3.4.64 \(\int e^{\tanh ^{-1}(x)} (1+x) \, dx\) [364]

Optimal. Leaf size=47 \[ -\frac {3}{2} \sqrt {1-x} \sqrt {1+x}-\frac {1}{2} \sqrt {1-x} (1+x)^{3/2}+\frac {3 \text {ArcSin}(x)}{2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6264, 52, 41, 222} \begin {gather*} \frac {3 \text {ArcSin}(x)}{2}-\frac {1}{2} \sqrt {1-x} (x+1)^{3/2}-\frac {3}{2} \sqrt {1-x} \sqrt {x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*(1 + x),x]

[Out]

(-3*Sqrt[1 - x]*Sqrt[1 + x])/2 - (Sqrt[1 - x]*(1 + x)^(3/2))/2 + (3*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 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 222

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 6264

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

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Mathematica [A]
time = 0.01, size = 37, normalized size = 0.79 \begin {gather*} -\frac {1}{2} (4+x) \sqrt {1-x^2}-3 \text {ArcSin}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[x]*(1 + x),x]

[Out]

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

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Maple [A]
time = 1.25, size = 29, normalized size = 0.62

method	result	size
risch	\(\frac {\left (x +4\right ) \left (x^{2}-1\right )}{2 \sqrt {-x^{2}+1}}+\frac {3 \arcsin \left (x \right )}{2}\)	\(25\)
default	\(-\frac {x \sqrt {-x^{2}+1}}{2}+\frac {3 \arcsin \left (x \right )}{2}-2 \sqrt {-x^{2}+1}\)	\(29\)
trager	\(\left (-\frac {x}{2}-2\right ) \sqrt {-x^{2}+1}+\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+1}+x \right )}{2}\)	\(44\)
meijerg	\(\arcsin \left (x \right )-\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-x^{2}+1}}{\sqrt {\pi }}+\frac {i \left (i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{2 \sqrt {\pi }}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^2/(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 28, normalized size = 0.60 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 1} x - 2 \, \sqrt {-x^{2} + 1} + \frac {3}{2} \, \arcsin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 33, normalized size = 0.70 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 1} {\left (x + 4\right )} - 3 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.07, size = 27, normalized size = 0.57 \begin {gather*} - \frac {x \sqrt {1 - x^{2}}}{2} - 2 \sqrt {1 - x^{2}} + \frac {3 \operatorname {asin}{\left (x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 19, normalized size = 0.40 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 1} {\left (x + 4\right )} + \frac {3}{2} \, \arcsin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.03, size = 21, normalized size = 0.45 \begin {gather*} \frac {3\,\mathrm {asin}\left (x\right )}{2}-\left (\frac {x}{2}+2\right )\,\sqrt {1-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^2/(1 - x^2)^(1/2),x)

[Out]

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

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