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

Optimal. Leaf size=191 \[ -\frac {15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {15 b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]

[Out]

-15/4*b^2*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b*x-arctanh(tanh(b*x+a)))^(7/2)
+3/4*b/x/arctanh(tanh(b*x+a))^(5/2)-3/4*b^2/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(5/2)-1/2/x^2/arct
anh(tanh(b*x+a))^(3/2)+5/4*b^2/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(3/2)-15/4*b^2/(b*x-arctanh(t
anh(b*x+a)))^3/arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ -\frac {15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {15 b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*b^2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b
*x]])^(7/2)) + (3*b)/(4*x*ArcTanh[Tanh[a + b*x]]^(5/2)) - (3*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Ta
nh[a + b*x]]^(5/2)) - 1/(2*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)) + (5*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^2*Arc
Tanh[Tanh[a + b*x]]^(3/2)) - (15*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^3*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^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{4} (3 b) \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac {3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {1}{8} \left (15 b^2\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=\frac {3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\left (15 b^2\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {\left (15 b^2\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {\left (15 b^2\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {15 b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 115, normalized size = 0.60 \[ \frac {1}{4} \left (\frac {9 b x \tanh ^{-1}(\tanh (a+b x))-2 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2}{x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-15*b^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[
a + b*x]])^(7/2) + (8*b^2*x^2 + 9*b*x*ArcTanh[Tanh[a + b*x]] - 2*ArcTanh[Tanh[a + b*x]]^2)/(x^2*Sqrt[ArcTanh[T
anh[a + b*x]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3))/4

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fricas [A]  time = 0.63, size = 189, normalized size = 0.99 \[ \left [\frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{8 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, \frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{4 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^3*x^3 + a*b^2*x^2)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(15*a*b^2*x^2 + 5*a^2*
b*x - 2*a^3)*sqrt(b*x + a))/(a^4*b*x^3 + a^5*x^2), 1/4*(15*(b^3*x^3 + a*b^2*x^2)*sqrt(-a)*arctan(sqrt(b*x + a)
*sqrt(-a)/a) + (15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x + a))/(a^4*b*x^3 + a^5*x^2)]

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giac [A]  time = 0.18, size = 80, normalized size = 0.42 \[ \frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3}} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2*b^2/(sqrt(b*x + a)*a^3) + 1/4*(7*(b*x + a)^(3/2)*b^
2 - 9*sqrt(b*x + a)*a*b^2)/(a^3*b^2*x^2)

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maple [A]  time = 0.16, size = 131, normalized size = 0.69 \[ 2 b^{2} \left (\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {\frac {\frac {7 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{8}+\left (-\frac {9 \arctanh \left (\tanh \left (b x +a \right )\right )}{8}+\frac {9 b x}{8}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{2} x^{2}}-\frac {15 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{8 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*(1/(arctanh(tanh(b*x+a))-b*x)^3/arctanh(tanh(b*x+a))^(1/2)+1/(arctanh(tanh(b*x+a))-b*x)^3*((7/8*arctanh(
tanh(b*x+a))^(3/2)+(-9/8*arctanh(tanh(b*x+a))+9/8*b*x)*arctanh(tanh(b*x+a))^(1/2))/b^2/x^2-15/8/(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 \[ \int \frac {1}{x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.07, size = 1028, normalized size = 5.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*b^2*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)^7 + 84*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)^5 - 280*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)^4 + 560*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)^3 - 672*a^5*(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 - 128*a^7 - 14*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^6 + 448*a^6*(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)))*15i)/(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)^(7/2) - (2*(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))/(x^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) + ((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)*((14*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)^2 - (60*b^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)^3))/(x*(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))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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