3.200 \(\int \frac{x^2}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=57 \[ -\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{5/2}}-\frac{3}{2 b^2 x}+\frac{1}{2 b x \left (b+c x^2\right )} \]

[Out]

-3/(2*b^2*x) + 1/(2*b*x*(b + c*x^2)) - (3*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(5/2))

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Rubi [A]  time = 0.025463, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1584, 290, 325, 205} \[ -\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{5/2}}-\frac{3}{2 b^2 x}+\frac{1}{2 b x \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(b*x^2 + c*x^4)^2,x]

[Out]

-3/(2*b^2*x) + 1/(2*b*x*(b + c*x^2)) - (3*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(5/2))

Rule 1584

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

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 (b x^2+c x^4\right )^2} \, dx &=\int \frac{1}{x^2 \left (b+c x^2\right )^2} \, dx\\ &=\frac{1}{2 b x \left (b+c x^2\right )}+\frac{3 \int \frac{1}{x^2 \left (b+c x^2\right )} \, dx}{2 b}\\ &=-\frac{3}{2 b^2 x}+\frac{1}{2 b x \left (b+c x^2\right )}-\frac{(3 c) \int \frac{1}{b+c x^2} \, dx}{2 b^2}\\ &=-\frac{3}{2 b^2 x}+\frac{1}{2 b x \left (b+c x^2\right )}-\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.036817, size = 54, normalized size = 0.95 \[ -\frac{c x}{2 b^2 \left (b+c x^2\right )}-\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{5/2}}-\frac{1}{b^2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b*x^2 + c*x^4)^2,x]

[Out]

-(1/(b^2*x)) - (c*x)/(2*b^2*(b + c*x^2)) - (3*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(5/2))

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Maple [A]  time = 0.053, size = 46, normalized size = 0.8 \begin{align*} -{\frac{1}{{b}^{2}x}}-{\frac{cx}{2\,{b}^{2} \left ( c{x}^{2}+b \right ) }}-{\frac{3\,c}{2\,{b}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^2/x-1/2*c/b^2*x/(c*x^2+b)-3/2*c/b^2/(b*c)^(1/2)*arctan(x*c/(b*c)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49902, size = 288, normalized size = 5.05 \begin{align*} \left [-\frac{6 \, c x^{2} - 3 \,{\left (c x^{3} + b x\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} - 2 \, b x \sqrt{-\frac{c}{b}} - b}{c x^{2} + b}\right ) + 4 \, b}{4 \,{\left (b^{2} c x^{3} + b^{3} x\right )}}, -\frac{3 \, c x^{2} + 3 \,{\left (c x^{3} + b x\right )} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right ) + 2 \, b}{2 \,{\left (b^{2} c x^{3} + b^{3} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(6*c*x^2 - 3*(c*x^3 + b*x)*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)) + 4*b)/(b^2*c*x^3
+ b^3*x), -1/2*(3*c*x^2 + 3*(c*x^3 + b*x)*sqrt(c/b)*arctan(x*sqrt(c/b)) + 2*b)/(b^2*c*x^3 + b^3*x)]

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Sympy [A]  time = 0.525042, size = 90, normalized size = 1.58 \begin{align*} \frac{3 \sqrt{- \frac{c}{b^{5}}} \log{\left (- \frac{b^{3} \sqrt{- \frac{c}{b^{5}}}}{c} + x \right )}}{4} - \frac{3 \sqrt{- \frac{c}{b^{5}}} \log{\left (\frac{b^{3} \sqrt{- \frac{c}{b^{5}}}}{c} + x \right )}}{4} - \frac{2 b + 3 c x^{2}}{2 b^{3} x + 2 b^{2} c x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(-c/b**5)*log(-b**3*sqrt(-c/b**5)/c + x)/4 - 3*sqrt(-c/b**5)*log(b**3*sqrt(-c/b**5)/c + x)/4 - (2*b + 3*
c*x**2)/(2*b**3*x + 2*b**2*c*x**3)

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Giac [A]  time = 1.26842, size = 63, normalized size = 1.11 \begin{align*} -\frac{3 \, c \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} b^{2}} - \frac{3 \, c x^{2} + 2 \, b}{2 \,{\left (c x^{3} + b x\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*c*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2) - 1/2*(3*c*x^2 + 2*b)/((c*x^3 + b*x)*b^2)