3.213 \(\int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=201 \[ -\frac {35 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac {5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {35 b}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4} \]

[Out]

-35/4*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)))^(9/2)+35/12
/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^3+7/4/b/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^2+5/4/b^2/x^(7/2)/(b*x-arctanh(
tanh(b*x+a)))-1/2/b/x^(5/2)/arctanh(tanh(b*x+a))^2+5/4/b^2/x^(7/2)/arctanh(tanh(b*x+a))+35/4*b/(b*x-arctanh(ta
nh(b*x+a)))^4/x^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2162} \[ \frac {5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {35 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac {35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {35 b}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

(-35*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b*x]])^
(9/2)) + (35*b)/(4*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^4) + 35/(12*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3
) + 7/(4*b*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + 5/(4*b^2*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(2
*b*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2) + 5/(4*b^2*x^(7/2)*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]

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^{5/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac {5 \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {35 \int \frac {1}{x^{9/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac {5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {35 \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {35 \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {(35 b) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {35 b}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (35 b^2\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=-\frac {35 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac {35 b}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 156, normalized size = 0.78 \[ \frac {1}{12} \left (\frac {105 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}}+\frac {6 b^2 \sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}+\frac {33 b^2 \sqrt {x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}+\frac {80 b x-8 \tanh ^{-1}(\tanh (a+b x))}{x^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

((105*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]
])^(9/2) + (80*b*x - 8*ArcTanh[Tanh[a + b*x]])/(x^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (33*b^2*Sqrt[x]
)/(ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (6*b^2*Sqrt[x])/(ArcTanh[Tanh[a + b*x]]^2*(-(
b*x) + ArcTanh[Tanh[a + b*x]])^3))/12

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fricas [A]  time = 0.54, size = 250, normalized size = 1.24 \[ \left [\frac {105 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, -\frac {105 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

[1/24*(105*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) +
2*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), -1/12*(10
5*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (105*b^3*x^3 + 175*a*b^2*x^2
+ 56*a^2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)]

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giac [A]  time = 0.43, size = 71, normalized size = 0.35 \[ \frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} + \frac {2 \, {\left (9 \, b x - a\right )}}{3 \, a^{4} x^{\frac {3}{2}}} + \frac {11 \, b^{3} x^{\frac {3}{2}} + 13 \, a b^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

35/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + 2/3*(9*b*x - a)/(a^4*x^(3/2)) + 1/4*(11*b^3*x^(3/2) + 1
3*a*b^2*sqrt(x))/((b*x + a)^2*a^4)

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maple [A]  time = 0.29, size = 207, normalized size = 1.03 \[ -\frac {2}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {3}{2}}}+\frac {6 b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {x}}+\frac {11 b^{3} x^{\frac {3}{2}}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {13 b^{2} a \sqrt {x}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {13 b^{2} \sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {35 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(arctanh(tanh(b*x+a))-b*x)^3/x^(3/2)+6/(arctanh(tanh(b*x+a))-b*x)^4*b/x^(1/2)+11/4/(arctanh(tanh(b*x+a))-
b*x)^4*b^3/arctanh(tanh(b*x+a))^2*x^(3/2)+13/4/(arctanh(tanh(b*x+a))-b*x)^4*b^2/arctanh(tanh(b*x+a))^2*a*x^(1/
2)+13/4/(arctanh(tanh(b*x+a))-b*x)^4*b^2/arctanh(tanh(b*x+a))^2*x^(1/2)*(arctanh(tanh(b*x+a))-b*x-a)+35/4/(arc
tanh(tanh(b*x+a))-b*x)^4*b^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))

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maxima [A]  time = 0.43, size = 86, normalized size = 0.43 \[ \frac {105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}}{12 \, {\left (a^{4} b^{2} x^{\frac {7}{2}} + 2 \, a^{5} b x^{\frac {5}{2}} + a^{6} x^{\frac {3}{2}}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

1/12*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)/(a^4*b^2*x^(7/2) + 2*a^5*b*x^(5/2) + a^6*x^(3/2)) + 35
/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4)

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mupad [B]  time = 2.50, size = 1362, normalized size = 6.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*atanh(tanh(a + b*x))^3),x)

[Out]

(x^(1/2)*((2*(2*b*(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) - 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)))/(3*(2*a*b - b*(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))*(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)) + (56*b^2*x)/(3*(2*a*b - b*(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))*(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*b*x^2 - x*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + 2*b*x))^2 - (x^(1/2)*((280*b)/(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)^3) - (280*b^2*x)/(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*e
xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4))/(2*b*x^2 - x*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - l
og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)) + (70*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)^8 + 112*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)^6 - 448*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)^5 + 1120*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)^4 - 1792*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)^3 + 1792*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)^2 + 256*a^8 - 16*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)^7 - 1024*a^7*(2*a - log((2*exp(2*a)*ex
p(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)^(9/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {5}{2}} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/atanh(tanh(b*x+a))**3,x)

[Out]

Integral(1/(x**(5/2)*atanh(tanh(a + b*x))**3), x)

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