3.148 \(\int \frac{1}{x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\)

Optimal. Leaf size=212 \[ \frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

(5*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
]])^(7/2)) - b^2/(8*x*ArcTanh[Tanh[a + b*x]]^(5/2)) + b^3/(8*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b
*x]]^(5/2)) + b/(12*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)) - (5*b^3)/(24*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[T
anh[a + b*x]]^(3/2)) - 1/(3*x^3*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^3*Sq
rt[ArcTanh[Tanh[a + b*x]]])

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Rubi [A]  time = 0.150657, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ \frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*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
]])^(7/2)) - b^2/(8*x*ArcTanh[Tanh[a + b*x]]^(5/2)) + b^3/(8*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b
*x]]^(5/2)) + b/(12*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)) - (5*b^3)/(24*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[T
anh[a + b*x]]^(3/2)) - 1/(3*x^3*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^3*Sq
rt[ArcTanh[Tanh[a + b*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]))

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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx &=-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{1}{6} b \int \frac{1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{1}{8} b^2 \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{1}{16} \left (5 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{\left (5 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 )}\\ &=-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (5 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 )^2}\\ &=-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (5 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 )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{5 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 )^{7/2}}-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}

Mathematica [A]  time = 0.107486, size = 117, normalized size = 0.55 \[ \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-26 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+33 b^2 x^2\right )}{24 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(8*(-(b*x) + ArcTanh[Tanh[
a + b*x]])^(7/2)) + (Sqrt[ArcTanh[Tanh[a + b*x]]]*(33*b^2*x^2 - 26*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh
[a + b*x]]^2))/(24*x^3*(b*x - ArcTanh[Tanh[a + b*x]])^3)

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Maple [A]  time = 0.148, size = 200, normalized size = 0.9 \begin{align*} 2\,{b}^{3} \left ( 2/3\,{\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ){b}^{3}{x}^{3}}}+10/3\,{\frac{1}{-4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx} \left ({\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ){b}^{2}{x}^{2}}}+3\,{\frac{1}{-4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx} \left ( 2\,{\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) bx}}-2\,{\frac{1}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}{\it Artanh} \left ({\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}} \right ) } \right ) } \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3*(2/3*arctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/b^3/x^3+10/3/(-4*arctanh(tanh(b*x+a))+4*
b*x)*(arctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/b^2/x^2+3/(-4*arctanh(tanh(b*x+a))+4*b*x)*(2*
arctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/b/x-2/(-4*arctanh(tanh(b*x+a))+4*b*x)/(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \sqrt{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.15512, size = 356, normalized size = 1.68 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{3} x^{3} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{48 \, a^{4} x^{3}}, -\frac{15 \, \sqrt{-a} b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{24 \, a^{4} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*sqrt(a)*b^3*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^2*x^2 - 10*a^2*b*x + 8*a^3)
*sqrt(b*x + a))/(a^4*x^3), -1/24*(15*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^2*x^2 - 10*a^
2*b*x + 8*a^3)*sqrt(b*x + a))/(a^4*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14601, size = 113, normalized size = 0.53 \begin{align*} -\frac{\frac{15 \, b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{15 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} - 40 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} + 33 \, \sqrt{b x + a} a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(15*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (15*(b*x + a)^(5/2)*b^4 - 40*(b*x + a)^(3/2)*a*b
^4 + 33*sqrt(b*x + a)*a^2*b^4)/(a^3*b^3*x^3))/b