3.2.28 \(\int e^{\frac {2}{3} \tanh ^{-1}(x)} x^2 \, dx\) [128]

Optimal. Leaf size=133 \[ -\frac {11}{27} (1-x)^{2/3} \sqrt [3]{1+x}-\frac {1}{9} (1-x)^{2/3} (1+x)^{4/3}-\frac {1}{3} (1-x)^{2/3} x (1+x)^{4/3}+\frac {22 \text {ArcTan}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-x}}{\sqrt {3} \sqrt [3]{1+x}}\right )}{27 \sqrt {3}}+\frac {11}{81} \log (1+x)+\frac {11}{27} \log \left (1+\frac {\sqrt [3]{1-x}}{\sqrt [3]{1+x}}\right ) \]

[Out]

-11/27*(1-x)^(2/3)*(1+x)^(1/3)-1/9*(1-x)^(2/3)*(1+x)^(4/3)-1/3*(1-x)^(2/3)*x*(1+x)^(4/3)+11/81*ln(1+x)+11/27*l
n(1+(1-x)^(1/3)/(1+x)^(1/3))-22/81*arctan(-1/3*3^(1/2)+2/3*(1-x)^(1/3)/(1+x)^(1/3)*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6261, 92, 81, 52, 62} \begin {gather*} \frac {22 \text {ArcTan}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-x}}{\sqrt {3} \sqrt [3]{x+1}}\right )}{27 \sqrt {3}}-\frac {1}{3} (1-x)^{2/3} x (x+1)^{4/3}-\frac {1}{9} (1-x)^{2/3} (x+1)^{4/3}-\frac {11}{27} (1-x)^{2/3} \sqrt [3]{x+1}+\frac {11}{81} \log (x+1)+\frac {11}{27} \log \left (\frac {\sqrt [3]{1-x}}{\sqrt [3]{x+1}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((2*ArcTanh[x])/3)*x^2,x]

[Out]

(-11*(1 - x)^(2/3)*(1 + x)^(1/3))/27 - ((1 - x)^(2/3)*(1 + x)^(4/3))/9 - ((1 - x)^(2/3)*x*(1 + x)^(4/3))/3 + (
22*ArcTan[1/Sqrt[3] - (2*(1 - x)^(1/3))/(Sqrt[3]*(1 + x)^(1/3))])/(27*Sqrt[3]) + (11*Log[1 + x])/81 + (11*Log[
1 + (1 - x)^(1/3)/(1 + x)^(1/3)])/27

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 62

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-d/b, 3]}, Simp[Sqrt[
3]*(q/d)*ArcTan[1/Sqrt[3] - 2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], x] + (Simp[3*(q/(2*d))*Log[q*((a
+ b*x)^(1/3)/(c + d*x)^(1/3)) + 1], x] + Simp[(q/(2*d))*Log[c + d*x], x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && NegQ[d/b]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 6261

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rubi steps

\begin {align*} \int e^{\frac {2}{3} \tanh ^{-1}(x)} x^2 \, dx &=\int \frac {x^2 \sqrt [3]{1+x}}{\sqrt [3]{1-x}} \, dx\\ &=-\frac {1}{3} (1-x)^{2/3} x (1+x)^{4/3}-\frac {1}{3} \int \frac {\left (-1-\frac {2 x}{3}\right ) \sqrt [3]{1+x}}{\sqrt [3]{1-x}} \, dx\\ &=-\frac {1}{9} (1-x)^{2/3} (1+x)^{4/3}-\frac {1}{3} (1-x)^{2/3} x (1+x)^{4/3}+\frac {11}{27} \int \frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x}} \, dx\\ &=-\frac {11}{27} (1-x)^{2/3} \sqrt [3]{1+x}-\frac {1}{9} (1-x)^{2/3} (1+x)^{4/3}-\frac {1}{3} (1-x)^{2/3} x (1+x)^{4/3}+\frac {22}{81} \int \frac {1}{\sqrt [3]{1-x} (1+x)^{2/3}} \, dx\\ &=-\frac {11}{27} (1-x)^{2/3} \sqrt [3]{1+x}-\frac {1}{9} (1-x)^{2/3} (1+x)^{4/3}-\frac {1}{3} (1-x)^{2/3} x (1+x)^{4/3}+\frac {22 \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-x}}{\sqrt {3} \sqrt [3]{1+x}}\right )}{27 \sqrt {3}}+\frac {11}{81} \log (1+x)+\frac {11}{27} \log \left (1+\frac {\sqrt [3]{1-x}}{\sqrt [3]{1+x}}\right )\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 59, normalized size = 0.44 \begin {gather*} -\frac {1}{18} (1-x)^{2/3} \left (2 \sqrt [3]{1+x} \left (1+4 x+3 x^2\right )+11 \sqrt [3]{2} \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};\frac {1-x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/18*((1 - x)^(2/3)*(2*(1 + x)^(1/3)*(1 + 4*x + 3*x^2) + 11*2^(1/3)*Hypergeometric2F1[-1/3, 2/3, 5/3, (1 - x)
/2]))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (\frac {1+x}{\sqrt {-x^{2}+1}}\right )^{\frac {2}{3}} x^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 159, normalized size = 1.20 \begin {gather*} \frac {1}{27} \, {\left (9 \, x^{3} + 3 \, x^{2} + 2 \, x - 14\right )} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + \frac {22}{81} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {22}{81} \, \log \left (\left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + 1\right ) - \frac {11}{81} \, \log \left (-\frac {{\left (x - 1\right )} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} - x + \sqrt {-x^{2} + 1} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + 1}{x - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*(9*x^3 + 3*x^2 + 2*x - 14)*(-sqrt(-x^2 + 1)/(x - 1))^(2/3) + 22/81*sqrt(3)*arctan(2/3*sqrt(3)*(-sqrt(-x^2
+ 1)/(x - 1))^(2/3) - 1/3*sqrt(3)) + 22/81*log((-sqrt(-x^2 + 1)/(x - 1))^(2/3) + 1) - 11/81*log(-((x - 1)*(-s
qrt(-x^2 + 1)/(x - 1))^(2/3) - x + sqrt(-x^2 + 1)*(-sqrt(-x^2 + 1)/(x - 1))^(1/3) + 1)/(x - 1))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\left (\frac {x+1}{\sqrt {1-x^2}}\right )}^{2/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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