3.166 \(\int \frac {1}{x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=278 \[ \frac {105 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {105 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{11/2}}-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}} \]

[Out]

105/8*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)))^(11/2
)-35/24*b^2/x/arctanh(tanh(b*x+a))^(9/2)+35/24*b^3/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(9/2)+5/12*
b/x^2/arctanh(tanh(b*x+a))^(7/2)-15/8*b^3/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(7/2)-1/3/x^3/arct
anh(tanh(b*x+a))^(5/2)+21/8*b^3/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(5/2)-35/8*b^3/(b*x-arctanh(
tanh(b*x+a)))^4/arctanh(tanh(b*x+a))^(3/2)+105/8*b^3/(b*x-arctanh(tanh(b*x+a)))^5/arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ \frac {105 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {105 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{11/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

(105*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]])^(11/2)) - (35*b^2)/(24*x*ArcTanh[Tanh[a + b*x]]^(9/2)) + (35*b^3)/(24*(b*x - ArcTanh[Tanh[a + b*x]])*Arc
Tanh[Tanh[a + b*x]]^(9/2)) + (5*b)/(12*x^2*ArcTanh[Tanh[a + b*x]]^(7/2)) - (15*b^3)/(8*(b*x - ArcTanh[Tanh[a +
b*x]])^2*ArcTanh[Tanh[a + b*x]]^(7/2)) - 1/(3*x^3*ArcTanh[Tanh[a + b*x]]^(5/2)) + (21*b^3)/(8*(b*x - ArcTanh[
Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(5/2)) - (35*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^4*ArcTanh[Tanh[a
+ b*x]]^(3/2)) + (105*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^5*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))^{5/2}} \, dx &=-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{6} (5 b) \int \frac {1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {1}{24} \left (35 b^2\right ) \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx\\ &=-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{16} \left (105 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{11/2}} \, dx\\ &=-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {\left (105 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {\left (105 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 )^2}\\ &=-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {\left (105 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 )^3}\\ &=-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\left (105 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 )^4}\\ &=-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {105 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (105 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 )^4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {105 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 )^{11/2}}-\frac {35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac {5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {105 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 150, normalized size = 0.54 \[ \frac {1}{24} \left (\frac {315 b^3 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{11/2}}+\frac {208 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+165 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-50 b x \tanh ^{-1}(\tanh (a+b x))^3+8 \tanh ^{-1}(\tanh (a+b x))^4-16 b^4 x^4}{x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \tanh ^{-1}(\tanh (a+b x))^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

((315*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[
a + b*x]])^(11/2) + (-16*b^4*x^4 + 208*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 165*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 -
50*b*x*ArcTanh[Tanh[a + b*x]]^3 + 8*ArcTanh[Tanh[a + b*x]]^4)/(x^3*(b*x - ArcTanh[Tanh[a + b*x]])^5*ArcTanh[T
anh[a + b*x]]^(3/2)))/24

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fricas [A]  time = 0.51, size = 277, normalized size = 1.00 \[ \left [\frac {315 \, {\left (b^{5} x^{5} + 2 \, a b^{4} x^{4} + a^{2} b^{3} x^{3}\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x + a}}{48 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}, -\frac {315 \, {\left (b^{5} x^{5} + 2 \, a b^{4} x^{4} + a^{2} b^{3} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x + a}}{24 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

[1/48*(315*(b^5*x^5 + 2*a*b^4*x^4 + a^2*b^3*x^3)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(315
*a*b^4*x^4 + 420*a^2*b^3*x^3 + 63*a^3*b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4
+ a^8*x^3), -1/24*(315*(b^5*x^5 + 2*a*b^4*x^4 + a^2*b^3*x^3)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (315*
a*b^4*x^4 + 420*a^2*b^3*x^3 + 63*a^3*b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4 +
a^8*x^3)]

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giac [A]  time = 0.22, size = 115, normalized size = 0.41 \[ -\frac {105 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{5}} - \frac {315 \, {\left (b x + a\right )}^{4} b^{3} - 840 \, {\left (b x + a\right )}^{3} a b^{3} + 693 \, {\left (b x + a\right )}^{2} a^{2} b^{3} - 144 \, {\left (b x + a\right )} a^{3} b^{3} - 16 \, a^{4} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )}^{3} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-105/8*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5) - 1/24*(315*(b*x + a)^4*b^3 - 840*(b*x + a)^3*a*b^3 +
693*(b*x + a)^2*a^2*b^3 - 144*(b*x + a)*a^3*b^3 - 16*a^4*b^3)/(((b*x + a)^(3/2) - sqrt(b*x + a)*a)^3*a^5)

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maple [A]  time = 0.16, size = 211, normalized size = 0.76 \[ 2 b^{3} \left (-\frac {1}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {\frac {\frac {41 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16}+\left (-\frac {35 \arctanh \left (\tanh \left (b x +a \right )\right )}{6}+\frac {35 b x}{6}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (\frac {55 a^{2}}{16}+\frac {55 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8}+\frac {55 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{16}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{3} x^{3}}-\frac {105 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

2*b^3*(-1/3/(arctanh(tanh(b*x+a))-b*x)^4/arctanh(tanh(b*x+a))^(3/2)-4/(arctanh(tanh(b*x+a))-b*x)^5/arctanh(tan
h(b*x+a))^(1/2)-1/(arctanh(tanh(b*x+a))-b*x)^5*((41/16*arctanh(tanh(b*x+a))^(5/2)+(-35/6*arctanh(tanh(b*x+a))+
35/6*b*x)*arctanh(tanh(b*x+a))^(3/2)+(55/16*a^2+55/8*a*(arctanh(tanh(b*x+a))-b*x-a)+55/16*(arctanh(tanh(b*x+a)
)-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3-105/16/(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 \[ \int \frac {1}{x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(x^4*arctanh(tanh(b*x + a))^(5/2)), x)

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mupad [B]  time = 6.42, size = 2359, normalized size = 8.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*atanh(tanh(a + b*x))^(5/2)),x)

[Out]

(8*(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)^3) - ((
(140*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 - (840*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)*(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)))) - ((log((2*e
xp(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*((((1232*b^4)/
(9*(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) - (80*b^3*(2*b*(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) - 7*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)))/(9*(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) - (952*b^3)/(9*(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) + (128*b^2*(2*b*(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) - 7*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)))/(9*(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)) - (68*b*(2*b*(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) - 7*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)))/(9*(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^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) + (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)*ex
p(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*ex
p(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)^11 + 220*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)^9 - 1320*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)^8 + 5280*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)^7 - 14784*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)^6 + 29
568*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)^5 - 42240*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)^4 + 42240*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)*ex
p(2*b*x) + 1)) + 2*b*x)^3 - 28160*a^9*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(e
xp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 - 2048*a^11 - 22*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)^10 + 11264*a^10*(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)) - l
og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2)))*210i)/(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 \[ \int \frac {1}{x^{4} \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Integral(1/(x**4*atanh(tanh(a + b*x))**(5/2)), x)

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