3.214 \(\int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=228 \[ -\frac {63 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{11/2}}+\frac {7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {63 b^2}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac {21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

-63/4*b^(5/2)*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b*x-arctanh(tanh(b*x+a)))^(11/2)+21/4
*b/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^4+63/20/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^3+9/4/b/x^(7/2)/(b*x-arctanh(
tanh(b*x+a)))^2+7/4/b^2/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))-1/2/b/x^(7/2)/arctanh(tanh(b*x+a))^2+7/4/b^2/x^(9/2
)/arctanh(tanh(b*x+a))+63/4*b^2/(b*x-arctanh(tanh(b*x+a)))^5/x^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 228, 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, 2162} \[ \frac {7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {63 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{11/2}}+\frac {63 b^2}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac {21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

(-63*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b*x]])^
(11/2)) + (63*b^2)/(4*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^5) + (21*b)/(4*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*
x]])^4) + 63/(20*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + 9/(4*b*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2)
+ 7/(4*b^2*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(2*b*x^(7/2)*ArcTanh[Tanh[a + b*x]]^2) + 7/(4*b^2*x^(9/
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^{7/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac {7 \int \frac {1}{x^{9/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {63 \int \frac {1}{x^{11/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac {7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {63 \int \frac {1}{x^{9/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {63 \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {(63 b) \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (63 b^2\right ) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^4}\\ &=\frac {63 b^2}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac {21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (63 b^3\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 )^4}\\ &=-\frac {63 b^{5/2} \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 )^{11/2}}+\frac {63 b^2}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac {21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 174, normalized size = 0.76 \[ \frac {1}{20} \left (-\frac {315 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{11/2}}+\frac {75 b^3 \sqrt {x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \tanh ^{-1}(\tanh (a+b x))}-\frac {10 b^3 \sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}+\frac {8 \left (-7 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2+36 b^2 x^2\right )}{x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

((75*b^3*Sqrt[x])/((b*x - ArcTanh[Tanh[a + b*x]])^5*ArcTanh[Tanh[a + b*x]]) - (315*b^(5/2)*ArcTan[(Sqrt[b]*Sqr
t[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(11/2) - (10*b^3*Sqrt[x])/(Arc
Tanh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (8*(36*b^2*x^2 - 7*b*x*ArcTanh[Tanh[a + b*x]] + A
rcTanh[Tanh[a + b*x]]^2))/(x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^5))/20

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fricas [A]  time = 0.65, size = 276, normalized size = 1.21 \[ \left [\frac {315 \, {\left (b^{4} x^{5} + 2 \, a b^{3} x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (315 \, b^{4} x^{4} + 525 \, a b^{3} x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {x}}{40 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, \frac {315 \, {\left (b^{4} x^{5} + 2 \, a b^{3} x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (315 \, b^{4} x^{4} + 525 \, a b^{3} x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {x}}{20 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

[1/40*(315*(b^4*x^5 + 2*a*b^3*x^4 + a^2*b^2*x^3)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a))
- 2*(315*b^4*x^4 + 525*a*b^3*x^3 + 168*a^2*b^2*x^2 - 24*a^3*b*x + 8*a^4)*sqrt(x))/(a^5*b^2*x^5 + 2*a^6*b*x^4 +
a^7*x^3), 1/20*(315*(b^4*x^5 + 2*a*b^3*x^4 + a^2*b^2*x^3)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (315*b^
4*x^4 + 525*a*b^3*x^3 + 168*a^2*b^2*x^2 - 24*a^3*b*x + 8*a^4)*sqrt(x))/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)]

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giac [A]  time = 0.20, size = 80, normalized size = 0.35 \[ -\frac {63 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} - \frac {15 \, b^{4} x^{\frac {3}{2}} + 17 \, a b^{3} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{5}} - \frac {2 \, {\left (30 \, b^{2} x^{2} - 5 \, a b x + a^{2}\right )}}{5 \, a^{5} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

-63/4*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - 1/4*(15*b^4*x^(3/2) + 17*a*b^3*sqrt(x))/((b*x + a)^2*a
^5) - 2/5*(30*b^2*x^2 - 5*a*b*x + a^2)/(a^5*x^(5/2))

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maple [A]  time = 0.33, size = 229, normalized size = 1.00 \[ -\frac {15 b^{4} x^{\frac {3}{2}}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {17 b^{3} a \sqrt {x}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {17 b^{3} \sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {63 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {5}{2}}}-\frac {12 b^{2}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \sqrt {x}}+\frac {2 b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/arctanh(tanh(b*x+a))^3,x)

[Out]

-15/4/(arctanh(tanh(b*x+a))-b*x)^5*b^4/arctanh(tanh(b*x+a))^2*x^(3/2)-17/4/(arctanh(tanh(b*x+a))-b*x)^5*b^3/ar
ctanh(tanh(b*x+a))^2*a*x^(1/2)-17/4/(arctanh(tanh(b*x+a))-b*x)^5*b^3/arctanh(tanh(b*x+a))^2*x^(1/2)*(arctanh(t
anh(b*x+a))-b*x-a)-63/4/(arctanh(tanh(b*x+a))-b*x)^5*b^3/((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))-2/5/(arctanh(tanh(b*x+a))-b*x)^3/x^(5/2)-12/(arctanh(tanh(b*x+a))-b*x)^
5*b^2/x^(1/2)+2/(arctanh(tanh(b*x+a))-b*x)^4*b/x^(3/2)

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maxima [A]  time = 0.43, size = 97, normalized size = 0.43 \[ -\frac {315 \, b^{4} x^{4} + 525 \, a b^{3} x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x + 8 \, a^{4}}{20 \, {\left (a^{5} b^{2} x^{\frac {9}{2}} + 2 \, a^{6} b x^{\frac {7}{2}} + a^{7} x^{\frac {5}{2}}\right )}} - \frac {63 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-1/20*(315*b^4*x^4 + 525*a*b^3*x^3 + 168*a^2*b^2*x^2 - 24*a^3*b*x + 8*a^4)/(a^5*b^2*x^(9/2) + 2*a^6*b*x^(7/2)
+ a^7*x^(5/2)) - 63/4*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5)

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mupad [B]  time = 1.95, size = 2151, normalized size = 9.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*atanh(tanh(a + b*x))^3),x)

[Out]

16/(5*x^(5/2)*(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) - (x*((((448*b^4)/(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)^3) - (16*b^3*(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)^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))/(2*b) - (112*b^3)/((2*a*b - b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex
p(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) + (8*b^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)))/((2*a*b - b*(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))*(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*b*(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)))/((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))/(x^(3/2)*(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)))^2) - ((336*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)^4 - (1008*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)^5)/(x^(1/2)*(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)))) + (252*2^(1/2)*b^(5/2)*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)^10 + 180*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)^8 - 960*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)^7 + 3360*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)^6 - 8064*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)^5 + 13440*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)^4 - 15360*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)^3 + 11520*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)^2 + 1024*a^10 - 20*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)^9 - 512
0*a^9*(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*(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)^(11/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {7}{2}} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(7/2)/atanh(tanh(b*x+a))**3,x)

[Out]

Integral(1/(x**(7/2)*atanh(tanh(a + b*x))**3), x)

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