Integrand size = 27, antiderivative size = 227 \[ \int e^{-3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {4 \sqrt {c-a^2 c x^2}}{a^4 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2 x \sqrt {c-a^2 c x^2}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 x^2 \sqrt {c-a^2 c x^2}}{3 a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 x^3 \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^4 \sqrt {c-a^2 c x^2}}{5 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 \sqrt {c-a^2 c x^2} \log (1+a x)}{a^5 \sqrt {1-\frac {1}{a^2 x^2}} x} \] Output:
4*(-a^2*c*x^2+c)^(1/2)/a^4/(1-1/a^2/x^2)^(1/2)-2*x*(-a^2*c*x^2+c)^(1/2)/a^ 3/(1-1/a^2/x^2)^(1/2)+4/3*x^2*(-a^2*c*x^2+c)^(1/2)/a^2/(1-1/a^2/x^2)^(1/2) -3/4*x^3*(-a^2*c*x^2+c)^(1/2)/a/(1-1/a^2/x^2)^(1/2)+1/5*x^4*(-a^2*c*x^2+c) ^(1/2)/(1-1/a^2/x^2)^(1/2)-4*(-a^2*c*x^2+c)^(1/2)*ln(a*x+1)/a^5/(1-1/a^2/x ^2)^(1/2)/x
Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.38 \[ \int e^{-3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (\frac {4 x}{a^4}-\frac {2 x^2}{a^3}+\frac {4 x^3}{3 a^2}-\frac {3 x^4}{4 a}+\frac {x^5}{5}-\frac {4 \log (1+a x)}{a^5}\right )}{\sqrt {1-\frac {1}{a^2 x^2}} x} \] Input:
Integrate[(x^3*Sqrt[c - a^2*c*x^2])/E^(3*ArcCoth[a*x]),x]
Output:
(Sqrt[c - a^2*c*x^2]*((4*x)/a^4 - (2*x^2)/a^3 + (4*x^3)/(3*a^2) - (3*x^4)/ (4*a) + x^5/5 - (4*Log[1 + a*x])/a^5))/(Sqrt[1 - 1/(a^2*x^2)]*x)
Time = 0.87 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.38, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, 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.
\(\displaystyle \int x^3 \sqrt {c-a^2 c x^2} e^{-3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^4dx}{x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {x^3 (1-a x)^2}{a x+1}dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \left (a x^4-3 x^3+\frac {4 x^2}{a}-\frac {4 x}{a^2}-\frac {4}{a^3 (a x+1)}+\frac {4}{a^3}\right )dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (-\frac {4 \log (a x+1)}{a^4}+\frac {4 x}{a^3}-\frac {2 x^2}{a^2}+\frac {a x^5}{5}+\frac {4 x^3}{3 a}-\frac {3 x^4}{4}\right )}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(x^3*Sqrt[c - a^2*c*x^2])/E^(3*ArcCoth[a*x]),x]
Output:
(Sqrt[c - a^2*c*x^2]*((4*x)/a^3 - (2*x^2)/a^2 + (4*x^3)/(3*a) - (3*x^4)/4 + (a*x^5)/5 - (4*Log[1 + a*x])/a^4))/(a*Sqrt[1 - 1/(a^2*x^2)]*x)
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 = 92, normalized size of antiderivative = 0.41
method | result | size |
default | \(-\frac {\left (-12 a^{5} x^{5}+45 a^{4} x^{4}-80 a^{3} x^{3}+120 a^{2} x^{2}-240 a x +240 \ln \left (a x +1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{60 a^{4} \left (a x -1\right )^{2}}\) | \(92\) |
Input:
int(x^3*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBO SE)
Output:
-1/60*(-12*a^5*x^5+45*a^4*x^4-80*a^3*x^3+120*a^2*x^2-240*a*x+240*ln(a*x+1) )*(-c*(a^2*x^2-1))^(1/2)*(a*x+1)*((a*x-1)/(a*x+1))^(3/2)/a^4/(a*x-1)^2
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.26 \[ \int e^{-3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {{\left (12 \, a^{5} x^{5} - 45 \, a^{4} x^{4} + 80 \, a^{3} x^{3} - 120 \, a^{2} x^{2} + 240 \, a x - 240 \, \log \left (a x + 1\right )\right )} \sqrt {-a^{2} c}}{60 \, a^{5}} \] Input:
integrate(x^3*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="f ricas")
Output:
1/60*(12*a^5*x^5 - 45*a^4*x^4 + 80*a^3*x^3 - 120*a^2*x^2 + 240*a*x - 240*l og(a*x + 1))*sqrt(-a^2*c)/a^5
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\text {Timed out} \] Input:
integrate(x**3*(-a**2*c*x**2+c)**(1/2)*((a*x-1)/(a*x+1))**(3/2),x)
Output:
Timed out
\[ \int e^{-3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\int { \sqrt {-a^{2} c x^{2} + c} x^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(x^3*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="m axima")
Output:
integrate(sqrt(-a^2*c*x^2 + c)*x^3*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.44 \[ \int e^{-3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=-\frac {1}{60} \, \sqrt {-c} {\left (\frac {240 \, \log \left ({\left | a x + 1 \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{a^{4}} - \frac {12 \, a^{6} x^{5} \mathrm {sgn}\left (a x + 1\right ) - 45 \, a^{5} x^{4} \mathrm {sgn}\left (a x + 1\right ) + 80 \, a^{4} x^{3} \mathrm {sgn}\left (a x + 1\right ) - 120 \, a^{3} x^{2} \mathrm {sgn}\left (a x + 1\right ) + 240 \, a^{2} x \mathrm {sgn}\left (a x + 1\right )}{a^{5}}\right )} \] Input:
integrate(x^3*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="g iac")
Output:
-1/60*sqrt(-c)*(240*log(abs(a*x + 1))*sgn(a*x + 1)/a^4 - (12*a^6*x^5*sgn(a *x + 1) - 45*a^5*x^4*sgn(a*x + 1) + 80*a^4*x^3*sgn(a*x + 1) - 120*a^3*x^2* sgn(a*x + 1) + 240*a^2*x*sgn(a*x + 1))/a^5)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\int x^3\,\sqrt {c-a^2\,c\,x^2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \] Input:
int(x^3*(c - a^2*c*x^2)^(1/2)*((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int(x^3*(c - a^2*c*x^2)^(1/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.24 \[ \int e^{-3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c}\, i \left (240 \,\mathrm {log}\left (a x +1\right )-12 a^{5} x^{5}+45 a^{4} x^{4}-80 a^{3} x^{3}+120 a^{2} x^{2}-240 a x +167\right )}{60 a^{4}} \] Input:
int(x^3*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x)
Output:
(sqrt(c)*i*(240*log(a*x + 1) - 12*a**5*x**5 + 45*a**4*x**4 - 80*a**3*x**3 + 120*a**2*x**2 - 240*a*x + 167))/(60*a**4)