Integrand size = 11, antiderivative size = 82 \[ \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx=3 \sqrt {1-\frac {1}{x^2}} x+\frac {15}{8} \sqrt {1-\frac {1}{x^2}} x^2+\sqrt {1-\frac {1}{x^2}} x^3+\frac {1}{4} \sqrt {1-\frac {1}{x^2}} x^4+\frac {15}{8} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \] Output:
3*(1-1/x^2)^(1/2)*x+15/8*(1-1/x^2)^(1/2)*x^2+(1-1/x^2)^(1/2)*x^3+1/4*(1-1/ x^2)^(1/2)*x^4+15/8*arctanh((1-1/x^2)^(1/2))
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.63 \[ \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx=\frac {1}{8} \sqrt {1-\frac {1}{x^2}} x \left (24+15 x+8 x^2+2 x^3\right )+\frac {15}{8} \log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \] Input:
Integrate[E^ArcCoth[x]*x*(1 + x)^2,x]
Output:
(Sqrt[1 - x^(-2)]*x*(24 + 15*x + 8*x^2 + 2*x^3))/8 + (15*Log[(1 + Sqrt[1 - x^(-2)])*x])/8
Time = 0.48 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.76, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {6729, 107, 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.
\(\displaystyle \int x (x+1)^2 e^{\coth ^{-1}(x)} \, dx\) |
\(\Big \downarrow \) 6729 |
\(\displaystyle -\int \frac {\left (1+\frac {1}{x}\right )^{5/2} x^5}{\sqrt {1-\frac {1}{x}}}d\frac {1}{x}\) |
\(\Big \downarrow \) 107 |
\(\displaystyle \frac {1}{4} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{7/2} x^4-\frac {3}{4} \int \frac {\left (1+\frac {1}{x}\right )^{5/2} x^4}{\sqrt {1-\frac {1}{x}}}d\frac {1}{x}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {1}{4} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{7/2} x^4-\frac {3}{4} \left (\frac {5}{3} \int \frac {\left (1+\frac {1}{x}\right )^{3/2} x^3}{\sqrt {1-\frac {1}{x}}}d\frac {1}{x}-\frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3\right )\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {1}{4} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{7/2} x^4-\frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \int \frac {\sqrt {1+\frac {1}{x}} x^2}{\sqrt {1-\frac {1}{x}}}d\frac {1}{x}-\frac {1}{2} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{3/2} x^2\right )-\frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3\right )\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {1}{4} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{7/2} x^4-\frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\int \frac {x}{\sqrt {1-\frac {1}{x}} \sqrt {1+\frac {1}{x}}}d\frac {1}{x}-\sqrt {1-\frac {1}{x}} \sqrt {\frac {1}{x}+1} x\right )-\frac {1}{2} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{3/2} x^2\right )-\frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3\right )\) |
\(\Big \downarrow \) 103 |
\(\displaystyle \frac {1}{4} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{7/2} x^4-\frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (-\int \frac {1}{1-\frac {1}{x^2}}d\left (\sqrt {1-\frac {1}{x}} \sqrt {1+\frac {1}{x}}\right )-\sqrt {1-\frac {1}{x}} \sqrt {\frac {1}{x}+1} x\right )-\frac {1}{2} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{3/2} x^2\right )-\frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{7/2} x^4-\frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (-\text {arctanh}\left (\sqrt {1-\frac {1}{x}} \sqrt {\frac {1}{x}+1}\right )-\sqrt {1-\frac {1}{x}} \sqrt {\frac {1}{x}+1} x\right )-\frac {1}{2} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{3/2} x^2\right )-\frac {1}{3} \sqrt {1-\frac {1}{x}} \left (\frac {1}{x}+1\right )^{5/2} x^3\right )\) |
Input:
Int[E^ArcCoth[x]*x*(1 + x)^2,x]
Output:
(Sqrt[1 - x^(-1)]*(1 + x^(-1))^(7/2)*x^4)/4 - (3*(-1/3*(Sqrt[1 - x^(-1)]*( 1 + x^(-1))^(5/2)*x^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) - ArcTanh[Sqrt[1 - x^(-1)]* Sqrt[1 + x^(-1)]]))/2))/3))/4
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[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_.))*((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(p _.), x_Symbol] :> Simp[(-d^p)*(e*x)^m*(1/x)^m Subst[Int[((1 + c*(x/d))^p* ((1 + x/a)^(n/2)/x^(m + p + 2)))/(1 - x/a)^(n/2), x], x, 1/x], x] /; FreeQ[ {a, c, d, e, m, n}, x] && EqQ[a^2*c^2 - d^2, 0] && IntegerQ[p]
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85
method | result | size |
risch | \(\frac {\left (2 x^{3}+8 x^{2}+15 x +24\right ) \left (x -1\right )}{8 \sqrt {\frac {x -1}{1+x}}}+\frac {15 \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{8 \sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\) | \(70\) |
trager | \(\frac {\left (1+x \right ) \left (2 x^{3}+8 x^{2}+15 x +24\right ) \sqrt {-\frac {1-x}{1+x}}}{8}-\frac {15 \ln \left (-\sqrt {-\frac {1-x}{1+x}}\, x -\sqrt {-\frac {1-x}{1+x}}+x \right )}{8}\) | \(74\) |
default | \(\frac {\left (x -1\right ) \left (2 x \left (x^{2}-1\right )^{\frac {3}{2}}+8 \left (\left (x -1\right ) \left (1+x \right )\right )^{\frac {3}{2}}+17 x \sqrt {x^{2}-1}+32 \sqrt {x^{2}-1}+15 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{8 \sqrt {\frac {x -1}{1+x}}\, \sqrt {\left (x -1\right ) \left (1+x \right )}}\) | \(79\) |
Input:
int(1/((x-1)/(1+x))^(1/2)*x*(1+x)^2,x,method=_RETURNVERBOSE)
Output:
1/8*(2*x^3+8*x^2+15*x+24)*(x-1)/((x-1)/(1+x))^(1/2)+15/8*ln(x+(x^2-1)^(1/2 ))/((x-1)/(1+x))^(1/2)/(1+x)*((x-1)*(1+x))^(1/2)
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx=\frac {1}{8} \, {\left (2 \, x^{4} + 10 \, x^{3} + 23 \, x^{2} + 39 \, x + 24\right )} \sqrt {\frac {x - 1}{x + 1}} + \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*x*(1+x)^2,x, algorithm="fricas")
Output:
1/8*(2*x^4 + 10*x^3 + 23*x^2 + 39*x + 24)*sqrt((x - 1)/(x + 1)) + 15/8*log (sqrt((x - 1)/(x + 1)) + 1) - 15/8*log(sqrt((x - 1)/(x + 1)) - 1)
\[ \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx=\int \frac {x \left (x + 1\right )^{2}}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \] Input:
integrate(1/((x-1)/(1+x))**(1/2)*x*(1+x)**2,x)
Output:
Integral(x*(x + 1)**2/sqrt((x - 1)/(x + 1)), x)
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (66) = 132\).
Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.68 \[ \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx=\frac {15 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {7}{2}} - 55 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{2}} + 73 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} - 49 \, \sqrt {\frac {x - 1}{x + 1}}}{4 \, {\left (\frac {4 \, {\left (x - 1\right )}}{x + 1} - \frac {6 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {4 \, {\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - \frac {{\left (x - 1\right )}^{4}}{{\left (x + 1\right )}^{4}} - 1\right )}} + \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*x*(1+x)^2,x, algorithm="maxima")
Output:
1/4*(15*((x - 1)/(x + 1))^(7/2) - 55*((x - 1)/(x + 1))^(5/2) + 73*((x - 1) /(x + 1))^(3/2) - 49*sqrt((x - 1)/(x + 1)))/(4*(x - 1)/(x + 1) - 6*(x - 1) ^2/(x + 1)^2 + 4*(x - 1)^3/(x + 1)^3 - (x - 1)^4/(x + 1)^4 - 1) + 15/8*log (sqrt((x - 1)/(x + 1)) + 1) - 15/8*log(sqrt((x - 1)/(x + 1)) - 1)
Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87 \[ \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx=\frac {1}{8} \, {\left ({\left (2 \, x {\left (\frac {x}{\mathrm {sgn}\left (x + 1\right )} + \frac {4}{\mathrm {sgn}\left (x + 1\right )}\right )} + \frac {15}{\mathrm {sgn}\left (x + 1\right )}\right )} x + \frac {24}{\mathrm {sgn}\left (x + 1\right )}\right )} \sqrt {x^{2} - 1} - \frac {15 \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{8 \, \mathrm {sgn}\left (x + 1\right )} \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*x*(1+x)^2,x, algorithm="giac")
Output:
1/8*((2*x*(x/sgn(x + 1) + 4/sgn(x + 1)) + 15/sgn(x + 1))*x + 24/sgn(x + 1) )*sqrt(x^2 - 1) - 15/8*log(abs(-x + sqrt(x^2 - 1)))/sgn(x + 1)
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.44 \[ \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx=\frac {15\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )}{4}+\frac {\frac {49\,\sqrt {\frac {x-1}{x+1}}}{4}-\frac {73\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{4}+\frac {55\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{4}-\frac {15\,{\left (\frac {x-1}{x+1}\right )}^{7/2}}{4}}{\frac {6\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {4\,\left (x-1\right )}{x+1}-\frac {4\,{\left (x-1\right )}^3}{{\left (x+1\right )}^3}+\frac {{\left (x-1\right )}^4}{{\left (x+1\right )}^4}+1} \] Input:
int((x*(x + 1)^2)/((x - 1)/(x + 1))^(1/2),x)
Output:
(15*atanh(((x - 1)/(x + 1))^(1/2)))/4 + ((49*((x - 1)/(x + 1))^(1/2))/4 - (73*((x - 1)/(x + 1))^(3/2))/4 + (55*((x - 1)/(x + 1))^(5/2))/4 - (15*((x - 1)/(x + 1))^(7/2))/4)/((6*(x - 1)^2)/(x + 1)^2 - (4*(x - 1))/(x + 1) - ( 4*(x - 1)^3)/(x + 1)^3 + (x - 1)^4/(x + 1)^4 + 1)
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx=\frac {\sqrt {x +1}\, \sqrt {x -1}\, x^{3}}{4}+\sqrt {x +1}\, \sqrt {x -1}\, x^{2}+\frac {15 \sqrt {x +1}\, \sqrt {x -1}\, x}{8}+3 \sqrt {x +1}\, \sqrt {x -1}+\frac {15 \,\mathrm {log}\left (\frac {\sqrt {x -1}+\sqrt {x +1}}{\sqrt {2}}\right )}{4} \] Input:
int(1/((x-1)/(1+x))^(1/2)*x*(1+x)^2,x)
Output:
(2*sqrt(x + 1)*sqrt(x - 1)*x**3 + 8*sqrt(x + 1)*sqrt(x - 1)*x**2 + 15*sqrt (x + 1)*sqrt(x - 1)*x + 24*sqrt(x + 1)*sqrt(x - 1) + 30*log((sqrt(x - 1) + sqrt(x + 1))/sqrt(2)))/8