3.71 \(\int (1+\coth (x))^{5/2} \, dx\)

Optimal. Leaf size=45 \[ -\frac {2}{3} (\coth (x)+1)^{3/2}-4 \sqrt {\coth (x)+1}+4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\coth (x)+1}}{\sqrt {2}}\right ) \]

[Out]

-2/3*(1+coth(x))^(3/2)+4*arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))*2^(1/2)-4*(1+coth(x))^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3478, 3480, 206} \[ -\frac {2}{3} (\coth (x)+1)^{3/2}-4 \sqrt {\coth (x)+1}+4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\coth (x)+1}}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Coth[x])^(5/2),x]

[Out]

4*Sqrt[2]*ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]] - 4*Sqrt[1 + Coth[x]] - (2*(1 + Coth[x])^(3/2))/3

Rule 206

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

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int (1+\coth (x))^{5/2} \, dx &=-\frac {2}{3} (1+\coth (x))^{3/2}+2 \int (1+\coth (x))^{3/2} \, dx\\ &=-4 \sqrt {1+\coth (x)}-\frac {2}{3} (1+\coth (x))^{3/2}+4 \int \sqrt {1+\coth (x)} \, dx\\ &=-4 \sqrt {1+\coth (x)}-\frac {2}{3} (1+\coth (x))^{3/2}+8 \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1+\coth (x)}\right )\\ &=4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1+\coth (x)}}{\sqrt {2}}\right )-4 \sqrt {1+\coth (x)}-\frac {2}{3} (1+\coth (x))^{3/2}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 92, normalized size = 2.04 \[ -\frac {2 \sinh (x) (\coth (x)+1)^{5/2} \left (\cosh (x) \sqrt {i (\coth (x)+1)}+\sinh (x) \left (7 \sqrt {i (\coth (x)+1)}-(6-6 i) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {i (\coth (x)+1)}\right )\right )\right )}{3 \sqrt {i (\coth (x)+1)} (\sinh (x)+\cosh (x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Coth[x])^(5/2),x]

[Out]

(-2*(1 + Coth[x])^(5/2)*Sinh[x]*(Cosh[x]*Sqrt[I*(1 + Coth[x])] + ((-6 + 6*I)*ArcTan[(1/2 + I/2)*Sqrt[I*(1 + Co
th[x])]] + 7*Sqrt[I*(1 + Coth[x])])*Sinh[x]))/(3*Sqrt[I*(1 + Coth[x])]*(Cosh[x] + Sinh[x])^2)

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fricas [B]  time = 0.41, size = 259, normalized size = 5.76 \[ -\frac {2 \, {\left (2 \, \sqrt {2} {\left (4 \, \sqrt {2} \cosh \relax (x)^{3} + 12 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{2} + 4 \, \sqrt {2} \sinh \relax (x)^{3} + 3 \, {\left (4 \, \sqrt {2} \cosh \relax (x)^{2} - \sqrt {2}\right )} \sinh \relax (x) - 3 \, \sqrt {2} \cosh \relax (x)\right )} \sqrt {\frac {\sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} - 3 \, {\left (\sqrt {2} \cosh \relax (x)^{4} + 4 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{3} + \sqrt {2} \sinh \relax (x)^{4} + 2 \, {\left (3 \, \sqrt {2} \cosh \relax (x)^{2} - \sqrt {2}\right )} \sinh \relax (x)^{2} - 2 \, \sqrt {2} \cosh \relax (x)^{2} + 4 \, {\left (\sqrt {2} \cosh \relax (x)^{3} - \sqrt {2} \cosh \relax (x)\right )} \sinh \relax (x) + \sqrt {2}\right )} \log \left (2 \, \sqrt {2} \sqrt {\frac {\sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} + 2 \, \cosh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x) + 2 \, \sinh \relax (x)^{2} - 1\right )\right )}}{3 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + 2 \, {\left (3 \, \cosh \relax (x)^{2} - 1\right )} \sinh \relax (x)^{2} - 2 \, \cosh \relax (x)^{2} + 4 \, {\left (\cosh \relax (x)^{3} - \cosh \relax (x)\right )} \sinh \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^(5/2),x, algorithm="fricas")

