3.204 \(\int \frac{1}{x^2 (b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=81 \[ -\frac{7 c^2}{2 b^4 x}-\frac{7 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{9/2}}+\frac{7 c}{6 b^3 x^3}-\frac{7}{10 b^2 x^5}+\frac{1}{2 b x^5 \left (b+c x^2\right )} \]

[Out]

-7/(10*b^2*x^5) + (7*c)/(6*b^3*x^3) - (7*c^2)/(2*b^4*x) + 1/(2*b*x^5*(b + c*x^2)) - (7*c^(5/2)*ArcTan[(Sqrt[c]
*x)/Sqrt[b]])/(2*b^(9/2))

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Rubi [A]  time = 0.0454493, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1584, 290, 325, 205} \[ -\frac{7 c^2}{2 b^4 x}-\frac{7 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{9/2}}+\frac{7 c}{6 b^3 x^3}-\frac{7}{10 b^2 x^5}+\frac{1}{2 b x^5 \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(b*x^2 + c*x^4)^2),x]

[Out]

-7/(10*b^2*x^5) + (7*c)/(6*b^3*x^3) - (7*c^2)/(2*b^4*x) + 1/(2*b*x^5*(b + c*x^2)) - (7*c^(5/2)*ArcTan[(Sqrt[c]
*x)/Sqrt[b]])/(2*b^(9/2))

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{1}{x^2 \left (b x^2+c x^4\right )^2} \, dx &=\int \frac{1}{x^6 \left (b+c x^2\right )^2} \, dx\\ &=\frac{1}{2 b x^5 \left (b+c x^2\right )}+\frac{7 \int \frac{1}{x^6 \left (b+c x^2\right )} \, dx}{2 b}\\ &=-\frac{7}{10 b^2 x^5}+\frac{1}{2 b x^5 \left (b+c x^2\right )}-\frac{(7 c) \int \frac{1}{x^4 \left (b+c x^2\right )} \, dx}{2 b^2}\\ &=-\frac{7}{10 b^2 x^5}+\frac{7 c}{6 b^3 x^3}+\frac{1}{2 b x^5 \left (b+c x^2\right )}+\frac{\left (7 c^2\right ) \int \frac{1}{x^2 \left (b+c x^2\right )} \, dx}{2 b^3}\\ &=-\frac{7}{10 b^2 x^5}+\frac{7 c}{6 b^3 x^3}-\frac{7 c^2}{2 b^4 x}+\frac{1}{2 b x^5 \left (b+c x^2\right )}-\frac{\left (7 c^3\right ) \int \frac{1}{b+c x^2} \, dx}{2 b^4}\\ &=-\frac{7}{10 b^2 x^5}+\frac{7 c}{6 b^3 x^3}-\frac{7 c^2}{2 b^4 x}+\frac{1}{2 b x^5 \left (b+c x^2\right )}-\frac{7 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0441998, size = 80, normalized size = 0.99 \[ -\frac{c^3 x}{2 b^4 \left (b+c x^2\right )}-\frac{3 c^2}{b^4 x}-\frac{7 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{9/2}}+\frac{2 c}{3 b^3 x^3}-\frac{1}{5 b^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(b*x^2 + c*x^4)^2),x]

[Out]

-1/(5*b^2*x^5) + (2*c)/(3*b^3*x^3) - (3*c^2)/(b^4*x) - (c^3*x)/(2*b^4*(b + c*x^2)) - (7*c^(5/2)*ArcTan[(Sqrt[c
]*x)/Sqrt[b]])/(2*b^(9/2))

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Maple [A]  time = 0.055, size = 70, normalized size = 0.9 \begin{align*} -{\frac{1}{5\,{b}^{2}{x}^{5}}}-3\,{\frac{{c}^{2}}{{b}^{4}x}}+{\frac{2\,c}{3\,{b}^{3}{x}^{3}}}-{\frac{{c}^{3}x}{2\,{b}^{4} \left ( c{x}^{2}+b \right ) }}-{\frac{7\,{c}^{3}}{2\,{b}^{4}}\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(1/x^2/(c*x^4+b*x^2)^2,x)

[Out]

-1/5/b^2/x^5-3*c^2/b^4/x+2/3*c/b^3/x^3-1/2*c^3/b^4*x/(c*x^2+b)-7/2*c^3/b^4/(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(1/x^2/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.48718, size = 423, normalized size = 5.22 \begin{align*} \left [-\frac{210 \, c^{3} x^{6} + 140 \, b c^{2} x^{4} - 28 \, b^{2} c x^{2} + 12 \, b^{3} - 105 \,{\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} - 2 \, b x \sqrt{-\frac{c}{b}} - b}{c x^{2} + b}\right )}{60 \,{\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, -\frac{105 \, c^{3} x^{6} + 70 \, b c^{2} x^{4} - 14 \, b^{2} c x^{2} + 6 \, b^{3} + 105 \,{\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right )}{30 \,{\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

[-1/60*(210*c^3*x^6 + 140*b*c^2*x^4 - 28*b^2*c*x^2 + 12*b^3 - 105*(c^3*x^7 + b*c^2*x^5)*sqrt(-c/b)*log((c*x^2
- 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^4*c*x^7 + b^5*x^5), -1/30*(105*c^3*x^6 + 70*b*c^2*x^4 - 14*b^2*c*x^2
+ 6*b^3 + 105*(c^3*x^7 + b*c^2*x^5)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^4*c*x^7 + b^5*x^5)]

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Sympy [A]  time = 0.848313, size = 126, normalized size = 1.56 \begin{align*} \frac{7 \sqrt{- \frac{c^{5}}{b^{9}}} \log{\left (- \frac{b^{5} \sqrt{- \frac{c^{5}}{b^{9}}}}{c^{3}} + x \right )}}{4} - \frac{7 \sqrt{- \frac{c^{5}}{b^{9}}} \log{\left (\frac{b^{5} \sqrt{- \frac{c^{5}}{b^{9}}}}{c^{3}} + x \right )}}{4} - \frac{6 b^{3} - 14 b^{2} c x^{2} + 70 b c^{2} x^{4} + 105 c^{3} x^{6}}{30 b^{5} x^{5} + 30 b^{4} c x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(c*x**4+b*x**2)**2,x)

[Out]

7*sqrt(-c**5/b**9)*log(-b**5*sqrt(-c**5/b**9)/c**3 + x)/4 - 7*sqrt(-c**5/b**9)*log(b**5*sqrt(-c**5/b**9)/c**3
+ x)/4 - (6*b**3 - 14*b**2*c*x**2 + 70*b*c**2*x**4 + 105*c**3*x**6)/(30*b**5*x**5 + 30*b**4*c*x**7)

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Giac [A]  time = 1.25199, size = 95, normalized size = 1.17 \begin{align*} -\frac{7 \, c^{3} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} b^{4}} - \frac{c^{3} x}{2 \,{\left (c x^{2} + b\right )} b^{4}} - \frac{45 \, c^{2} x^{4} - 10 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{4} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

-7/2*c^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) - 1/2*c^3*x/((c*x^2 + b)*b^4) - 1/15*(45*c^2*x^4 - 10*b*c*x^2 +
3*b^2)/(b^4*x^5)