Integrand size = 25, antiderivative size = 184 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{8 a^4 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 a^4 c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a^4 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{8 a^4 c^2 \sqrt {c-a^2 c x^2}} \] Output:
1/8*(-a^2*x^2+1)^(1/2)/a^4/c^2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)-1/2*(-a^2*x ^2+1)^(1/2)/a^4/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+1/8*(-a^2*x^2+1)^(1/2)/a ^4/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)+3/8*(-a^2*x^2+1)^(1/2)*arctanh(a*x)/a^ 4/c^2/(-a^2*c*x^2+c)^(1/2)
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.46 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-2-a x+5 a^2 x^2+3 (-1+a x)^2 (1+a x) \text {arctanh}(a x)\right )}{8 a^4 c^2 (-1+a x)^2 (1+a x) \sqrt {c-a^2 c x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x^3)/(c - a^2*c*x^2)^(5/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(-2 - a*x + 5*a^2*x^2 + 3*(-1 + a*x)^2*(1 + a*x)*ArcTan h[a*x]))/(8*a^4*c^2*(-1 + a*x)^2*(1 + a*x)*Sqrt[c - a^2*c*x^2])
Time = 0.82 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 99, 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.
\(\displaystyle \int \frac {x^3 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^3}{(1-a x)^3 (a x+1)^2}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {1}{2 a^3 (a x-1)^2}-\frac {1}{8 a^3 (a x+1)^2}-\frac {1}{4 a^3 (a x-1)^3}-\frac {3}{8 a^3 \left (a^2 x^2-1\right )}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {3 \text {arctanh}(a x)}{8 a^4}-\frac {1}{2 a^4 (1-a x)}+\frac {1}{8 a^4 (a x+1)}+\frac {1}{8 a^4 (1-a x)^2}\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
Input:
Int[(E^ArcTanh[a*x]*x^3)/(c - a^2*c*x^2)^(5/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(1/(8*a^4*(1 - a*x)^2) - 1/(2*a^4*(1 - a*x)) + 1/(8*a^4 *(1 + a*x)) + (3*ArcTanh[a*x])/(8*a^4)))/(c^2*Sqrt[c - a^2*c*x^2])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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]
Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {\left (3 \ln \left (a x +1\right ) x^{3} a^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}-3 \ln \left (a x +1\right ) x^{2} a^{2}+3 a^{2} \ln \left (a x -1\right ) x^{2}+10 a^{2} x^{2}-3 \ln \left (a x +1\right ) x a +3 a \ln \left (a x -1\right ) x -2 a x +3 \ln \left (a x +1\right )-3 \ln \left (a x -1\right )-4\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{16 \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (a x -1\right )^{2} c^{3} a^{4}}\) | \(155\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVE RBOSE)
Output:
1/16*(3*ln(a*x+1)*x^3*a^3-3*a^3*ln(a*x-1)*x^3-3*ln(a*x+1)*x^2*a^2+3*a^2*ln (a*x-1)*x^2+10*a^2*x^2-3*ln(a*x+1)*x*a+3*a*ln(a*x-1)*x-2*a*x+3*ln(a*x+1)-3 *ln(a*x-1)-4)/(-a^2*x^2+1)^(1/2)/(a*x+1)/(a*x-1)^2*(-c*(a^2*x^2-1))^(1/2)/ c^3/a^4
Time = 0.17 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.51 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) - 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} - 3 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{32 \, {\left (a^{9} c^{3} x^{5} - a^{8} c^{3} x^{4} - 2 \, a^{7} c^{3} x^{3} + 2 \, a^{6} c^{3} x^{2} + a^{5} c^{3} x - a^{4} c^{3}\right )}}, \frac {3 \, {\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) - 2 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} - 3 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{16 \, {\left (a^{9} c^{3} x^{5} - a^{8} c^{3} x^{4} - 2 \, a^{7} c^{3} x^{3} + 2 \, a^{6} c^{3} x^{2} + a^{5} c^{3} x - a^{4} c^{3}\right )}}\right ] \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm ="fricas")
Output:
[1/32*(3*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*sqrt(c)*log (-(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a^2*c*x^2 - 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x ^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)) - 4*(2*a^3*x^3 + 3*a^2*x^2 - 3*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^ 2 + 1))/(a^9*c^3*x^5 - a^8*c^3*x^4 - 2*a^7*c^3*x^3 + 2*a^6*c^3*x^2 + a^5*c ^3*x - a^4*c^3), 1/16*(3*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*sqrt(-c)* x/(a^4*c*x^4 - c)) - 2*(2*a^3*x^3 + 3*a^2*x^2 - 3*a*x)*sqrt(-a^2*c*x^2 + c )*sqrt(-a^2*x^2 + 1))/(a^9*c^3*x^5 - a^8*c^3*x^4 - 2*a^7*c^3*x^3 + 2*a^6*c ^3*x^2 + a^5*c^3*x - a^4*c^3)]
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**3/(-a**2*c*x**2+c)**(5/2),x)
Output:
Integral(x**3*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1 ))**(5/2)), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{3}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm ="maxima")
Output:
a*integrate(-x^4/((a^4*c^(5/2)*x^4 - 2*a^2*c^(5/2)*x^2 + c^(5/2))*(a*x + 1 )*(a*x - 1)), x) + 1/4/(a^8*c^(5/2)*x^4 - 2*a^6*c^(5/2)*x^2 + a^4*c^(5/2)) + 1/2/(a^6*c^(5/2)*x^2 - a^4*c^(5/2))
Exception generated. \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm ="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((x^3*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((x^3*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)), x)
Time = 0.14 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-3 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+3 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+3 \,\mathrm {log}\left (a x -1\right ) a x -3 \,\mathrm {log}\left (a x -1\right )+3 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-3 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-3 \,\mathrm {log}\left (a x +1\right ) a x +3 \,\mathrm {log}\left (a x +1\right )+10 a^{3} x^{3}-12 a x +6\right )}{16 a^{4} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*( - 3*log(a*x - 1)*a**3*x**3 + 3*log(a*x - 1)*a**2*x**2 + 3*log(a *x - 1)*a*x - 3*log(a*x - 1) + 3*log(a*x + 1)*a**3*x**3 - 3*log(a*x + 1)*a **2*x**2 - 3*log(a*x + 1)*a*x + 3*log(a*x + 1) + 10*a**3*x**3 - 12*a*x + 6 ))/(16*a**4*c**3*(a**3*x**3 - a**2*x**2 - a*x + 1))