Integrand size = 12, antiderivative size = 169 \[ \int e^{2 \text {sech}^{-1}(a x)} x^2 \, dx=\frac {(1+a x) \left (1-\sqrt {\frac {1-a x}{1+a x}}\right ) \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )}{2 a^3}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^2 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}{6 a^3}+\frac {(1+a x)^3 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^4}{12 a^3}-\frac {2 \arctan \left (\sqrt {\frac {1-a x}{1+a x}}\right )}{a^3} \] Output:
1/2*(a*x+1)*(1-((-a*x+1)/(a*x+1))^(1/2))*(1+((-a*x+1)/(a*x+1))^(1/2))/a^3- 1/6*((-a*x+1)/(a*x+1))^(1/2)*(a*x+1)^2*(1+((-a*x+1)/(a*x+1))^(1/2))^3/a^3+ 1/12*(a*x+1)^3*(1+((-a*x+1)/(a*x+1))^(1/2))^4/a^3-2*arctan(((-a*x+1)/(a*x+ 1))^(1/2))/a^3
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.51 \[ \int e^{2 \text {sech}^{-1}(a x)} x^2 \, dx=\frac {2 x}{a^2}-\frac {x^3}{3}+\sqrt {\frac {1-a x}{1+a x}} \left (\frac {x}{a^2}+\frac {x^2}{a}\right )+\frac {i \log \left (-2 i a x+2 \sqrt {\frac {1-a x}{1+a x}} (1+a x)\right )}{a^3} \] Input:
Integrate[E^(2*ArcSech[a*x])*x^2,x]
Output:
(2*x)/a^2 - x^3/3 + Sqrt[(1 - a*x)/(1 + a*x)]*(x/a^2 + x^2/a) + (I*Log[(-2 *I)*a*x + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)])/a^3
Time = 1.07 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6891, 7268, 531, 27, 490, 487, 216}
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.
| \(\displaystyle \int x^2 e^{2 \text {sech}^{-1}(a x)} \, dx\) |
| \(\Big \downarrow \) 6891 |
| \(\displaystyle \int x^2 \left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}\right )^2dx\) |
| \(\Big \downarrow \) 7268 |
| \(\displaystyle -\frac {4 \int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}{\left (\frac {1-a x}{a x+1}+1\right )^4}d\sqrt {\frac {1-a x}{a x+1}}}{a^3}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle -\frac {4 \left (-\frac {1}{6} \int -\frac {4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}{\left (\frac {1-a x}{a x+1}+1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}-\frac {\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}{6 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {4 \left (\frac {2}{3} \int \frac {\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}{\left (\frac {1-a x}{a x+1}+1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}-\frac {\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}{6 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
| \(\Big \downarrow \) 490 |
| \(\displaystyle -\frac {4 \left (\frac {2}{3} \left (\frac {3}{4} \int \frac {\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}{\left (\frac {1-a x}{a x+1}+1\right )^2}d\sqrt {\frac {1-a x}{a x+1}}+\frac {\sqrt {\frac {1-a x}{a x+1}} \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}{4 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )-\frac {\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}{6 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
| \(\Big \downarrow \) 487 |
| \(\displaystyle -\frac {4 \left (\frac {2}{3} \left (\frac {3}{4} \left (\int \frac {1}{\frac {1-a x}{a x+1}+1}d\sqrt {\frac {1-a x}{a x+1}}-\frac {\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}{2 \left (\frac {1-a x}{a x+1}+1\right )}\right )+\frac {\sqrt {\frac {1-a x}{a x+1}} \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}{4 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )-\frac {\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}{6 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {4 \left (\frac {2}{3} \left (\frac {3}{4} \left (\arctan \left (\sqrt {\frac {1-a x}{a x+1}}\right )-\frac {\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}{2 \left (\frac {1-a x}{a x+1}+1\right )}\right )+\frac {\sqrt {\frac {1-a x}{a x+1}} \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}{4 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )-\frac {\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}{6 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
Input:
Int[E^(2*ArcSech[a*x])*x^2,x]
Output:
(-4*(-1/6*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^4/(1 + (1 - a*x)/(1 + a*x))^3 + (2*((Sqrt[(1 - a*x)/(1 + a*x)]*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^3)/(4*(1 + (1 - a*x)/(1 + a*x))^2) + (3*(-1/2*((1 - Sqrt[(1 - a*x)/(1 + a*x)])*(1 + S qrt[(1 - a*x)/(1 + a*x)]))/(1 + (1 - a*x)/(1 + a*x)) + ArcTan[Sqrt[(1 - a* x)/(1 + a*x)]]))/4))/3))/a^3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n - 1)*(a*d - b*c*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[(2*p + 3)*((b*c^2 + a*d^2)/(2*a*b*(p + 1))) Int[(c + d*x)^(n - 2)*( a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[n + 2*p + 2, 0] && LtQ[p, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-x)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] - Simp[c*(n/(2*a*( p + 1))) Int[(c + d*x)^(n - 1)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b , c, d, n, p}, x] && EqQ[n + 2*p + 3, 0] && LtQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b *x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
