3.30 \(\int (1+\coth ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=50 \[ -\frac{1}{2} \coth (x) \sqrt{\coth ^2(x)+1}+2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \coth (x)}{\sqrt{\coth ^2(x)+1}}\right )-\frac{5}{2} \sinh ^{-1}(\coth (x)) \]

[Out]

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

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Rubi [A]  time = 0.0392872, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3661, 416, 523, 215, 377, 206} \[ -\frac{1}{2} \coth (x) \sqrt{\coth ^2(x)+1}+2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \coth (x)}{\sqrt{\coth ^2(x)+1}}\right )-\frac{5}{2} \sinh ^{-1}(\coth (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Coth[x]^2)^(3/2),x]

[Out]

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

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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

Mathematica [B]  time = 0.330443, size = 116, normalized size = 2.32 \[ -\frac{1}{8} \left (\coth ^2(x)+1\right )^{3/2} \text{sech}^2(2 x) \left (\sinh (4 x)+16 \sinh ^3(x) \sqrt{\cosh (2 x)} \tanh ^{-1}\left (\frac{\cosh (x)}{\sqrt{\cosh (2 x)}}\right )+4 \sinh ^3(x) \left (\sqrt{-\cosh (2 x)} \tan ^{-1}\left (\frac{\cosh (x)}{\sqrt{-\cosh (2 x)}}\right )-4 \sqrt{2} \sqrt{\cosh (2 x)} \log \left (\sqrt{2} \cosh (x)+\sqrt{\cosh (2 x)}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Coth[x]^2)^(3/2),x]

[Out]

-((1 + Coth[x]^2)^(3/2)*Sech[2*x]^2*(16*ArcTanh[Cosh[x]/Sqrt[Cosh[2*x]]]*Sqrt[Cosh[2*x]]*Sinh[x]^3 + 4*(ArcTan
[Cosh[x]/Sqrt[-Cosh[2*x]]]*Sqrt[-Cosh[2*x]] - 4*Sqrt[2]*Sqrt[Cosh[2*x]]*Log[Sqrt[2]*Cosh[x] + Sqrt[Cosh[2*x]]]
)*Sinh[x]^3 + Sinh[4*x]))/8

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Maple [B]  time = 0.018, size = 158, normalized size = 3.2 \begin{align*}{\frac{1}{6} \left ( \left ( 1+{\rm coth} \left (x\right ) \right ) ^{2}-2\,{\rm coth} \left (x\right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{\rm coth} \left (x\right )}{4}\sqrt{ \left ( 1+{\rm coth} \left (x\right ) \right ) ^{2}-2\,{\rm coth} \left (x\right )}}-{\frac{5\,{\it Arcsinh} \left ({\rm coth} \left (x\right ) \right ) }{2}}+\sqrt{ \left ( 1+{\rm coth} \left (x\right ) \right ) ^{2}-2\,{\rm coth} \left (x\right )}-\sqrt{2}{\it Artanh} \left ({\frac{ \left ( 2-2\,{\rm coth} \left (x\right ) \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ( 1+{\rm coth} \left (x\right ) \right ) ^{2}-2\,{\rm coth} \left (x\right )}}}} \right ) -{\frac{1}{6} \left ( \left ({\rm coth} \left (x\right )-1 \right ) ^{2}+2\,{\rm coth} \left (x\right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{\rm coth} \left (x\right )}{4}\sqrt{ \left ({\rm coth} \left (x\right )-1 \right ) ^{2}+2\,{\rm coth} \left (x\right )}}-\sqrt{ \left ({\rm coth} \left (x\right )-1 \right ) ^{2}+2\,{\rm coth} \left (x\right )}+\sqrt{2}{\it Artanh} \left ({\frac{ \left ( 2\,{\rm coth} \left (x\right )+2 \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ({\rm coth} \left (x\right )-1 \right ) ^{2}+2\,{\rm coth} \left (x\right )}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+coth(x)^2)^(3/2),x)

[Out]

1/6*((1+coth(x))^2-2*coth(x))^(3/2)-1/4*coth(x)*((1+coth(x))^2-2*coth(x))^(1/2)-5/2*arcsinh(coth(x))+((1+coth(
x))^2-2*coth(x))^(1/2)-2^(1/2)*arctanh(1/4*(2-2*coth(x))*2^(1/2)/((1+coth(x))^2-2*coth(x))^(1/2))-1/6*((coth(x
)-1)^2+2*coth(x))^(3/2)-1/4*coth(x)*((coth(x)-1)^2+2*coth(x))^(1/2)-((coth(x)-1)^2+2*coth(x))^(1/2)+2^(1/2)*ar
ctanh(1/4*(2*coth(x)+2)*2^(1/2)/((coth(x)-1)^2+2*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 )^{2} + 1\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((coth(x)^2 + 1)^(3/2), x)

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Fricas [B]  time = 2.51501, size = 3526, normalized size = 70.52 \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)^2)^(3/2),x, algorithm="fricas")

