3.365 \(\int e^{\tanh ^{-1}(x)} x (1+x)^2 \, dx\)

Optimal. Leaf size=87 \[ -\frac{1}{4} \sqrt{1-x} (x+1)^{7/2}-\frac{1}{4} \sqrt{1-x} (x+1)^{5/2}-\frac{5}{8} \sqrt{1-x} (x+1)^{3/2}-\frac{15}{8} \sqrt{1-x} \sqrt{x+1}+\frac{15}{8} \sin ^{-1}(x) \]

[Out]

(-15*Sqrt[1 - x]*Sqrt[1 + x])/8 - (5*Sqrt[1 - x]*(1 + x)^(3/2))/8 - (Sqrt[1 - x]*(1 + x)^(5/2))/4 - (Sqrt[1 -
x]*(1 + x)^(7/2))/4 + (15*ArcSin[x])/8

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Rubi [A]  time = 0.0482521, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {6129, 80, 50, 41, 216} \[ -\frac{1}{4} \sqrt{1-x} (x+1)^{7/2}-\frac{1}{4} \sqrt{1-x} (x+1)^{5/2}-\frac{5}{8} \sqrt{1-x} (x+1)^{3/2}-\frac{15}{8} \sqrt{1-x} \sqrt{x+1}+\frac{15}{8} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*x*(1 + x)^2,x]

[Out]

(-15*Sqrt[1 - x]*Sqrt[1 + x])/8 - (5*Sqrt[1 - x]*(1 + x)^(3/2))/8 - (Sqrt[1 - x]*(1 + x)^(5/2))/4 - (Sqrt[1 -
x]*(1 + x)^(7/2))/4 + (15*ArcSin[x])/8

Rule 6129

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

Rule 80

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 50

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 41

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(x)} x (1+x)^2 \, dx &=\int \frac{x (1+x)^{5/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{4} \sqrt{1-x} (1+x)^{7/2}+\frac{3}{4} \int \frac{(1+x)^{5/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{4} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{7/2}+\frac{5}{4} \int \frac{(1+x)^{3/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{5}{8} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{7/2}+\frac{15}{8} \int \frac{\sqrt{1+x}}{\sqrt{1-x}} \, dx\\ &=-\frac{15}{8} \sqrt{1-x} \sqrt{1+x}-\frac{5}{8} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{7/2}+\frac{15}{8} \int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=-\frac{15}{8} \sqrt{1-x} \sqrt{1+x}-\frac{5}{8} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{7/2}+\frac{15}{8} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-\frac{15}{8} \sqrt{1-x} \sqrt{1+x}-\frac{5}{8} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{4} \sqrt{1-x} (1+x)^{7/2}+\frac{15}{8} \sin ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0310331, size = 51, normalized size = 0.59 \[ \frac{1}{8} \left (-\sqrt{1-x^2} \left (2 x^3+8 x^2+15 x+24\right )-30 \sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[x]*x*(1 + x)^2,x]

[Out]

(-(Sqrt[1 - x^2]*(24 + 15*x + 8*x^2 + 2*x^3)) - 30*ArcSin[Sqrt[1 - x]/Sqrt[2]])/8

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Maple [A]  time = 0.035, size = 57, normalized size = 0.7 \begin{align*} -{\frac{{x}^{3}}{4}\sqrt{-{x}^{2}+1}}-{\frac{15\,x}{8}\sqrt{-{x}^{2}+1}}+{\frac{15\,\arcsin \left ( x \right ) }{8}}-{x}^{2}\sqrt{-{x}^{2}+1}-3\,\sqrt{-{x}^{2}+1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^3*(-x^2+1)^(1/2)-15/8*x*(-x^2+1)^(1/2)+15/8*arcsin(x)-x^2*(-x^2+1)^(1/2)-3*(-x^2+1)^(1/2)

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Maxima [A]  time = 1.43023, size = 76, normalized size = 0.87 \begin{align*} -\frac{1}{4} \, \sqrt{-x^{2} + 1} x^{3} - \sqrt{-x^{2} + 1} x^{2} - \frac{15}{8} \, \sqrt{-x^{2} + 1} x - 3 \, \sqrt{-x^{2} + 1} + \frac{15}{8} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(-x^2 + 1)*x^3 - sqrt(-x^2 + 1)*x^2 - 15/8*sqrt(-x^2 + 1)*x - 3*sqrt(-x^2 + 1) + 15/8*arcsin(x)

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Fricas [A]  time = 1.86501, size = 117, normalized size = 1.34 \begin{align*} -\frac{1}{8} \,{\left (2 \, x^{3} + 8 \, x^{2} + 15 \, x + 24\right )} \sqrt{-x^{2} + 1} - \frac{15}{4} \, \arctan \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*x^3 + 8*x^2 + 15*x + 24)*sqrt(-x^2 + 1) - 15/4*arctan((sqrt(-x^2 + 1) - 1)/x)

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Sympy [A]  time = 0.865044, size = 54, normalized size = 0.62 \begin{align*} - \frac{x^{3} \sqrt{1 - x^{2}}}{4} - x^{2} \sqrt{1 - x^{2}} - \frac{15 x \sqrt{1 - x^{2}}}{8} - 3 \sqrt{1 - x^{2}} + \frac{15 \operatorname{asin}{\left (x \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3*sqrt(1 - x**2)/4 - x**2*sqrt(1 - x**2) - 15*x*sqrt(1 - x**2)/8 - 3*sqrt(1 - x**2) + 15*asin(x)/8

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Giac [A]  time = 1.28611, size = 38, normalized size = 0.44 \begin{align*} -\frac{1}{8} \,{\left ({\left (2 \,{\left (x + 4\right )} x + 15\right )} x + 24\right )} \sqrt{-x^{2} + 1} + \frac{15}{8} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((2*(x + 4)*x + 15)*x + 24)*sqrt(-x^2 + 1) + 15/8*arcsin(x)