3.56 \(\int \frac {1}{(1+\tanh (x))^{5/2}} \, dx\)

Optimal. Leaf size=61 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{4 \sqrt {2}}-\frac {1}{4 \sqrt {\tanh (x)+1}}-\frac {1}{6 (\tanh (x)+1)^{3/2}}-\frac {1}{5 (\tanh (x)+1)^{5/2}} \]

[Out]

1/8*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(1/2)-1/4/(1+tanh(x))^(1/2)-1/5/(1+tanh(x))^(5/2)-1/6/(1+tanh(x))
^(3/2)

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, 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 = {3479, 3480, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{4 \sqrt {2}}-\frac {1}{4 \sqrt {\tanh (x)+1}}-\frac {1}{6 (\tanh (x)+1)^{3/2}}-\frac {1}{5 (\tanh (x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tanh[x])^(-5/2),x]

[Out]

ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/(4*Sqrt[2]) - 1/(5*(1 + Tanh[x])^(5/2)) - 1/(6*(1 + Tanh[x])^(3/2)) - 1/(4*
Sqrt[1 + Tanh[x]])

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 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(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] && LtQ[n
, 0]

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 \frac {1}{(1+\tanh (x))^{5/2}} \, dx &=-\frac {1}{5 (1+\tanh (x))^{5/2}}+\frac {1}{2} \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx\\ &=-\frac {1}{5 (1+\tanh (x))^{5/2}}-\frac {1}{6 (1+\tanh (x))^{3/2}}+\frac {1}{4} \int \frac {1}{\sqrt {1+\tanh (x)}} \, dx\\ &=-\frac {1}{5 (1+\tanh (x))^{5/2}}-\frac {1}{6 (1+\tanh (x))^{3/2}}-\frac {1}{4 \sqrt {1+\tanh (x)}}+\frac {1}{8} \int \sqrt {1+\tanh (x)} \, dx\\ &=-\frac {1}{5 (1+\tanh (x))^{5/2}}-\frac {1}{6 (1+\tanh (x))^{3/2}}-\frac {1}{4 \sqrt {1+\tanh (x)}}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1+\tanh (x)}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )}{4 \sqrt {2}}-\frac {1}{5 (1+\tanh (x))^{5/2}}-\frac {1}{6 (1+\tanh (x))^{3/2}}-\frac {1}{4 \sqrt {1+\tanh (x)}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 62, normalized size = 1.02 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {(\sinh (2 x)-\cosh (2 x)) (20 \sinh (2 x)+26 \cosh (2 x)+11)}{60 \sqrt {\tanh (x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tanh[x])^(-5/2),x]

[Out]

ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/(4*Sqrt[2]) + ((-Cosh[2*x] + Sinh[2*x])*(11 + 26*Cosh[2*x] + 20*Sinh[2*x]))
/(60*Sqrt[1 + Tanh[x]])

