Integrand size = 43, antiderivative size = 134 \[ \int \frac {2+3 x^2+5 x^4}{x^5 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}}{4 x^4}+\frac {63 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}}{16 x^2}-\frac {1223 \sqrt {-1+9 x^2} \arctan \left (\frac {\sqrt {2+x^2}}{\sqrt {2} \sqrt {-1+9 x^2}}\right )}{16 \sqrt {2} \sqrt {-1+3 x} \sqrt {1+3 x}} \] Output:
1/4*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2)^(1/2)/x^4+63/16*(-1+3*x)^(1/2)*(1 +3*x)^(1/2)*(x^2+2)^(1/2)/x^2-1223/32*2^(1/2)*(9*x^2-1)^(1/2)*arctan(1/2*( x^2+2)^(1/2)*2^(1/2)/(9*x^2-1)^(1/2))/(-1+3*x)^(1/2)/(1+3*x)^(1/2)
Time = 10.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.65 \[ \int \frac {2+3 x^2+5 x^4}{x^5 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {2 \sqrt {2+x^2} \left (-4-27 x^2+567 x^4\right )-1223 x^4 \sqrt {-2+18 x^2} \arctan \left (\frac {\sqrt {2+x^2}}{\sqrt {-2+18 x^2}}\right )}{32 x^4 \sqrt {-1+3 x} \sqrt {1+3 x}} \] Input:
Integrate[(2 + 3*x^2 + 5*x^4)/(x^5*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x]*Sqrt[2 + x ^2]),x]
Output:
(2*Sqrt[2 + x^2]*(-4 - 27*x^2 + 567*x^4) - 1223*x^4*Sqrt[-2 + 18*x^2]*ArcT an[Sqrt[2 + x^2]/Sqrt[-2 + 18*x^2]])/(32*x^4*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x])
Time = 1.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2038, 7293, 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.
\(\displaystyle \int \frac {5 x^4+3 x^2+2}{x^5 \sqrt {3 x-1} \sqrt {3 x+1} \sqrt {x^2+2}} \, dx\) |
\(\Big \downarrow \) 2038 |
\(\displaystyle \frac {\sqrt {9 x^2-1} \int \frac {5 x^4+3 x^2+2}{x^5 \sqrt {x^2+2} \sqrt {9 x^2-1}}dx}{\sqrt {3 x-1} \sqrt {3 x+1}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\sqrt {9 x^2-1} \int \left (\frac {5}{x \sqrt {x^2+2} \sqrt {9 x^2-1}}+\frac {3}{x^3 \sqrt {x^2+2} \sqrt {9 x^2-1}}+\frac {2}{x^5 \sqrt {x^2+2} \sqrt {9 x^2-1}}\right )dx}{\sqrt {3 x-1} \sqrt {3 x+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {9 x^2-1} \left (-\frac {1223 \arctan \left (\frac {\sqrt {x^2+2}}{\sqrt {2} \sqrt {9 x^2-1}}\right )}{16 \sqrt {2}}+\frac {63 \sqrt {x^2+2} \sqrt {9 x^2-1}}{16 x^2}+\frac {\sqrt {x^2+2} \sqrt {9 x^2-1}}{4 x^4}\right )}{\sqrt {3 x-1} \sqrt {3 x+1}}\) |
Input:
Int[(2 + 3*x^2 + 5*x^4)/(x^5*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x]*Sqrt[2 + x^2]),x ]
Output:
(Sqrt[-1 + 9*x^2]*((Sqrt[2 + x^2]*Sqrt[-1 + 9*x^2])/(4*x^4) + (63*Sqrt[2 + x^2]*Sqrt[-1 + 9*x^2])/(16*x^2) - (1223*ArcTan[Sqrt[2 + x^2]/(Sqrt[2]*Sqr t[-1 + 9*x^2])])/(16*Sqrt[2])))/(Sqrt[-1 + 3*x]*Sqrt[1 + 3*x])
Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p _)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_), x_Symbol] :> Simp[(a1 + b1*x^(n/2)) ^FracPart[p]*((a2 + b2*x^(n/2))^FracPart[p]/(a1*a2 + b1*b2*x^n)^FracPart[p] ) Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x] && EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && !(Eq Q[n, 2] && IGtQ[q, 0])
Time = 0.87 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}\, \left (63 x^{2}+4\right )}{16 x^{4}}+\frac {1223 \sqrt {2}\, \arctan \left (\frac {\left (17 x^{2}-4\right ) \sqrt {2}}{4 \sqrt {9 x^{4}+17 x^{2}-2}}\right ) \sqrt {\left (-1+3 x \right ) \left (1+3 x \right ) \left (x^{2}+2\right )}}{64 \sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\) | \(106\) |
elliptic | \(\frac {\sqrt {\left (x^{2}+2\right ) \left (9 x^{2}-1\right )}\, \left (\frac {\sqrt {9 x^{4}+17 x^{2}-2}}{4 x^{4}}+\frac {63 \sqrt {9 x^{4}+17 x^{2}-2}}{16 x^{2}}+\frac {1223 \sqrt {2}\, \arctan \left (\frac {\left (17 x^{2}-4\right ) \sqrt {2}}{4 \sqrt {9 x^{4}+17 x^{2}-2}}\right )}{64}\right )}{\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\) | \(109\) |
