3.229 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{9/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

4/35*b*arctanh(tanh(b*x+a))^(5/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/7*arctanh(tanh(b*x+a))^(5/2)/x^(7/2)/
(b*x-arctanh(tanh(b*x+a)))

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Rubi [A]  time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(9/2),x]

[Out]

(4*b*ArcTanh[Tanh[a + b*x]]^(5/2))/(35*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(
5/2))/(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]

Rule 2171

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] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{9/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(2 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx}{7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{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 = 48, normalized size = 0.67 \[ \frac {2 \left (7 b x-5 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 x^{7/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(9/2),x]

[Out]

(2*(7*b*x - 5*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(5/2))/(35*x^(7/2)*(-(b*x) + ArcTanh[Tanh[a + b*x
]])^2)

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fricas [A]  time = 0.50, size = 45, normalized size = 0.62 \[ \frac {2 \, {\left (2 \, b^{3} x^{3} - a b^{2} x^{2} - 8 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x + a}}{35 \, a^{2} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(9/2),x, algorithm="fricas")

[Out]

2/35*(2*b^3*x^3 - a*b^2*x^2 - 8*a^2*b*x - 5*a^3)*sqrt(b*x + a)/(a^2*x^(7/2))

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giac [A]  time = 0.16, size = 59, normalized size = 0.82 \[ \frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{7}}{a^{2}} - \frac {7 \, \sqrt {2} b^{7}}{a}\right )} {\left (b x + a\right )}^{\frac {5}{2}} b}{35 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(9/2),x, algorithm="giac")

[Out]

1/35*sqrt(2)*(2*sqrt(2)*(b*x + a)*b^7/a^2 - 7*sqrt(2)*b^7/a)*(b*x + a)^(5/2)*b/(((b*x + a)*b - a*b)^(7/2)*abs(
b))

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maple [A]  time = 0.25, size = 59, normalized size = 0.82 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}+\frac {4 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{35 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(3/2)/x^(9/2),x)

[Out]

-2/7/(arctanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(5/2)+4/35*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(5/2)
*arctanh(tanh(b*x+a))^(5/2)

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maxima [A]  time = 0.43, size = 34, normalized size = 0.47 \[ \frac {2 \, {\left (2 \, b^{2} x^{2} - 3 \, a b x - 5 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{35 \, a^{2} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(9/2),x, algorithm="maxima")

[Out]

2/35*(2*b^2*x^2 - 3*a*b*x - 5*a^2)*(b*x + a)^(3/2)/(a^2*x^(7/2))

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mupad [B]  time = 1.54, size = 228, normalized size = 3.17 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{7}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{7}-\frac {6\,b\,x}{35}+\frac {4\,b^2\,x^2}{35\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {16\,b^3\,x^3}{35\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(3/2)/x^(9/2),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))/7 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/7 - (6*b*x)/35 + (4*b^2*
x^2)/(35*(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))
+ (16*b^3*x^3)/(35*(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)))/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))**(3/2)/x**(9/2),x)

[Out]

Timed out

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