\(\int \frac {x^2}{(-2 a+3 x^2) (-a+3 x^2)^{3/4}} \, dx\) [1077]

Optimal result

Integrand size = 28, antiderivative size = 85 \[ \int \frac {x^2}{\left (-2 a+3 x^2\right ) \left (-a+3 x^2\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{a} \sqrt [4]{-a+3 x^2}}\right )}{3 \sqrt {6} \sqrt [4]{a}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{a} \sqrt [4]{-a+3 x^2}}\right )}{3 \sqrt {6} \sqrt [4]{a}} \]

[Out]

1/18*arctan(1/2*x*6^(1/2)/a^(1/4)/(3*x^2-a)^(1/4))/a^(1/4)*6^(1/2)-1/18*arctanh(1/2*x*6^(1/2)/a^(1/4)/(3*x^2-a
)^(1/4))/a^(1/4)*6^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {453} \[ \int \frac {x^2}{\left (-2 a+3 x^2\right ) \left (-a+3 x^2\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{a} \sqrt [4]{3 x^2-a}}\right )}{3 \sqrt {6} \sqrt [4]{a}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{a} \sqrt [4]{3 x^2-a}}\right )}{3 \sqrt {6} \sqrt [4]{a}} \]

[In]

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

[Out]

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

Rule 453

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[(-b/(Sqrt[2]*a*d*Rt[-b^2/a,
4]^3))*ArcTan[(Rt[-b^2/a, 4]*x)/(Sqrt[2]*(a + b*x^2)^(1/4))], x] + Simp[(b/(Sqrt[2]*a*d*Rt[-b^2/a, 4]^3))*ArcT
anh[(Rt[-b^2/a, 4]*x)/(Sqrt[2]*(a + b*x^2)^(1/4))], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && Neg
Q[b^2/a]

Rubi steps \begin{align*} \text {integral}& = \frac {\tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{a} \sqrt [4]{-a+3 x^2}}\right )}{3 \sqrt {6} \sqrt [4]{a}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{a} \sqrt [4]{-a+3 x^2}}\right )}{3 \sqrt {6} \sqrt [4]{a}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\left (-2 a+3 x^2\right ) \left (-a+3 x^2\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{a} \sqrt [4]{-a+3 x^2}}\right )-\text {arctanh}\left (\frac {\sqrt {\frac {2}{3}} \sqrt [4]{a} \sqrt [4]{-a+3 x^2}}{x}\right )}{3 \sqrt {6} \sqrt [4]{a}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.65 \[ \int \frac {x^2}{\left (-2 a+3 x^2\right ) \left (-a+3 x^2\right )^{3/4}} \, dx=-\frac {\left (\frac {1}{36}\right )^{\frac {1}{4}} \log \left (\frac {\frac {3 \, \left (\frac {1}{36}\right )^{\frac {1}{4}} x}{a^{\frac {1}{4}}} + {\left (3 \, x^{2} - a\right )}^{\frac {1}{4}}}{x}\right )}{6 \, a^{\frac {1}{4}}} + \frac {\left (\frac {1}{36}\right )^{\frac {1}{4}} \log \left (-\frac {\frac {3 \, \left (\frac {1}{36}\right )^{\frac {1}{4}} x}{a^{\frac {1}{4}}} - {\left (3 \, x^{2} - a\right )}^{\frac {1}{4}}}{x}\right )}{6 \, a^{\frac {1}{4}}} - \frac {i \, \left (\frac {1}{36}\right )^{\frac {1}{4}} \log \left (\frac {\frac {3 i \, \left (\frac {1}{36}\right )^{\frac {1}{4}} x}{a^{\frac {1}{4}}} + {\left (3 \, x^{2} - a\right )}^{\frac {1}{4}}}{x}\right )}{6 \, a^{\frac {1}{4}}} + \frac {i \, \left (\frac {1}{36}\right )^{\frac {1}{4}} \log \left (\frac {-\frac {3 i \, \left (\frac {1}{36}\right )^{\frac {1}{4}} x}{a^{\frac {1}{4}}} + {\left (3 \, x^{2} - a\right )}^{\frac {1}{4}}}{x}\right )}{6 \, a^{\frac {1}{4}}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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