Integrand size = 25, antiderivative size = 312 \[ \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {x \sqrt {1-a^2 x^2}}{a^6 c^2 \sqrt {c-a^2 c x^2}}-\frac {x^2 \sqrt {1-a^2 x^2}}{2 a^5 c^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a^7 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2}}{4 a^7 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a^7 c^2 (1+a x) \sqrt {c-a^2 c x^2}}-\frac {39 \sqrt {1-a^2 x^2} \log (1-a x)}{16 a^7 c^2 \sqrt {c-a^2 c x^2}}-\frac {9 \sqrt {1-a^2 x^2} \log (1+a x)}{16 a^7 c^2 \sqrt {c-a^2 c x^2}} \] Output:
-x*(-a^2*x^2+1)^(1/2)/a^6/c^2/(-a^2*c*x^2+c)^(1/2)-1/2*x^2*(-a^2*x^2+1)^(1 /2)/a^5/c^2/(-a^2*c*x^2+c)^(1/2)+1/8*(-a^2*x^2+1)^(1/2)/a^7/c^2/(-a*x+1)^2 /(-a^2*c*x^2+c)^(1/2)-5/4*(-a^2*x^2+1)^(1/2)/a^7/c^2/(-a*x+1)/(-a^2*c*x^2+ c)^(1/2)-1/8*(-a^2*x^2+1)^(1/2)/a^7/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)-39/16 *(-a^2*x^2+1)^(1/2)*ln(-a*x+1)/a^7/c^2/(-a^2*c*x^2+c)^(1/2)-9/16*(-a^2*x^2 +1)^(1/2)*ln(a*x+1)/a^7/c^2/(-a^2*c*x^2+c)^(1/2)
Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.31 \[ \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (2 \left (-8 a x-4 a^2 x^2+\frac {1}{(-1+a x)^2}+\frac {10}{-1+a x}-\frac {1}{1+a x}\right )-39 \log (1-a x)-9 \log (1+a x)\right )}{16 a^7 c^2 \sqrt {c-a^2 c x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x^6)/(c - a^2*c*x^2)^(5/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(2*(-8*a*x - 4*a^2*x^2 + (-1 + a*x)^(-2) + 10/(-1 + a*x ) - (1 + a*x)^(-1)) - 39*Log[1 - a*x] - 9*Log[1 + a*x]))/(16*a^7*c^2*Sqrt[ c - a^2*c*x^2])
Time = 0.88 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.39, 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.
| \(\displaystyle \int \frac {x^6 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6703 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 6700 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^6}{(1-a x)^3 (a x+1)^2}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {x}{a^5}-\frac {39}{16 a^6 (a x-1)}-\frac {9}{16 a^6 (a x+1)}-\frac {5}{4 a^6 (a x-1)^2}+\frac {1}{8 a^6 (a x+1)^2}-\frac {1}{4 a^6 (a x-1)^3}-\frac {1}{a^6}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (-\frac {5}{4 a^7 (1-a x)}-\frac {1}{8 a^7 (a x+1)}+\frac {1}{8 a^7 (1-a x)^2}-\frac {39 \log (1-a x)}{16 a^7}-\frac {9 \log (a x+1)}{16 a^7}-\frac {x}{a^6}-\frac {x^2}{2 a^5}\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
Input:
Int[(E^ArcTanh[a*x]*x^6)/(c - a^2*c*x^2)^(5/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(-(x/a^6) - x^2/(2*a^5) + 1/(8*a^7*(1 - a*x)^2) - 5/(4* a^7*(1 - a*x)) - 1/(8*a^7*(1 + a*x)) - (39*Log[1 - a*x])/(16*a^7) - (9*Log [1 + a*x])/(16*a^7)))/(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 = 179, normalized size of antiderivative = 0.57
| method | result | size |
| default | \(-\frac {\left (8 a^{5} x^{5}+8 a^{4} x^{4}+9 \ln \left (a x +1\right ) x^{3} a^{3}+39 a^{3} \ln \left (a x -1\right ) x^{3}-24 a^{3} x^{3}-9 \ln \left (a x +1\right ) x^{2} a^{2}-39 a^{2} \ln \left (a x -1\right ) x^{2}-26 a^{2} x^{2}-9 \ln \left (a x +1\right ) x a -39 a \ln \left (a x -1\right ) x +10 a x +9 \ln \left (a x +1\right )+39 \ln \left (a x -1\right )+20\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{16 \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (a x -1\right )^{2} c^{3} a^{7}}\) | \(179\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^6/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVE RBOSE)
