Integrand size = 24, antiderivative size = 124 \[ \int \frac {x^2}{\left (2+3 x^2\right )^{3/4} \left (4+3 x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt {3} x \sqrt [4]{2+3 x^2}}{2+\sqrt {2} \sqrt {2+3 x^2}}\right )}{3 \sqrt [4]{2} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {2\ 2^{3/4}-2 \sqrt [4]{2} \sqrt {2+3 x^2}}{2 \sqrt {3} x \sqrt [4]{2+3 x^2}}\right )}{3 \sqrt [4]{2} \sqrt {3}} \] Output:
1/18*arctan(2^(1/4)*3^(1/2)*x*(3*x^2+2)^(1/4)/(2+2^(1/2)*(3*x^2+2)^(1/2))) *2^(3/4)*3^(1/2)+1/18*arctanh(1/6*(2*2^(3/4)-2*2^(1/4)*(3*x^2+2)^(1/2))*3^ (1/2)/x/(3*x^2+2)^(1/4))*2^(3/4)*3^(1/2)
Time = 1.76 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\left (2+3 x^2\right )^{3/4} \left (4+3 x^2\right )} \, dx=-\frac {\arctan \left (\frac {-3 \sqrt {2} x^2+4 \sqrt {2+3 x^2}}{2\ 2^{3/4} \sqrt {3} x \sqrt [4]{2+3 x^2}}\right )+\text {arctanh}\left (\frac {2 \sqrt {3} x \sqrt [4]{4+6 x^2}}{3 x^2+2 \sqrt {4+6 x^2}}\right )}{6 \sqrt [4]{2} \sqrt {3}} \] Input:
Integrate[x^2/((2 + 3*x^2)^(3/4)*(4 + 3*x^2)),x]
Output:
-1/6*(ArcTan[(-3*Sqrt[2]*x^2 + 4*Sqrt[2 + 3*x^2])/(2*2^(3/4)*Sqrt[3]*x*(2 + 3*x^2)^(1/4))] + ArcTanh[(2*Sqrt[3]*x*(4 + 6*x^2)^(1/4))/(3*x^2 + 2*Sqrt [4 + 6*x^2])])/(2^(1/4)*Sqrt[3])
Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.04, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {350}
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.
\(\displaystyle \int \frac {x^2}{\left (3 x^2+2\right )^{3/4} \left (3 x^2+4\right )} \, dx\) |
\(\Big \downarrow \) 350 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {2\ 2^{3/4}-2 \sqrt [4]{2} \sqrt {3 x^2+2}}{2 \sqrt {3} x \sqrt [4]{3 x^2+2}}\right )}{3 \sqrt [4]{2} \sqrt {3}}-\frac {\arctan \left (\frac {2 \sqrt [4]{2} \sqrt {3 x^2+2}+2\ 2^{3/4}}{2 \sqrt {3} x \sqrt [4]{3 x^2+2}}\right )}{3 \sqrt [4]{2} \sqrt {3}}\) |
Input:
Int[x^2/((2 + 3*x^2)^(3/4)*(4 + 3*x^2)),x]
Output:
-1/3*ArcTan[(2*2^(3/4) + 2*2^(1/4)*Sqrt[2 + 3*x^2])/(2*Sqrt[3]*x*(2 + 3*x^ 2)^(1/4))]/(2^(1/4)*Sqrt[3]) + ArcTanh[(2*2^(3/4) - 2*2^(1/4)*Sqrt[2 + 3*x ^2])/(2*Sqrt[3]*x*(2 + 3*x^2)^(1/4))]/(3*2^(1/4)*Sqrt[3])
Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] : > Simp[(-b/(a*d*Rt[b^2/a, 4]^3))*ArcTan[(b + Rt[b^2/a, 4]^2*Sqrt[a + b*x^2] )/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))], x] + Simp[(b/(a*d*Rt[b^2/a, 4]^3)) *ArcTanh[(b - Rt[b^2/a, 4]^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2) ^(1/4))], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[b^2/a ]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.13 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.49
method | result | size |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right )^{2}\right ) \ln \left (\frac {\left (3 x^{2}+2\right )^{\frac {3}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right )^{2}\right )+9 \sqrt {3 x^{2}+2}\, x +3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right )^{2} x -6 \left (3 x^{2}+2\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right )^{2}\right )}{3 x^{2}+4}\right )}{18}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right ) \ln \left (\frac {\left (3 x^{2}+2\right )^{\frac {3}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right )^{3}+9 \sqrt {3 x^{2}+2}\, x -3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right )^{2} x +6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+18\right ) \left (3 x^{2}+2\right )^{\frac {1}{4}}}{3 x^{2}+4}\right )}{18}\) | \(185\) |
