3.219 \(\int \frac{1}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=100 \[ -\frac{63 c^2}{8 b^5 x}-\frac{63 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{11/2}}+\frac{21 c}{8 b^4 x^3}+\frac{9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac{63}{40 b^3 x^5}+\frac{1}{4 b x^5 \left (b+c x^2\right )^2} \]

[Out]

-63/(40*b^3*x^5) + (21*c)/(8*b^4*x^3) - (63*c^2)/(8*b^5*x) + 1/(4*b*x^5*(b + c*x^2)^2) + 9/(8*b^2*x^5*(b + c*x
^2)) - (63*c^(5/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(11/2))

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Rubi [A]  time = 0.0491003, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {1593, 290, 325, 205} \[ -\frac{63 c^2}{8 b^5 x}-\frac{63 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{11/2}}+\frac{21 c}{8 b^4 x^3}+\frac{9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac{63}{40 b^3 x^5}+\frac{1}{4 b x^5 \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(-3),x]

[Out]

-63/(40*b^3*x^5) + (21*c)/(8*b^4*x^3) - (63*c^2)/(8*b^5*x) + 1/(4*b*x^5*(b + c*x^2)^2) + 9/(8*b^2*x^5*(b + c*x
^2)) - (63*c^(5/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(11/2))

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 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}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{1}{x^6 \left (b+c x^2\right )^3} \, dx\\ &=\frac{1}{4 b x^5 \left (b+c x^2\right )^2}+\frac{9 \int \frac{1}{x^6 \left (b+c x^2\right )^2} \, dx}{4 b}\\ &=\frac{1}{4 b x^5 \left (b+c x^2\right )^2}+\frac{9}{8 b^2 x^5 \left (b+c x^2\right )}+\frac{63 \int \frac{1}{x^6 \left (b+c x^2\right )} \, dx}{8 b^2}\\ &=-\frac{63}{40 b^3 x^5}+\frac{1}{4 b x^5 \left (b+c x^2\right )^2}+\frac{9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac{(63 c) \int \frac{1}{x^4 \left (b+c x^2\right )} \, dx}{8 b^3}\\ &=-\frac{63}{40 b^3 x^5}+\frac{21 c}{8 b^4 x^3}+\frac{1}{4 b x^5 \left (b+c x^2\right )^2}+\frac{9}{8 b^2 x^5 \left (b+c x^2\right )}+\frac{\left (63 c^2\right ) \int \frac{1}{x^2 \left (b+c x^2\right )} \, dx}{8 b^4}\\ &=-\frac{63}{40 b^3 x^5}+\frac{21 c}{8 b^4 x^3}-\frac{63 c^2}{8 b^5 x}+\frac{1}{4 b x^5 \left (b+c x^2\right )^2}+\frac{9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac{\left (63 c^3\right ) \int \frac{1}{b+c x^2} \, dx}{8 b^5}\\ &=-\frac{63}{40 b^3 x^5}+\frac{21 c}{8 b^4 x^3}-\frac{63 c^2}{8 b^5 x}+\frac{1}{4 b x^5 \left (b+c x^2\right )^2}+\frac{9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac{63 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.053974, size = 90, normalized size = 0.9 \[ -\frac{168 b^2 c^2 x^4-24 b^3 c x^2+8 b^4+525 b c^3 x^6+315 c^4 x^8}{40 b^5 x^5 \left (b+c x^2\right )^2}-\frac{63 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(-3),x]

[Out]

-(8*b^4 - 24*b^3*c*x^2 + 168*b^2*c^2*x^4 + 525*b*c^3*x^6 + 315*c^4*x^8)/(40*b^5*x^5*(b + c*x^2)^2) - (63*c^(5/
2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(11/2))

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Maple [A]  time = 0.056, size = 89, normalized size = 0.9 \begin{align*} -{\frac{1}{5\,{b}^{3}{x}^{5}}}-6\,{\frac{{c}^{2}}{{b}^{5}x}}+{\frac{c}{{b}^{4}{x}^{3}}}-{\frac{15\,{c}^{4}{x}^{3}}{8\,{b}^{5} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{17\,{c}^{3}x}{8\,{b}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{63\,{c}^{3}}{8\,{b}^{5}}\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/(c*x^4+b*x^2)^3,x)

