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

Optimal. Leaf size=124 \[ -\frac {3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {3 b \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ -\frac {3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {3 b \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^
(5/2) - 1/(x*ArcTanh[Tanh[a + b*x]]^(3/2)) + b/((b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2)) -
(3*b)/((b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])

Rule 2161

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

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 91, normalized size = 0.73 \[ \frac {3 b \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}}-\frac {\tanh ^{-1}(\tanh (a+b x))+2 b x}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 151, normalized size = 1.22 \[ \left [\frac {3 \, {\left (b^{2} x^{2} + a b x\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b x + a^{2}\right )} \sqrt {b x + a}}{2 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}, -\frac {3 \, {\left (b^{2} x^{2} + a b x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x + a^{2}\right )} \sqrt {b x + a}}{a^{3} b x^{2} + a^{4} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(b^2*x^2 + a*b*x)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(3*a*b*x + a^2)*sqrt(b*x +
a))/(a^3*b*x^2 + a^4*x), -(3*(b^2*x^2 + a*b*x)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*a*b*x + a^2)*sqr
t(b*x + a))/(a^3*b*x^2 + a^4*x)]

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giac [A]  time = 0.22, size = 64, normalized size = 0.52 \[ -\frac {3 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {3 \, {\left (b x + a\right )} b - 2 \, a b}{{\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 105, normalized size = 0.85 \[ 2 b \left (-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{2 b x}-\frac {3 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.86, size = 807, normalized size = 6.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^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)*(4/(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) - (24*b*x)/(lo
g(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))/(2*b*x^2
- 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)) + (2^
(1/2)*b*log((((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)^(1/2
)*2i + 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^(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)^5 + 40*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)*ex
p(2*b*x) + 1)) + 2*b*x)^3 - 80*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 - 32*a^5 - 10*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)^4 + 80*a^4*(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))*1i)/(2*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)^(1/2)))*6i)/(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 \[ \int \frac {1}{x^{2} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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