3.139 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=221 \[ \frac {3 b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {3 b^5 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac {b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{64 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4} \]

[Out]

3/128*b^5*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)
+1/128*b^4/x/arctanh(tanh(b*x+a))^(3/2)-1/128*b^5/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)-1/8*b*
arctanh(tanh(b*x+a))^(3/2)/x^4-1/5*arctanh(tanh(b*x+a))^(5/2)/x^5-1/64*b^3/x^2/arctanh(tanh(b*x+a))^(1/2)+3/12
8*b^5/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(1/2)-1/16*b^2*arctanh(tanh(b*x+a))^(1/2)/x^3

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Rubi [A]  time = 0.17, antiderivative size = 221, 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} \[ -\frac {b^3}{64 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}+\frac {3 b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {3 b^5 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^6,x]

[Out]

(3*b^5*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(128*(b*x - ArcTanh[Tanh[a + b
*x]])^(5/2)) + b^4/(128*x*ArcTanh[Tanh[a + b*x]]^(3/2)) - b^5/(128*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh
[a + b*x]]^(3/2)) - b^3/(64*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (3*b^5)/(128*(b*x - ArcTanh[Tanh[a + b*x]])^2*
Sqrt[ArcTanh[Tanh[a + b*x]]]) - (b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(16*x^3) - (b*ArcTanh[Tanh[a + b*x]]^(3/2))
/(8*x^4) - ArcTanh[Tanh[a + b*x]]^(5/2)/(5*x^5)

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 {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^6} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}+\frac {1}{2} b \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^5} \, dx\\ &=-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}+\frac {1}{16} \left (3 b^2\right ) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^4} \, dx\\ &=-\frac {b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}+\frac {1}{32} b^3 \int \frac {1}{x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac {b^3}{64 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}-\frac {1}{128} b^4 \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac {b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{64 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}+\frac {1}{256} \left (3 b^5\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac {b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{64 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}-\frac {\left (3 b^5\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{256 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{64 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {3 b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}-\frac {\left (3 b^5\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{256 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {3 b^5 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac {b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{64 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {3 b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac {b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 150, normalized size = 0.68 \[ \frac {1}{640} \left (-\frac {15 b^5 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (10 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+8 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-176 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+15 b^4 x^4\right )}{x^5 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^6,x]

[Out]

((-15*b^5*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[
a + b*x]])^(5/2) - (Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^4*x^4 + 10*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 8*b^2*x^2*A
rcTanh[Tanh[a + b*x]]^2 - 176*b*x*ArcTanh[Tanh[a + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(x^5*(-(b*x) + Arc
Tanh[Tanh[a + b*x]])^2))/640

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fricas [A]  time = 0.84, size = 189, normalized size = 0.86 \[ \left [\frac {15 \, \sqrt {a} b^{5} x^{5} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt {b x + a}}{1280 \, a^{3} x^{5}}, \frac {15 \, \sqrt {-a} b^{5} x^{5} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt {b x + a}}{640 \, a^{3} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/1280*(15*sqrt(a)*b^5*x^5*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(15*a*b^4*x^4 - 10*a^2*b^3*x^3 -
248*a^3*b^2*x^2 - 336*a^4*b*x - 128*a^5)*sqrt(b*x + a))/(a^3*x^5), 1/640*(15*sqrt(-a)*b^5*x^5*arctan(sqrt(b*x
+ a)*sqrt(-a)/a) + (15*a*b^4*x^4 - 10*a^2*b^3*x^3 - 248*a^3*b^2*x^2 - 336*a^4*b*x - 128*a^5)*sqrt(b*x + a))/(a
^3*x^5)]

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giac [A]  time = 0.31, size = 123, normalized size = 0.56 \[ \frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} b^{6} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {\sqrt {2} {\left (15 \, {\left (b x + a\right )}^{\frac {9}{2}} b^{6} - 70 \, {\left (b x + a\right )}^{\frac {7}{2}} a b^{6} - 128 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{6} + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{6} - 15 \, \sqrt {b x + a} a^{4} b^{6}\right )}}{a^{2} b^{5} x^{5}}\right )}}{1280 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^6,x, algorithm="giac")

[Out]

1/1280*sqrt(2)*(15*sqrt(2)*b^6*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + sqrt(2)*(15*(b*x + a)^(9/2)*b^6
- 70*(b*x + a)^(7/2)*a*b^6 - 128*(b*x + a)^(5/2)*a^2*b^6 + 70*(b*x + a)^(3/2)*a^3*b^6 - 15*sqrt(b*x + a)*a^4*
b^6)/(a^2*b^5*x^5))/b

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maple [A]  time = 0.18, size = 262, normalized size = 1.19 \[ 2 b^{5} \left (\frac {\frac {3 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{256 \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}-\frac {7 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{128 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{10}+\left (\frac {7 \arctanh \left (\tanh \left (b x +a \right )\right )}{128}-\frac {7 b x}{128}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (-\frac {3 a^{2}}{256}-\frac {3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{128}-\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{256}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{5} x^{5}}-\frac {3 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{256 \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^6,x)

[Out]

2*b^5*((3/256/(a^2+2*a*(arctanh(tanh(b*x+a))-b*x-a)+(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(9/2)
-7/128/(arctanh(tanh(b*x+a))-b*x)*arctanh(tanh(b*x+a))^(7/2)-1/10*arctanh(tanh(b*x+a))^(5/2)+(7/128*arctanh(ta
nh(b*x+a))-7/128*b*x)*arctanh(tanh(b*x+a))^(3/2)+(-3/256*a^2-3/128*a*(arctanh(tanh(b*x+a))-b*x-a)-3/256*(arcta
nh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^5/x^5-3/256/(a^2+2*a*(arctanh(tanh(b*x+a))-b*x-a)+(arc
tanh(tanh(b*x+a))-b*x-a)^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 \[ \int \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^6,x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^6, x)

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mupad [B]  time = 6.51, size = 1292, normalized size = 5.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^6,x)

[Out]

(3*b^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))/
(32*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) -
((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)^3)/(4*x^5*(5*log
(2/(exp(2*a)*exp(2*b*x) + 1)) - 5*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 10*b*x)) + (b^3*(lo
g((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))/(16*x^2*(2
*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 4*b*x)) + (2^(1
/2)*b^5*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))*1024i)/(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)))*3i)/(64*(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)) - (93*b^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)*(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))/(80*x^3*(3*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 3*log
((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x)) + (21*b*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e
xp(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)^2)/(20*x^4*(4*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)) + 8*b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2)/x**6,x)

[Out]

Integral(atanh(tanh(a + b*x))**(5/2)/x**6, x)

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