\(\int \frac {x^2}{(a+3 x^2)^{3/4} (2 a+3 x^2)} \, dx\) [1057]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 26, antiderivative size = 120 \[ \int \frac {x^2}{\left (a+3 x^2\right )^{3/4} \left (2 a+3 x^2\right )} \, dx=-\frac {\arctan \left (\frac {a^{3/4} \left (1+\frac {\sqrt {a+3 x^2}}{\sqrt {a}}\right )}{\sqrt {3} x \sqrt [4]{a+3 x^2}}\right )}{3 \sqrt {3} \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {a^{3/4} \left (1-\frac {\sqrt {a+3 x^2}}{\sqrt {a}}\right )}{\sqrt {3} x \sqrt [4]{a+3 x^2}}\right )}{3 \sqrt {3} \sqrt [4]{a}} \]

[Out]

-1/9*arctan(1/3*a^(3/4)*(1+(3*x^2+a)^(1/2)/a^(1/2))/x/(3*x^2+a)^(1/4)*3^(1/2))/a^(1/4)*3^(1/2)+1/9*arctanh(1/3
*a^(3/4)*(1-(3*x^2+a)^(1/2)/a^(1/2))/x/(3*x^2+a)^(1/4)*3^(1/2))/a^(1/4)*3^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {452} \[ \int \frac {x^2}{\left (a+3 x^2\right )^{3/4} \left (2 a+3 x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {a^{3/4} \left (1-\frac {\sqrt {a+3 x^2}}{\sqrt {a}}\right )}{\sqrt {3} x \sqrt [4]{a+3 x^2}}\right )}{3 \sqrt {3} \sqrt [4]{a}}-\frac {\arctan \left (\frac {a^{3/4} \left (\frac {\sqrt {a+3 x^2}}{\sqrt {a}}+1\right )}{\sqrt {3} x \sqrt [4]{a+3 x^2}}\right )}{3 \sqrt {3} \sqrt [4]{a}} \]

[In]

Int[x^2/((a + 3*x^2)^(3/4)*(2*a + 3*x^2)),x]

[Out]

-1/3*ArcTan[(a^(3/4)*(1 + Sqrt[a + 3*x^2]/Sqrt[a]))/(Sqrt[3]*x*(a + 3*x^2)^(1/4))]/(Sqrt[3]*a^(1/4)) + ArcTanh
[(a^(3/4)*(1 - Sqrt[a + 3*x^2]/Sqrt[a]))/(Sqrt[3]*x*(a + 3*x^2)^(1/4))]/(3*Sqrt[3]*a^(1/4))

Rule 452

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))*Ar
cTan[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^{-1}\left (\frac {a^{3/4} \left (1+\frac {\sqrt {a+3 x^2}}{\sqrt {a}}\right )}{\sqrt {3} x \sqrt [4]{a+3 x^2}}\right )}{3 \sqrt {3} \sqrt [4]{a}}+\frac {\tanh ^{-1}\left (\frac {a^{3/4} \left (1-\frac {\sqrt {a+3 x^2}}{\sqrt {a}}\right )}{\sqrt {3} x \sqrt [4]{a+3 x^2}}\right )}{3 \sqrt {3} \sqrt [4]{a}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.77 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.99 \[ \int \frac {x^2}{\left (a+3 x^2\right )^{3/4} \left (2 a+3 x^2\right )} \, dx=-\frac {\arctan \left (\frac {-3 x^2+2 \sqrt {a} \sqrt {a+3 x^2}}{2 \sqrt {3} \sqrt [4]{a} x \sqrt [4]{a+3 x^2}}\right )+\text {arctanh}\left (\frac {2 \sqrt {3} \sqrt [4]{a} x \sqrt [4]{a+3 x^2}}{3 x^2+2 \sqrt {a} \sqrt {a+3 x^2}}\right )}{6 \sqrt {3} \sqrt [4]{a}} \]

[In]

Integrate[x^2/((a + 3*x^2)^(3/4)*(2*a + 3*x^2)),x]

[Out]

-1/6*(ArcTan[(-3*x^2 + 2*Sqrt[a]*Sqrt[a + 3*x^2])/(2*Sqrt[3]*a^(1/4)*x*(a + 3*x^2)^(1/4))] + ArcTanh[(2*Sqrt[3
]*a^(1/4)*x*(a + 3*x^2)^(1/4))/(3*x^2 + 2*Sqrt[a]*Sqrt[a + 3*x^2])])/(Sqrt[3]*a^(1/4))

Maple [F]

\[\int \frac {x^{2}}{\left (3 x^{2}+a \right )^{\frac {3}{4}} \left (3 x^{2}+2 a \right )}d x\]

[In]

int(x^2/(3*x^2+a)^(3/4)/(3*x^2+2*a),x)

