3.373 \(\int e^{\tanh ^{-1}(x)} (1-x)^{3/2} x \, dx\)

Optimal. Leaf size=34 \[ -\frac{2}{7} (x+1)^{7/2}+\frac{6}{5} (x+1)^{5/2}-\frac{4}{3} (x+1)^{3/2} \]

[Out]

(-4*(1 + x)^(3/2))/3 + (6*(1 + x)^(5/2))/5 - (2*(1 + x)^(7/2))/7

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Rubi [A]  time = 0.0620389, antiderivative size = 47, normalized size of antiderivative = 1.38, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {6128, 795, 627, 43} \[ -\frac{2}{7} \sqrt{1-x} \left (1-x^2\right )^{3/2}+\frac{2}{35} (x+1)^{5/2}-\frac{4}{21} (x+1)^{3/2} \]

Warning: Unable to verify antiderivative.

[In]

Int[E^ArcTanh[x]*(1 - x)^(3/2)*x,x]

[Out]

(-4*(1 + x)^(3/2))/21 + (2*(1 + x)^(5/2))/35 - (2*Sqrt[1 - x]*(1 - x^2)^(3/2))/7

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && IntegerQ[2*p]

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(x)} (1-x)^{3/2} x \, dx &=\int \sqrt{1-x} x \sqrt{1-x^2} \, dx\\ &=-\frac{2}{7} \sqrt{1-x} \left (1-x^2\right )^{3/2}-\frac{1}{7} \int \sqrt{1-x} \sqrt{1-x^2} \, dx\\ &=-\frac{2}{7} \sqrt{1-x} \left (1-x^2\right )^{3/2}-\frac{1}{7} \int (1-x) \sqrt{1+x} \, dx\\ &=-\frac{2}{7} \sqrt{1-x} \left (1-x^2\right )^{3/2}-\frac{1}{7} \int \left (2 \sqrt{1+x}-(1+x)^{3/2}\right ) \, dx\\ &=-\frac{4}{21} (1+x)^{3/2}+\frac{2}{35} (1+x)^{5/2}-\frac{2}{7} \sqrt{1-x} \left (1-x^2\right )^{3/2}\\ \end{align*}

Mathematica [A]  time = 0.0098989, size = 21, normalized size = 0.62 \[ -\frac{2}{105} (x+1)^{3/2} \left (15 x^2-33 x+22\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[x]*(1 - x)^(3/2)*x,x]

[Out]

(-2*(1 + x)^(3/2)*(22 - 33*x + 15*x^2))/105

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Maple [A]  time = 0.028, size = 34, normalized size = 1. \begin{align*} -{\frac{2\, \left ( 15\,{x}^{2}-33\,x+22 \right ) \left ( 1+x \right ) ^{2}}{105}\sqrt{1-x}{\frac{1}{\sqrt{-{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)/(-x^2+1)^(1/2)*(1-x)^(3/2)*x,x)

[Out]

-2/105*(1+x)^2*(15*x^2-33*x+22)*(1-x)^(1/2)/(-x^2+1)^(1/2)

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Maxima [B]  time = 0.971683, size = 65, normalized size = 1.91 \begin{align*} -\frac{2 \,{\left (15 \, x^{4} - 24 \, x^{3} + 13 \, x^{2} - 52 \, x - 104\right )}}{105 \, \sqrt{x + 1}} - \frac{2 \,{\left (x^{3} - 2 \, x^{2} + 3 \, x + 6\right )}}{5 \, \sqrt{x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(3/2)*x,x, algorithm="maxima")

[Out]

-2/105*(15*x^4 - 24*x^3 + 13*x^2 - 52*x - 104)/sqrt(x + 1) - 2/5*(x^3 - 2*x^2 + 3*x + 6)/sqrt(x + 1)

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Fricas [A]  time = 1.72861, size = 99, normalized size = 2.91 \begin{align*} \frac{2 \,{\left (15 \, x^{3} - 18 \, x^{2} - 11 \, x + 22\right )} \sqrt{-x^{2} + 1} \sqrt{-x + 1}}{105 \,{\left (x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(3/2)*x,x, algorithm="fricas")

[Out]

2/105*(15*x^3 - 18*x^2 - 11*x + 22)*sqrt(-x^2 + 1)*sqrt(-x + 1)/(x - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (1 - x\right )^{\frac{3}{2}} \left (x + 1\right )}{\sqrt{- \left (x - 1\right ) \left (x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-x**2+1)**(1/2)*(1-x)**(3/2)*x,x)

[Out]

Integral(x*(1 - x)**(3/2)*(x + 1)/sqrt(-(x - 1)*(x + 1)), x)

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Giac [A]  time = 1.2457, size = 36, normalized size = 1.06 \begin{align*} -\frac{2}{7} \,{\left (x + 1\right )}^{\frac{7}{2}} + \frac{6}{5} \,{\left (x + 1\right )}^{\frac{5}{2}} - \frac{4}{3} \,{\left (x + 1\right )}^{\frac{3}{2}} + \frac{16}{105} \, \sqrt{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-x^2+1)^(1/2)*(1-x)^(3/2)*x,x, algorithm="giac")

[Out]

-2/7*(x + 1)^(7/2) + 6/5*(x + 1)^(5/2) - 4/3*(x + 1)^(3/2) + 16/105*sqrt(2)