3.257 \(\int \frac{x^2 (3 a+b x^2)}{a^2+2 a b x^2+b^2 x^4+c^2 x^6} \, dx\)

Optimal. Leaf size=19 \[ \frac{\tan ^{-1}\left (\frac{c x^3}{a+b x^2}\right )}{c} \]

[Out]

ArcTan[(c*x^3)/(a + b*x^2)]/c

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Rubi [A]  time = 0.104799, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {2094, 205} \[ \frac{\tan ^{-1}\left (\frac{c x^3}{a+b x^2}\right )}{c} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(3*a + b*x^2))/(a^2 + 2*a*b*x^2 + b^2*x^4 + c^2*x^6),x]

[Out]

ArcTan[(c*x^3)/(a + b*x^2)]/c

Rule 2094

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(n_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_
Symbol] :> Dist[(A^2*(m - n + 1))/(m + 1), Subst[Int[1/(a + A^2*b*(m - n + 1)^2*x^2), x], x, x^(m + 1)/(A*(m -
n + 1) + B*(m + 1)*x^n)], x] /; FreeQ[{a, b, c, d, A, B, m, n}, x] && EqQ[n2, 2*n] && EqQ[k, 2*(m + 1)] && Eq
Q[a*B^2*(m + 1)^2 - A^2*d*(m - n + 1)^2, 0] && EqQ[B*c*(m + 1) - 2*A*d*(m - n + 1), 0]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{x^2 \left (3 a+b x^2\right )}{a^2+2 a b x^2+b^2 x^4+c^2 x^6} \, dx &=\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a^2+9 a^2 c^2 x^2} \, dx,x,\frac{x^3}{3 a+3 b x^2}\right )\\ &=\frac{\tan ^{-1}\left (\frac{c x^3}{a+b x^2}\right )}{c}\\ \end{align*}

Mathematica [C]  time = 0.0452552, size = 87, normalized size = 4.58 \[ \frac{1}{2} \text{RootSum}\left [2 \text{$\#$1}^2 a b+\text{$\#$1}^4 b^2+\text{$\#$1}^6 c^2+a^2\& ,\frac{\text{$\#$1}^3 b \log (x-\text{$\#$1})+3 \text{$\#$1} a \log (x-\text{$\#$1})}{2 \text{$\#$1}^2 b^2+3 \text{$\#$1}^4 c^2+2 a b}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(3*a + b*x^2))/(a^2 + 2*a*b*x^2 + b^2*x^4 + c^2*x^6),x]

[Out]

RootSum[a^2 + 2*a*b*#1^2 + b^2*#1^4 + c^2*#1^6 & , (3*a*Log[x - #1]*#1 + b*Log[x - #1]*#1^3)/(2*a*b + 2*b^2*#1
^2 + 3*c^2*#1^4) & ]/2

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Maple [C]  time = 0.107, size = 75, normalized size = 4. \begin{align*}{\frac{1}{2}\sum _{{\it \_R}={\it RootOf} \left ({c}^{2}{{\it \_Z}}^{6}+{b}^{2}{{\it \_Z}}^{4}+2\,ab{{\it \_Z}}^{2}+{a}^{2} \right ) }{\frac{ \left ({{\it \_R}}^{4}b+3\,{{\it \_R}}^{2}a \right ) \ln \left ( x-{\it \_R} \right ) }{3\,{{\it \_R}}^{5}{c}^{2}+2\,{{\it \_R}}^{3}{b}^{2}+2\,{\it \_R}\,ab}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^2+3*a)/(c^2*x^6+b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

1/2*sum((_R^4*b+3*_R^2*a)/(3*_R^5*c^2+2*_R^3*b^2+2*_R*a*b)*ln(x-_R),_R=RootOf(_Z^6*c^2+_Z^4*b^2+2*_Z^2*a*b+a^2
))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + 3 \, a\right )} x^{2}}{c^{2} x^{6} + b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+3*a)/(c^2*x^6+b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + 3*a)*x^2/(c^2*x^6 + b^2*x^4 + 2*a*b*x^2 + a^2), x)

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Fricas [B]  time = 1.41602, size = 171, normalized size = 9. \begin{align*} \frac{\arctan \left (\frac{c x}{b}\right ) - \arctan \left (\frac{b c^{2} x^{5} + a b^{2} x +{\left (b^{3} - a c^{2}\right )} x^{3}}{a^{2} c}\right ) + \arctan \left (\frac{b c^{2} x^{3} +{\left (b^{3} - a c^{2}\right )} x}{a b c}\right )}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+3*a)/(c^2*x^6+b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

(arctan(c*x/b) - arctan((b*c^2*x^5 + a*b^2*x + (b^3 - a*c^2)*x^3)/(a^2*c)) + arctan((b*c^2*x^3 + (b^3 - a*c^2)
*x)/(a*b*c)))/c

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Sympy [C]  time = 1.04768, size = 44, normalized size = 2.32 \begin{align*} \frac{- \frac{i \log{\left (- \frac{i a}{c} - \frac{i b x^{2}}{c} + x^{3} \right )}}{2} + \frac{i \log{\left (\frac{i a}{c} + \frac{i b x^{2}}{c} + x^{3} \right )}}{2}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**2+3*a)/(c**2*x**6+b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

(-I*log(-I*a/c - I*b*x**2/c + x**3)/2 + I*log(I*a/c + I*b*x**2/c + x**3)/2)/c

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Giac [B]  time = 3.98667, size = 117, normalized size = 6.16 \begin{align*} \frac{\arctan \left (\frac{c x}{b}\right ) + \arctan \left (-\frac{b c^{2} x^{5} + b^{3} x^{3} - a c^{2} x^{3} + a b^{2} x}{a^{2} c}\right ) - \arctan \left (-\frac{b c^{2} x^{3} + b^{3} x - a c^{2} x}{a b c}\right )}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+3*a)/(c^2*x^6+b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

(arctan(c*x/b) + arctan(-(b*c^2*x^5 + b^3*x^3 - a*c^2*x^3 + a*b^2*x)/(a^2*c)) - arctan(-(b*c^2*x^3 + b^3*x - a
*c^2*x)/(a*b*c)))/c