3.4.63 \(\int e^{\tanh ^{-1}(x)} x (1+x) \, dx\) [363]

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

[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 = 0.02, 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 = {6264, 81, 52, 41, 222} \begin {gather*} \text {ArcSin}(x)-\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} \end {gather*}

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 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 81

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 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)} 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 = 0.02, size = 42, normalized size = 0.69 \begin {gather*} -\frac {1}{3} \sqrt {1-x^2} \left (5+3 x+x^2\right )-2 \text {ArcSin}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right ) \end {gather*}

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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Maple [A]
time = 1.57, size = 41, normalized size = 0.67

method	result	size
risch	\(\frac {\left (x^{2}+3 x +5\right ) \left (x^{2}-1\right )}{3 \sqrt {-x^{2}+1}}+\arcsin \left (x \right )\)	\(28\)
default	\(-\frac {x^{2} \sqrt {-x^{2}+1}}{3}-\frac {5 \sqrt {-x^{2}+1}}{3}-x \sqrt {-x^{2}+1}+\arcsin \left (x \right )\)	\(41\)
trager	\(\left (-\frac {1}{3} x^{2}-x -\frac {5}{3}\right ) \sqrt {-x^{2}+1}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+1}+x \right )\)	\(48\)
meijerg	\(-\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-x^{2}+1}}{2 \sqrt {\pi }}+\frac {i \left (i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{\sqrt {\pi }}+\frac {\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 x^{2}+8\right ) \sqrt {-x^{2}+1}}{6}}{2 \sqrt {\pi }}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.46, size = 40, normalized size = 0.66 \begin {gather*} -\frac {1}{3} \, \sqrt {-x^{2} + 1} x^{2} - \sqrt {-x^{2} + 1} x - \frac {5}{3} \, \sqrt {-x^{2} + 1} + \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,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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Fricas [A]
time = 0.40, size = 38, normalized size = 0.62 \begin {gather*} -\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 ) \end {gather*}

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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Sympy [A]
time = 0.11, size = 37, normalized size = 0.61 \begin {gather*} - \frac {x^{2} \sqrt {1 - x^{2}}}{3} - x \sqrt {1 - x^{2}} - \frac {5 \sqrt {1 - x^{2}}}{3} + \operatorname {asin}{\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,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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Giac [A]
time = 0.41, size = 21, normalized size = 0.34 \begin {gather*} -\frac {1}{3} \, {\left ({\left (x + 3\right )} x + 5\right )} \sqrt {-x^{2} + 1} + \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,x, algorithm="giac")

[Out]

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

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

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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