Integrand size = 11, antiderivative size = 65 \[ \int e^{\text {arctanh}(x)} x (1+x)^2 \, dx=-4 \sqrt {1-x^2}-\frac {15}{8} x \sqrt {1-x^2}-\frac {1}{4} x^3 \sqrt {1-x^2}+\left (1-x^2\right )^{3/2}+\frac {15 \arcsin (x)}{8} \] Output:
-4*(-x^2+1)^(1/2)-15/8*x*(-x^2+1)^(1/2)-1/4*x^3*(-x^2+1)^(1/2)+(-x^2+1)^(3 /2)+15/8*arcsin(x)
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int e^{\text {arctanh}(x)} x (1+x)^2 \, dx=\frac {1}{8} \left (-\sqrt {1-x^2} \left (24+15 x+8 x^2+2 x^3\right )-30 \arcsin \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )\right ) \] Input:
Integrate[E^ArcTanh[x]*x*(1 + x)^2,x]
Output:
(-(Sqrt[1 - x^2]*(24 + 15*x + 8*x^2 + 2*x^3)) - 30*ArcSin[Sqrt[1 - x]/Sqrt [2]])/8
Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.40, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {6679, 90, 60, 60, 50, 223}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x (x+1)^2 e^{\text {arctanh}(x)} \, dx\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \int \frac {x (x+1)^{5/2}}{\sqrt {1-x}}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {3}{4} \int \frac {(x+1)^{5/2}}{\sqrt {1-x}}dx-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3}{4} \left (\frac {5}{3} \int \frac {(x+1)^{3/2}}{\sqrt {1-x}}dx-\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}\right )-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \int \frac {\sqrt {x+1}}{\sqrt {1-x}}dx-\frac {1}{2} \sqrt {1-x} (x+1)^{3/2}\right )-\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}\right )-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\) |
\(\Big \downarrow \) 50 |
\(\displaystyle \frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {1-x^2}}dx-\sqrt {1-x^2}\right )-\frac {1}{2} \sqrt {1-x} (x+1)^{3/2}\right )-\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}\right )-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\arcsin (x)-\sqrt {1-x^2}\right )-\frac {1}{2} \sqrt {1-x} (x+1)^{3/2}\right )-\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}\right )-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\) |
Input:
Int[E^ArcTanh[x]*x*(1 + x)^2,x]
Output:
-1/4*(Sqrt[1 - x]*(1 + x)^(7/2)) + (3*(-1/3*(Sqrt[1 - x]*(1 + x)^(5/2)) + (5*(-1/2*(Sqrt[1 - x]*(1 + x)^(3/2)) + (3*(-Sqrt[1 - x^2] + ArcSin[x]))/2) )/3))/4
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a *c + b*d*x^2)^m/(2*d*m), x] + Simp[a Int[(a*c + b*d*x^2)^n, x], x] /; Fre eQ[{a, b, c, d, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 1] && GtQ[m, 0 ] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(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]
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]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Simp[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])
Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.57
method | result | size |
risch | \(\frac {\left (2 x^{3}+8 x^{2}+15 x +24\right ) \left (x^{2}-1\right )}{8 \sqrt {-x^{2}+1}}+\frac {15 \arcsin \left (x \right )}{8}\) | \(37\) |
trager | \(\left (-\frac {1}{4} x^{3}-x^{2}-\frac {15}{8} x -3\right ) \sqrt {-x^{2}+1}+\frac {15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+1}+x \right )}{8}\) | \(54\) |
