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]

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

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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 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 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]

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])

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*}

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Mathematica [A]  time = 0.02, 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]

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

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fricas [A]  time = 0.43, size = 40, normalized size = 0.60 \[ -\frac {1}{6} \, {\left (2 \, x^{2} + 9 \, x + 22\right )} \sqrt {-x^{2} + 1} - 5 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

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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giac [A]  time = 0.18, size = 25, normalized size = 0.37 \[ -\frac {1}{6} \, {\left ({\left (2 \, x + 9\right )} x + 22\right )} \sqrt {-x^{2} + 1} + \frac {5}{2} \, \arcsin \relax (x) \]

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)

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maple [A]  time = 0.03, size = 43, normalized size = 0.64 \[ -\frac {x^{2} \sqrt {-x^{2}+1}}{3}-\frac {11 \sqrt {-x^{2}+1}}{3}-\frac {3 x \sqrt {-x^{2}+1}}{2}+\frac {5 \arcsin \relax (x )}{2} \]

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 = 0.49, size = 42, normalized size = 0.63 \[ -\frac {1}{3} \, \sqrt {-x^{2} + 1} x^{2} - \frac {3}{2} \, \sqrt {-x^{2} + 1} x - \frac {11}{3} \, \sqrt {-x^{2} + 1} + \frac {5}{2} \, \arcsin \relax (x) \]

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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mupad [B]  time = 0.03, size = 26, normalized size = 0.39 \[ \frac {5\,\mathrm {asin}\relax (x)}{2}-\sqrt {1-x^2}\,\left (\frac {x^2}{3}+\frac {3\,x}{2}+\frac {11}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^3/(1 - x^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 0.33, size = 44, normalized size = 0.66 \[ - \frac {x^{2} \sqrt {1 - x^{2}}}{3} - \frac {3 x \sqrt {1 - x^{2}}}{2} - \frac {11 \sqrt {1 - x^{2}}}{3} + \frac {5 \operatorname {asin}{\relax (x )}}{2} \]

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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