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

Optimal. Leaf size=99 \[ -\frac{16 x^3}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{32 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^5}-\frac{2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0660405, 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} \[ -\frac{16 x^3}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{32 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^5}-\frac{2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^4)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (16*x^3)/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (32*x^2*Sqrt[Arc
Tanh[Tanh[a + b*x]]])/b^3 - (128*x*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b^4) + (256*ArcTanh[Tanh[a + b*x]]^(5/2))/
(15*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))^{5/2}} \, dx &=-\frac{2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{8 \int \frac{x^3}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 b}\\ &=-\frac{2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{16 x^3}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 \int \frac{x^2}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b^2}\\ &=-\frac{2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{16 x^3}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{32 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{64 \int x \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=-\frac{2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{16 x^3}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{32 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac{128 \int \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b^4}\\ &=-\frac{2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{16 x^3}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{32 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac{128 \operatorname{Subst}\left (\int x^{3/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^5}\\ &=-\frac{2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{16 x^3}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{32 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^5}\\ \end{align*}

Mathematica [A]  time = 0.0480728, size = 83, normalized size = 0.84 \[ -\frac{2 \left (40 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))-240 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+320 b x \tanh ^{-1}(\tanh (a+b x))^3-128 \tanh ^{-1}(\tanh (a+b x))^4+5 b^4 x^4\right )}{15 b^5 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*b^4*x^4 + 40*b^3*x^3*ArcTanh[Tanh[a + b*x]] - 240*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 320*b*x*ArcTanh[Ta
nh[a + b*x]]^3 - 128*ArcTanh[Tanh[a + b*x]]^4))/(15*b^5*ArcTanh[Tanh[a + b*x]]^(3/2))

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Maple [B]  time = 0.045, size = 295, normalized size = 3. \begin{align*} 2\,{\frac{1}{{b}^{5}} \left ( 1/5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}-4/3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}a-4/3\, \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 ) +6\,\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{a}^{2}+12\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }+6\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-{\frac{-4\,{a}^{3}-12\,{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) -12\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}-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 ) }}}-1/3\,{\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}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.78818, size = 86, normalized size = 0.87 \begin{align*} \frac{2 \,{\left (3 \, b^{5} x^{5} - 5 \, a b^{4} x^{4} + 40 \, a^{2} b^{3} x^{3} + 240 \, a^{3} b^{2} x^{2} + 320 \, a^{4} b x + 128 \, a^{5}\right )}}{15 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b^5*x^5 - 5*a*b^4*x^4 + 40*a^2*b^3*x^3 + 240*a^3*b^2*x^2 + 320*a^4*b*x + 128*a^5)/((b*x + a)^(5/2)*b^5
)

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Fricas [A]  time = 2.07703, size = 161, normalized size = 1.63 \begin{align*} \frac{2 \,{\left (3 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} + 192 \, a^{3} b x + 128 \, a^{4}\right )} \sqrt{b x + a}}{15 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b^4*x^4 - 8*a*b^3*x^3 + 48*a^2*b^2*x^2 + 192*a^3*b*x + 128*a^4)*sqrt(b*x + a)/(b^7*x^2 + 2*a*b^6*x + a
^2*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{atanh}^{\frac{5}{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))**(5/2),x)

[Out]

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

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Giac [A]  time = 1.14978, size = 101, normalized size = 1.02 \begin{align*} \frac{2 \,{\left (12 \,{\left (b x + a\right )} a^{3} - a^{4}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{5}} + \frac{2 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{20} - 20 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{20} + 90 \, \sqrt{b x + a} a^{2} b^{20}\right )}}{15 \, b^{25}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(12*(b*x + a)*a^3 - a^4)/((b*x + a)^(3/2)*b^5) + 2/15*(3*(b*x + a)^(5/2)*b^20 - 20*(b*x + a)^(3/2)*a*b^20
+ 90*sqrt(b*x + a)*a^2*b^20)/b^25