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

Optimal. Leaf size=99 \[ \frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^5}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac{2 x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b} \]

[Out]

(2*x^4*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (16*x^3*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b^2) + (32*x^2*ArcTanh[Tanh[
a + b*x]]^(5/2))/(5*b^3) - (128*x*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^4) + (256*ArcTanh[Tanh[a + b*x]]^(9/2))/
(315*b^5)

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Rubi [A]  time = 0.0653643, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2157, 30} \[ \frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^5}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac{2 x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^4*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (16*x^3*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b^2) + (32*x^2*ArcTanh[Tanh[
a + b*x]]^(5/2))/(5*b^3) - (128*x*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^4) + (256*ArcTanh[Tanh[a + b*x]]^(9/2))/
(315*b^5)

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 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{x^4}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac{2 x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{8 \int x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac{2 x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{16 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{b^2}\\ &=\frac{2 x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac{64 \int x \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b^3}\\ &=\frac{2 x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac{128 \int \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{35 b^4}\\ &=\frac{2 x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac{128 \operatorname{Subst}\left (\int x^{7/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{35 b^5}\\ &=\frac{2 x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^5}\\ \end{align*}

Mathematica [A]  time = 0.0427971, size = 83, normalized size = 0.84 \[ \frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-840 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+1008 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-576 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+315 b^4 x^4\right )}{315 b^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(315*b^4*x^4 - 840*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 1008*b^2*x^2*ArcTanh[Tanh[
a + b*x]]^2 - 576*b*x*ArcTanh[Tanh[a + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(315*b^5)

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Maple [A]  time = 0.061, size = 153, normalized size = 1.6 \begin{align*} 2\,{\frac{1/9\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{9/2}+1/7\, \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}+1/5\, \left ( 2\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}+ \left ( -2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +2\,bx \right ) ^{2} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}+2/3\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2} \left ( -2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +2\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{{b}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.79655, size = 86, normalized size = 0.87 \begin{align*} \frac{2 \,{\left (35 \, b^{5} x^{5} - 5 \, a b^{4} x^{4} + 8 \, a^{2} b^{3} x^{3} - 16 \, a^{3} b^{2} x^{2} + 64 \, a^{4} b x + 128 \, a^{5}\right )}}{315 \, \sqrt{b x + a} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^5*x^5 - 5*a*b^4*x^4 + 8*a^2*b^3*x^3 - 16*a^3*b^2*x^2 + 64*a^4*b*x + 128*a^5)/(sqrt(b*x + a)*b^5)

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Fricas [A]  time = 2.08191, size = 126, normalized size = 1.27 \begin{align*} \frac{2 \,{\left (35 \, b^{4} x^{4} - 40 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} - 64 \, a^{3} b x + 128 \, a^{4}\right )} \sqrt{b x + a}}{315 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^4*x^4 - 40*a*b^3*x^3 + 48*a^2*b^2*x^2 - 64*a^3*b*x + 128*a^4)*sqrt(b*x + a)/b^5

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{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(x**4/atanh(tanh(b*x+a))**(1/2),x)

[Out]

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

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Giac [A]  time = 1.16333, size = 82, normalized size = 0.83 \begin{align*} \frac{2 \,{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 180 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 378 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 420 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3} + 315 \, \sqrt{b x + a} a^{4}\right )}}{315 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sq
rt(b*x + a)*a^4)/b^5