Integrand size = 25, antiderivative size = 211 \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4}{a \left (c-a^2 c x^2\right )^{3/2}}+\frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^5}{2 \left (c-a^2 c x^2\right )^{3/2}}+\frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{2 a^2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac {7 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (1-a x)}{4 a^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (1+a x)}{4 a^2 \left (c-a^2 c x^2\right )^{3/2}} \] Output:
(1-1/a^2/x^2)^(3/2)*x^4/a/(-a^2*c*x^2+c)^(3/2)+1/2*(1-1/a^2/x^2)^(3/2)*x^5 /(-a^2*c*x^2+c)^(3/2)+1/2*(1-1/a^2/x^2)^(3/2)*x^3/a^2/(-a*x+1)/(-a^2*c*x^2 +c)^(3/2)+7/4*(1-1/a^2/x^2)^(3/2)*x^3*ln(-a*x+1)/a^2/(-a^2*c*x^2+c)^(3/2)+ 1/4*(1-1/a^2/x^2)^(3/2)*x^3*ln(a*x+1)/a^2/(-a^2*c*x^2+c)^(3/2)
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.39 \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \left (2 \left (\frac {2 x}{a}+x^2+\frac {1}{a^2-a^3 x}\right )+\frac {7 \log (1-a x)}{a^2}+\frac {\log (1+a x)}{a^2}\right )}{4 \left (c-a^2 c x^2\right )^{3/2}} \] Input:
Integrate[(E^ArcCoth[a*x]*x^4)/(c - a^2*c*x^2)^(3/2),x]
Output:
((1 - 1/(a^2*x^2))^(3/2)*x^3*(2*((2*x)/a + x^2 + (a^2 - a^3*x)^(-1)) + (7* Log[1 - a*x])/a^2 + Log[1 + a*x]/a^2))/(4*(c - a^2*c*x^2)^(3/2))
Time = 0.83 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.45, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6746, 6747, 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.
\(\displaystyle \int \frac {x^4 e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \frac {e^{\coth ^{-1}(a x)} x}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \frac {x^4}{(1-a x)^2 (a x+1)}dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \left (\frac {x}{a^3}+\frac {7}{4 a^4 (a x-1)}+\frac {1}{4 a^4 (a x+1)}+\frac {1}{2 a^4 (a x-1)^2}+\frac {1}{a^4}\right )dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (\frac {1}{2 a^5 (1-a x)}+\frac {7 \log (1-a x)}{4 a^5}+\frac {\log (a x+1)}{4 a^5}+\frac {x}{a^4}+\frac {x^2}{2 a^3}\right )}{\left (c-a^2 c x^2\right )^{3/2}}\) |
Input:
Int[(E^ArcCoth[a*x]*x^4)/(c - a^2*c*x^2)^(3/2),x]
Output:
(a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3*(x/a^4 + x^2/(2*a^3) + 1/(2*a^5*(1 - a*x) ) + (7*Log[1 - a*x])/(4*a^5) + Log[1 + a*x]/(4*a^5)))/(c - a^2*c*x^2)^(3/2 )
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^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p) Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && !IntegerQ[p]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[c^p/a^(2*p) Int[(u/x^(2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !Inte gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
Time = 0.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.50
method | result | size |
default | \(\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (2 a^{3} x^{3}+2 a^{2} x^{2}+\ln \left (a x +1\right ) x a +7 a \ln \left (a x -1\right ) x -4 a x -\ln \left (a x +1\right )-7 \ln \left (a x -1\right )-2\right )}{4 \sqrt {\frac {a x -1}{a x +1}}\, \left (a^{2} x^{2}-1\right ) c^{2} a^{5}}\) | \(106\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVER BOSE)
Output:
1/4/((a*x-1)/(a*x+1))^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(2*a^3*x^3+2*a^2*x^2+ln (a*x+1)*x*a+7*a*ln(a*x-1)*x-4*a*x-ln(a*x+1)-7*ln(a*x-1)-2)/(a^2*x^2-1)/c^2 /a^5
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.36 \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {{\left (2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 4 \, a x + {\left (a x - 1\right )} \log \left (a x + 1\right ) + 7 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 2\right )} \sqrt {-a^{2} c}}{4 \, {\left (a^{7} c^{2} x - a^{6} c^{2}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),x, algorithm= "fricas")
Output:
1/4*(2*a^3*x^3 + 2*a^2*x^2 - 4*a*x + (a*x - 1)*log(a*x + 1) + 7*(a*x - 1)* log(a*x - 1) - 2)*sqrt(-a^2*c)/(a^7*c^2*x - a^6*c^2)
\[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^{4}}{\sqrt {\frac {a x - 1}{a x + 1}} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)*x**4/(-a**2*c*x**2+c)**(3/2),x)
Output:
Integral(x**4/(sqrt((a*x - 1)/(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(3/2)), x)
\[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),x, algorithm= "maxima")
Output:
integrate(x^4/((-a^2*c*x^2 + c)^(3/2)*sqrt((a*x - 1)/(a*x + 1))), x)
Exception generated. \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),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. \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \] Input:
int(x^4/((c - a^2*c*x^2)^(3/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
int(x^4/((c - a^2*c*x^2)^(3/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)
Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, i \left (-\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )-\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x +\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )-14 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x +14 \,\mathrm {log}\left (\sqrt {-a x +1}\right )-2 a^{3} x^{3}-2 a^{2} x^{2}+10 a x -4\right )}{4 a^{5} c^{2} \left (a x -1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*i*( - log(sqrt( - a*x + 1) - sqrt(2))*a*x + log(sqrt( - a*x + 1) - sqrt(2)) - log(sqrt( - a*x + 1) + sqrt(2))*a*x + log(sqrt( - a*x + 1) + sqrt(2)) - 14*log(sqrt( - a*x + 1))*a*x + 14*log(sqrt( - a*x + 1)) - 2*a** 3*x**3 - 2*a**2*x**2 + 10*a*x - 4))/(4*a**5*c**2*(a*x - 1))