3.237 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{9/2}} \, dx\)

Optimal. Leaf size=35 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

2/7*arctanh(tanh(b*x+a))^(7/2)/x^(7/2)/(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} \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(9/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*x^(7/2)*(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))^{5/2}}{x^{9/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 34, normalized size = 0.97 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{x^{7/2} \left (7 b x-7 \tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(9/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(7/2))/(x^(7/2)*(7*b*x - 7*ArcTanh[Tanh[a + b*x]]))

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fricas [A]  time = 0.42, size = 42, normalized size = 1.20 \[ -\frac {2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt {b x + a}}{7 \, a x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(9/2),x, algorithm="fricas")

[Out]

-2/7*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*x + a)/(a*x^(7/2))

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giac [A]  time = 0.18, size = 33, normalized size = 0.94 \[ -\frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{8}}{7 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}} a {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(9/2),x, algorithm="giac")

[Out]

-2/7*(b*x + a)^(7/2)*b^8/(((b*x + a)*b - a*b)^(7/2)*a*abs(b))

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maple [A]  time = 0.25, size = 29, normalized size = 0.83 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^(9/2),x)

[Out]

-2/7/(arctanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(7/2)

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maxima [A]  time = 0.42, size = 15, normalized size = 0.43 \[ -\frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}}}{7 \, a x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(9/2),x, algorithm="maxima")

[Out]

-2/7*(b*x + a)^(7/2)/(a*x^(7/2))

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mupad [B]  time = 1.65, size = 396, normalized size = 11.31 \[ -\frac {\sqrt {2}\,\sqrt {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}\,\left (\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{14\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-14\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+28\,b\,x}-\frac {{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{14\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-14\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+28\,b\,x}+\frac {3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{14\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-14\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+28\,b\,x}-\frac {3\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{14\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-14\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+28\,b\,x}\right )}{2\,x^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^(9/2),x)

[Out]

-(2^(1/2)*(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)))^(1/2)*(l
og(1/(exp(2*a)*exp(2*b*x) + 1))^3/(14*log(1/(exp(2*a)*exp(2*b*x) + 1)) - 14*log((exp(2*a)*exp(2*b*x))/(exp(2*a
)*exp(2*b*x) + 1)) + 28*b*x) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3/(14*log(1/(exp(2*a)*exp(
2*b*x) + 1)) - 14*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 28*b*x) + (3*log(1/(exp(2*a)*exp(2*b*
x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2)/(14*log(1/(exp(2*a)*exp(2*b*x) + 1)) - 14*log
((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 28*b*x) - (3*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log((exp(2
*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/(14*log(1/(exp(2*a)*exp(2*b*x) + 1)) - 14*log((exp(2*a)*exp(2*b*x)
)/(exp(2*a)*exp(2*b*x) + 1)) + 28*b*x)))/(2*x^(7/2))

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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))**(5/2)/x**(9/2),x)

[Out]

Timed out

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