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

Optimal. Leaf size=67 \[ -\frac{1}{3} \sqrt{1-x} (x+1)^{5/2}-\frac{5}{6} \sqrt{1-x} (x+1)^{3/2}-\frac{5}{2} \sqrt{1-x} \sqrt{x+1}+\frac{5}{2} \sin ^{-1}(x) \]

[Out]

(-5*Sqrt[1 - x]*Sqrt[1 + x])/2 - (5*Sqrt[1 - x]*(1 + x)^(3/2))/6 - (Sqrt[1 - x]*(1 + x)^(5/2))/3 + (5*ArcSin[x
])/2

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Rubi [A]  time = 0.0290799, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6129, 50, 41, 216} \[ -\frac{1}{3} \sqrt{1-x} (x+1)^{5/2}-\frac{5}{6} \sqrt{1-x} (x+1)^{3/2}-\frac{5}{2} \sqrt{1-x} \sqrt{x+1}+\frac{5}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*(1 + x)^2,x]

[Out]

(-5*Sqrt[1 - x]*Sqrt[1 + x])/2 - (5*Sqrt[1 - x]*(1 + x)^(3/2))/6 - (Sqrt[1 - x]*(1 + x)^(5/2))/3 + (5*ArcSin[x
])/2

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 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)} (1+x)^2 \, dx &=\int \frac{(1+x)^{5/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{3} \sqrt{1-x} (1+x)^{5/2}+\frac{5}{3} \int \frac{(1+x)^{3/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{5}{6} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{3} \sqrt{1-x} (1+x)^{5/2}+\frac{5}{2} \int \frac{\sqrt{1+x}}{\sqrt{1-x}} \, dx\\ &=-\frac{5}{2} \sqrt{1-x} \sqrt{1+x}-\frac{5}{6} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{3} \sqrt{1-x} (1+x)^{5/2}+\frac{5}{2} \int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=-\frac{5}{2} \sqrt{1-x} \sqrt{1+x}-\frac{5}{6} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{3} \sqrt{1-x} (1+x)^{5/2}+\frac{5}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-\frac{5}{2} \sqrt{1-x} \sqrt{1+x}-\frac{5}{6} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{3} \sqrt{1-x} (1+x)^{5/2}+\frac{5}{2} \sin ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0216032, size = 44, normalized size = 0.66 \[ -\frac{1}{6} \sqrt{1-x^2} \left (2 x^2+9 x+22\right )-5 \sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[x]*(1 + x)^2,x]

[Out]

-(Sqrt[1 - x^2]*(22 + 9*x + 2*x^2))/6 - 5*ArcSin[Sqrt[1 - x]/Sqrt[2]]

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Maple [A]  time = 0.045, size = 43, normalized size = 0.6 \begin{align*} -{\frac{{x}^{2}}{3}\sqrt{-{x}^{2}+1}}-{\frac{11}{3}\sqrt{-{x}^{2}+1}}-{\frac{3\,x}{2}\sqrt{-{x}^{2}+1}}+{\frac{5\,\arcsin \left ( x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^2*(-x^2+1)^(1/2)-11/3*(-x^2+1)^(1/2)-3/2*x*(-x^2+1)^(1/2)+5/2*arcsin(x)

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Maxima [A]  time = 1.44777, size = 57, normalized size = 0.85 \begin{align*} -\frac{1}{3} \, \sqrt{-x^{2} + 1} x^{2} - \frac{3}{2} \, \sqrt{-x^{2} + 1} x - \frac{11}{3} \, \sqrt{-x^{2} + 1} + \frac{5}{2} \, \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, algorithm="maxima")

[Out]

-1/3*sqrt(-x^2 + 1)*x^2 - 3/2*sqrt(-x^2 + 1)*x - 11/3*sqrt(-x^2 + 1) + 5/2*arcsin(x)

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Fricas [A]  time = 1.73717, size = 101, normalized size = 1.51 \begin{align*} -\frac{1}{6} \,{\left (2 \, x^{2} + 9 \, x + 22\right )} \sqrt{-x^{2} + 1} - 5 \, \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, algorithm="fricas")

[Out]

-1/6*(2*x^2 + 9*x + 22)*sqrt(-x^2 + 1) - 5*arctan((sqrt(-x^2 + 1) - 1)/x)

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Sympy [A]  time = 0.418913, size = 44, normalized size = 0.66 \begin{align*} - \frac{x^{2} \sqrt{1 - x^{2}}}{3} - \frac{3 x \sqrt{1 - x^{2}}}{2} - \frac{11 \sqrt{1 - x^{2}}}{3} + \frac{5 \operatorname{asin}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2*sqrt(1 - x**2)/3 - 3*x*sqrt(1 - x**2)/2 - 11*sqrt(1 - x**2)/3 + 5*asin(x)/2

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Giac [A]  time = 1.17571, size = 34, normalized size = 0.51 \begin{align*} -\frac{1}{6} \,{\left ({\left (2 \, x + 9\right )} x + 22\right )} \sqrt{-x^{2} + 1} + \frac{5}{2} \, \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, algorithm="giac")

[Out]

-1/6*((2*x + 9)*x + 22)*sqrt(-x^2 + 1) + 5/2*arcsin(x)