Integrand size = 29, antiderivative size = 153 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt [4]{1-a^2 x^2}}{a^3 c \sqrt {1-a x} \sqrt [4]{c-a^2 c x^2}}+\frac {2 \sqrt {1-a x} \sqrt [4]{1-a^2 x^2}}{a^3 c \sqrt [4]{c-a^2 c x^2}}-\frac {\sqrt [4]{1-a^2 x^2} \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a^3 c \sqrt [4]{c-a^2 c x^2}} \] Output:
(-a^2*x^2+1)^(1/4)/a^3/c/(-a*x+1)^(1/2)/(-a^2*c*x^2+c)^(1/4)+2*(-a*x+1)^(1 /2)*(-a^2*x^2+1)^(1/4)/a^3/c/(-a^2*c*x^2+c)^(1/4)-1/2*(-a^2*x^2+1)^(1/4)*a rctanh(1/2*(-a*x+1)^(1/2)*2^(1/2))*2^(1/2)/a^3/c/(-a^2*c*x^2+c)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.46 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt [4]{1-a^2 x^2} \left (2-2 a x+\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1-a x)\right )\right )}{a^3 c \sqrt {1-a x} \sqrt [4]{c-a^2 c x^2}} \] Input:
Integrate[(E^(ArcTanh[a*x]/2)*x^2)/(c - a^2*c*x^2)^(5/4),x]
Output:
((1 - a^2*x^2)^(1/4)*(2 - 2*a*x + Hypergeometric2F1[-1/2, 1, 1/2, (1 - a*x )/2]))/(a^3*c*Sqrt[1 - a*x]*(c - a^2*c*x^2)^(1/4))
Time = 0.55 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.59, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {6703, 6700, 98, 2009}
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.
\(\displaystyle \int \frac {x^2 e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (1-a^2 x^2\right )^{5/4}}dx}{c \sqrt [4]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \int \frac {x^2}{(1-a x)^{3/2} (a x+1)}dx}{c \sqrt [4]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 98 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \int \left (-\frac {1}{a^2 \left (a^2 x^2-1\right ) \sqrt {1-a x}}-\frac {1}{a^2 \sqrt {1-a x}}\right )dx}{c \sqrt [4]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \left (-\frac {\text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a^3}+\frac {2 \sqrt {1-a x}}{a^3}+\frac {1}{a^3 \sqrt {1-a x}}\right )}{c \sqrt [4]{c-a^2 c x^2}}\) |
Input:
Int[(E^(ArcTanh[a*x]/2)*x^2)/(c - a^2*c*x^2)^(5/4),x]
Output:
((1 - a^2*x^2)^(1/4)*(1/(a^3*Sqrt[1 - a*x]) + (2*Sqrt[1 - a*x])/a^3 - ArcT anh[Sqrt[1 - a*x]/Sqrt[2]]/(Sqrt[2]*a^3)))/(c*(c - a^2*c*x^2)^(1/4))
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar t[p]) Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0]) && !I ntegerQ[n/2]
\[\int \frac {\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, x^{2}}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{4}}}d x\]
Input:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(5/4),x)
Output:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(5/4),x)
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\text {Timed out} \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(5/4),x, a lgorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\text {Timed out} \] Input:
integrate(((a*x+1)/(-a**2*x**2+1)**(1/2))**(1/2)*x**2/(-a**2*c*x**2+c)**(5 /4),x)
Output:
Timed out
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\int { \frac {x^{2} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(5/4),x, a lgorithm="maxima")
Output:
integrate(x^2*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(5/4), x )
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\int { \frac {x^{2} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(5/4),x, a lgorithm="giac")
Output:
integrate(x^2*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(5/4), x )
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\int \frac {x^2\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{{\left (c-a^2\,c\,x^2\right )}^{5/4}} \,d x \] Input:
int((x^2*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2))/(c - a^2*c*x^2)^(5/4),x)
Output:
int((x^2*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2))/(c - a^2*c*x^2)^(5/4), x)
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.39 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {a x +1}}{\sqrt {2}}\right )}{2}\right )\right )-5 \sqrt {-a x +1}\, \sqrt {2}-4 a x +6}{2 c^{\frac {5}{4}} \sqrt {-a x +1}\, a^{3}} \] Input:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(5/4),x)
Output:
(c**(3/4)*(sqrt( - a*x + 1)*sqrt(2)*log(tan(asin(sqrt(a*x + 1)/sqrt(2))/2) ) - 5*sqrt( - a*x + 1)*sqrt(2) - 4*a*x + 6))/(2*sqrt( - a*x + 1)*a**3*c**2 )