3.149 \(\int \frac{x^4}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=95 \[ \frac{16 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac{256 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^5}-\frac{2 x^4}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

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

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Rubi [A]  time = 0.0671925, antiderivative size = 95, 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} \[ \frac{16 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac{256 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^5}-\frac{2 x^4}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(-2*x^4)/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (16*x^3*Sqrt[ArcTanh[Tanh[a + b*x]]])/b^2 - (32*x^2*ArcTanh[Tanh[a
+ b*x]]^(3/2))/b^3 + (128*x*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b^4) - (256*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*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}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac{2 x^4}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{8 \int \frac{x^3}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b}\\ &=-\frac{2 x^4}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{48 \int x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=-\frac{2 x^4}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac{64 \int x \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{b^3}\\ &=-\frac{2 x^4}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac{128 \int \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b^4}\\ &=-\frac{2 x^4}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac{128 \operatorname{Subst}\left (\int x^{5/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^5}\\ &=-\frac{2 x^4}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac{256 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^5}\\ \end{align*}

Mathematica [A]  time = 0.0435768, size = 83, normalized size = 0.87 \[ -\frac{2 \left (-280 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+560 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-448 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+35 b^4 x^4\right )}{35 b^5 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(-2*(35*b^4*x^4 - 280*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 560*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 448*b*x*ArcTanh[
Tanh[a + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(35*b^5*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Maple [B]  time = 0.042, size = 319, normalized size = 3.4 \begin{align*} 2\,{\frac{1}{{b}^{5}} \left ( 1/7\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}-4/5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}a-4/5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}{a}^{2}+4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}-4\,\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{a}^{3}-12\,{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-12\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-{\frac{{a}^{4}+4\,{a}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +6\,{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}+4\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}+ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{4}}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.78609, size = 86, normalized size = 0.91 \begin{align*} \frac{2 \,{\left (5 \, b^{5} x^{5} - 3 \, a b^{4} x^{4} + 8 \, a^{2} b^{3} x^{3} - 48 \, a^{3} b^{2} x^{2} - 192 \, a^{4} b x - 128 \, a^{5}\right )}}{35 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/35*(5*b^5*x^5 - 3*a*b^4*x^4 + 8*a^2*b^3*x^3 - 48*a^3*b^2*x^2 - 192*a^4*b*x - 128*a^5)/((b*x + a)^(3/2)*b^5)

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Fricas [A]  time = 2.34281, size = 138, normalized size = 1.45 \begin{align*} \frac{2 \,{\left (5 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 16 \, a^{2} b^{2} x^{2} - 64 \, a^{3} b x - 128 \, a^{4}\right )} \sqrt{b x + a}}{35 \,{\left (b^{6} x + a b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/35*(5*b^4*x^4 - 8*a*b^3*x^3 + 16*a^2*b^2*x^2 - 64*a^3*b*x - 128*a^4)*sqrt(b*x + a)/(b^6*x + a*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{atanh}^{\frac{3}{2}}{\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))**(3/2),x)

[Out]

Integral(x**4/atanh(tanh(a + b*x))**(3/2), x)

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Giac [A]  time = 1.16118, size = 104, normalized size = 1.09 \begin{align*} -\frac{2 \, a^{4}}{\sqrt{b x + a} b^{5}} + \frac{2 \,{\left (5 \,{\left (b x + a\right )}^{\frac{7}{2}} b^{30} - 28 \,{\left (b x + a\right )}^{\frac{5}{2}} a b^{30} + 70 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2} b^{30} - 140 \, \sqrt{b x + a} a^{3} b^{30}\right )}}{35 \, b^{35}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

-2*a^4/(sqrt(b*x + a)*b^5) + 2/35*(5*(b*x + a)^(7/2)*b^30 - 28*(b*x + a)^(5/2)*a*b^30 + 70*(b*x + a)^(3/2)*a^2
*b^30 - 140*sqrt(b*x + a)*a^3*b^30)/b^35