Integrand size = 27, antiderivative size = 93 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {(1+a x)^2}{3 a^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac {5 (1+a x)}{3 a^3 c \sqrt {c-a^2 c x^2}}+\frac {\arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^3 c^{3/2}} \] Output:
1/3*(a*x+1)^2/a^3/(-a^2*c*x^2+c)^(3/2)-5/3*(a*x+1)/a^3/c/(-a^2*c*x^2+c)^(1 /2)+arctan(a*c^(1/2)*x/(-a^2*c*x^2+c)^(1/2))/a^3/c^(3/2)
Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\frac {(-4+5 a x) \sqrt {c-a^2 c x^2}}{(-1+a x)^2}-3 \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )}{3 a^3 c^2} \] Input:
Integrate[(E^(2*ArcTanh[a*x])*x^2)/(c - a^2*c*x^2)^(3/2),x]
Output:
(((-4 + 5*a*x)*Sqrt[c - a^2*c*x^2])/(-1 + a*x)^2 - 3*Sqrt[c]*ArcTan[(a*x*S qrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/(3*a^3*c^2)
Time = 0.42 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6701, 529, 27, 665, 27, 224, 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.
\(\displaystyle \int \frac {x^2 e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6701 |
\(\displaystyle c \int \frac {x^2 (a x+1)^2}{\left (c-a^2 c x^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 529 |
\(\displaystyle c \left (\frac {(a x+1)^2}{3 a^3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\int \frac {(a x+1) (3 a x+2)}{a^2 \left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (\frac {(a x+1)^2}{3 a^3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\int \frac {(a x+1) (3 a x+2)}{\left (c-a^2 c x^2\right )^{3/2}}dx}{3 a^2 c}\right )\) |
\(\Big \downarrow \) 665 |
\(\displaystyle c \left (\frac {(a x+1)^2}{3 a^3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\frac {5 (a x+1)}{a c \sqrt {c-a^2 c x^2}}-\frac {\int \frac {3 a}{\sqrt {c-a^2 c x^2}}dx}{a c}}{3 a^2 c}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (\frac {(a x+1)^2}{3 a^3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\frac {5 (a x+1)}{a c \sqrt {c-a^2 c x^2}}-\frac {3 \int \frac {1}{\sqrt {c-a^2 c x^2}}dx}{c}}{3 a^2 c}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle c \left (\frac {(a x+1)^2}{3 a^3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\frac {5 (a x+1)}{a c \sqrt {c-a^2 c x^2}}-\frac {3 \int \frac {1}{\frac {a^2 c x^2}{c-a^2 c x^2}+1}d\frac {x}{\sqrt {c-a^2 c x^2}}}{c}}{3 a^2 c}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle c \left (\frac {(a x+1)^2}{3 a^3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\frac {5 (a x+1)}{a c \sqrt {c-a^2 c x^2}}-\frac {3 \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a c^{3/2}}}{3 a^2 c}\right )\) |
Input:
Int[(E^(2*ArcTanh[a*x])*x^2)/(c - a^2*c*x^2)^(3/2),x]
Output:
c*((1 + a*x)^2/(3*a^3*c*(c - a^2*c*x^2)^(3/2)) - ((5*(1 + a*x))/(a*c*Sqrt[ c - a^2*c*x^2]) - (3*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(a*c^(3/2) ))/(3*a^2*c))
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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ (2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b* x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* c^2 + a*d^2, 0]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2)^(3/2), x_Symbol] :> Simp[(-2^(m - 1))*d^(m - 2)*(e*f + d*g)^n*((d + e*x)/(c*e^(n - 1)*Sqrt[a + c*x^2])), x] + Simp[1/(c*e^(n - 2)) Int[Expand ToSum[(2^(m - 1)*d^(m - 1)*(e*f + d*g)^n - e^n*(d + e*x)^(m - 1)*(f + g*x)^ n)/(d - e*x), x]/Sqrt[a + c*x^2], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^(n/2) Int[x^m*(c + d*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ [c, 0]) && IGtQ[n/2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(188\) vs. \(2(79)=158\).
