Integrand size = 12, antiderivative size = 75 \[ \int e^{-\text {sech}^{-1}(a x)} x^2 \, dx=-\frac {x}{a^2}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^3}{3 a^3}+\frac {(1+a x)^2 \left (3+4 \sqrt {\frac {1-a x}{1+a x}}\right )}{6 a^3} \]
-x/a^2-1/3*(a*x+1)^3*((-a*x+1)/(a*x+1))^(1/2)/a^3+1/6*(a*x+1)^2*(3+4*((-a* x+1)/(a*x+1))^(1/2))/a^3
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.64 \[ \int e^{-\text {sech}^{-1}(a x)} x^2 \, dx=\frac {3 a^2 x^2-2 (-1+a x) \sqrt {\frac {1-a x}{1+a x}} (1+a x)^2}{6 a^3} \]
Time = 0.71 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.56, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6891, 7268, 25, 2335, 27, 2345, 27, 241}
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.
\(\displaystyle \int x^2 e^{-\text {sech}^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6891 |
\(\displaystyle \int \frac {x^2}{\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}}dx\) |
\(\Big \downarrow \) 7268 |
\(\displaystyle \frac {4 \int -\frac {\sqrt {\frac {1-a x}{a x+1}} \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}{\left (\frac {1-a x}{a x+1}+1\right )^4}d\sqrt {\frac {1-a x}{a x+1}}}{a^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {4 \int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}{\left (\frac {1-a x}{a x+1}+1\right )^4}d\sqrt {\frac {1-a x}{a x+1}}}{a^3}\) |
\(\Big \downarrow \) 2335 |
\(\displaystyle \frac {4 \left (\frac {1}{6} \int \frac {2 \left (3 \left (\frac {1-a x}{a x+1}\right )^{3/2}-3 \sqrt {\frac {1-a x}{a x+1}}-\frac {6 (1-a x)}{a x+1}+2\right )}{\left (\frac {1-a x}{a x+1}+1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}-\frac {2 \sqrt {\frac {1-a x}{a x+1}}}{3 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \left (\frac {1}{3} \int \frac {3 \left (\frac {1-a x}{a x+1}\right )^{3/2}-3 \sqrt {\frac {1-a x}{a x+1}}-\frac {6 (1-a x)}{a x+1}+2}{\left (\frac {1-a x}{a x+1}+1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}-\frac {2 \sqrt {\frac {1-a x}{a x+1}}}{3 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {4 \left (\frac {1}{3} \left (\frac {4 \sqrt {\frac {1-a x}{a x+1}}+3}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}-\frac {1}{4} \int -\frac {12 \sqrt {\frac {1-a x}{a x+1}}}{\left (\frac {1-a x}{a x+1}+1\right )^2}d\sqrt {\frac {1-a x}{a x+1}}\right )-\frac {2 \sqrt {\frac {1-a x}{a x+1}}}{3 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \left (\frac {1}{3} \left (3 \int \frac {\sqrt {\frac {1-a x}{a x+1}}}{\left (\frac {1-a x}{a x+1}+1\right )^2}d\sqrt {\frac {1-a x}{a x+1}}+\frac {4 \sqrt {\frac {1-a x}{a x+1}}+3}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )-\frac {2 \sqrt {\frac {1-a x}{a x+1}}}{3 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {4 \left (\frac {1}{3} \left (\frac {4 \sqrt {\frac {1-a x}{a x+1}}+3}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}-\frac {3}{2 \left (\frac {1-a x}{a x+1}+1\right )}\right )-\frac {2 \sqrt {\frac {1-a x}{a x+1}}}{3 \left (\frac {1-a x}{a x+1}+1\right )^3}\right )}{a^3}\) |
(4*((-2*Sqrt[(1 - a*x)/(1 + a*x)])/(3*(1 + (1 - a*x)/(1 + a*x))^3) + ((3 + 4*Sqrt[(1 - a*x)/(1 + a*x)])/(2*(1 + (1 - a*x)/(1 + a*x))^2) - 3/(2*(1 + (1 - a*x)/(1 + a*x))))/3))/a^3
3.1.78.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq , a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] + Simp[c/(2*a*b*(p + 1)) Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
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 3 vs. order 2.
Time = 0.29 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.59
method | result | size |
default | \(\frac {\left (a x +1\right ) \left (3 a^{6} x^{6} \left (\frac {a x +1}{a x}\right )^{\frac {3}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}}+3 \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}} \ln \left (a^{2} x^{2}\right ) \left (\frac {a x +1}{a x}\right )^{\frac {3}{2}} x^{4} a^{4}-3 \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}} \sqrt {\frac {a x +1}{a x}}\, \ln \left (a^{2} x^{2}\right ) a^{4} x^{4}+2 a^{7} x^{7}-3 x^{3} \ln \left (a^{2} x^{2}\right ) \sqrt {\frac {a x +1}{a x}}\, \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}} a^{3}-2 a^{6} x^{6}-6 a^{5} x^{5}+6 a^{4} x^{4}+6 a^{3} x^{3}-6 a^{2} x^{2}-2 a x +2\right )}{6 x^{5} a^{8} \left (\frac {a x +1}{a x}\right )^{\frac {5}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}}}\) | \(269\) |
1/6*(a*x+1)/x^5*(3*a^6*x^6*((a*x+1)/a/x)^(3/2)*(-(a*x-1)/a/x)^(5/2)+3*(-(a *x-1)/a/x)^(5/2)*ln(a^2*x^2)*((a*x+1)/a/x)^(3/2)*x^4*a^4-3*(-(a*x-1)/a/x)^ (5/2)*((a*x+1)/a/x)^(1/2)*ln(a^2*x^2)*a^4*x^4+2*a^7*x^7-3*x^3*ln(a^2*x^2)* ((a*x+1)/a/x)^(1/2)*(-(a*x-1)/a/x)^(5/2)*a^3-2*a^6*x^6-6*a^5*x^5+6*a^4*x^4 +6*a^3*x^3-6*a^2*x^2-2*a*x+2)/a^8/((a*x+1)/a/x)^(5/2)/(-(a*x-1)/a/x)^(5/2)
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72 \[ \int e^{-\text {sech}^{-1}(a x)} x^2 \, dx=\frac {3 \, a x^{2} - 2 \, {\left (a^{2} x^{3} - x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}}{6 \, a^{2}} \]
\[ \int e^{-\text {sech}^{-1}(a x)} x^2 \, dx=a \int \frac {x^{3}}{a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + 1}\, dx \]
\[ \int e^{-\text {sech}^{-1}(a x)} x^2 \, dx=\int { \frac {x^{2}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}} \,d x } \]
\[ \int e^{-\text {sech}^{-1}(a x)} x^2 \, dx=\int { \frac {x^{2}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}} \,d x } \]
Time = 4.89 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.76 \[ \int e^{-\text {sech}^{-1}(a x)} x^2 \, dx=\frac {x^2}{2\,a}+\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {x}{3\,a^2}+\frac {1}{3\,a^3}-\frac {x^3}{3}-\frac {x^2}{3\,a}\right )}{\sqrt {\frac {1}{a\,x}+1}} \]