\(\int \frac {x^2}{2 a+2 b+x^4} \, dx\) [713]

Optimal result

Integrand size = 16, antiderivative size = 79 \[ \int \frac {x^2}{2 a+2 b+x^4} \, dx=\frac {\arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{-a-b}}\right )}{2 \sqrt [4]{2} \sqrt [4]{-a-b}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{-a-b}}\right )}{2 \sqrt [4]{2} \sqrt [4]{-a-b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {304, 209, 212} \[ \int \frac {x^2}{2 a+2 b+x^4} \, dx=\frac {\arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{-a-b}}\right )}{2 \sqrt [4]{2} \sqrt [4]{-a-b}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{-a-b}}\right )}{2 \sqrt [4]{2} \sqrt [4]{-a-b}} \]

[In]

Int[x^2/(2*a + 2*b + x^4),x]

[Out]

ArcTan[x/(2^(1/4)*(-a - b)^(1/4))]/(2*2^(1/4)*(-a - b)^(1/4)) - ArcTanh[x/(2^(1/4)*(-a - b)^(1/4))]/(2*2^(1/4)
*(-a - b)^(1/4))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.62 \[ \int \frac {x^2}{2 a+2 b+x^4} \, dx=\frac {-2 \arctan \left (1-\frac {\sqrt [4]{2} x}{\sqrt [4]{a+b}}\right )+2 \arctan \left (1+\frac {\sqrt [4]{2} x}{\sqrt [4]{a+b}}\right )+\log \left (2 \sqrt {a+b}-2 \sqrt [4]{2} \sqrt [4]{a+b} x+\sqrt {2} x^2\right )-\log \left (2 \sqrt {a+b}+2 \sqrt [4]{2} \sqrt [4]{a+b} x+\sqrt {2} x^2\right )}{4\ 2^{3/4} \sqrt [4]{a+b}} \]

[In]

Integrate[x^2/(2*a + 2*b + x^4),x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.92 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.34

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+2 a +2 b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{4}\)	\(27\)
default	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (2 a +2 b \right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {2 a +2 b}}{x^{2}+\left (2 a +2 b \right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {2 a +2 b}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (2 a +2 b \right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (2 a +2 b \right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (2 a +2 b \right )^{\frac {1}{4}}}\)	\(113\)

[In]

int(x^2/(x^4+2*a+2*b),x,method=_RETURNVERBOSE)

[Out]

1/4*sum(1/_R*ln(x-_R),_R=RootOf(_Z^4+2*a+2*b))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.84 \[ \int \frac {x^2}{2 a+2 b+x^4} \, dx=\frac {1}{4} \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \left (-\frac {1}{a + b}\right )^{\frac {1}{4}} \log \left (2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (a + b\right )} \left (-\frac {1}{a + b}\right )^{\frac {3}{4}} + x\right ) - \frac {1}{4} \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \left (-\frac {1}{a + b}\right )^{\frac {1}{4}} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (a + b\right )} \left (-\frac {1}{a + b}\right )^{\frac {3}{4}} + x\right ) + \frac {1}{4} i \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \left (-\frac {1}{a + b}\right )^{\frac {1}{4}} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (i \, a + i \, b\right )} \left (-\frac {1}{a + b}\right )^{\frac {3}{4}} + x\right ) - \frac {1}{4} i \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \left (-\frac {1}{a + b}\right )^{\frac {1}{4}} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (-i \, a - i \, b\right )} \left (-\frac {1}{a + b}\right )^{\frac {3}{4}} + x\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.37 \[ \int \frac {x^2}{2 a+2 b+x^4} \, dx=\operatorname {RootSum} {\left (t^{4} \cdot \left (512 a + 512 b\right ) + 1, \left ( t \mapsto t \log {\left (128 t^{3} a + 128 t^{3} b + x \right )} \right )\right )} \]

[In]

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

[Out]

RootSum(_t**4*(512*a + 512*b) + 1, Lambda(_t, _t*log(128*_t**3*a + 128*_t**3*b + x)))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (61) = 122\).