default | \(-3 \sqrt {-x^{2}+1}-\frac {15 x \sqrt {-x^{2}+1}}{8}-x^{2} \sqrt {-x^{2}+1}-\frac {x^{3} \sqrt {-x^{2}+1}}{4}+\frac {15 \arcsin \left (x \right )}{8}\) | \(57\) |
meijerg | \(-\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-x^{2}+1}}{2 \sqrt {\pi }}+\frac {3 i \left (i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{2 \sqrt {\pi }}+\frac {2 \sqrt {\pi }-\frac {\sqrt {\pi }\, \left (4 x^{2}+8\right ) \sqrt {-x^{2}+1}}{4}}{\sqrt {\pi }}-\frac {i \left (-\frac {i \sqrt {\pi }\, x \left (10 x^{2}+15\right ) \sqrt {-x^{2}+1}}{20}+\frac {3 i \sqrt {\pi }\, \arcsin \left (x \right )}{4}\right )}{2 \sqrt {\pi }}\) | \(128\) |
Input:
int((1+x)^3/(-x^2+1)^(1/2)*x,x,method=_RETURNVERBOSE)
Output:
1/8*(2*x^3+8*x^2+15*x+24)*(x^2-1)/(-x^2+1)^(1/2)+15/8*arcsin(x)
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int e^{\text {arctanh}(x)} x (1+x)^2 \, dx=-\frac {1}{8} \, {\left (2 \, x^{3} + 8 \, x^{2} + 15 \, x + 24\right )} \sqrt {-x^{2} + 1} - \frac {15}{4} \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \] Input:
integrate((1+x)^3/(-x^2+1)^(1/2)*x,x, algorithm="fricas")
Output:
-1/8*(2*x^3 + 8*x^2 + 15*x + 24)*sqrt(-x^2 + 1) - 15/4*arctan((sqrt(-x^2 + 1) - 1)/x)
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int e^{\text {arctanh}(x)} x (1+x)^2 \, dx=- \frac {x^{3} \sqrt {1 - x^{2}}}{4} - x^{2} \sqrt {1 - x^{2}} - \frac {15 x \sqrt {1 - x^{2}}}{8} - 3 \sqrt {1 - x^{2}} + \frac {15 \operatorname {asin}{\left (x \right )}}{8} \] Input:
integrate((1+x)**3/(-x**2+1)**(1/2)*x,x)
Output:
-x**3*sqrt(1 - x**2)/4 - x**2*sqrt(1 - x**2) - 15*x*sqrt(1 - x**2)/8 - 3*s qrt(1 - x**2) + 15*asin(x)/8
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int e^{\text {arctanh}(x)} x (1+x)^2 \, dx=-\frac {1}{4} \, \sqrt {-x^{2} + 1} x^{3} - \sqrt {-x^{2} + 1} x^{2} - \frac {15}{8} \, \sqrt {-x^{2} + 1} x - 3 \, \sqrt {-x^{2} + 1} + \frac {15}{8} \, \arcsin \left (x\right ) \] Input:
integrate((1+x)^3/(-x^2+1)^(1/2)*x,x, algorithm="maxima")
Output:
-1/4*sqrt(-x^2 + 1)*x^3 - sqrt(-x^2 + 1)*x^2 - 15/8*sqrt(-x^2 + 1)*x - 3*s qrt(-x^2 + 1) + 15/8*arcsin(x)
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.43 \[ \int e^{\text {arctanh}(x)} x (1+x)^2 \, dx=-\frac {1}{8} \, {\left ({\left (2 \, {\left (x + 4\right )} x + 15\right )} x + 24\right )} \sqrt {-x^{2} + 1} + \frac {15}{8} \, \arcsin \left (x\right ) \] Input:
integrate((1+x)^3/(-x^2+1)^(1/2)*x,x, algorithm="giac")
Output:
-1/8*((2*(x + 4)*x + 15)*x + 24)*sqrt(-x^2 + 1) + 15/8*arcsin(x)
Time = 14.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.45 \[ \int e^{\text {arctanh}(x)} x (1+x)^2 \, dx=\frac {15\,\mathrm {asin}\left (x\right )}{8}-\sqrt {1-x^2}\,\left (\frac {x^3}{4}+x^2+\frac {15\,x}{8}+3\right ) \] Input:
int((x*(x + 1)^3)/(1 - x^2)^(1/2),x)
Output:
(15*asin(x))/8 - (1 - x^2)^(1/2)*((15*x)/8 + x^2 + x^3/4 + 3)
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82 \[ \int e^{\text {arctanh}(x)} x (1+x)^2 \, dx=\frac {15 \mathit {asin} \left (x \right )}{8}-\frac {\sqrt {-x^{2}+1}\, x^{3}}{4}-\sqrt {-x^{2}+1}\, x^{2}-\frac {15 \sqrt {-x^{2}+1}\, x}{8}-3 \sqrt {-x^{2}+1}+3 \] Input:
int((1+x)^3/(-x^2+1)^(1/2)*x,x)
Output:
(15*asin(x) - 2*sqrt( - x**2 + 1)*x**3 - 8*sqrt( - x**2 + 1)*x**2 - 15*sqr t( - x**2 + 1)*x - 24*sqrt( - x**2 + 1) + 24)/8