3.228 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=35 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

2/5*arctanh(tanh(b*x+a))^(5/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2167} \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(7/2),x]

[Out]

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

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 34, normalized size = 0.97 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{5/2} \left (5 b x-5 \tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(7/2),x]

[Out]

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

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fricas [A]  time = 0.44, size = 31, normalized size = 0.89 \[ -\frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b x + a}}{5 \, a x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(7/2),x, algorithm="fricas")

[Out]

-2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)/(a*x^(5/2))

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giac [A]  time = 0.15, size = 33, normalized size = 0.94 \[ -\frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{6}}{5 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}} a {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(7/2),x, algorithm="giac")

[Out]

-2/5*(b*x + a)^(5/2)*b^6/(((b*x + a)*b - a*b)^(5/2)*a*abs(b))

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maple [A]  time = 0.26, size = 29, normalized size = 0.83 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(3/2)/x^(7/2),x)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(5/2)

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maxima [A]  time = 0.44, size = 15, normalized size = 0.43 \[ -\frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}}}{5 \, a x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(7/2),x, algorithm="maxima")

[Out]

-2/5*(b*x + a)^(5/2)/(a*x^(5/2))

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mupad [B]  time = 1.51, size = 332, normalized size = 9.49 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{5\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-5\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+10\,b\,x}+\frac {4\,b^2\,x^2}{5\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-5\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+10\,b\,x}-\frac {4\,b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{5\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-5\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+10\,b\,x}\right )}{x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(3/2)/x^(7/2),x)

[Out]

((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)*((log(2
/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2/(5*log(2/(exp(
2*a)*exp(2*b*x) + 1)) - 5*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 10*b*x) + (4*b^2*x^2)/(5*lo
g(2/(exp(2*a)*exp(2*b*x) + 1)) - 5*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 10*b*x) - (4*b*x*(
log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(5*log(2/(
exp(2*a)*exp(2*b*x) + 1)) - 5*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 10*b*x)))/x^(5/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(3/2)/x**(7/2),x)

[Out]

Timed out

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