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fricas [B]  time = 0.70, size = 266, normalized size = 4.36 \[ -\frac {2 \, \sqrt {2} {\left (23 \, \sqrt {2} \cosh \relax (x)^{4} + 92 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{3} + 23 \, \sqrt {2} \sinh \relax (x)^{4} + {\left (138 \, \sqrt {2} \cosh \relax (x)^{2} + 11 \, \sqrt {2}\right )} \sinh \relax (x)^{2} + 11 \, \sqrt {2} \cosh \relax (x)^{2} + 2 \, {\left (46 \, \sqrt {2} \cosh \relax (x)^{3} + 11 \, \sqrt {2} \cosh \relax (x)\right )} \sinh \relax (x) + 3 \, \sqrt {2}\right )} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} - 15 \, {\left (\sqrt {2} \cosh \relax (x)^{5} + 5 \, \sqrt {2} \cosh \relax (x)^{4} \sinh \relax (x) + 10 \, \sqrt {2} \cosh \relax (x)^{3} \sinh \relax (x)^{2} + 10 \, \sqrt {2} \cosh \relax (x)^{2} \sinh \relax (x)^{3} + 5 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{4} + \sqrt {2} \sinh \relax (x)^{5}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} - 1\right )}{240 \, {\left (\cosh \relax (x)^{5} + 5 \, \cosh \relax (x)^{4} \sinh \relax (x) + 10 \, \cosh \relax (x)^{3} \sinh \relax (x)^{2} + 10 \, \cosh \relax (x)^{2} \sinh \relax (x)^{3} + 5 \, \cosh \relax (x) \sinh \relax (x)^{4} + \sinh \relax (x)^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(2*sqrt(2)*(23*sqrt(2)*cosh(x)^4 + 92*sqrt(2)*cosh(x)*sinh(x)^3 + 23*sqrt(2)*sinh(x)^4 + (138*sqrt(2)*c
osh(x)^2 + 11*sqrt(2))*sinh(x)^2 + 11*sqrt(2)*cosh(x)^2 + 2*(46*sqrt(2)*cosh(x)^3 + 11*sqrt(2)*cosh(x))*sinh(x
) + 3*sqrt(2))*sqrt(cosh(x)/(cosh(x) - sinh(x))) - 15*(sqrt(2)*cosh(x)^5 + 5*sqrt(2)*cosh(x)^4*sinh(x) + 10*sq
rt(2)*cosh(x)^3*sinh(x)^2 + 10*sqrt(2)*cosh(x)^2*sinh(x)^3 + 5*sqrt(2)*cosh(x)*sinh(x)^4 + sqrt(2)*sinh(x)^5)*
log(-2*sqrt(2)*sqrt(cosh(x)/(cosh(x) - sinh(x)))*(cosh(x) + sinh(x)) - 2*cosh(x)^2 - 4*cosh(x)*sinh(x) - 2*sin
h(x)^2 - 1))/(cosh(x)^5 + 5*cosh(x)^4*sinh(x) + 10*cosh(x)^3*sinh(x)^2 + 10*cosh(x)^2*sinh(x)^3 + 5*cosh(x)*si
nh(x)^4 + sinh(x)^5)

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giac [B]  time = 0.15, size = 140, normalized size = 2.30 \[ -\frac {1}{240} \, \sqrt {2} {\left (\frac {2 \, {\left (45 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{4} - 45 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} + 35 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 15 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 15 \, e^{\left (2 \, x\right )} + 3\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{5}} + 15 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right ) - 46\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*sqrt(2)*(2*(45*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^4 - 45*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^3 + 35*(s
qrt(e^(4*x) + e^(2*x)) - e^(2*x))^2 - 15*sqrt(e^(4*x) + e^(2*x)) + 15*e^(2*x) + 3)/(sqrt(e^(4*x) + e^(2*x)) -
e^(2*x))^5 + 15*log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1) - 46)

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maple [A]  time = 0.06, size = 43, normalized size = 0.70 \[ \frac {\arctanh \left (\frac {\sqrt {1+\tanh \relax (x )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{8}-\frac {1}{4 \sqrt {1+\tanh \relax (x )}}-\frac {1}{5 \left (1+\tanh \relax (x )\right )^{\frac {5}{2}}}-\frac {1}{6 \left (1+\tanh \relax (x )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(1/2)-1/4/(1+tanh(x))^(1/2)-1/5/(1+tanh(x))^(5/2)-1/6/(1+tanh(x))
^(3/2)

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maxima [A]  time = 0.41, size = 79, normalized size = 1.30 \[ -\frac {1}{120} \, \sqrt {2} {\left (\frac {5}{e^{\left (-2 \, x\right )} + 1} + \frac {15}{{\left (e^{\left (-2 \, x\right )} + 1\right )}^{2}} + 3\right )} {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}} - \frac {1}{16} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}{\sqrt {2} + \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*sqrt(2)*(5/(e^(-2*x) + 1) + 15/(e^(-2*x) + 1)^2 + 3)*(e^(-2*x) + 1)^(5/2) - 1/16*sqrt(2)*log(-(sqrt(2)
- sqrt(2)/sqrt(e^(-2*x) + 1))/(sqrt(2) + sqrt(2)/sqrt(e^(-2*x) + 1)))

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mupad [B]  time = 0.12, size = 40, normalized size = 0.66 \[ \frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\relax (x)+1}}{2}\right )}{8}-\frac {\frac {\mathrm {tanh}\relax (x)}{6}+\frac {{\left (\mathrm {tanh}\relax (x)+1\right )}^2}{4}+\frac {11}{30}}{{\left (\mathrm {tanh}\relax (x)+1\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tanh(x) + 1)^(5/2),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2))/8 - (tanh(x)/6 + (tanh(x) + 1)^2/4 + 11/30)/(tanh(x) + 1)^(5/
2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((tanh(x) + 1)**(-5/2), x)

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