Output:
-1/16*(8*a^5*x^5+8*a^4*x^4+9*ln(a*x+1)*x^3*a^3+39*a^3*ln(a*x-1)*x^3-24*a^3 *x^3-9*ln(a*x+1)*x^2*a^2-39*a^2*ln(a*x-1)*x^2-26*a^2*x^2-9*ln(a*x+1)*x*a-3 9*a*ln(a*x-1)*x+10*a*x+9*ln(a*x+1)+39*ln(a*x-1)+20)/(-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^7
\[ \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{6}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^6/(-a^2*c*x^2+c)^(5/2),x, algorithm ="fricas")
Output:
integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^6/(a^7*c^3*x^7 - a^6*c^ 3*x^6 - 3*a^5*c^3*x^5 + 3*a^4*c^3*x^4 + 3*a^3*c^3*x^3 - 3*a^2*c^3*x^2 - a* c^3*x + c^3), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^{6} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**6/(-a**2*c*x**2+c)**(5/2),x)
Output:
Integral(x**6*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1 ))**(5/2)), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{6}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^6/(-a^2*c*x^2+c)^(5/2),x, algorithm ="maxima")
Output:
integrate((a*x + 1)*x^6/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)), x)
Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.38 \[ \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a \sqrt {c} x^{2} + 2 \, \sqrt {c} x}{2 \, a^{6} c^{3}} - \frac {9 \, \log \left ({\left | a x + 1 \right |}\right )}{8 \, a^{7} c^{\frac {5}{2}}} - \frac {39 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, a^{7} c^{\frac {5}{2}}} + \frac {9 \, a^{2} x^{2} + 3 \, a x - 10}{4 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} a^{7} c^{\frac {5}{2}}} - \frac {a^{5} c^{\frac {5}{2}} x^{2} + 2 \, a^{4} c^{\frac {5}{2}} x}{2 \, a^{10} c^{5}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^6/(-a^2*c*x^2+c)^(5/2),x, algorithm ="giac")
Output:
-1/2*(a*sqrt(c)*x^2 + 2*sqrt(c)*x)/(a^6*c^3) - 9/8*log(abs(a*x + 1))/(a^7* c^(5/2)) - 39/8*log(abs(a*x - 1))/(a^7*c^(5/2)) + 1/4*(9*a^2*x^2 + 3*a*x - 10)/((a*x + 1)*(a*x - 1)^2*a^7*c^(5/2)) - 1/2*(a^5*c^(5/2)*x^2 + 2*a^4*c^ (5/2)*x)/(a^10*c^5)
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^6\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((x^6*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((x^6*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.50 \[ \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-39 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+39 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+39 \,\mathrm {log}\left (a x -1\right ) a x -39 \,\mathrm {log}\left (a x -1\right )-9 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+9 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+9 \,\mathrm {log}\left (a x +1\right ) a x -9 \,\mathrm {log}\left (a x +1\right )-8 a^{5} x^{5}-8 a^{4} x^{4}+50 a^{3} x^{3}-36 a x +6\right )}{16 a^{7} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^6/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*( - 39*log(a*x - 1)*a**3*x**3 + 39*log(a*x - 1)*a**2*x**2 + 39*lo g(a*x - 1)*a*x - 39*log(a*x - 1) - 9*log(a*x + 1)*a**3*x**3 + 9*log(a*x + 1)*a**2*x**2 + 9*log(a*x + 1)*a*x - 9*log(a*x + 1) - 8*a**5*x**5 - 8*a**4* x**4 + 50*a**3*x**3 - 36*a*x + 6))/(16*a**7*c**3*(a**3*x**3 - a**2*x**2 - a*x + 1))