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

Optimal. Leaf size=59 \[ -\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{5/2}}-\frac{3}{4 a^2 x^2}+\frac{1}{4 a x^2 \left (a+c x^4\right )} \]

[Out]

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

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Rubi [A]  time = 0.0314795, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {275, 290, 325, 205} \[ -\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{5/2}}-\frac{3}{4 a^2 x^2}+\frac{1}{4 a x^2 \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + c*x^4)^2),x]

[Out]

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [A]  time = 0.0531938, size = 94, normalized size = 1.59 \[ \frac{-\frac{\sqrt{a} \left (2 a+3 c x^4\right )}{x^2 \left (a+c x^4\right )}+3 \sqrt{c} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{4 a^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + c*x^4)^2),x]

[Out]

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

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Maple [A]  time = 0.012, size = 50, normalized size = 0.9 \begin{align*} -{\frac{c{x}^{2}}{4\,{a}^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{3\,c}{4\,{a}^{2}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{1}{2\,{a}^{2}{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(c*x^4+a)^2,x)

[Out]

-1/4/a^2*c*x^2/(c*x^4+a)-3/4/a^2*c/(a*c)^(1/2)*arctan(x^2*c/(a*c)^(1/2))-1/2/a^2/x^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^3/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.75888, size = 312, normalized size = 5.29 \begin{align*} \left [-\frac{6 \, c x^{4} - 3 \,{\left (c x^{6} + a x^{2}\right )} \sqrt{-\frac{c}{a}} \log \left (\frac{c x^{4} - 2 \, a x^{2} \sqrt{-\frac{c}{a}} - a}{c x^{4} + a}\right ) + 4 \, a}{8 \,{\left (a^{2} c x^{6} + a^{3} x^{2}\right )}}, -\frac{3 \, c x^{4} - 3 \,{\left (c x^{6} + a x^{2}\right )} \sqrt{\frac{c}{a}} \arctan \left (\frac{a \sqrt{\frac{c}{a}}}{c x^{2}}\right ) + 2 \, a}{4 \,{\left (a^{2} c x^{6} + a^{3} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(6*c*x^4 - 3*(c*x^6 + a*x^2)*sqrt(-c/a)*log((c*x^4 - 2*a*x^2*sqrt(-c/a) - a)/(c*x^4 + a)) + 4*a)/(a^2*c*
x^6 + a^3*x^2), -1/4*(3*c*x^4 - 3*(c*x^6 + a*x^2)*sqrt(c/a)*arctan(a*sqrt(c/a)/(c*x^2)) + 2*a)/(a^2*c*x^6 + a^
3*x^2)]

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Sympy [A]  time = 1.20302, size = 95, normalized size = 1.61 \begin{align*} \frac{3 \sqrt{- \frac{c}{a^{5}}} \log{\left (- \frac{a^{3} \sqrt{- \frac{c}{a^{5}}}}{c} + x^{2} \right )}}{8} - \frac{3 \sqrt{- \frac{c}{a^{5}}} \log{\left (\frac{a^{3} \sqrt{- \frac{c}{a^{5}}}}{c} + x^{2} \right )}}{8} - \frac{2 a + 3 c x^{4}}{4 a^{3} x^{2} + 4 a^{2} c x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(-c/a**5)*log(-a**3*sqrt(-c/a**5)/c + x**2)/8 - 3*sqrt(-c/a**5)*log(a**3*sqrt(-c/a**5)/c + x**2)/8 - (2*
a + 3*c*x**4)/(4*a**3*x**2 + 4*a**2*c*x**6)

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Giac [A]  time = 1.12101, size = 69, normalized size = 1.17 \begin{align*} -\frac{3 \, c \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{4 \, \sqrt{a c} a^{2}} - \frac{3 \, c x^{4} + 2 \, a}{4 \,{\left (c x^{6} + a x^{2}\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*c*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*a^2) - 1/4*(3*c*x^4 + 2*a)/((c*x^6 + a*x^2)*a^2)