3.157 \(\int \frac {1}{x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=245 \[ -\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {35 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac {5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]

[Out]

-35/8*b^3*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b*x-arctanh(tanh(b*x+a)))^(9/2)
-5/8*b^2/x/arctanh(tanh(b*x+a))^(7/2)+5/8*b^3/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(7/2)+1/4*b/x^2/
arctanh(tanh(b*x+a))^(5/2)-7/8*b^3/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(5/2)-1/3/x^3/arctanh(tan
h(b*x+a))^(3/2)+35/24*b^3/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(3/2)-35/8*b^3/(b*x-arctanh(tanh(b
*x+a)))^4/arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ -\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {35 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(-35*b^3*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(8*(b*x - ArcTanh[Tanh[a + b
*x]])^(9/2)) - (5*b^2)/(8*x*ArcTanh[Tanh[a + b*x]]^(7/2)) + (5*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[
Tanh[a + b*x]]^(7/2)) + b/(4*x^2*ArcTanh[Tanh[a + b*x]]^(5/2)) - (7*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^2*A
rcTanh[Tanh[a + b*x]]^(5/2)) - 1/(3*x^3*ArcTanh[Tanh[a + b*x]]^(3/2)) + (35*b^3)/(24*(b*x - ArcTanh[Tanh[a + b
*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2)) - (35*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^4*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^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{2} b \int \frac {1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {1}{8} \left (5 b^2\right ) \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=-\frac {5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{16} \left (35 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx\\ &=-\frac {5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {\left (35 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\left (35 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {\left (35 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=-\frac {5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {\left (35 b^3\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {35 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac {5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 133, normalized size = 0.54 \[ \frac {35 b^3 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}}-\frac {87 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-38 b x \tanh ^{-1}(\tanh (a+b x))^2+8 \tanh ^{-1}(\tanh (a+b x))^3+48 b^3 x^3}{24 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(35*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(8*(-(b*x) + ArcTanh[Tanh
[a + b*x]])^(9/2)) - (48*b^3*x^3 + 87*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 38*b*x*ArcTanh[Tanh[a + b*x]]^2 + 8*Arc
Tanh[Tanh[a + b*x]]^3)/(24*x^3*Sqrt[ArcTanh[Tanh[a + b*x]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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fricas [A]  time = 0.58, size = 211, normalized size = 0.86 \[ \left [\frac {105 \, {\left (b^{4} x^{4} + a b^{3} x^{3}\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x + a}}{48 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}, -\frac {105 \, {\left (b^{4} x^{4} + a b^{3} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x + a}}{24 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

[1/48*(105*(b^4*x^4 + a*b^3*x^3)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(105*a*b^3*x^3 + 35*
a^2*b^2*x^2 - 14*a^3*b*x + 8*a^4)*sqrt(b*x + a))/(a^5*b*x^4 + a^6*x^3), -1/24*(105*(b^4*x^4 + a*b^3*x^3)*sqrt(
-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (105*a*b^3*x^3 + 35*a^2*b^2*x^2 - 14*a^3*b*x + 8*a^4)*sqrt(b*x + a))/(a
^5*b*x^4 + a^6*x^3)]

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giac [A]  time = 0.21, size = 95, normalized size = 0.39 \[ -\frac {35 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{4}} - \frac {2 \, b^{3}}{\sqrt {b x + a} a^{4}} - \frac {57 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} - 136 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} + 87 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, a^{4} b^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

-35/8*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) - 2*b^3/(sqrt(b*x + a)*a^4) - 1/24*(57*(b*x + a)^(5/2)
*b^3 - 136*(b*x + a)^(3/2)*a*b^3 + 87*sqrt(b*x + a)*a^2*b^3)/(a^4*b^3*x^3)

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maple [A]  time = 0.16, size = 186, normalized size = 0.76 \[ 2 b^{3} \left (-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {\frac {\frac {19 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16}+\left (-\frac {17 \arctanh \left (\tanh \left (b x +a \right )\right )}{6}+\frac {17 b x}{6}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (\frac {29 a^{2}}{16}+\frac {29 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8}+\frac {29 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{16}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{3} x^{3}}-\frac {35 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

2*b^3*(-1/(arctanh(tanh(b*x+a))-b*x)^4/arctanh(tanh(b*x+a))^(1/2)-1/(arctanh(tanh(b*x+a))-b*x)^4*((19/16*arcta
nh(tanh(b*x+a))^(5/2)+(-17/6*arctanh(tanh(b*x+a))+17/6*b*x)*arctanh(tanh(b*x+a))^(3/2)+(29/16*a^2+29/8*a*(arct
anh(tanh(b*x+a))-b*x-a)+29/16*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3-35/16/(arcta
nh(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 \[ \int \frac {1}{x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(x^4*arctanh(tanh(b*x + a))^(3/2)), x)

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mupad [B]  time = 5.09, size = 1258, normalized size = 5.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((38*b^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)
^3 - (140*b^3*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)^4)*(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*(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)))) - (4*(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))/(3*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) + (2^(1/2)*b^
3*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)^9 + 144*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)^7 - 672*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)^6 + 2016*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)^5 - 4032*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)^4 + 5376*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)^3 - 4608*a^7*(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 - 512*a^9 - 18*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)^8 + 2304*a^8*(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)))*35i)/(log(2/(ex
p(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)^(9/2) - (22*b*(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))/(3*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)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Integral(1/(x**4*atanh(tanh(a + b*x))**(3/2)), x)

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