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

Optimal. Leaf size=101 \[ -\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {2 b}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2163, 2162} \[ -\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {2 b}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(5/2
) + (2*b)/(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^2) + 2/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]]))

Rule 2162

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-2*ArcTanh[Sqrt
[v]/Rt[-((b*u - a*v)/a), 2]])/(a*Rt[-((b*u - a*v)/a), 2]), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /;
PiecewiseLinearQ[u, v, x]

Rule 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 89, normalized size = 0.88 \[ \frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}}+\frac {2 \left (4 b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 x^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^
(5/2) + (2*(4*b*x - ArcTanh[Tanh[a + b*x]]))/(3*x^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)

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fricas [A]  time = 0.54, size = 118, normalized size = 1.17 \[ \left [\frac {3 \, b x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (3 \, b x - a\right )} \sqrt {x}}{3 \, a^{2} x^{2}}, -\frac {2 \, {\left (3 \, b x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (3 \, b x - a\right )} \sqrt {x}\right )}}{3 \, a^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*b*x^2*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(3*b*x - a)*sqrt(x))/(a^2*x^2),
-2/3*(3*b*x^2*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (3*b*x - a)*sqrt(x))/(a^2*x^2)]

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giac [A]  time = 0.19, size = 41, normalized size = 0.41 \[ \frac {2 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (3 \, b x - a\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 98, normalized size = 0.97 \[ \frac {2 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {2}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {2 b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2/(arctanh(tanh(b*x+a))-b*x)^2/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a)
)-b*x)*b)^(1/2))-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)+2/(arctanh(tanh(b*x+a))-b*x)^2*b/x^(1/2)

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maxima [A]  time = 0.43, size = 41, normalized size = 0.41 \[ \frac {2 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (3 \, b x - a\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.83, size = 642, normalized size = 6.36 \[ \frac {4}{3\,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 )}+\frac {8\,b}{\sqrt {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 )}^2}+\frac {4\,\sqrt {2}\,b^{3/2}\,\ln \left (\frac {\sqrt {b}\,\sqrt {\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}\,\left (\sqrt {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 )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\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}+2\,\sqrt {2}\,b\,x\right )\,\left ({\left (2\,a-\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 )+2\,b\,x\right )}^4+24\,a^2\,{\left (2\,a-\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 )+2\,b\,x\right )}^2+16\,a^4-8\,a\,{\left (2\,a-\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 )+2\,b\,x\right )}^3-32\,a^3\,\left (2\,a-\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 )+2\,b\,x\right )\right )}{2\,\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 )}\right )}{{\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 )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/(3*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)) + (8*b)/(x^(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) + (4*2^(1/2)*b^(3/2)*log((b^(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)^(1/2)*(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) - 4*b^(1/2)*x^(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)^(1/2) + 2*2^(1/2)*b*x)*((2*a - 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*b*x)^4 + 24*a^2*(2*a - 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*b*x)^2 + 16*a^4 - 8*a*(2*a - 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*b*x)^3 - 32*a^3*(2*a -
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*b*x)))/(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))))))/(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)^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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