\(\int \frac {x}{2 (a+b)+x^4} \, dx\) [712]

Optimal result

Integrand size = 13, antiderivative size = 33 \[ \int \frac {x}{2 (a+b)+x^4} \, dx=\frac {\arctan \left (\frac {x^2}{\sqrt {2} \sqrt {a+b}}\right )}{2 \sqrt {2} \sqrt {a+b}} \]

[Out]

1/4*arctan(1/2*x^2*2^(1/2)/(a+b)^(1/2))*2^(1/2)/(a+b)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {281, 209} \[ \int \frac {x}{2 (a+b)+x^4} \, dx=\frac {\arctan \left (\frac {x^2}{\sqrt {2} \sqrt {a+b}}\right )}{2 \sqrt {2} \sqrt {a+b}} \]

[In]

Int[x/(2*(a + b) + x^4),x]

[Out]

ArcTan[x^2/(Sqrt[2]*Sqrt[a + b])]/(2*Sqrt[2]*Sqrt[a + b])

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {x}{2 (a+b)+x^4} \, dx=\frac {\arctan \left (\frac {x^2}{\sqrt {2} \sqrt {a+b}}\right )}{2 \sqrt {2} \sqrt {a+b}} \]

[In]

Integrate[x/(2*(a + b) + x^4),x]

[Out]

ArcTan[x^2/(Sqrt[2]*Sqrt[a + b])]/(2*Sqrt[2]*Sqrt[a + b])

Maple [A] (verified)

Time = 4.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\arctan \left (\frac {x^{2}}{\sqrt {2 a +2 b}}\right )}{2 \sqrt {2 a +2 b}}\)	\(26\)
risch	\(-\frac {\ln \left (x^{2} \sqrt {-2 a -2 b}-2 a -2 b \right )}{4 \sqrt {-2 a -2 b}}+\frac {\ln \left (x^{2} \sqrt {-2 a -2 b}+2 a +2 b \right )}{4 \sqrt {-2 a -2 b}}\)	\(66\)

[In]

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

[Out]

1/2/(2*a+2*b)^(1/2)*arctan(x^2/(2*a+2*b)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.76 \[ \int \frac {x}{2 (a+b)+x^4} \, dx=\left [-\frac {\sqrt {-2 \, a - 2 \, b} \log \left (\frac {x^{4} - 2 \, \sqrt {-2 \, a - 2 \, b} x^{2} - 2 \, a - 2 \, b}{x^{4} + 2 \, a + 2 \, b}\right )}{8 \, {\left (a + b\right )}}, \frac {\sqrt {2 \, a + 2 \, b} \arctan \left (\frac {\sqrt {2 \, a + 2 \, b} x^{2}}{2 \, {\left (a + b\right )}}\right )}{4 \, {\left (a + b\right )}}\right ] \]

[In]

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

[Out]

[-1/8*sqrt(-2*a - 2*b)*log((x^4 - 2*sqrt(-2*a - 2*b)*x^2 - 2*a - 2*b)/(x^4 + 2*a + 2*b))/(a + b), 1/4*sqrt(2*a
+ 2*b)*arctan(1/2*sqrt(2*a + 2*b)*x^2/(a + b))/(a + b)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (31) = 62\).

Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.33 \[ \int \frac {x}{2 (a+b)+x^4} \, dx=- \frac {\sqrt {2} \sqrt {- \frac {1}{a + b}} \log {\left (- \sqrt {2} a \sqrt {- \frac {1}{a + b}} - \sqrt {2} b \sqrt {- \frac {1}{a + b}} + x^{2} \right )}}{8} + \frac {\sqrt {2} \sqrt {- \frac {1}{a + b}} \log {\left (\sqrt {2} a \sqrt {- \frac {1}{a + b}} + \sqrt {2} b \sqrt {- \frac {1}{a + b}} + x^{2} \right )}}{8} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {x}{2 (a+b)+x^4} \, dx=\frac {\arctan \left (\frac {x^{2}}{\sqrt {2 \, a + 2 \, b}}\right )}{2 \, \sqrt {2 \, a + 2 \, b}} \]

[In]

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

[Out]

1/2*arctan(x^2/sqrt(2*a + 2*b))/sqrt(2*a + 2*b)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {x}{2 (a+b)+x^4} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} x^{2}}{2 \, \sqrt {a + b}}\right )}{4 \, \sqrt {a + b}} \]

[In]

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

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*x^2/sqrt(a + b))/sqrt(a + b)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {x}{2 (a+b)+x^4} \, dx=\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2\,\sqrt {a+b}}{2\,a+2\,b}\right )}{4\,\sqrt {a+b}} \]

[In]

int(x/(2*a + 2*b + x^4),x)

[Out]

(2^(1/2)*atan((2^(1/2)*x^2*(a + b)^(1/2))/(2*a + 2*b)))/(4*(a + b)^(1/2))