Time = 0.29 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.27 \[ \int \frac {x^2}{2 a+2 b+x^4} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}\right )}}{2 \, {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}}\right )}{4 \, {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}} + \frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}\right )}}{2 \, {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}}\right )}{4 \, {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (x^{2} + \sqrt {2} {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}} x + \sqrt {2 \, a + 2 \, b}\right )}{8 \, {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (x^{2} - \sqrt {2} {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}} x + \sqrt {2 \, a + 2 \, b}\right )}{8 \, {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(2*a + 2*b)^(1/4))/(2*a + 2*b)^(1/4))/(2*a + 2*b)^(1/4) + 1/4*sq
rt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(2*a + 2*b)^(1/4))/(2*a + 2*b)^(1/4))/(2*a + 2*b)^(1/4) - 1/8*sqrt(2)*
log(x^2 + sqrt(2)*(2*a + 2*b)^(1/4)*x + sqrt(2*a + 2*b))/(2*a + 2*b)^(1/4) + 1/8*sqrt(2)*log(x^2 - sqrt(2)*(2*
a + 2*b)^(1/4)*x + sqrt(2*a + 2*b))/(2*a + 2*b)^(1/4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (61) = 122\).

Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.77 \[ \int \frac {x^2}{2 a+2 b+x^4} \, dx=\frac {{\left (2 \, a + 2 \, b\right )}^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}\right )}}{2 \, {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}}\right )}{4 \, {\left (\sqrt {2} a + \sqrt {2} b\right )}} + \frac {{\left (2 \, a + 2 \, b\right )}^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}\right )}}{2 \, {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}}}\right )}{4 \, {\left (\sqrt {2} a + \sqrt {2} b\right )}} - \frac {{\left (2 \, a + 2 \, b\right )}^{\frac {3}{4}} \log \left (x^{2} + \sqrt {2} {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}} x + \sqrt {2 \, a + 2 \, b}\right )}{8 \, {\left (\sqrt {2} a + \sqrt {2} b\right )}} + \frac {{\left (2 \, a + 2 \, b\right )}^{\frac {3}{4}} \log \left (x^{2} - \sqrt {2} {\left (2 \, a + 2 \, b\right )}^{\frac {1}{4}} x + \sqrt {2 \, a + 2 \, b}\right )}{8 \, {\left (\sqrt {2} a + \sqrt {2} b\right )}} \]

[In]

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

[Out]

1/4*(2*a + 2*b)^(3/4)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(2*a + 2*b)^(1/4))/(2*a + 2*b)^(1/4))/(sqrt(2)*a + sqr
t(2)*b) + 1/4*(2*a + 2*b)^(3/4)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(2*a + 2*b)^(1/4))/(2*a + 2*b)^(1/4))/(sqrt(
2)*a + sqrt(2)*b) - 1/8*(2*a + 2*b)^(3/4)*log(x^2 + sqrt(2)*(2*a + 2*b)^(1/4)*x + sqrt(2*a + 2*b))/(sqrt(2)*a
+ sqrt(2)*b) + 1/8*(2*a + 2*b)^(3/4)*log(x^2 - sqrt(2)*(2*a + 2*b)^(1/4)*x + sqrt(2*a + 2*b))/(sqrt(2)*a + sqr
t(2)*b)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.67 \[ \int \frac {x^2}{2 a+2 b+x^4} \, dx=\frac {8^{1/4}\,\mathrm {atan}\left (\frac {8^{1/4}\,x}{2\,{\left (-a-b\right )}^{1/4}}\right )-8^{1/4}\,\mathrm {atanh}\left (\frac {8^{1/4}\,x}{2\,{\left (-a-b\right )}^{1/4}}\right )}{4\,{\left (-a-b\right )}^{1/4}} \]

[In]

int(x^2/(2*a + 2*b + x^4),x)

[Out]

(8^(1/4)*atan((8^(1/4)*x)/(2*(- a - b)^(1/4))) - 8^(1/4)*atanh((8^(1/4)*x)/(2*(- a - b)^(1/4))))/(4*(- a - b)^
(1/4))