Input:
int(x^2/(3*x^2+2)^(3/4)/(3*x^2+4),x,method=_RETURNVERBOSE)
Output:
-1/18*RootOf(_Z^2+RootOf(_Z^4+18)^2)*ln(((3*x^2+2)^(3/4)*RootOf(_Z^4+18)^2 *RootOf(_Z^2+RootOf(_Z^4+18)^2)+9*(3*x^2+2)^(1/2)*x+3*RootOf(_Z^4+18)^2*x- 6*(3*x^2+2)^(1/4)*RootOf(_Z^2+RootOf(_Z^4+18)^2))/(3*x^2+4))+1/18*RootOf(_ Z^4+18)*ln(((3*x^2+2)^(3/4)*RootOf(_Z^4+18)^3+9*(3*x^2+2)^(1/2)*x-3*RootOf (_Z^4+18)^2*x+6*RootOf(_Z^4+18)*(3*x^2+2)^(1/4))/(3*x^2+4))
Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.20 \[ \int \frac {x^2}{\left (2+3 x^2\right )^{3/4} \left (4+3 x^2\right )} \, dx=-\frac {1}{108} \cdot 18^{\frac {3}{4}} \arctan \left (\frac {3 \, x + 2 \cdot 18^{\frac {1}{4}} {\left (3 \, x^{2} + 2\right )}^{\frac {1}{4}}}{3 \, x}\right ) - \frac {1}{108} \cdot 18^{\frac {3}{4}} \arctan \left (-\frac {3 \, x - 2 \cdot 18^{\frac {1}{4}} {\left (3 \, x^{2} + 2\right )}^{\frac {1}{4}}}{3 \, x}\right ) - \frac {1}{216} \cdot 18^{\frac {3}{4}} \log \left (\frac {9 \, \sqrt {2} x^{2} + 2 \cdot 18^{\frac {3}{4}} {\left (3 \, x^{2} + 2\right )}^{\frac {1}{4}} x + 12 \, \sqrt {3 \, x^{2} + 2}}{x^{2}}\right ) + \frac {1}{216} \cdot 18^{\frac {3}{4}} \log \left (\frac {9 \, \sqrt {2} x^{2} - 2 \cdot 18^{\frac {3}{4}} {\left (3 \, x^{2} + 2\right )}^{\frac {1}{4}} x + 12 \, \sqrt {3 \, x^{2} + 2}}{x^{2}}\right ) \] Input:
integrate(x^2/(3*x^2+2)^(3/4)/(3*x^2+4),x, algorithm="fricas")
Output:
-1/108*18^(3/4)*arctan(1/3*(3*x + 2*18^(1/4)*(3*x^2 + 2)^(1/4))/x) - 1/108 *18^(3/4)*arctan(-1/3*(3*x - 2*18^(1/4)*(3*x^2 + 2)^(1/4))/x) - 1/216*18^( 3/4)*log((9*sqrt(2)*x^2 + 2*18^(3/4)*(3*x^2 + 2)^(1/4)*x + 12*sqrt(3*x^2 + 2))/x^2) + 1/216*18^(3/4)*log((9*sqrt(2)*x^2 - 2*18^(3/4)*(3*x^2 + 2)^(1/ 4)*x + 12*sqrt(3*x^2 + 2))/x^2)
\[ \int \frac {x^2}{\left (2+3 x^2\right )^{3/4} \left (4+3 x^2\right )} \, dx=\int \frac {x^{2}}{\left (3 x^{2} + 2\right )^{\frac {3}{4}} \cdot \left (3 x^{2} + 4\right )}\, dx \] Input:
integrate(x**2/(3*x**2+2)**(3/4)/(3*x**2+4),x)
Output:
Integral(x**2/((3*x**2 + 2)**(3/4)*(3*x**2 + 4)), x)
\[ \int \frac {x^2}{\left (2+3 x^2\right )^{3/4} \left (4+3 x^2\right )} \, dx=\int { \frac {x^{2}}{{\left (3 \, x^{2} + 4\right )} {\left (3 \, x^{2} + 2\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(x^2/(3*x^2+2)^(3/4)/(3*x^2+4),x, algorithm="maxima")
Output:
integrate(x^2/((3*x^2 + 4)*(3*x^2 + 2)^(3/4)), x)
\[ \int \frac {x^2}{\left (2+3 x^2\right )^{3/4} \left (4+3 x^2\right )} \, dx=\int { \frac {x^{2}}{{\left (3 \, x^{2} + 4\right )} {\left (3 \, x^{2} + 2\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(x^2/(3*x^2+2)^(3/4)/(3*x^2+4),x, algorithm="giac")
Output:
integrate(x^2/((3*x^2 + 4)*(3*x^2 + 2)^(3/4)), x)
Timed out. \[ \int \frac {x^2}{\left (2+3 x^2\right )^{3/4} \left (4+3 x^2\right )} \, dx=\int \frac {x^2}{{\left (3\,x^2+2\right )}^{3/4}\,\left (3\,x^2+4\right )} \,d x \] Input:
int(x^2/((3*x^2 + 2)^(3/4)*(3*x^2 + 4)),x)
Output:
int(x^2/((3*x^2 + 2)^(3/4)*(3*x^2 + 4)), x)
\[ \int \frac {x^2}{\left (2+3 x^2\right )^{3/4} \left (4+3 x^2\right )} \, dx=\int \frac {x^{2}}{3 \left (3 x^{2}+2\right )^{\frac {3}{4}} x^{2}+4 \left (3 x^{2}+2\right )^{\frac {3}{4}}}d x \] Input:
int(x^2/(3*x^2+2)^(3/4)/(3*x^2+4),x)
Output:
int(x**2/(3*(3*x**2 + 2)**(3/4)*x**2 + 4*(3*x**2 + 2)**(3/4)),x)