Integrand size = 25, antiderivative size = 217 \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{4 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {11 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}} \] Output:
-1/8*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)+3/4*(1-1/a^2/ x^2)^(5/2)*x^5/(-a*x+1)/(-a^2*c*x^2+c)^(5/2)+1/8*(1-1/a^2/x^2)^(5/2)*x^5/( a*x+1)/(-a^2*c*x^2+c)^(5/2)+11/16*(1-1/a^2/x^2)^(5/2)*x^5*ln(-a*x+1)/(-a^2 *c*x^2+c)^(5/2)+5/16*(1-1/a^2/x^2)^(5/2)*x^5*ln(a*x+1)/(-a^2*c*x^2+c)^(5/2 )
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.39 \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (-\frac {2 \left (-6+3 a x+5 a^2 x^2\right )}{(-1+a x)^2 (1+a x)}+11 \log (1-a x)+5 \log (1+a x)\right )}{16 \left (c-a^2 c x^2\right )^{5/2}} \] Input:
Integrate[(E^ArcCoth[a*x]*x^4)/(c - a^2*c*x^2)^(5/2),x]
Output:
((1 - 1/(a^2*x^2))^(5/2)*x^5*((-2*(-6 + 3*a*x + 5*a^2*x^2))/((-1 + a*x)^2* (1 + a*x)) + 11*Log[1 - a*x] + 5*Log[1 + a*x]))/(16*(c - a^2*c*x^2)^(5/2))
Time = 0.92 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.50, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6746, 6747, 25, 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.
\(\displaystyle \int \frac {x^4 e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int -\frac {x^4}{(1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {x^4}{(1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \left (-\frac {5}{16 a^4 (a x+1)}+\frac {1}{8 a^4 (a x+1)^2}-\frac {11}{16 a^4 (a x-1)}-\frac {3}{4 a^4 (a x-1)^2}-\frac {1}{4 a^4 (a x-1)^3}\right )dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (-\frac {3}{4 a^5 (1-a x)}-\frac {1}{8 a^5 (a x+1)}+\frac {1}{8 a^5 (1-a x)^2}-\frac {11 \log (1-a x)}{16 a^5}-\frac {5 \log (a x+1)}{16 a^5}\right )}{\left (c-a^2 c x^2\right )^{5/2}}\) |
Input:
Int[(E^ArcCoth[a*x]*x^4)/(c - a^2*c*x^2)^(5/2),x]
Output:
-((a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(1/(8*a^5*(1 - a*x)^2) - 3/(4*a^5*(1 - a*x)) - 1/(8*a^5*(1 + a*x)) - (11*Log[1 - a*x])/(16*a^5) - (5*Log[1 + a*x] )/(16*a^5)))/(c - a^2*c*x^2)^(5/2))
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 = 0.15 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.78
method | result | size |
default | \(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (5 \ln \left (a x +1\right ) x^{3} a^{3}+11 a^{3} \ln \left (a x -1\right ) x^{3}-5 \ln \left (a x +1\right ) x^{2} a^{2}-11 a^{2} \ln \left (a x -1\right ) x^{2}-10 a^{2} x^{2}-5 \ln \left (a x +1\right ) x a -11 a \ln \left (a x -1\right ) x -6 a x +5 \ln \left (a x +1\right )+11 \ln \left (a x -1\right )+12\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a^{5} \left (a x +1\right )}\) | \(169\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVER BOSE)
Output:
-1/16/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(5*ln(a*x+1)* x^3*a^3+11*a^3*ln(a*x-1)*x^3-5*ln(a*x+1)*x^2*a^2-11*a^2*ln(a*x-1)*x^2-10*a ^2*x^2-5*ln(a*x+1)*x*a-11*a*ln(a*x-1)*x-6*a*x+5*ln(a*x+1)+11*ln(a*x-1)+12) /(a^2*x^2-1)/c^3/a^5/(a*x+1)
Time = 0.12 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (10 \, a^{2} x^{2} + 6 \, a x - 5 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) - 11 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 12\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{9} c^{3} x^{3} - a^{8} c^{3} x^{2} - a^{7} c^{3} x + a^{6} c^{3}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x, algorithm= "fricas")
Output:
1/16*(10*a^2*x^2 + 6*a*x - 5*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) - 11*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x - 1) - 12)*sqrt(-a^2*c)/(a^9*c^3* x^3 - a^8*c^3*x^2 - a^7*c^3*x + a^6*c^3)
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)*x**4/(-a**2*c*x**2+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x, algorithm= "maxima")
Output:
integrate(x^4/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)
Exception generated. \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(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. \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^4}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \] Input:
int(x^4/((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
int(x^4/((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)
Time = 0.16 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\coth ^{-1}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, i \left (5 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}-5 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}-5 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +5 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )+5 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}-5 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}-5 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x +5 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )+22 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}-22 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}-22 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x +22 \,\mathrm {log}\left (\sqrt {-a x +1}\right )+5 a^{3} x^{3}-15 a^{2} x^{2}-11 a x +17\right )}{16 a^{5} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*i*(5*log(sqrt( - a*x + 1) - sqrt(2))*a**3*x**3 - 5*log(sqrt( - a* x + 1) - sqrt(2))*a**2*x**2 - 5*log(sqrt( - a*x + 1) - sqrt(2))*a*x + 5*lo g(sqrt( - a*x + 1) - sqrt(2)) + 5*log(sqrt( - a*x + 1) + sqrt(2))*a**3*x** 3 - 5*log(sqrt( - a*x + 1) + sqrt(2))*a**2*x**2 - 5*log(sqrt( - a*x + 1) + sqrt(2))*a*x + 5*log(sqrt( - a*x + 1) + sqrt(2)) + 22*log(sqrt( - a*x + 1 ))*a**3*x**3 - 22*log(sqrt( - a*x + 1))*a**2*x**2 - 22*log(sqrt( - a*x + 1 ))*a*x + 22*log(sqrt( - a*x + 1)) + 5*a**3*x**3 - 15*a**2*x**2 - 11*a*x + 17))/(16*a**5*c**3*(a**3*x**3 - a**2*x**2 - a*x + 1))