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

Optimal. Leaf size=87 \[ \frac{35 c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{9/2}}+\frac{7}{8 b^2 x^3 \left (b+c x^2\right )}+\frac{35 c}{8 b^4 x}-\frac{35}{24 b^3 x^3}+\frac{1}{4 b x^3 \left (b+c x^2\right )^2} \]

[Out]

-35/(24*b^3*x^3) + (35*c)/(8*b^4*x) + 1/(4*b*x^3*(b + c*x^2)^2) + 7/(8*b^2*x^3*(b + c*x^2)) + (35*c^(3/2)*ArcT
an[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(9/2))

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Rubi [A]  time = 0.0418952, antiderivative size = 87, 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} \[ \frac{35 c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{9/2}}+\frac{7}{8 b^2 x^3 \left (b+c x^2\right )}+\frac{35 c}{8 b^4 x}-\frac{35}{24 b^3 x^3}+\frac{1}{4 b x^3 \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(b*x^2 + c*x^4)^3,x]

[Out]

-35/(24*b^3*x^3) + (35*c)/(8*b^4*x) + 1/(4*b*x^3*(b + c*x^2)^2) + 7/(8*b^2*x^3*(b + c*x^2)) + (35*c^(3/2)*ArcT
an[(Sqrt[c]*x)/Sqrt[b]])/(8*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{x^2}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{1}{x^4 \left (b+c x^2\right )^3} \, dx\\ &=\frac{1}{4 b x^3 \left (b+c x^2\right )^2}+\frac{7 \int \frac{1}{x^4 \left (b+c x^2\right )^2} \, dx}{4 b}\\ &=\frac{1}{4 b x^3 \left (b+c x^2\right )^2}+\frac{7}{8 b^2 x^3 \left (b+c x^2\right )}+\frac{35 \int \frac{1}{x^4 \left (b+c x^2\right )} \, dx}{8 b^2}\\ &=-\frac{35}{24 b^3 x^3}+\frac{1}{4 b x^3 \left (b+c x^2\right )^2}+\frac{7}{8 b^2 x^3 \left (b+c x^2\right )}-\frac{(35 c) \int \frac{1}{x^2 \left (b+c x^2\right )} \, dx}{8 b^3}\\ &=-\frac{35}{24 b^3 x^3}+\frac{35 c}{8 b^4 x}+\frac{1}{4 b x^3 \left (b+c x^2\right )^2}+\frac{7}{8 b^2 x^3 \left (b+c x^2\right )}+\frac{\left (35 c^2\right ) \int \frac{1}{b+c x^2} \, dx}{8 b^4}\\ &=-\frac{35}{24 b^3 x^3}+\frac{35 c}{8 b^4 x}+\frac{1}{4 b x^3 \left (b+c x^2\right )^2}+\frac{7}{8 b^2 x^3 \left (b+c x^2\right )}+\frac{35 c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0435254, size = 79, normalized size = 0.91 \[ \frac{56 b^2 c x^2-8 b^3+175 b c^2 x^4+105 c^3 x^6}{24 b^4 x^3 \left (b+c x^2\right )^2}+\frac{35 c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b*x^2 + c*x^4)^3,x]

[Out]

(-8*b^3 + 56*b^2*c*x^2 + 175*b*c^2*x^4 + 105*c^3*x^6)/(24*b^4*x^3*(b + c*x^2)^2) + (35*c^(3/2)*ArcTan[(Sqrt[c]
*x)/Sqrt[b]])/(8*b^(9/2))

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Maple [A]  time = 0.056, size = 79, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,{b}^{3}{x}^{3}}}+3\,{\frac{c}{{b}^{4}x}}+{\frac{11\,{c}^{3}{x}^{3}}{8\,{b}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{13\,{c}^{2}x}{8\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{35\,{c}^{2}}{8\,{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(x^2/(c*x^4+b*x^2)^3,x)

[Out]

-1/3/b^3/x^3+3*c/b^4/x+11/8/b^4*c^3/(c*x^2+b)^2*x^3+13/8/b^3*c^2/(c*x^2+b)^2*x+35/8/b^4*c^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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80281, size = 504, normalized size = 5.79 \begin{align*} \left [\frac{210 \, c^{3} x^{6} + 350 \, b c^{2} x^{4} + 112 \, b^{2} c x^{2} - 16 \, b^{3} + 105 \,{\left (c^{3} x^{7} + 2 \, b c^{2} x^{5} + b^{2} c x^{3}\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} + 2 \, b x \sqrt{-\frac{c}{b}} - b}{c x^{2} + b}\right )}{48 \,{\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}, \frac{105 \, c^{3} x^{6} + 175 \, b c^{2} x^{4} + 56 \, b^{2} c x^{2} - 8 \, b^{3} + 105 \,{\left (c^{3} x^{7} + 2 \, b c^{2} x^{5} + b^{2} c x^{3}\right )} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right )}{24 \,{\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(210*c^3*x^6 + 350*b*c^2*x^4 + 112*b^2*c*x^2 - 16*b^3 + 105*(c^3*x^7 + 2*b*c^2*x^5 + b^2*c*x^3)*sqrt(-c/
b)*log((c*x^2 + 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3), 1/24*(105*c^3*x^6 +
175*b*c^2*x^4 + 56*b^2*c*x^2 - 8*b^3 + 105*(c^3*x^7 + 2*b*c^2*x^5 + b^2*c*x^3)*sqrt(c/b)*arctan(x*sqrt(c/b)))
/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3)]

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Sympy [A]  time = 0.927996, size = 138, normalized size = 1.59 \begin{align*} - \frac{35 \sqrt{- \frac{c^{3}}{b^{9}}} \log{\left (- \frac{b^{5} \sqrt{- \frac{c^{3}}{b^{9}}}}{c^{2}} + x \right )}}{16} + \frac{35 \sqrt{- \frac{c^{3}}{b^{9}}} \log{\left (\frac{b^{5} \sqrt{- \frac{c^{3}}{b^{9}}}}{c^{2}} + x \right )}}{16} + \frac{- 8 b^{3} + 56 b^{2} c x^{2} + 175 b c^{2} x^{4} + 105 c^{3} x^{6}}{24 b^{6} x^{3} + 48 b^{5} c x^{5} + 24 b^{4} c^{2} x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-35*sqrt(-c**3/b**9)*log(-b**5*sqrt(-c**3/b**9)/c**2 + x)/16 + 35*sqrt(-c**3/b**9)*log(b**5*sqrt(-c**3/b**9)/c
**2 + x)/16 + (-8*b**3 + 56*b**2*c*x**2 + 175*b*c**2*x**4 + 105*c**3*x**6)/(24*b**6*x**3 + 48*b**5*c*x**5 + 24
*b**4*c**2*x**7)

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Giac [A]  time = 1.22641, size = 96, normalized size = 1.1 \begin{align*} \frac{35 \, c^{2} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} b^{4}} + \frac{11 \, c^{3} x^{3} + 13 \, b c^{2} x}{8 \,{\left (c x^{2} + b\right )}^{2} b^{4}} + \frac{9 \, c x^{2} - b}{3 \, b^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8*c^2*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) + 1/8*(11*c^3*x^3 + 13*b*c^2*x)/((c*x^2 + b)^2*b^4) + 1/3*(9*c*
x^2 - b)/(b^4*x^3)