a + 2*b*x) + 1)) - (3*exp(a + b*x))/(b*(exp(2*a + 2*b*x) - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*coth(b*x+a)**3,x)

[Out]

exp(a)*Integral(exp(b*x)*coth(a + b*x)**3, x)

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