[Out]

1/4*(2*(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*(cosh(x)^
8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (28*cosh(x)^2 + 3)*sinh(x)^6 + 3*cosh(x)^6 + 2*(28*cosh(x)^3 + 9*cosh(x)
)*sinh(x)^5 + 5*(14*cosh(x)^4 + 9*cosh(x)^2 + 1)*sinh(x)^4 + 5*cosh(x)^4 + 4*(14*cosh(x)^5 + 15*cosh(x)^3 + 5*
cosh(x))*sinh(x)^3 + (28*cosh(x)^6 + 45*cosh(x)^4 + 30*cosh(x)^2 + 4)*sinh(x)^2 + 4*cosh(x)^2 + 2*(4*cosh(x)^7
+ 9*cosh(x)^5 + 10*cosh(x)^3 + 4*cosh(x))*sinh(x) + (sqrt(2)*cosh(x)^6 + 6*sqrt(2)*cosh(x)*sinh(x)^5 + sqrt(2
)*sinh(x)^6 + 3*(5*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^4 + 3*sqrt(2)*cosh(x)^4 + 4*(5*sqrt(2)*cosh(x)^3 + 3*s
qrt(2)*cosh(x))*sinh(x)^3 + (15*sqrt(2)*cosh(x)^4 + 18*sqrt(2)*cosh(x)^2 + 4*sqrt(2))*sinh(x)^2 + 4*sqrt(2)*co
sh(x)^2 + 2*(3*sqrt(2)*cosh(x)^5 + 6*sqrt(2)*cosh(x)^3 + 4*sqrt(2)*cosh(x))*sinh(x) + 4*sqrt(2))*sqrt((cosh(x)
^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4)/(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x
)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6)) + 2*(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*(cosh(x)^4 + 4*cosh(x
)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 - 1)*sinh(x)^2 - cosh(x)^2 + 2*(2*cosh(x)^3 - cosh(x))*sinh(x) + (sqrt(
2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2 - sqrt(2))*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^
2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) - 5*(cosh(x)^4 + 4*cosh(
x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 4*(cosh(x)^3 - cosh(x))*sinh(x) + 1)*
log((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 2*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x)
+ sinh(x)^2)) + 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) + 5*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x
)^4 + 2*(3*cosh(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 4*(cosh(x)^3 - cosh(x))*sinh(x) + 1)*log((cosh(x)^2 + 2*co
sh(x)*sinh(x) + sinh(x)^2 - 2*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 1)/(
cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) - 4*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt((cosh(x)^
2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\coth ^{2}{\left (x \right )} + 1\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x)**2)**(3/2),x)

[Out]

Integral((coth(x)**2 + 1)**(3/2), x)

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Giac [B]  time = 1.22714, size = 358, normalized size = 7.16 \begin{align*} \frac{1}{4} \, \sqrt{2}{\left (5 \, \sqrt{2} \log \left (\frac{{\left | -2 \, \sqrt{2} + 2 \, \sqrt{e^{\left (4 \, x\right )} + 1} - 2 \, e^{\left (2 \, x\right )} + 2 \right |}}{2 \,{\left (\sqrt{2} + \sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right )}}\right ) \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) + 4 \, \log \left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right ) \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) - 4 \, \log \left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right ) \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) - 4 \, \log \left (-\sqrt{e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1\right ) \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) - \frac{4 \,{\left (3 \,{\left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{3} \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) +{\left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{2} \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) -{\left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )} \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) + \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right )\right )}}{{\left ({\left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{2} + 2 \, \sqrt{e^{\left (4 \, x\right )} + 1} - 2 \, e^{\left (2 \, x\right )} - 1\right )}^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x)^2)^(3/2),x, algorithm="giac")

[Out]

1/4*sqrt(2)*(5*sqrt(2)*log(1/2*abs(-2*sqrt(2) + 2*sqrt(e^(4*x) + 1) - 2*e^(2*x) + 2)/(sqrt(2) + sqrt(e^(4*x) +
1) - e^(2*x) + 1))*sgn(e^(2*x) - 1) + 4*log(sqrt(e^(4*x) + 1) - e^(2*x) + 1)*sgn(e^(2*x) - 1) - 4*log(sqrt(e^
(4*x) + 1) - e^(2*x))*sgn(e^(2*x) - 1) - 4*log(-sqrt(e^(4*x) + 1) + e^(2*x) + 1)*sgn(e^(2*x) - 1) - 4*(3*(sqrt
(e^(4*x) + 1) - e^(2*x))^3*sgn(e^(2*x) - 1) + (sqrt(e^(4*x) + 1) - e^(2*x))^2*sgn(e^(2*x) - 1) - (sqrt(e^(4*x)
+ 1) - e^(2*x))*sgn(e^(2*x) - 1) + sgn(e^(2*x) - 1))/((sqrt(e^(4*x) + 1) - e^(2*x))^2 + 2*sqrt(e^(4*x) + 1) -
2*e^(2*x) - 1)^2)