Integrand size = 25, antiderivative size = 113 \[ \int e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx=-\frac {3 \sqrt {c-\frac {c}{a^2 x^2}} x^2}{\sqrt {1-a^2 x^2}}+\frac {a \sqrt {c-\frac {c}{a^2 x^2}} x^3}{2 \sqrt {1-a^2 x^2}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} x \log (1+a x)}{a \sqrt {1-a^2 x^2}} \] Output:
-3*(c-c/a^2/x^2)^(1/2)*x^2/(-a^2*x^2+1)^(1/2)+1/2*a*(c-c/a^2/x^2)^(1/2)*x^ 3/(-a^2*x^2+1)^(1/2)+4*(c-c/a^2/x^2)^(1/2)*x*ln(a*x+1)/a/(-a^2*x^2+1)^(1/2 )
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.48 \[ \int e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left (-3 x+\frac {a x^2}{2}+\frac {4 \log (1+a x)}{a}\right )}{\sqrt {1-a^2 x^2}} \] Input:
Integrate[(Sqrt[c - c/(a^2*x^2)]*x)/E^(3*ArcTanh[a*x]),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*x*(-3*x + (a*x^2)/2 + (4*Log[1 + a*x])/a))/Sqrt[1 - a^2*x^2]
Time = 0.66 (sec) , antiderivative size = 54, 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 = {6710, 6690, 49, 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.
\(\displaystyle \int x e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} \, dx\) |
\(\Big \downarrow \) 6710 |
\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int e^{-3 \text {arctanh}(a x)} \sqrt {1-a^2 x^2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^2}{a x+1}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \left (a x+\frac {4}{a x+1}-3\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {a x^2}{2}+\frac {4 \log (a x+1)}{a}-3 x\right )}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[(Sqrt[c - c/(a^2*x^2)]*x)/E^(3*ArcTanh[a*x]),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*x*(-3*x + (a*x^2)/2 + (4*Log[1 + a*x])/a))/Sqrt[1 - a^2*x^2]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p Int[(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a , c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[x^(2*p)*((c + d/x^2)^p/(1 - a^2*x^2)^p) Int[(u/x^(2*p))*(1 - a ^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !IntegerQ[p] && !IntegerQ[n/2]
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.51
method | result | size |
default | \(\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \left (a^{2} x^{2}-6 a x +8 \ln \left (a x +1\right )\right )}{2 \sqrt {-a^{2} x^{2}+1}\, a}\) | \(58\) |
Input:
int((c-c/a^2/x^2)^(1/2)*x/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x,method=_RETURNVER BOSE)
Output:
1/2*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x*(a^2*x^2-6*a*x+8*ln(a*x+1))/(-a^2*x^2+ 1)^(1/2)/a
Time = 0.14 (sec) , antiderivative size = 387, normalized size of antiderivative = 3.42 \[ \int e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx=\left [\frac {4 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \log \left (\frac {a^{6} c x^{6} + 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} - 4 \, a c x + {\left (a^{5} x^{5} + 4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} + 4 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 2 \, c}{a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1}\right ) - {\left (a^{3} x^{3} - 6 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{2} - a^{2}\right )}}, -\frac {8 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {c} \arctan \left (\frac {{\left (a^{2} x^{2} + 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{3} c x^{3} + 2 \, a^{2} c x^{2} - a c x - 2 \, c}\right ) + {\left (a^{3} x^{3} - 6 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{2} - a^{2}\right )}}\right ] \] Input:
integrate((c-c/a^2/x^2)^(1/2)*x/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm= "fricas")
Output:
[1/2*(4*(a^2*x^2 - 1)*sqrt(-c)*log((a^6*c*x^6 + 4*a^5*c*x^5 + 5*a^4*c*x^4 - 4*a^2*c*x^2 - 4*a*c*x + (a^5*x^5 + 4*a^4*x^4 + 6*a^3*x^3 + 4*a^2*x^2)*sq rt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - 2*c)/(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)) - (a^3*x^3 - 6*a^2*x^2)*sqrt(-a^2*x^2 + 1)*sqrt( (a^2*c*x^2 - c)/(a^2*x^2)))/(a^4*x^2 - a^2), -1/2*(8*(a^2*x^2 - 1)*sqrt(c) *arctan((a^2*x^2 + 2*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^3*c*x^3 + 2*a^2*c*x^2 - a*c*x - 2*c)) + (a^3*x^3 - 6*a^2 *x^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^4*x^2 - a^2)]
\[ \int e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx=\int \frac {x \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )}}{\left (a x + 1\right )^{3}}\, dx \] Input:
integrate((c-c/a**2/x**2)**(1/2)*x/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)
Output:
Integral(x*(-(a*x - 1)*(a*x + 1))**(3/2)*sqrt(-c*(-1 + 1/(a*x))*(1 + 1/(a* x)))/(a*x + 1)**3, x)
\[ \int e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {c - \frac {c}{a^{2} x^{2}}} x}{{\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(1/2)*x/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm= "maxima")
Output:
integrate((-a^2*x^2 + 1)^(3/2)*sqrt(c - c/(a^2*x^2))*x/(a*x + 1)^3, x)
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.41 \[ \int e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx=\frac {1}{2} \, \sqrt {-c} {\left (\frac {8 \, \log \left ({\left | a x + 1 \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{3}} + \frac {a^{5} x^{2} \mathrm {sgn}\left (x\right ) - 6 \, a^{4} x \mathrm {sgn}\left (x\right )}{a^{6}}\right )} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(1/2)*x/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm= "giac")
Output:
1/2*sqrt(-c)*(8*log(abs(a*x + 1))*sgn(x)/a^3 + (a^5*x^2*sgn(x) - 6*a^4*x*s gn(x))/a^6)*abs(a)
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx=\int \frac {x\,\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \] Input:
int((x*(c - c/(a^2*x^2))^(1/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3,x)
Output:
int((x*(c - c/(a^2*x^2))^(1/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3, x)
Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.26 \[ \int e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx=\frac {\sqrt {c}\, i \left (-8 \,\mathrm {log}\left (a x +1\right )-a^{2} x^{2}+6 a x \right )}{2 a^{2}} \] Input:
int((c-c/a^2/x^2)^(1/2)*x/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
Output:
(sqrt(c)*i*( - 8*log(a*x + 1) - a**2*x**2 + 6*a*x))/(2*a**2)