Integrand size = 27, antiderivative size = 186 \[ \int e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=-\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} x}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 \sqrt {c-\frac {c}{a^2 x^2}} x^2}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^3}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^4}{4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1+a x)}{a^4 \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
-4*(c-c/a^2/x^2)^(1/2)*x/a^3/(1-1/a^2/x^2)^(1/2)+2*(c-c/a^2/x^2)^(1/2)*x^2 /a^2/(1-1/a^2/x^2)^(1/2)-(c-c/a^2/x^2)^(1/2)*x^3/a/(1-1/a^2/x^2)^(1/2)+1/4 *(c-c/a^2/x^2)^(1/2)*x^4/(1-1/a^2/x^2)^(1/2)+4*(c-c/a^2/x^2)^(1/2)*ln(a*x+ 1)/a^4/(1-1/a^2/x^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.38 \[ \int e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {4 x}{a^3}+\frac {2 x^2}{a^2}-\frac {x^3}{a}+\frac {x^4}{4}+\frac {4 \log (1+a x)}{a^4}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}} \] Input:
Integrate[(Sqrt[c - c/(a^2*x^2)]*x^3)/E^(3*ArcCoth[a*x]),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*((-4*x)/a^3 + (2*x^2)/a^2 - x^3/a + x^4/4 + (4*Log[ 1 + a*x])/a^4))/Sqrt[1 - 1/(a^2*x^2)]
Time = 0.91 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.39, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6751, 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.
\(\displaystyle \int x^3 \sqrt {c-\frac {c}{a^2 x^2}} e^{-3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^3dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {x^2 (1-a x)^2}{a x+1}dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \left (a x^3-3 x^2+\frac {4 x}{a}+\frac {4}{a^2 (a x+1)}-\frac {4}{a^2}\right )dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {4 \log (a x+1)}{a^3}-\frac {4 x}{a^2}+\frac {a x^4}{4}+\frac {2 x^2}{a}-x^3\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(Sqrt[c - c/(a^2*x^2)]*x^3)/E^(3*ArcCoth[a*x]),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*((-4*x)/a^2 + (2*x^2)/a - x^3 + (a*x^4)/4 + (4*Log[ 1 + a*x])/a^3))/(a*Sqrt[1 - 1/(a^2*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^(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]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[c^IntPart[p]*((c + d/x^2)^FracPart[p]/(1 - 1/(a^2*x^2))^FracPart [p]) Int[u*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !IntegerQ[n/2] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {\left (a^{4} x^{4}-4 a^{3} x^{3}+8 a^{2} x^{2}-16 a x +16 \ln \left (a x +1\right )\right ) x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{4 a^{3} \left (a x -1\right )^{2}}\) | \(89\) |
Input:
int((c-c/a^2/x^2)^(1/2)*x^3*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOS E)
Output:
1/4*(a^4*x^4-4*a^3*x^3+8*a^2*x^2-16*a*x+16*ln(a*x+1))*x*(c*(a^2*x^2-1)/a^2 /x^2)^(1/2)*(a*x+1)*((a*x-1)/(a*x+1))^(3/2)/a^3/(a*x-1)^2
Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.26 \[ \int e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 16 \, a x + 16 \, \log \left (a x + 1\right )\right )} \sqrt {a^{2} c}}{4 \, a^{5}} \] Input:
integrate((c-c/a^2/x^2)^(1/2)*x^3*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fr icas")
Output:
1/4*(a^4*x^4 - 4*a^3*x^3 + 8*a^2*x^2 - 16*a*x + 16*log(a*x + 1))*sqrt(a^2* c)/a^5
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(1/2)*x**3*((a*x-1)/(a*x+1))**(3/2),x)
Output:
Timed out
\[ \int e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\int { \sqrt {c - \frac {c}{a^{2} x^{2}}} x^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(1/2)*x^3*((a*x-1)/(a*x+1))^(3/2),x, algorithm="ma xima")
Output:
integrate(sqrt(c - c/(a^2*x^2))*x^3*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.51 \[ \int e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {1}{4} \, \sqrt {c} {\left (\frac {16 \, \log \left ({\left | a x + 1 \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{5}} + \frac {a^{11} x^{4} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 4 \, a^{10} x^{3} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + 8 \, a^{9} x^{2} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 16 \, a^{8} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{12}}\right )} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(1/2)*x^3*((a*x-1)/(a*x+1))^(3/2),x, algorithm="gi ac")
Output:
1/4*sqrt(c)*(16*log(abs(a*x + 1))*sgn(a*x + 1)*sgn(x)/a^5 + (a^11*x^4*sgn( a*x + 1)*sgn(x) - 4*a^10*x^3*sgn(a*x + 1)*sgn(x) + 8*a^9*x^2*sgn(a*x + 1)* sgn(x) - 16*a^8*x*sgn(a*x + 1)*sgn(x))/a^12)*abs(a)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\int x^3\,\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \] Input:
int(x^3*(c - c/(a^2*x^2))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int(x^3*(c - c/(a^2*x^2))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.24 \[ \int e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {\sqrt {c}\, \left (16 \,\mathrm {log}\left (a x +1\right )+a^{4} x^{4}-4 a^{3} x^{3}+8 a^{2} x^{2}-16 a x +11\right )}{4 a^{4}} \] Input:
int((c-c/a^2/x^2)^(1/2)*x^3*((a*x-1)/(a*x+1))^(3/2),x)
Output:
(sqrt(c)*(16*log(a*x + 1) + a**4*x**4 - 4*a**3*x**3 + 8*a**2*x**2 - 16*a*x + 11))/(4*a**4)