[Out]

-2/3*(2*sqrt(2)*(4*sqrt(2)*cosh(x)^3 + 12*sqrt(2)*cosh(x)*sinh(x)^2 + 4*sqrt(2)*sinh(x)^3 + 3*(4*sqrt(2)*cosh(
x)^2 - sqrt(2))*sinh(x) - 3*sqrt(2)*cosh(x))*sqrt(sinh(x)/(cosh(x) - sinh(x))) - 3*(sqrt(2)*cosh(x)^4 + 4*sqrt
(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)^4 + 2*(3*sqrt(2)*cosh(x)^2 - sqrt(2))*sinh(x)^2 - 2*sqrt(2)*cosh(x)^2
+ 4*(sqrt(2)*cosh(x)^3 - sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(2*sqrt(2)*sqrt(sinh(x)/(cosh(x) - sinh(x)))*(
cosh(x) + sinh(x)) + 2*cosh(x)^2 + 4*cosh(x)*sinh(x) + 2*sinh(x)^2 - 1))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + si
nh(x)^4 + 2*(3*cosh(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 4*(cosh(x)^3 - cosh(x))*sinh(x) + 1)

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giac [B]  time = 0.15, size = 112, normalized size = 2.49 \[ -\frac {2}{3} \, \sqrt {2} {\left (\frac {2 \, {\left (6 \, {\left (\sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} + 9 \, \sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 9 \, e^{\left (2 \, x\right )} + 4\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} + 1\right )}^{3}} + 3 \, \log \left ({\left | 2 \, \sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 2 \, e^{\left (2 \, x\right )} + 1 \right |}\right )\right )} \mathrm {sgn}\left (e^{\left (2 \, x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^(5/2),x, algorithm="giac")

[Out]

-2/3*sqrt(2)*(2*(6*(sqrt(e^(4*x) - e^(2*x)) - e^(2*x))^2 + 9*sqrt(e^(4*x) - e^(2*x)) - 9*e^(2*x) + 4)/(sqrt(e^
(4*x) - e^(2*x)) - e^(2*x) + 1)^3 + 3*log(abs(2*sqrt(e^(4*x) - e^(2*x)) - 2*e^(2*x) + 1)))*sgn(e^(2*x) - 1)

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maple [A]  time = 0.06, size = 35, normalized size = 0.78 \[ -\frac {2 \left (1+\coth \relax (x )\right )^{\frac {3}{2}}}{3}+4 \arctanh \left (\frac {\sqrt {1+\coth \relax (x )}\, \sqrt {2}}{2}\right ) \sqrt {2}-4 \sqrt {1+\coth \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+coth(x))^(5/2),x)

[Out]

-2/3*(1+coth(x))^(3/2)+4*arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))*2^(1/2)-4*(1+coth(x))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\coth \relax (x) + 1\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^(5/2),x, algorithm="maxima")

[Out]

integrate((coth(x) + 1)^(5/2), x)

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mupad [B]  time = 1.19, size = 54, normalized size = 1.20 \[ \sqrt {8}\,\ln \left (-2\,\sqrt {8}\,\sqrt {\mathrm {coth}\relax (x)+1}-8\right )-\frac {2\,{\left (\mathrm {coth}\relax (x)+1\right )}^{3/2}}{3}-2\,\sqrt {2}\,\ln \left (4\,\sqrt {2}\,\sqrt {\mathrm {coth}\relax (x)+1}-8\right )-4\,\sqrt {\mathrm {coth}\relax (x)+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((coth(x) + 1)^(5/2),x)

[Out]

8^(1/2)*log(- 2*8^(1/2)*(coth(x) + 1)^(1/2) - 8) - (2*(coth(x) + 1)^(3/2))/3 - 2*2^(1/2)*log(4*2^(1/2)*(coth(x
) + 1)^(1/2) - 8) - 4*(coth(x) + 1)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\coth {\relax (x )} + 1\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))**(5/2),x)

[Out]

Integral((coth(x) + 1)**(5/2), x)

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