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

Optimal. Leaf size=135 \[ -\frac {7 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{9/2}}+\frac {7 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}+\frac {7 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^3}-\frac {x^{7/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {7 x^{5/2}}{5 b^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2159, 2162} \[ \frac {7 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^3}+\frac {7 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {7 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{9/2}}-\frac {x^{7/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {7 x^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

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

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Mathematica [A]  time = 0.21, size = 144, normalized size = 1.07 \[ -\frac {7 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{9/2}}+\frac {\sqrt {x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}{b^4 \tanh ^{-1}(\tanh (a+b x))}+\frac {6 \sqrt {x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{b^4}-\frac {4 x^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{3 b^3}+\frac {2 x^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2))/(5*b^2) - (4*x^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/(3*b^3) + (6*Sqrt[x]*(-(b*x) + ArcTanh[Tan
h[a + b*x]])^2)/b^4 - (7*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tan
h[a + b*x]])^(5/2))/b^(9/2) + (Sqrt[x]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3)/(b^4*ArcTanh[Tanh[a + b*x]])

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fricas [A]  time = 0.41, size = 188, normalized size = 1.39 \[ \left [\frac {105 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} + 70 \, a^{2} b x + 105 \, a^{3}\right )} \sqrt {x}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {105 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} + 70 \, a^{2} b x + 105 \, a^{3}\right )} \sqrt {x}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(105*(a^2*b*x + a^3)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(6*b^3*x^3 - 14*a*
b^2*x^2 + 70*a^2*b*x + 105*a^3)*sqrt(x))/(b^5*x + a*b^4), -1/15*(105*(a^2*b*x + a^3)*sqrt(a/b)*arctan(b*sqrt(x
)*sqrt(a/b)/a) - (6*b^3*x^3 - 14*a*b^2*x^2 + 70*a^2*b*x + 105*a^3)*sqrt(x))/(b^5*x + a*b^4)]

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giac [A]  time = 0.16, size = 76, normalized size = 0.56 \[ -\frac {7 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {a^{3} \sqrt {x}}{{\left (b x + a\right )} b^{4}} + \frac {2 \, {\left (3 \, b^{8} x^{\frac {5}{2}} - 10 \, a b^{7} x^{\frac {3}{2}} + 45 \, a^{2} b^{6} \sqrt {x}\right )}}{15 \, b^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*a^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + a^3*sqrt(x)/((b*x + a)*b^4) + 2/15*(3*b^8*x^(5/2) - 10*a*
b^7*x^(3/2) + 45*a^2*b^6*sqrt(x))/b^10

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maple [B]  time = 0.27, size = 452, normalized size = 3.35 \[ \frac {2 x^{\frac {5}{2}}}{5 b^{2}}-\frac {4 x^{\frac {3}{2}} a}{3 b^{3}}-\frac {4 x^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3 b^{3}}+\frac {6 a^{2} \sqrt {x}}{b^{4}}+\frac {12 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{b^{4}}+\frac {6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{b^{4}}+\frac {\sqrt {x}\, a^{3}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {3 \sqrt {x}\, a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {3 \sqrt {x}\, a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {7 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{3}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {21 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {21 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {7 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{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(x^(7/2)/arctanh(tanh(b*x+a))^2,x)

[Out]

2/5*x^(5/2)/b^2-4/3/b^3*x^(3/2)*a-4/3/b^3*x^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)+6/b^4*a^2*x^(1/2)+12/b^4*a*(arc
tanh(tanh(b*x+a))-b*x-a)*x^(1/2)+6/b^4*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)+1/b^4*x^(1/2)/arctanh(tanh(b*x+a
))*a^3+3/b^4*x^(1/2)/arctanh(tanh(b*x+a))*a^2*(arctanh(tanh(b*x+a))-b*x-a)+3/b^4*x^(1/2)/arctanh(tanh(b*x+a))*
a*(arctanh(tanh(b*x+a))-b*x-a)^2+1/b^4*x^(1/2)/arctanh(tanh(b*x+a))*(arctanh(tanh(b*x+a))-b*x-a)^3-7/b^4/((arc
tanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2))*a^3-21/b^4/((arctanh(ta
nh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2))*a^2*(arctanh(tanh(b*x+a))-b*x-
a)-21/b^4/((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))*a*(arcta
nh(tanh(b*x+a))-b*x-a)^2-7/b^4/((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))*(arctanh(tanh(b*x+a))-b*x-a)^3

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maxima [A]  time = 0.42, size = 75, normalized size = 0.56 \[ \frac {6 \, b^{3} x^{\frac {7}{2}} - 14 \, a b^{2} x^{\frac {5}{2}} + 70 \, a^{2} b x^{\frac {3}{2}} + 105 \, a^{3} \sqrt {x}}{15 \, {\left (b^{5} x + a b^{4}\right )}} - \frac {7 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(6*b^3*x^(7/2) - 14*a*b^2*x^(5/2) + 70*a^2*b*x^(3/2) + 105*a^3*sqrt(x))/(b^5*x + a*b^4) - 7*a^3*arctan(b*
sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4)

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mupad [B]  time = 1.61, size = 523, normalized size = 3.87 \[ \frac {2\,x^{5/2}}{5\,b^2}+\frac {2\,x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{3\,b^3}+\frac {3\,\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2\,b^4}+\frac {7\,\sqrt {2}\,\ln \left (\frac {64\,b^{19/2}\,\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}+2\,\sqrt {2}\,b\,x\right )}{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^{5/2}}{16\,b^{9/2}}-\frac {\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4\,b^4\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/atanh(tanh(a + b*x))^2,x)

[Out]

(2*x^(5/2))/(5*b^2) + (2*x^(3/2)*(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*b^3) + (3*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)^2)/(2*b^4) + (7*2^(1/2)*log((64*b^(19/2)*(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))/(
(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)*e
xp(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)))*(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))/(16*b^(9/2)) - (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)^3)/(4*b^4
*(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(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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