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]

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

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Rubi [A]  time = 0.06, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.52, size = 38, normalized size = 1.12 \[ \frac {2 \, {\left (15 \, x^{3} - 18 \, x^{2} - 11 \, x + 22\right )} \sqrt {-x^{2} + 1} \sqrt {-x + 1}}{105 \, {\left (x - 1\right )}} \]

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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giac [A]  time = 0.43, size = 27, normalized size = 0.79 \[ -\frac {2}{7} \, {\left (x + 1\right )}^{\frac {7}{2}} + \frac {6}{5} \, {\left (x + 1\right )}^{\frac {5}{2}} - \frac {4}{3} \, {\left (x + 1\right )}^{\frac {3}{2}} + \frac {16}{105} \, \sqrt {2} \]

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)

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maple [A]  time = 0.03, size = 34, normalized size = 1.00 \[ -\frac {2 \left (1+x \right )^{2} \left (15 x^{2}-33 x +22\right ) \sqrt {1-x}}{105 \sqrt {-x^{2}+1}} \]

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 = 0.38, size = 48, normalized size = 1.41 \[ -\frac {2 \, {\left (15 \, x^{4} - 24 \, x^{3} + 13 \, x^{2} - 52 \, x - 104\right )}}{105 \, \sqrt {x + 1}} - \frac {2 \, {\left (x^{3} - 2 \, x^{2} + 3 \, x + 6\right )}}{5 \, \sqrt {x + 1}} \]

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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mupad [B]  time = 0.92, size = 33, normalized size = 0.97 \[ \frac {2\,\sqrt {1-x^2}\,\left (-15\,x^3+18\,x^2+11\,x-22\right )}{105\,\sqrt {1-x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (1 - x\right )^{\frac {3}{2}} \left (x + 1\right )}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \]

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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