3.162 \(\int \frac {1}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=18 \[ -\frac {2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2157, 30} \[ -\frac {2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(-5/2),x]

[Out]

-2/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \[ -\frac {2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(-5/2),x]

[Out]

-2/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2))

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fricas [B]  time = 0.52, size = 31, normalized size = 1.72 \[ -\frac {2 \, \sqrt {b x + a}}{3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(b*x + a)/(b^3*x^2 + 2*a*b^2*x + a^2*b)

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giac [A]  time = 0.23, size = 12, normalized size = 0.67 \[ -\frac {2}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/((b*x + a)^(3/2)*b)

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maple [A]  time = 0.03, size = 15, normalized size = 0.83 \[ -\frac {2}{3 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

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

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maxima [A]  time = 0.50, size = 12, normalized size = 0.67 \[ -\frac {2}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/((b*x + a)^(3/2)*b)

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mupad [B]  time = 1.37, size = 103, normalized size = 5.72 \[ -\frac {8\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{3\,b\,{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/atanh(tanh(a + b*x))^(5/2),x)

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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