\(\int \frac {x^2}{(2-b x^2)^{3/4} (4-b x^2)} \, dx\) [1056]

Optimal result

Integrand size = 26, antiderivative size = 119 \[ \int \frac {x^2}{\left (2-b x^2\right )^{3/4} \left (4-b x^2\right )} \, dx=\frac {\arctan \left (\frac {2-\sqrt {2} \sqrt {2-b x^2}}{\sqrt [4]{2} \sqrt {b} x \sqrt [4]{2-b x^2}}\right )}{\sqrt [4]{2} b^{3/2}}-\frac {\text {arctanh}\left (\frac {2+\sqrt {2} \sqrt {2-b x^2}}{\sqrt [4]{2} \sqrt {b} x \sqrt [4]{2-b x^2}}\right )}{\sqrt [4]{2} b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 119, 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} \[ \int \frac {x^2}{\left (2-b x^2\right )^{3/4} \left (4-b x^2\right )} \, dx=\frac {\arctan \left (\frac {2-\sqrt {2} \sqrt {2-b x^2}}{\sqrt [4]{2} \sqrt {b} x \sqrt [4]{2-b x^2}}\right )}{\sqrt [4]{2} b^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {2-b x^2}+2}{\sqrt [4]{2} \sqrt {b} x \sqrt [4]{2-b x^2}}\right )}{\sqrt [4]{2} b^{3/2}} \]

[In]

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

[Out]

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

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 {2-\sqrt {2} \sqrt {2-b x^2}}{\sqrt [4]{2} \sqrt {b} x \sqrt [4]{2-b x^2}}\right )}{\sqrt [4]{2} b^{3/2}}-\frac {\tanh ^{-1}\left (\frac {2+\sqrt {2} \sqrt {2-b x^2}}{\sqrt [4]{2} \sqrt {b} x \sqrt [4]{2-b x^2}}\right )}{\sqrt [4]{2} b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.83 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int \frac {x^2}{\left (2-b x^2\right )^{3/4} \left (4-b x^2\right )} \, dx=-\frac {\arctan \left (\frac {-2^{3/4} b x^2+4 \sqrt [4]{2} \sqrt {2-b x^2}}{4 \sqrt {b} x \sqrt [4]{2-b x^2}}\right )+\text {arctanh}\left (\frac {2 \sqrt {b} x \sqrt [4]{4-2 b x^2}}{b x^2+2 \sqrt {4-2 b x^2}}\right )}{2 \sqrt [4]{2} b^{3/2}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.50 \[ \int \frac {x^2}{\left (2-b x^2\right )^{3/4} \left (4-b x^2\right )} \, dx=-\frac {1}{2} \, \left (\frac {1}{8}\right )^{\frac {1}{4}} \left (-\frac {1}{b^{6}}\right )^{\frac {1}{4}} \log \left (\frac {\left (\frac {1}{8}\right )^{\frac {1}{4}} b^{2} x \left (-\frac {1}{b^{6}}\right )^{\frac {1}{4}} + {\left (-b x^{2} + 2\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \left (\frac {1}{8}\right )^{\frac {1}{4}} \left (-\frac {1}{b^{6}}\right )^{\frac {1}{4}} \log \left (-\frac {\left (\frac {1}{8}\right )^{\frac {1}{4}} b^{2} x \left (-\frac {1}{b^{6}}\right )^{\frac {1}{4}} - {\left (-b x^{2} + 2\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} i \, \left (\frac {1}{8}\right )^{\frac {1}{4}} \left (-\frac {1}{b^{6}}\right )^{\frac {1}{4}} \log \left (\frac {i \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b^{2} x \left (-\frac {1}{b^{6}}\right )^{\frac {1}{4}} + {\left (-b x^{2} + 2\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} i \, \left (\frac {1}{8}\right )^{\frac {1}{4}} \left (-\frac {1}{b^{6}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b^{2} x \left (-\frac {1}{b^{6}}\right )^{\frac {1}{4}} + {\left (-b x^{2} + 2\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

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

[Out]

-1/2*(1/8)^(1/4)*(-1/b^6)^(1/4)*log(((1/8)^(1/4)*b^2*x*(-1/b^6)^(1/4) + (-b*x^2 + 2)^(1/4))/x) + 1/2*(1/8)^(1/
4)*(-1/b^6)^(1/4)*log(-((1/8)^(1/4)*b^2*x*(-1/b^6)^(1/4) - (-b*x^2 + 2)^(1/4))/x) - 1/2*I*(1/8)^(1/4)*(-1/b^6)
^(1/4)*log((I*(1/8)^(1/4)*b^2*x*(-1/b^6)^(1/4) + (-b*x^2 + 2)^(1/4))/x) + 1/2*I*(1/8)^(1/4)*(-1/b^6)^(1/4)*log
((-I*(1/8)^(1/4)*b^2*x*(-1/b^6)^(1/4) + (-b*x^2 + 2)^(1/4))/x)

Sympy [F]

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

[In]

integrate(x**2/(-b*x**2+2)**(3/4)/(-b*x**2+4),x)

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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