\(\int \frac {b x+d x^3}{2+3 x^4} \, dx\) [161]

Optimal result

Integrand size = 19, antiderivative size = 36 \[ \int \frac {b x+d x^3}{2+3 x^4} \, dx=\frac {b \arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}}+\frac {1}{12} d \log \left (2+3 x^4\right ) \]

[Out]

1/12*d*ln(3*x^4+2)+1/12*b*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1607, 1262, 649, 209, 266} \[ \int \frac {b x+d x^3}{2+3 x^4} \, dx=\frac {b \arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}}+\frac {1}{12} d \log \left (3 x^4+2\right ) \]

[In]

Int[(b*x + d*x^3)/(2 + 3*x^4),x]

[Out]

(b*ArcTan[Sqrt[3/2]*x^2])/(2*Sqrt[6]) + (d*Log[2 + 3*x^4])/12

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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 1262

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (b+d x^2\right )}{2+3 x^4} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {b+d x}{2+3 x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} b \text {Subst}\left (\int \frac {1}{2+3 x^2} \, dx,x,x^2\right )+\frac {1}{2} d \text {Subst}\left (\int \frac {x}{2+3 x^2} \, dx,x,x^2\right ) \\ & = \frac {b \tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}}+\frac {1}{12} d \log \left (2+3 x^4\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.81 \[ \int \frac {b x+d x^3}{2+3 x^4} \, dx=\frac {1}{24} \left (i \sqrt {6} b+2 d\right ) \log \left (\sqrt {6}-3 i x^2\right )+\frac {1}{24} \left (-i \sqrt {6} b+2 d\right ) \log \left (\sqrt {6}+3 i x^2\right ) \]

[In]

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

[Out]

((I*Sqrt[6]*b + 2*d)*Log[Sqrt[6] - (3*I)*x^2])/24 + (((-I)*Sqrt[6]*b + 2*d)*Log[Sqrt[6] + (3*I)*x^2])/24

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {d \ln \left (3 x^{4}+2\right )}{12}+\frac {b \arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{12}\)	\(28\)
risch	\(\frac {d \ln \left (9 x^{4}+6\right )}{12}+\frac {b \arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{12}\)	\(28\)
meijerg	\(\frac {d \ln \left (\frac {3 x^{4}}{2}+1\right )}{12}+\frac {\sqrt {6}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, x^{2}}{2}\right )}{12}\)	\(31\)

[In]

int((d*x^3+b*x)/(3*x^4+2),x,method=_RETURNVERBOSE)

[Out]

1/12*d*ln(3*x^4+2)+1/12*b*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int \frac {b x+d x^3}{2+3 x^4} \, dx=\frac {1}{12} \, \sqrt {6} b \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) + \frac {1}{12} \, d \log \left (3 \, x^{4} + 2\right ) \]

[In]

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

[Out]

1/12*sqrt(6)*b*arctan(1/2*sqrt(6)*x^2) + 1/12*d*log(3*x^4 + 2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.47 \[ \int \frac {b x+d x^3}{2+3 x^4} \, dx=\left (- \frac {\sqrt {6} i b}{24} + \frac {d}{12}\right ) \log {\left (x^{2} - \frac {\sqrt {6} i}{3} \right )} + \left (\frac {\sqrt {6} i b}{24} + \frac {d}{12}\right ) \log {\left (x^{2} + \frac {\sqrt {6} i}{3} \right )} \]

[In]

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

[Out]

(-sqrt(6)*I*b/24 + d/12)*log(x**2 - sqrt(6)*I/3) + (sqrt(6)*I*b/24 + d/12)*log(x**2 + sqrt(6)*I/3)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (27) = 54\).

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 3.14 \[ \int \frac {b x+d x^3}{2+3 x^4} \, dx=-\frac {1}{12} \, \sqrt {3} \sqrt {2} b \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x + 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) + \frac {1}{12} \, \sqrt {3} \sqrt {2} b \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x - 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) + \frac {1}{12} \, d \log \left (\sqrt {3} x^{2} + 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) + \frac {1}{12} \, d \log \left (\sqrt {3} x^{2} - 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) \]

[In]

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

[Out]

-1/12*sqrt(3)*sqrt(2)*b*arctan(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x + 3^(1/4)*2^(3/4))) + 1/12*sqrt(3)*sqrt(2)*b*a
rctan(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x - 3^(1/4)*2^(3/4))) + 1/12*d*log(sqrt(3)*x^2 + 3^(1/4)*2^(3/4)*x + sqrt
(2)) + 1/12*d*log(sqrt(3)*x^2 - 3^(1/4)*2^(3/4)*x + sqrt(2))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (27) = 54\).

Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.58 \[ \int \frac {b x+d x^3}{2+3 x^4} \, dx=-\frac {1}{12} \, \sqrt {6} b \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{12} \, \sqrt {6} b \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{12} \, d \log \left (x^{2} + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) + \frac {1}{12} \, d \log \left (x^{2} - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) \]

[In]

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

[Out]

-1/12*sqrt(6)*b*arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x + sqrt(2)*(2/3)^(1/4))) + 1/12*sqrt(6)*b*arctan(3/4*sqrt(2
)*(2/3)^(3/4)*(2*x - sqrt(2)*(2/3)^(1/4))) + 1/12*d*log(x^2 + sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3)) + 1/12*d*log(
x^2 - sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \frac {b x+d x^3}{2+3 x^4} \, dx=\frac {d\,\ln \left (x^4+\frac {2}{3}\right )}{12}+\frac {\sqrt {6}\,b\,\mathrm {atan}\left (\frac {\sqrt {6}\,x^2}{2}\right )}{12} \]

[In]

int((b*x + d*x^3)/(3*x^4 + 2),x)

[Out]

(d*log(x^4 + 2/3))/12 + (6^(1/2)*b*atan((6^(1/2)*x^2)/2))/12