Int[E^(ArcSech[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[(1 - u)/(1 + u)] + (1/u)*Sqrt[(1 - u)/(1 + u)])^n, x] /; FreeQ[m, x] && Integer Q[n]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfQuotientOfLinears [u, x]}, Simp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/ls t[[2]])], x] /; !FalseQ[lst]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.62
| method | result | size |
| default | \(\frac {-\frac {1}{3} a^{2} x^{3}+x}{a^{2}}+\frac {\sqrt {-\frac {a x -1}{a x}}\, x \sqrt {\frac {a x +1}{a x}}\, \left (x \sqrt {-a^{2} x^{2}+1}\, \operatorname {csgn}\left (a \right ) a +\arctan \left (\frac {\operatorname {csgn}\left (a \right ) a x}{\sqrt {-a^{2} x^{2}+1}}\right )\right ) \operatorname {csgn}\left (a \right )}{a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {x}{a^{2}}\) | \(105\) |
Input:
int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*x^2,x,method=_RETURNVERBOSE )
Output:
1/a^2*(-1/3*a^2*x^3+x)+1/a^2*(-(a*x-1)/a/x)^(1/2)*x*((a*x+1)/a/x)^(1/2)*(x *(-a^2*x^2+1)^(1/2)*csgn(a)*a+arctan(csgn(a)*a*x/(-a^2*x^2+1)^(1/2)))/(-a^ 2*x^2+1)^(1/2)*csgn(a)+x/a^2
Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.51 \[ \int e^{2 \text {sech}^{-1}(a x)} x^2 \, dx=-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 6 \, a x + 3 \, \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right )}{3 \, a^{3}} \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*x^2,x, algorithm="fri cas")
Output:
-1/3*(a^3*x^3 - 3*a^2*x^2*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 6 *a*x + 3*arctan(sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x))))/a^3
\[ \int e^{2 \text {sech}^{-1}(a x)} x^2 \, dx=\frac {\int 2\, dx + \int \left (- a^{2} x^{2}\right )\, dx + \int 2 a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}\, dx}{a^{2}} \] Input:
integrate((1/a/x+(-1+1/a/x)**(1/2)*(1+1/a/x)**(1/2))**2*x**2,x)
Output:
(Integral(2, x) + Integral(-a**2*x**2, x) + Integral(2*a*x*sqrt(-1 + 1/(a* x))*sqrt(1 + 1/(a*x)), x))/a**2
\[ \int e^{2 \text {sech}^{-1}(a x)} x^2 \, dx=\int { x^{2} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2} \,d x } \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*x^2,x, algorithm="max ima")
Output:
2*x/a^2 + 2*integrate(sqrt(a*x + 1)*sqrt(-a*x + 1), x)/a^2 - integrate(x^2 , x)
\[ \int e^{2 \text {sech}^{-1}(a x)} x^2 \, dx=\int { x^{2} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2} \,d x } \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*x^2,x, algorithm="gia c")
Output:
integrate(x^2*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))^2, x)
Time = 34.07 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.49 \[ \int e^{2 \text {sech}^{-1}(a x)} x^2 \, dx=\frac {\frac {1{}\mathrm {i}}{16\,a^3}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{8\,a^3\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{16\,a^3\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}}{\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}}-\frac {x^3\,\left (\frac {a^2}{3}-\frac {2}{x^2}\right )}{a^2}+\frac {\left (\ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\right )\,2{}\mathrm {i}}{a^3}+\frac {\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\,1{}\mathrm {i}}{a^3}-\frac {\ln \left (\frac {2\,a\,\sqrt {\frac {a+\frac {1}{x}}{a}}-\frac {2}{x}+a\,\sqrt {-\frac {a-\frac {1}{x}}{a}}\,2{}\mathrm {i}}{2\,a+\frac {1}{x}-2\,a\,\sqrt {\frac {a+\frac {1}{x}}{a}}}\right )\,1{}\mathrm {i}}{a^3}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,a^3\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2} \] Input:
int(x^2*((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x))^2,x)
Output:
((log(((1/(a*x) - 1)^(1/2) - 1i)^2/((1/(a*x) + 1)^(1/2) - 1)^2 + 1) - log( ((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1)))*2i)/a^3 + (log(((1/ (a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1))*1i)/a^3 + (1i/(16*a^3) + (((1/(a*x) - 1)^(1/2) - 1i)^2*1i)/(8*a^3*((1/(a*x) + 1)^(1/2) - 1)^2) - ( ((1/(a*x) - 1)^(1/2) - 1i)^4*15i)/(16*a^3*((1/(a*x) + 1)^(1/2) - 1)^4))/(( (1/(a*x) - 1)^(1/2) - 1i)^2/((1/(a*x) + 1)^(1/2) - 1)^2 + (2*((1/(a*x) - 1 )^(1/2) - 1i)^4)/((1/(a*x) + 1)^(1/2) - 1)^4 + ((1/(a*x) - 1)^(1/2) - 1i)^ 6/((1/(a*x) + 1)^(1/2) - 1)^6) - (log((a*(-(a - 1/x)/a)^(1/2)*2i - 2/x + 2 *a*((a + 1/x)/a)^(1/2))/(2*a + 1/x - 2*a*((a + 1/x)/a)^(1/2)))*1i)/a^3 + ( ((1/(a*x) - 1)^(1/2) - 1i)^2*1i)/(16*a^3*((1/(a*x) + 1)^(1/2) - 1)^2) - (x ^3*(a^2/3 - 2/x^2))/a^2
Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.30 \[ \int e^{2 \text {sech}^{-1}(a x)} x^2 \, dx=\frac {-6 \mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )+3 \sqrt {a x +1}\, \sqrt {-a x +1}\, a x -a^{3} x^{3}+6 a x}{3 a^{3}} \] Input:
int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*x^2,x)
Output:
( - 6*asin(sqrt( - a*x + 1)/sqrt(2)) + 3*sqrt(a*x + 1)*sqrt( - a*x + 1)*a* x - a**3*x**3 + 6*a*x)/(3*a**3)