[Out]

int(x^2/(3*x^2+a)^(3/4)/(3*x^2+2*a),x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.37 \[ \int \frac {x^2}{\left (a+3 x^2\right )^{3/4} \left (2 a+3 x^2\right )} \, dx=-\frac {1}{6} \, \left (\frac {1}{36}\right )^{\frac {1}{4}} \left (-\frac {1}{a}\right )^{\frac {1}{4}} \log \left (\frac {3 \, \left (\frac {1}{36}\right )^{\frac {1}{4}} x \left (-\frac {1}{a}\right )^{\frac {1}{4}} + {\left (3 \, x^{2} + a\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{6} \, \left (\frac {1}{36}\right )^{\frac {1}{4}} \left (-\frac {1}{a}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, \left (\frac {1}{36}\right )^{\frac {1}{4}} x \left (-\frac {1}{a}\right )^{\frac {1}{4}} - {\left (3 \, x^{2} + a\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{6} i \, \left (\frac {1}{36}\right )^{\frac {1}{4}} \left (-\frac {1}{a}\right )^{\frac {1}{4}} \log \left (\frac {3 i \, \left (\frac {1}{36}\right )^{\frac {1}{4}} x \left (-\frac {1}{a}\right )^{\frac {1}{4}} + {\left (3 \, x^{2} + a\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{6} i \, \left (\frac {1}{36}\right )^{\frac {1}{4}} \left (-\frac {1}{a}\right )^{\frac {1}{4}} \log \left (\frac {-3 i \, \left (\frac {1}{36}\right )^{\frac {1}{4}} x \left (-\frac {1}{a}\right )^{\frac {1}{4}} + {\left (3 \, x^{2} + a\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

integrate(x^2/(3*x^2+a)^(3/4)/(3*x^2+2*a),x, algorithm="fricas")

[Out]

-1/6*(1/36)^(1/4)*(-1/a)^(1/4)*log((3*(1/36)^(1/4)*x*(-1/a)^(1/4) + (3*x^2 + a)^(1/4))/x) + 1/6*(1/36)^(1/4)*(
-1/a)^(1/4)*log(-(3*(1/36)^(1/4)*x*(-1/a)^(1/4) - (3*x^2 + a)^(1/4))/x) - 1/6*I*(1/36)^(1/4)*(-1/a)^(1/4)*log(
(3*I*(1/36)^(1/4)*x*(-1/a)^(1/4) + (3*x^2 + a)^(1/4))/x) + 1/6*I*(1/36)^(1/4)*(-1/a)^(1/4)*log((-3*I*(1/36)^(1
/4)*x*(-1/a)^(1/4) + (3*x^2 + a)^(1/4))/x)

Sympy [F]

\[ \int \frac {x^2}{\left (a+3 x^2\right )^{3/4} \left (2 a+3 x^2\right )} \, dx=\int \frac {x^{2}}{\left (a + 3 x^{2}\right )^{\frac {3}{4}} \cdot \left (2 a + 3 x^{2}\right )}\, dx \]

[In]

integrate(x**2/(3*x**2+a)**(3/4)/(3*x**2+2*a),x)

[Out]

Integral(x**2/((a + 3*x**2)**(3/4)*(2*a + 3*x**2)), x)

Maxima [F]

\[ \int \frac {x^2}{\left (a+3 x^2\right )^{3/4} \left (2 a+3 x^2\right )} \, dx=\int { \frac {x^{2}}{{\left (3 \, x^{2} + 2 \, a\right )} {\left (3 \, x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^2/(3*x^2+a)^(3/4)/(3*x^2+2*a),x, algorithm="maxima")

[Out]

integrate(x^2/((3*x^2 + 2*a)*(3*x^2 + a)^(3/4)), x)

Giac [F]

\[ \int \frac {x^2}{\left (a+3 x^2\right )^{3/4} \left (2 a+3 x^2\right )} \, dx=\int { \frac {x^{2}}{{\left (3 \, x^{2} + 2 \, a\right )} {\left (3 \, x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^2/(3*x^2+a)^(3/4)/(3*x^2+2*a),x, algorithm="giac")

[Out]

integrate(x^2/((3*x^2 + 2*a)*(3*x^2 + a)^(3/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (a+3 x^2\right )^{3/4} \left (2 a+3 x^2\right )} \, dx=\int \frac {x^2}{\left (3\,x^2+2\,a\right )\,{\left (3\,x^2+a\right )}^{3/4}} \,d x \]

[In]

int(x^2/((2*a + 3*x^2)*(a + 3*x^2)^(3/4)),x)

[Out]

int(x^2/((2*a + 3*x^2)*(a + 3*x^2)^(3/4)), x)