3.70 \(\int (1+\coth (x))^{7/2} \, dx\)

Optimal. Leaf size=57 \[ -\frac{2}{5} (\coth (x)+1)^{5/2}-\frac{4}{3} (\coth (x)+1)^{3/2}-8 \sqrt{\coth (x)+1}+8 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\coth (x)+1}}{\sqrt{2}}\right ) \]

[Out]

8*Sqrt[2]*ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]] - 8*Sqrt[1 + Coth[x]] - (4*(1 + Coth[x])^(3/2))/3 - (2*(1 + Coth[
x])^(5/2))/5

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Rubi [A]  time = 0.0419785, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3478, 3480, 206} \[ -\frac{2}{5} (\coth (x)+1)^{5/2}-\frac{4}{3} (\coth (x)+1)^{3/2}-8 \sqrt{\coth (x)+1}+8 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\coth (x)+1}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Coth[x])^(7/2),x]

[Out]

8*Sqrt[2]*ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]] - 8*Sqrt[1 + Coth[x]] - (4*(1 + Coth[x])^(3/2))/3 - (2*(1 + Coth[
x])^(5/2))/5

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]

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])

Rubi steps

\begin{align*} \int (1+\coth (x))^{7/2} \, dx &=-\frac{2}{5} (1+\coth (x))^{5/2}+2 \int (1+\coth (x))^{5/2} \, dx\\ &=-\frac{4}{3} (1+\coth (x))^{3/2}-\frac{2}{5} (1+\coth (x))^{5/2}+4 \int (1+\coth (x))^{3/2} \, dx\\ &=-8 \sqrt{1+\coth (x)}-\frac{4}{3} (1+\coth (x))^{3/2}-\frac{2}{5} (1+\coth (x))^{5/2}+8 \int \sqrt{1+\coth (x)} \, dx\\ &=-8 \sqrt{1+\coth (x)}-\frac{4}{3} (1+\coth (x))^{3/2}-\frac{2}{5} (1+\coth (x))^{5/2}+16 \operatorname{Subst}\left (\int \frac{1}{2-x^2} \, dx,x,\sqrt{1+\coth (x)}\right )\\ &=8 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1+\coth (x)}}{\sqrt{2}}\right )-8 \sqrt{1+\coth (x)}-\frac{4}{3} (1+\coth (x))^{3/2}-\frac{2}{5} (1+\coth (x))^{5/2}\\ \end{align*}

Mathematica [C]  time = 0.258514, size = 101, normalized size = 1.77 \[ -\frac{2 (\coth (x)+1)^{7/2} \left ((8 \sinh (2 x)+3) \sinh (x) \sqrt{i (\coth (x)+1)}+4 \sinh ^3(x) \left (19 \sqrt{i (\coth (x)+1)}-(15-15 i) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{i (\coth (x)+1)}\right )\right )\right )}{15 \sqrt{i (\coth (x)+1)} (\sinh (x)+\cosh (x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Coth[x])^(7/2),x]

[Out]

(-2*(1 + Coth[x])^(7/2)*(4*((-15 + 15*I)*ArcTan[(1/2 + I/2)*Sqrt[I*(1 + Coth[x])]] + 19*Sqrt[I*(1 + Coth[x])])
*Sinh[x]^3 + Sqrt[I*(1 + Coth[x])]*Sinh[x]*(3 + 8*Sinh[2*x])))/(15*Sqrt[I*(1 + Coth[x])]*(Cosh[x] + Sinh[x])^3
)

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Maple [A]  time = 0.016, size = 43, normalized size = 0.8 \begin{align*} -{\frac{4}{3} \left ( 1+{\rm coth} \left (x\right ) \right ) ^{{\frac{3}{2}}}}-{\frac{2}{5} \left ( 1+{\rm coth} \left (x\right ) \right ) ^{{\frac{5}{2}}}}+8\,{\it Artanh} \left ( 1/2\,\sqrt{1+{\rm coth} \left (x\right )}\sqrt{2} \right ) \sqrt{2}-8\,\sqrt{1+{\rm coth} \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+coth(x))^(7/2),x)

[Out]

-4/3*(1+coth(x))^(3/2)-2/5*(1+coth(x))^(5/2)+8*arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))*2^(1/2)-8*(1+coth(x))^(1
/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\coth \left (x\right ) + 1\right )}^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^(7/2),x, algorithm="maxima")

[Out]

integrate((coth(x) + 1)^(7/2), x)

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Fricas [B]  time = 2.41786, size = 1469, normalized size = 25.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^(7/2),x, algorithm="fricas")

[Out]

-4/15*(2*sqrt(2)*(23*sqrt(2)*cosh(x)^5 + 115*sqrt(2)*cosh(x)*sinh(x)^4 + 23*sqrt(2)*sinh(x)^5 + 5*(46*sqrt(2)*
cosh(x)^2 - 7*sqrt(2))*sinh(x)^3 - 35*sqrt(2)*cosh(x)^3 + 5*(46*sqrt(2)*cosh(x)^3 - 21*sqrt(2)*cosh(x))*sinh(x
)^2 + 5*(23*sqrt(2)*cosh(x)^4 - 21*sqrt(2)*cosh(x)^2 + 3*sqrt(2))*sinh(x) + 15*sqrt(2)*cosh(x))*sqrt(sinh(x)/(
cosh(x) - sinh(x))) - 15*(sqrt(2)*cosh(x)^6 + 6*sqrt(2)*cosh(x)*sinh(x)^5 + sqrt(2)*sinh(x)^6 + 3*(5*sqrt(2)*c
osh(x)^2 - sqrt(2))*sinh(x)^4 - 3*sqrt(2)*cosh(x)^4 + 4*(5*sqrt(2)*cosh(x)^3 - 3*sqrt(2)*cosh(x))*sinh(x)^3 +
3*(5*sqrt(2)*cosh(x)^4 - 6*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^2 + 3*sqrt(2)*cosh(x)^2 + 6*(sqrt(2)*cosh(x)^5
- 2*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)^6 + 6*cosh(x)*sinh(x)^5 + si
nh(x)^6 + 3*(5*cosh(x)^2 - 1)*sinh(x)^4 - 3*cosh(x)^4 + 4*(5*cosh(x)^3 - 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^4
- 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 - 2*cosh(x)^3 + cosh(x))*sinh(x) - 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 1.16966, size = 216, normalized size = 3.79 \begin{align*} -\frac{4}{15} \, \sqrt{2}{\left (\frac{2 \,{\left (45 \,{\left (\sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{4} + 135 \,{\left (\sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} + 170 \,{\left (\sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} + 100 \, \sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 100 \, e^{\left (2 \, x\right )} + 23\right )}}{{\left (\sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} + 1\right )}^{5}} + 15 \, \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^(7/2),x, algorithm="giac")

[Out]

-4/15*sqrt(2)*(2*(45*(sqrt(e^(4*x) - e^(2*x)) - e^(2*x))^4 + 135*(sqrt(e^(4*x) - e^(2*x)) - e^(2*x))^3 + 170*(
sqrt(e^(4*x) - e^(2*x)) - e^(2*x))^2 + 100*sqrt(e^(4*x) - e^(2*x)) - 100*e^(2*x) + 23)/(sqrt(e^(4*x) - e^(2*x)
) - e^(2*x) + 1)^5 + 15*log(abs(2*sqrt(e^(4*x) - e^(2*x)) - 2*e^(2*x) + 1)))*sgn(e^(2*x) - 1)