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

Optimal. Leaf size=68 \[ \frac{3 x^2}{16 a^2 \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{c}}+\frac{x^2}{8 a \left (a+c x^4\right )^2} \]

[Out]

x^2/(8*a*(a + c*x^4)^2) + (3*x^2)/(16*a^2*(a + c*x^4)) + (3*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(16*a^(5/2)*Sqrt[c]
)

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Rubi [A]  time = 0.0313811, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {275, 199, 205} \[ \frac{3 x^2}{16 a^2 \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{c}}+\frac{x^2}{8 a \left (a+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + c*x^4)^3,x]

[Out]

x^2/(8*a*(a + c*x^4)^2) + (3*x^2)/(16*a^2*(a + c*x^4)) + (3*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(16*a^(5/2)*Sqrt[c]
)

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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}{\left (a+c x^4\right )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\left (a+c x^2\right )^3} \, dx,x,x^2\right )\\ &=\frac{x^2}{8 a \left (a+c x^4\right )^2}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\left (a+c x^2\right )^2} \, dx,x,x^2\right )}{8 a}\\ &=\frac{x^2}{8 a \left (a+c x^4\right )^2}+\frac{3 x^2}{16 a^2 \left (a+c x^4\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{a+c x^2} \, dx,x,x^2\right )}{16 a^2}\\ &=\frac{x^2}{8 a \left (a+c x^4\right )^2}+\frac{3 x^2}{16 a^2 \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.0423269, size = 58, normalized size = 0.85 \[ \frac{1}{16} \left (\frac{5 a x^2+3 c x^6}{a^2 \left (a+c x^4\right )^2}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{a^{5/2} \sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + c*x^4)^3,x]

[Out]

((5*a*x^2 + 3*c*x^6)/(a^2*(a + c*x^4)^2) + (3*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(a^(5/2)*Sqrt[c]))/16

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Maple [A]  time = 0.006, size = 57, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}}{8\,a \left ( c{x}^{4}+a \right ) ^{2}}}+{\frac{3\,{x}^{2}}{16\,{a}^{2} \left ( c{x}^{4}+a \right ) }}+{\frac{3}{16\,{a}^{2}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^2/a/(c*x^4+a)^2+3/16*x^2/a^2/(c*x^4+a)+3/16/a^2/(a*c)^(1/2)*arctan(x^2*c/(a*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/(c*x^4+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.93816, size = 416, normalized size = 6.12 \begin{align*} \left [\frac{6 \, a c^{2} x^{6} + 10 \, a^{2} c x^{2} - 3 \,{\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \sqrt{-a c} \log \left (\frac{c x^{4} - 2 \, \sqrt{-a c} x^{2} - a}{c x^{4} + a}\right )}{32 \,{\left (a^{3} c^{3} x^{8} + 2 \, a^{4} c^{2} x^{4} + a^{5} c\right )}}, \frac{3 \, a c^{2} x^{6} + 5 \, a^{2} c x^{2} - 3 \,{\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c}}{c x^{2}}\right )}{16 \,{\left (a^{3} c^{3} x^{8} + 2 \, a^{4} c^{2} x^{4} + a^{5} c\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(6*a*c^2*x^6 + 10*a^2*c*x^2 - 3*(c^2*x^8 + 2*a*c*x^4 + a^2)*sqrt(-a*c)*log((c*x^4 - 2*sqrt(-a*c)*x^2 - a
)/(c*x^4 + a)))/(a^3*c^3*x^8 + 2*a^4*c^2*x^4 + a^5*c), 1/16*(3*a*c^2*x^6 + 5*a^2*c*x^2 - 3*(c^2*x^8 + 2*a*c*x^
4 + a^2)*sqrt(a*c)*arctan(sqrt(a*c)/(c*x^2)))/(a^3*c^3*x^8 + 2*a^4*c^2*x^4 + a^5*c)]

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Sympy [A]  time = 1.67901, size = 110, normalized size = 1.62 \begin{align*} - \frac{3 \sqrt{- \frac{1}{a^{5} c}} \log{\left (- a^{3} \sqrt{- \frac{1}{a^{5} c}} + x^{2} \right )}}{32} + \frac{3 \sqrt{- \frac{1}{a^{5} c}} \log{\left (a^{3} \sqrt{- \frac{1}{a^{5} c}} + x^{2} \right )}}{32} + \frac{5 a x^{2} + 3 c x^{6}}{16 a^{4} + 32 a^{3} c x^{4} + 16 a^{2} c^{2} x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(-1/(a**5*c))*log(-a**3*sqrt(-1/(a**5*c)) + x**2)/32 + 3*sqrt(-1/(a**5*c))*log(a**3*sqrt(-1/(a**5*c)) +
x**2)/32 + (5*a*x**2 + 3*c*x**6)/(16*a**4 + 32*a**3*c*x**4 + 16*a**2*c**2*x**8)

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Giac [A]  time = 1.1685, size = 66, normalized size = 0.97 \begin{align*} \frac{3 \, \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} a^{2}} + \frac{3 \, c x^{6} + 5 \, a x^{2}}{16 \,{\left (c x^{4} + a\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*a^2) + 1/16*(3*c*x^6 + 5*a*x^2)/((c*x^4 + a)^2*a^2)