[Out]

-1/5/b^3/x^5-6*c^2/b^5/x+c/b^4/x^3-15/8/b^5*c^4/(c*x^2+b)^2*x^3-17/8/b^4*c^3/(c*x^2+b)^2*x-63/8/b^5*c^3/(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/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55846, size = 560, normalized size = 5.6 \begin{align*} \left [-\frac{630 \, c^{4} x^{8} + 1050 \, b c^{3} x^{6} + 336 \, b^{2} c^{2} x^{4} - 48 \, b^{3} c x^{2} + 16 \, b^{4} - 315 \,{\left (c^{4} x^{9} + 2 \, b c^{3} x^{7} + b^{2} 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 )}{80 \,{\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}, -\frac{315 \, c^{4} x^{8} + 525 \, b c^{3} x^{6} + 168 \, b^{2} c^{2} x^{4} - 24 \, b^{3} c x^{2} + 8 \, b^{4} + 315 \,{\left (c^{4} x^{9} + 2 \, b c^{3} x^{7} + b^{2} c^{2} x^{5}\right )} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right )}{40 \,{\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

[-1/80*(630*c^4*x^8 + 1050*b*c^3*x^6 + 336*b^2*c^2*x^4 - 48*b^3*c*x^2 + 16*b^4 - 315*(c^4*x^9 + 2*b*c^3*x^7 +
b^2*c^2*x^5)*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5)
, -1/40*(315*c^4*x^8 + 525*b*c^3*x^6 + 168*b^2*c^2*x^4 - 24*b^3*c*x^2 + 8*b^4 + 315*(c^4*x^9 + 2*b*c^3*x^7 + b
^2*c^2*x^5)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5)]

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Sympy [A]  time = 1.75648, size = 150, normalized size = 1.5 \begin{align*} \frac{63 \sqrt{- \frac{c^{5}}{b^{11}}} \log{\left (- \frac{b^{6} \sqrt{- \frac{c^{5}}{b^{11}}}}{c^{3}} + x \right )}}{16} - \frac{63 \sqrt{- \frac{c^{5}}{b^{11}}} \log{\left (\frac{b^{6} \sqrt{- \frac{c^{5}}{b^{11}}}}{c^{3}} + x \right )}}{16} - \frac{8 b^{4} - 24 b^{3} c x^{2} + 168 b^{2} c^{2} x^{4} + 525 b c^{3} x^{6} + 315 c^{4} x^{8}}{40 b^{7} x^{5} + 80 b^{6} c x^{7} + 40 b^{5} c^{2} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**4+b*x**2)**3,x)

[Out]

63*sqrt(-c**5/b**11)*log(-b**6*sqrt(-c**5/b**11)/c**3 + x)/16 - 63*sqrt(-c**5/b**11)*log(b**6*sqrt(-c**5/b**11
)/c**3 + x)/16 - (8*b**4 - 24*b**3*c*x**2 + 168*b**2*c**2*x**4 + 525*b*c**3*x**6 + 315*c**4*x**8)/(40*b**7*x**
5 + 80*b**6*c*x**7 + 40*b**5*c**2*x**9)

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Giac [A]  time = 1.30243, size = 108, normalized size = 1.08 \begin{align*} -\frac{63 \, c^{3} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} b^{5}} - \frac{15 \, c^{4} x^{3} + 17 \, b c^{3} x}{8 \,{\left (c x^{2} + b\right )}^{2} b^{5}} - \frac{30 \, c^{2} x^{4} - 5 \, b c x^{2} + b^{2}}{5 \, b^{5} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-63/8*c^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^5) - 1/8*(15*c^4*x^3 + 17*b*c^3*x)/((c*x^2 + b)^2*b^5) - 1/5*(30*
c^2*x^4 - 5*b*c*x^2 + b^2)/(b^5*x^5)