Integrand size = 10, antiderivative size = 69 \[ \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx=\frac {11}{3} \sqrt {1-\frac {1}{x^2}} x+\frac {3}{2} \sqrt {1-\frac {1}{x^2}} x^2+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x^3+\frac {5}{2} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \] Output:
11/3*(1-1/x^2)^(1/2)*x+3/2*(1-1/x^2)^(1/2)*x^2+1/3*(1-1/x^2)^(1/2)*x^3+5/2 *arctanh((1-1/x^2)^(1/2))
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.68 \[ \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx=\frac {1}{6} \sqrt {1-\frac {1}{x^2}} x \left (22+9 x+2 x^2\right )+\frac {5}{2} \log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \] Input:
Integrate[E^ArcCoth[x]*(1 + x)^2,x]
Output:
(Sqrt[1 - x^(-2)]*x*(22 + 9*x + 2*x^2))/6 + (5*Log[(1 + Sqrt[1 - x^(-2)])* x])/2
Time = 0.42 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.62, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6725, 105, 105, 105, 103, 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.
\(\displaystyle \int (x+1)^2 e^{\coth ^{-1}(x)} \, dx\) |
\(\Big \downarrow \) 6725 |
\(\displaystyle -\int \frac {\left (1+\frac {1}{x}\right )^{5/2} x^4}{\sqrt {1-\frac {1}{x}}}d\frac {1}{x}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3-\frac {5}{3} \int \frac {\left (1+\frac {1}{x}\right )^{3/2} x^3}{\sqrt {1-\frac {1}{x}}}d\frac {1}{x}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3-\frac {5}{3} \left (\frac {3}{2} \int \frac {\sqrt {1+\frac {1}{x}} x^2}{\sqrt {1-\frac {1}{x}}}d\frac {1}{x}-\frac {1}{2} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{3/2} x^2\right )\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3-\frac {5}{3} \left (\frac {3}{2} \left (\int \frac {x}{\sqrt {1-\frac {1}{x}} \sqrt {1+\frac {1}{x}}}d\frac {1}{x}-\sqrt {1-\frac {1}{x}} \sqrt {\frac {1}{x}+1} x\right )-\frac {1}{2} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{3/2} x^2\right )\) |
\(\Big \downarrow \) 103 |
\(\displaystyle \frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3-\frac {5}{3} \left (\frac {3}{2} \left (-\int \frac {1}{1-\frac {1}{x^2}}d\left (\sqrt {1-\frac {1}{x}} \sqrt {1+\frac {1}{x}}\right )-\sqrt {1-\frac {1}{x}} \sqrt {\frac {1}{x}+1} x\right )-\frac {1}{2} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{3/2} x^2\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3-\frac {5}{3} \left (\frac {3}{2} \left (-\text {arctanh}\left (\sqrt {1-\frac {1}{x}} \sqrt {\frac {1}{x}+1}\right )-\sqrt {1-\frac {1}{x}} \sqrt {\frac {1}{x}+1} x\right )-\frac {1}{2} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{3/2} x^2\right )\) |
Input:
Int[E^ArcCoth[x]*(1 + x)^2,x]
Output:
(Sqrt[1 - x^(-1)]*(1 + x^(-1))^(5/2)*x^3)/3 - (5*(-1/2*(Sqrt[1 - x^(-1)]*( 1 + x^(-1))^(3/2)*x^2) + (3*(-(Sqrt[1 - x^(-1)]*Sqrt[1 + x^(-1)]*x) - ArcT anh[Sqrt[1 - x^(-1)]*Sqrt[1 + x^(-1)]]))/2))/3
Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_ ))), x_] :> Simp[b*f Subst[Int[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sq rt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[2*b*d *e - f*(b*c + a*d), 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
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^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[-d^p Subst[Int[((1 + c*(x/d))^p*((1 + x/a)^(n/2)/x^(p + 2)))/(1 - x/a )^(n/2), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && IntegerQ[p]
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94
method | result | size |
risch | \(\frac {\left (2 x^{2}+9 x +22\right ) \left (x -1\right )}{6 \sqrt {\frac {x -1}{1+x}}}+\frac {5 \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{2 \sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\) | \(65\) |
trager | \(\frac {\left (1+x \right ) \left (2 x^{2}+9 x +22\right ) \sqrt {-\frac {1-x}{1+x}}}{6}+\frac {5 \ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )}{2}\) | \(66\) |
default | \(\frac {\left (x -1\right ) \left (2 \left (\left (x -1\right ) \left (1+x \right )\right )^{\frac {3}{2}}+9 x \sqrt {x^{2}-1}+15 \ln \left (x +\sqrt {x^{2}-1}\right )+24 \sqrt {x^{2}-1}\right )}{6 \sqrt {\frac {x -1}{1+x}}\, \sqrt {\left (x -1\right ) \left (1+x \right )}}\) | \(69\) |
Input:
int(1/((x-1)/(1+x))^(1/2)*(1+x)^2,x,method=_RETURNVERBOSE)
Output:
1/6*(2*x^2+9*x+22)*(x-1)/((x-1)/(1+x))^(1/2)+5/2*ln(x+(x^2-1)^(1/2))/((x-1 )/(1+x))^(1/2)/(1+x)*((x-1)*(1+x))^(1/2)
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88 \[ \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx=\frac {1}{6} \, {\left (2 \, x^{3} + 11 \, x^{2} + 31 \, x + 22\right )} \sqrt {\frac {x - 1}{x + 1}} + \frac {5}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {5}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*(1+x)^2,x, algorithm="fricas")
Output:
1/6*(2*x^3 + 11*x^2 + 31*x + 22)*sqrt((x - 1)/(x + 1)) + 5/2*log(sqrt((x - 1)/(x + 1)) + 1) - 5/2*log(sqrt((x - 1)/(x + 1)) - 1)
\[ \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx=\int \frac {\left (x + 1\right )^{2}}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \] Input:
integrate(1/((x-1)/(1+x))**(1/2)*(1+x)**2,x)
Output:
Integral((x + 1)**2/sqrt((x - 1)/(x + 1)), x)
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (53) = 106\).
Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.62 \[ \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx=-\frac {15 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{2}} - 40 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} + 33 \, \sqrt {\frac {x - 1}{x + 1}}}{3 \, {\left (\frac {3 \, {\left (x - 1\right )}}{x + 1} - \frac {3 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {{\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - 1\right )}} + \frac {5}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {5}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*(1+x)^2,x, algorithm="maxima")
Output:
-1/3*(15*((x - 1)/(x + 1))^(5/2) - 40*((x - 1)/(x + 1))^(3/2) + 33*sqrt((x - 1)/(x + 1)))/(3*(x - 1)/(x + 1) - 3*(x - 1)^2/(x + 1)^2 + (x - 1)^3/(x + 1)^3 - 1) + 5/2*log(sqrt((x - 1)/(x + 1)) + 1) - 5/2*log(sqrt((x - 1)/(x + 1)) - 1)
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87 \[ \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx=\frac {1}{6} \, \sqrt {x^{2} - 1} {\left (x {\left (\frac {2 \, x}{\mathrm {sgn}\left (x + 1\right )} + \frac {9}{\mathrm {sgn}\left (x + 1\right )}\right )} + \frac {22}{\mathrm {sgn}\left (x + 1\right )}\right )} - \frac {5 \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{2 \, \mathrm {sgn}\left (x + 1\right )} \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*(1+x)^2,x, algorithm="giac")
Output:
1/6*sqrt(x^2 - 1)*(x*(2*x/sgn(x + 1) + 9/sgn(x + 1)) + 22/sgn(x + 1)) - 5/ 2*log(abs(-x + sqrt(x^2 - 1)))/sgn(x + 1)
Time = 13.60 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.36 \[ \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx=5\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {11\,\sqrt {\frac {x-1}{x+1}}-\frac {40\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{3}+5\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{\frac {3\,\left (x-1\right )}{x+1}-\frac {3\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}+\frac {{\left (x-1\right )}^3}{{\left (x+1\right )}^3}-1} \] Input:
int((x + 1)^2/((x - 1)/(x + 1))^(1/2),x)
Output:
5*atanh(((x - 1)/(x + 1))^(1/2)) - (11*((x - 1)/(x + 1))^(1/2) - (40*((x - 1)/(x + 1))^(3/2))/3 + 5*((x - 1)/(x + 1))^(5/2))/((3*(x - 1))/(x + 1) - (3*(x - 1)^2)/(x + 1)^2 + (x - 1)^3/(x + 1)^3 - 1)
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75 \[ \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx=\frac {\sqrt {x +1}\, \sqrt {x -1}\, x^{2}}{3}+\frac {3 \sqrt {x +1}\, \sqrt {x -1}\, x}{2}+\frac {11 \sqrt {x +1}\, \sqrt {x -1}}{3}+5 \,\mathrm {log}\left (\frac {\sqrt {x -1}+\sqrt {x +1}}{\sqrt {2}}\right ) \] Input:
int(1/((x-1)/(1+x))^(1/2)*(1+x)^2,x)
Output:
(2*sqrt(x + 1)*sqrt(x - 1)*x**2 + 9*sqrt(x + 1)*sqrt(x - 1)*x + 22*sqrt(x + 1)*sqrt(x - 1) + 30*log((sqrt(x - 1) + sqrt(x + 1))/sqrt(2)))/6