Time = 0.31 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.03
method | result | size |
default | \(-\frac {3 x}{a^{2} c \sqrt {-a^{2} c \,x^{2}+c}}+\frac {\arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{a^{2} c \sqrt {a^{2} c}}-\frac {2}{a^{3} c \sqrt {-a^{2} c \,x^{2}+c}}-\frac {2 \left (\frac {1}{3 a c \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}+\frac {-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c}{3 a \,c^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{a^{3}}\) | \(189\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^2/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVERBOS E)
Output:
-3*x/a^2/c/(-a^2*c*x^2+c)^(1/2)+1/a^2/c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2) *x/(-a^2*c*x^2+c)^(1/2))-2/a^3/c/(-a^2*c*x^2+c)^(1/2)-2/a^3*(1/3/a/c/(x-1/ a)/(-(x-1/a)^2*a^2*c-2*(x-1/a)*a*c)^(1/2)+1/3/a/c^2*(-2*(x-1/a)*a^2*c-2*a* c)/(-(x-1/a)^2*a^2*c-2*(x-1/a)*a*c)^(1/2))
Time = 0.09 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.31 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (5 \, a x - 4\right )}}{6 \, {\left (a^{5} c^{2} x^{2} - 2 \, a^{4} c^{2} x + a^{3} c^{2}\right )}}, -\frac {3 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (5 \, a x - 4\right )}}{3 \, {\left (a^{5} c^{2} x^{2} - 2 \, a^{4} c^{2} x + a^{3} c^{2}\right )}}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="fr icas")
Output:
[-1/6*(3*(a^2*x^2 - 2*a*x + 1)*sqrt(-c)*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^ 2 + c)*a*sqrt(-c)*x - c) - 2*sqrt(-a^2*c*x^2 + c)*(5*a*x - 4))/(a^5*c^2*x^ 2 - 2*a^4*c^2*x + a^3*c^2), -1/3*(3*(a^2*x^2 - 2*a*x + 1)*sqrt(c)*arctan(s qrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) - sqrt(-a^2*c*x^2 + c)*(5 *a*x - 4))/(a^5*c^2*x^2 - 2*a^4*c^2*x + a^3*c^2)]
\[ \int \frac {e^{2 \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=- \int \frac {x^{2}}{- a^{3} c x^{3} \sqrt {- a^{2} c x^{2} + c} + a^{2} c x^{2} \sqrt {- a^{2} c x^{2} + c} + a c x \sqrt {- a^{2} c x^{2} + c} - c \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {a x^{3}}{- a^{3} c x^{3} \sqrt {- a^{2} c x^{2} + c} + a^{2} c x^{2} \sqrt {- a^{2} c x^{2} + c} + a c x \sqrt {- a^{2} c x^{2} + c} - c \sqrt {- a^{2} c x^{2} + c}}\, dx \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*x**2/(-a**2*c*x**2+c)**(3/2),x)
Output:
-Integral(x**2/(-a**3*c*x**3*sqrt(-a**2*c*x**2 + c) + a**2*c*x**2*sqrt(-a* *2*c*x**2 + c) + a*c*x*sqrt(-a**2*c*x**2 + c) - c*sqrt(-a**2*c*x**2 + c)), x) - Integral(a*x**3/(-a**3*c*x**3*sqrt(-a**2*c*x**2 + c) + a**2*c*x**2*s qrt(-a**2*c*x**2 + c) + a*c*x*sqrt(-a**2*c*x**2 + c) - c*sqrt(-a**2*c*x**2 + c)), x)
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (79) = 158\).
Time = 0.19 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.47 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {1}{3} \, a {\left (\frac {a}{\sqrt {-a^{2} c x^{2} + c} a^{6} c x + \sqrt {-a^{2} c x^{2} + c} a^{5} c} - \frac {a}{\sqrt {-a^{2} c x^{2} + c} a^{6} c x - \sqrt {-a^{2} c x^{2} + c} a^{5} c} - \frac {1}{\sqrt {-a^{2} c x^{2} + c} a^{5} c x + \sqrt {-a^{2} c x^{2} + c} a^{4} c} - \frac {1}{\sqrt {-a^{2} c x^{2} + c} a^{5} c x - \sqrt {-a^{2} c x^{2} + c} a^{4} c} - \frac {5 \, x}{\sqrt {-a^{2} c x^{2} + c} a^{3} c} + \frac {3 \, \arcsin \left (a x\right )}{a^{4} c^{\frac {3}{2}}} - \frac {6}{\sqrt {-a^{2} c x^{2} + c} a^{4} c}\right )} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="ma xima")
Output:
1/3*a*(a/(sqrt(-a^2*c*x^2 + c)*a^6*c*x + sqrt(-a^2*c*x^2 + c)*a^5*c) - a/( sqrt(-a^2*c*x^2 + c)*a^6*c*x - sqrt(-a^2*c*x^2 + c)*a^5*c) - 1/(sqrt(-a^2* c*x^2 + c)*a^5*c*x + sqrt(-a^2*c*x^2 + c)*a^4*c) - 1/(sqrt(-a^2*c*x^2 + c) *a^5*c*x - sqrt(-a^2*c*x^2 + c)*a^4*c) - 5*x/(sqrt(-a^2*c*x^2 + c)*a^3*c) + 3*arcsin(a*x)/(a^4*c^(3/2)) - 6/(sqrt(-a^2*c*x^2 + c)*a^4*c))
Exception generated. \[ \int \frac {e^{2 \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="gi ac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {e^{2 \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int -\frac {x^2\,{\left (a\,x+1\right )}^2}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\left (a^2\,x^2-1\right )} \,d x \] Input:
int(-(x^2*(a*x + 1)^2)/((c - a^2*c*x^2)^(3/2)*(a^2*x^2 - 1)),x)
Output:
int(-(x^2*(a*x + 1)^2)/((c - a^2*c*x^2)^(3/2)*(a^2*x^2 - 1)), x)
Time = 0.16 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.23 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (3 \mathit {asin} \left (a x \right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-9 \mathit {asin} \left (a x \right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+9 \mathit {asin} \left (a x \right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-3 \mathit {asin} \left (a x \right )+2 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-12 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )+6\right )}{3 a^{3} c^{2} \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^2/(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*(3*asin(a*x)*tan(asin(a*x)/2)**3 - 9*asin(a*x)*tan(asin(a*x)/2)** 2 + 9*asin(a*x)*tan(asin(a*x)/2) - 3*asin(a*x) + 2*tan(asin(a*x)/2)**3 - 1 2*tan(asin(a*x)/2) + 6))/(3*a**3*c**2*(tan(asin(a*x)/2)**3 - 3*tan(asin(a* x)/2)**2 + 3*tan(asin(a*x)/2) - 1))