3.255 \(\int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac {16 b^2 \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 b}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

2/3/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(1/2)+8/3*b/(b*x-arctanh(tanh(b*x+a)))^2/x^(1/2)/a
rctanh(tanh(b*x+a))^(1/2)-16/3*b^2*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ -\frac {16 b^2 \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 b}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(-16*b^2*Sqrt[x])/(3*(b*x - ArcTanh[Tanh[a + b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (8*b)/(3*Sqrt[x]*(b*x -
ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + 2/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[Arc
Tanh[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 {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {(4 b) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {8 b}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (8 b^2\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {16 b^2 \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 b}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 64, normalized size = 0.58 \[ \frac {2 \left (-6 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2-3 b^2 x^2\right )}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(2*(-3*b^2*x^2 - 6*b*x*ArcTanh[Tanh[a + b*x]] + ArcTanh[Tanh[a + b*x]]^2))/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a +
b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]])

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fricas [A]  time = 0.58, size = 49, normalized size = 0.45 \[ \frac {2 \, {\left (8 \, b^{2} x^{2} + 4 \, a b x - a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/3*(8*b^2*x^2 + 4*a*b*x - a^2)*sqrt(b*x + a)*sqrt(x)/(a^3*b*x^3 + a^4*x^2)

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giac [A]  time = 0.17, size = 107, normalized size = 0.97 \[ \frac {2 \, b^{2} \sqrt {x}}{\sqrt {b x + a} a^{3}} - \frac {4 \, {\left (3 \, b^{\frac {3}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 12 \, a b^{\frac {3}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + 5 \, a^{2} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

2*b^2*sqrt(x)/(sqrt(b*x + a)*a^3) - 4/3*(3*b^(3/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 12*a*b^(3/2)*(sqrt(b)
*sqrt(x) - sqrt(b*x + a))^2 + 5*a^2*b^(3/2))/(((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^3*a^2)

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maple [A]  time = 0.25, size = 105, normalized size = 0.95 \[ -\frac {2}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {8 b \left (-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {2 b \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\right )}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)/arctanh(tanh(b*x+a))^(1/2)-8/3*b/(arctanh(tanh(b*x+a))-b*x)*(-1/(arcta
nh(tanh(b*x+a))-b*x)/x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-2*b/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)/arctanh(tanh(
b*x+a))^(1/2))

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maxima [A]  time = 0.42, size = 45, normalized size = 0.41 \[ \frac {2 \, {\left (8 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 3 \, a^{2} b x - a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/3*(8*b^3*x^3 + 12*a*b^2*x^2 + 3*a^2*b*x - a^3)/((b*x + a)^(3/2)*a^3*x^(3/2))

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mupad [B]  time = 1.69, size = 286, normalized size = 2.60 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {32\,x}{3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {4}{3\,b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\frac {128\,b\,x^2}{3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}\right )}{x^{5/2}-\frac {x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2\,b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*atanh(tanh(a + b*x))^(3/2)),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)*((32*x)
/(3*(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) + 4
/(3*b*(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)) - (
128*b*x^2)/(3*(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)^3)))/(x^(5/2) - (x^(3/2)*(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*b))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {5}{2}} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Integral(1/(x**(5/2)*atanh(tanh(a + b*x))**(3/2)), x)

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