Integrand size = 15, antiderivative size = 51 \[ \int \frac {e^{\text {arctanh}(x)} x}{(1-x)^{3/2}} \, dx=\frac {5 \sqrt {1+x}}{2}+\frac {(1+x)^{3/2}}{2 (1-x)}-\frac {5 \text {arctanh}\left (\frac {\sqrt {1+x}}{\sqrt {2}}\right )}{\sqrt {2}} \] Output:
5/2*(1+x)^(1/2)+(1+x)^(3/2)/(2-2*x)-5/2*2^(1/2)*arctanh(1/2*2^(1/2)*(1+x)^ (1/2))
Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\text {arctanh}(x)} x}{(1-x)^{3/2}} \, dx=\frac {\sqrt {1+x} (-3+2 x)}{-1+x}-\frac {5 \text {arctanh}\left (\frac {\sqrt {1+x}}{\sqrt {2}}\right )}{\sqrt {2}} \] Input:
Integrate[(E^ArcTanh[x]*x)/(1 - x)^(3/2),x]
Output:
(Sqrt[1 + x]*(-3 + 2*x))/(-1 + x) - (5*ArcTanh[Sqrt[1 + x]/Sqrt[2]])/Sqrt[ 2]
Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6679, 87, 60, 73, 219}
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.
\(\displaystyle \int \frac {x e^{\text {arctanh}(x)}}{(1-x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \int \frac {x \sqrt {x+1}}{(1-x)^2}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(x+1)^{3/2}}{2 (1-x)}-\frac {5}{4} \int \frac {\sqrt {x+1}}{1-x}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(x+1)^{3/2}}{2 (1-x)}-\frac {5}{4} \left (2 \int \frac {1}{(1-x) \sqrt {x+1}}dx-2 \sqrt {x+1}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(x+1)^{3/2}}{2 (1-x)}-\frac {5}{4} \left (4 \int \frac {1}{1-x}d\sqrt {x+1}-2 \sqrt {x+1}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(x+1)^{3/2}}{2 (1-x)}-\frac {5}{4} \left (2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )-2 \sqrt {x+1}\right )\) |
Input:
Int[(E^ArcTanh[x]*x)/(1 - x)^(3/2),x]
Output:
(1 + x)^(3/2)/(2*(1 - x)) - (5*(-2*Sqrt[1 + x] + 2*Sqrt[2]*ArcTanh[Sqrt[1 + x]/Sqrt[2]]))/4
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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 = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.43
method | result | size |
default | \(\frac {\sqrt {-x^{2}+1}\, \left (5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right ) x -5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right )-4 \sqrt {1+x}\, x +6 \sqrt {1+x}\right )}{2 \left (1-x \right )^{\frac {3}{2}} \sqrt {1+x}}\) | \(73\) |
risch | \(-\frac {\left (2 x^{2}-x -3\right ) \sqrt {\frac {\left (1-x \right ) \left (-x^{2}+1\right )}{\left (x -1\right )^{2}}}}{\sqrt {1+x}\, \sqrt {-x^{2}+1}\, \sqrt {1-x}}+\frac {5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right ) \sqrt {\frac {\left (1-x \right ) \left (-x^{2}+1\right )}{\left (x -1\right )^{2}}}\, \left (x -1\right )}{2 \sqrt {-x^{2}+1}\, \sqrt {1-x}}\) | \(110\) |
Input:
int((1+x)/(-x^2+1)^(1/2)*x/(1-x)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2*(-x^2+1)^(1/2)*(5*2^(1/2)*arctanh(1/2*(1+x)^(1/2)*2^(1/2))*x-5*2^(1/2) *arctanh(1/2*(1+x)^(1/2)*2^(1/2))-4*(1+x)^(1/2)*x+6*(1+x)^(1/2))/(1-x)^(3/ 2)/(1+x)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (36) = 72\).
Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.78 \[ \int \frac {e^{\text {arctanh}(x)} x}{(1-x)^{3/2}} \, dx=\frac {5 \, \sqrt {2} {\left (x^{2} - 2 \, x + 1\right )} \log \left (-\frac {x^{2} + 2 \, \sqrt {2} \sqrt {-x^{2} + 1} \sqrt {-x + 1} + 2 \, x - 3}{x^{2} - 2 \, x + 1}\right ) - 4 \, \sqrt {-x^{2} + 1} {\left (2 \, x - 3\right )} \sqrt {-x + 1}}{4 \, {\left (x^{2} - 2 \, x + 1\right )}} \] Input:
integrate((1+x)/(-x^2+1)^(1/2)*x/(1-x)^(3/2),x, algorithm="fricas")
Output:
1/4*(5*sqrt(2)*(x^2 - 2*x + 1)*log(-(x^2 + 2*sqrt(2)*sqrt(-x^2 + 1)*sqrt(- x + 1) + 2*x - 3)/(x^2 - 2*x + 1)) - 4*sqrt(-x^2 + 1)*(2*x - 3)*sqrt(-x + 1))/(x^2 - 2*x + 1)
\[ \int \frac {e^{\text {arctanh}(x)} x}{(1-x)^{3/2}} \, dx=\int \frac {x \left (x + 1\right )}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \left (1 - x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((1+x)/(-x**2+1)**(1/2)*x/(1-x)**(3/2),x)
Output:
Integral(x*(x + 1)/(sqrt(-(x - 1)*(x + 1))*(1 - x)**(3/2)), x)
\[ \int \frac {e^{\text {arctanh}(x)} x}{(1-x)^{3/2}} \, dx=\int { \frac {{\left (x + 1\right )} x}{\sqrt {-x^{2} + 1} {\left (-x + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((1+x)/(-x^2+1)^(1/2)*x/(1-x)^(3/2),x, algorithm="maxima")
Output:
integrate((x + 1)*x/(sqrt(-x^2 + 1)*(-x + 1)^(3/2)), x)
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\text {arctanh}(x)} x}{(1-x)^{3/2}} \, dx=\frac {5}{4} \, \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {x + 1}}{\sqrt {2} + \sqrt {x + 1}}\right ) + 2 \, \sqrt {x + 1} - \frac {\sqrt {x + 1}}{x - 1} \] Input:
integrate((1+x)/(-x^2+1)^(1/2)*x/(1-x)^(3/2),x, algorithm="giac")
Output:
5/4*sqrt(2)*log((sqrt(2) - sqrt(x + 1))/(sqrt(2) + sqrt(x + 1))) + 2*sqrt( x + 1) - sqrt(x + 1)/(x - 1)
Timed out. \[ \int \frac {e^{\text {arctanh}(x)} x}{(1-x)^{3/2}} \, dx=\int \frac {x\,\left (x+1\right )}{\sqrt {1-x^2}\,{\left (1-x\right )}^{3/2}} \,d x \] Input:
int((x*(x + 1))/((1 - x^2)^(1/2)*(1 - x)^(3/2)),x)
Output:
int((x*(x + 1))/((1 - x^2)^(1/2)*(1 - x)^(3/2)), x)
Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\text {arctanh}(x)} x}{(1-x)^{3/2}} \, dx=\frac {16 \sqrt {x +1}\, x -24 \sqrt {x +1}+20 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )\right ) x -20 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )\right )-17 \sqrt {2}\, x +17 \sqrt {2}}{8 x -8} \] Input:
int((1+x)/(-x^2+1)^(1/2)*x/(1-x)^(3/2),x)
Output:
(16*sqrt(x + 1)*x - 24*sqrt(x + 1) + 20*sqrt(2)*log(tan(asin(sqrt( - x + 1 )/sqrt(2))/2))*x - 20*sqrt(2)*log(tan(asin(sqrt( - x + 1)/sqrt(2))/2)) - 1 7*sqrt(2)*x + 17*sqrt(2))/(8*(x - 1))