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

Optimal. Leaf size=36 \[ \frac{b \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right )}{2 \sqrt{6}}+\frac{1}{12} d \log \left (3 x^4+2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0314551, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {1593, 1248, 635, 203, 260} \[ \frac{b \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right )}{2 \sqrt{6}}+\frac{1}{12} d \log \left (3 x^4+2\right ) \]

Antiderivative was successfully verified.

[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 1593

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]

Rule 1248

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 635

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 203

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

Rule 260

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]

Rubi steps

\begin{align*} \int \frac{b x+d x^3}{2+3 x^4} \, dx &=\int \frac{x \left (b+d x^2\right )}{2+3 x^4} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{b+d x}{2+3 x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} b \operatorname{Subst}\left (\int \frac{1}{2+3 x^2} \, dx,x,x^2\right )+\frac{1}{2} d \operatorname{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]  time = 0.0304642, size = 65, normalized size = 1.81 \[ \frac{1}{24} \left (2 d+i \sqrt{6} b\right ) \log \left (\sqrt{6}-3 i x^2\right )+\frac{1}{24} \left (2 d-i \sqrt{6} b\right ) \log \left (\sqrt{6}+3 i x^2\right ) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.042, size = 28, normalized size = 0.8 \begin{align*}{\frac{d\ln \left ( 3\,{x}^{4}+2 \right ) }{12}}+{\frac{b\sqrt{6}}{12}\arctan \left ({\frac{{x}^{2}\sqrt{6}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.46961, size = 153, normalized size = 4.25 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]  time = 1.47641, size = 86, normalized size = 2.39 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [C]  time = 0.305611, size = 53, normalized size = 1.47 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B]  time = 1.08778, size = 126, normalized size = 3.5 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))