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

Optimal. Leaf size=61 \[ -\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}-\frac {1}{3} \sqrt {1-x} (x+1)^{3/2}-\sqrt {1-x} \sqrt {x+1}+\sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6129, 80, 50, 41, 216} \[ -\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}-\frac {1}{3} \sqrt {1-x} (x+1)^{3/2}-\sqrt {1-x} \sqrt {x+1}+\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*x*(1 + x),x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.69 \[ -\frac {1}{3} \sqrt {1-x^2} \left (x^2+3 x+5\right )-2 \sin ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[x]*x*(1 + x),x]

[Out]

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

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fricas [A]  time = 0.60, size = 38, normalized size = 0.62 \[ -\frac {1}{3} \, {\left (x^{2} + 3 \, x + 5\right )} \sqrt {-x^{2} + 1} - 2 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 21, normalized size = 0.34 \[ -\frac {1}{3} \, {\left ({\left (x + 3\right )} x + 5\right )} \sqrt {-x^{2} + 1} + \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 41, normalized size = 0.67 \[ -\frac {x^{2} \sqrt {-x^{2}+1}}{3}-\frac {5 \sqrt {-x^{2}+1}}{3}-x \sqrt {-x^{2}+1}+\arcsin \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 40, normalized size = 0.66 \[ -\frac {1}{3} \, \sqrt {-x^{2} + 1} x^{2} - \sqrt {-x^{2} + 1} x - \frac {5}{3} \, \sqrt {-x^{2} + 1} + \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 22, normalized size = 0.36 \[ \mathrm {asin}\relax (x)-\sqrt {1-x^2}\,\left (\frac {x^2}{3}+x+\frac {5}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(x + 1)^2)/(1 - x^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 0.32, size = 37, normalized size = 0.61 \[ - \frac {x^{2} \sqrt {1 - x^{2}}}{3} - x \sqrt {1 - x^{2}} - \frac {5 \sqrt {1 - x^{2}}}{3} + \operatorname {asin}{\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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