default | \(\frac {\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}\, \left (1223 \sqrt {2}\, \arctan \left (\frac {\left (17 x^{2}-4\right ) \sqrt {2}}{4 \sqrt {9 x^{4}+17 x^{2}-2}}\right ) x^{4}-9312 \ln \left (\frac {17}{6}+3 x^{2}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) x^{4}+9312 \ln \left (6 x^{2}+\frac {17}{3}+2 \sqrt {9 x^{4}+17 x^{2}-2}\right ) x^{4}+252 \sqrt {9 x^{4}+17 x^{2}-2}\, x^{2}+16 \sqrt {9 x^{4}+17 x^{2}-2}\right )}{64 x^{4} \sqrt {9 x^{4}+17 x^{2}-2}}\) | \(168\) |
Input:
int((5*x^4+3*x^2+2)/x^5/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2),x,metho d=_RETURNVERBOSE)
Output:
1/16*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2)^(1/2)*(63*x^2+4)/x^4+1223/64*2^( 1/2)*arctan(1/4*(17*x^2-4)*2^(1/2)/(9*x^4+17*x^2-2)^(1/2))*((-1+3*x)*(1+3* x)*(x^2+2))^(1/2)/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.63 \[ \int \frac {2+3 x^2+5 x^4}{x^5 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {1223 \, \sqrt {2} x^{4} \arctan \left (-\frac {3}{2} \, \sqrt {2} x^{2} + \frac {1}{2} \, \sqrt {2} \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1}\right ) + 378 \, x^{4} + 2 \, {\left (63 \, x^{2} + 4\right )} \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1}}{32 \, x^{4}} \] Input:
integrate((5*x^4+3*x^2+2)/x^5/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2),x , algorithm="fricas")
Output:
1/32*(1223*sqrt(2)*x^4*arctan(-3/2*sqrt(2)*x^2 + 1/2*sqrt(2)*sqrt(x^2 + 2) *sqrt(3*x + 1)*sqrt(3*x - 1)) + 378*x^4 + 2*(63*x^2 + 4)*sqrt(x^2 + 2)*sqr t(3*x + 1)*sqrt(3*x - 1))/x^4
Timed out. \[ \int \frac {2+3 x^2+5 x^4}{x^5 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\text {Timed out} \] Input:
integrate((5*x**4+3*x**2+2)/x**5/(-1+3*x)**(1/2)/(1+3*x)**(1/2)/(x**2+2)** (1/2),x)
Output:
Timed out
\[ \int \frac {2+3 x^2+5 x^4}{x^5 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\int { \frac {5 \, x^{4} + 3 \, x^{2} + 2}{\sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} x^{5}} \,d x } \] Input:
integrate((5*x^4+3*x^2+2)/x^5/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2),x , algorithm="maxima")
Output:
integrate((5*x^4 + 3*x^2 + 2)/(sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1)*x ^5), x)
\[ \int \frac {2+3 x^2+5 x^4}{x^5 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\int { \frac {5 \, x^{4} + 3 \, x^{2} + 2}{\sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} x^{5}} \,d x } \] Input:
integrate((5*x^4+3*x^2+2)/x^5/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2),x , algorithm="giac")
Output:
integrate((5*x^4 + 3*x^2 + 2)/(sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1)*x ^5), x)
Timed out. \[ \int \frac {2+3 x^2+5 x^4}{x^5 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\int \frac {5\,x^4+3\,x^2+2}{x^5\,\sqrt {3\,x-1}\,\sqrt {3\,x+1}\,\sqrt {x^2+2}} \,d x \] Input:
int((3*x^2 + 5*x^4 + 2)/(x^5*(3*x - 1)^(1/2)*(3*x + 1)^(1/2)*(x^2 + 2)^(1/ 2)),x)
Output:
int((3*x^2 + 5*x^4 + 2)/(x^5*(3*x - 1)^(1/2)*(3*x + 1)^(1/2)*(x^2 + 2)^(1/ 2)), x)
Time = 1.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.65 \[ \int \frac {2+3 x^2+5 x^4}{x^5 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {-1223 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}\, \sqrt {2}}{18 x^{2}-2}\right ) x^{4}+126 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}\, x^{2}+8 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}}{32 x^{4}} \] Input:
int((5*x^4+3*x^2+2)/x^5/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2),x)
Output:
( - 1223*sqrt(2)*atan((sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2)*sqrt(2)) /(18*x**2 - 2))*x**4 + 126*sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2)*x**2 + 8*sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2))/(32*x**4)