3.673 \(\int \frac{x^9}{(a+c x^4)^3} \, dx\)

Optimal. Leaf size=68 \[ -\frac{3 x^2}{16 c^2 \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 \sqrt{a} c^{5/2}}-\frac{x^6}{8 c \left (a+c x^4\right )^2} \]

[Out]

-x^6/(8*c*(a + c*x^4)^2) - (3*x^2)/(16*c^2*(a + c*x^4)) + (3*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(16*Sqrt[a]*c^(5/2
))

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Rubi [A]  time = 0.035788, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {275, 288, 205} \[ -\frac{3 x^2}{16 c^2 \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 \sqrt{a} c^{5/2}}-\frac{x^6}{8 c \left (a+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(a + c*x^4)^3,x]

[Out]

-x^6/(8*c*(a + c*x^4)^2) - (3*x^2)/(16*c^2*(a + c*x^4)) + (3*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(16*Sqrt[a]*c^(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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& 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^9}{\left (a+c x^4\right )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{\left (a+c x^2\right )^3} \, dx,x,x^2\right )\\ &=-\frac{x^6}{8 c \left (a+c x^4\right )^2}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+c x^2\right )^2} \, dx,x,x^2\right )}{8 c}\\ &=-\frac{x^6}{8 c \left (a+c x^4\right )^2}-\frac{3 x^2}{16 c^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 c^2}\\ &=-\frac{x^6}{8 c \left (a+c x^4\right )^2}-\frac{3 x^2}{16 c^2 \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 \sqrt{a} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0498011, size = 58, normalized size = 0.85 \[ \frac{1}{16} \left (\frac{-3 a x^2-5 c x^6}{c^2 \left (a+c x^4\right )^2}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{\sqrt{a} c^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a + c*x^4)^3,x]

[Out]

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

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Maple [A]  time = 0.01, size = 52, normalized size = 0.8 \begin{align*}{\frac{1}{2\, \left ( c{x}^{4}+a \right ) ^{2}} \left ( -{\frac{5\,{x}^{6}}{8\,c}}-{\frac{3\,a{x}^{2}}{8\,{c}^{2}}} \right ) }+{\frac{3}{16\,{c}^{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^9/(c*x^4+a)^3,x)

[Out]

1/2*(-5/8*x^6/c-3/8*a*x^2/c^2)/(c*x^4+a)^2+3/16/c^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^9/(c*x^4+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74146, size = 419, normalized size = 6.16 \begin{align*} \left [-\frac{10 \, a c^{2} x^{6} + 6 \, 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 c^{5} x^{8} + 2 \, a^{2} c^{4} x^{4} + a^{3} c^{3}\right )}}, -\frac{5 \, a c^{2} x^{6} + 3 \, 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 c^{5} x^{8} + 2 \, a^{2} c^{4} x^{4} + a^{3} c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.46144, size = 114, normalized size = 1.68 \begin{align*} - \frac{3 \sqrt{- \frac{1}{a c^{5}}} \log{\left (- a c^{2} \sqrt{- \frac{1}{a c^{5}}} + x^{2} \right )}}{32} + \frac{3 \sqrt{- \frac{1}{a c^{5}}} \log{\left (a c^{2} \sqrt{- \frac{1}{a c^{5}}} + x^{2} \right )}}{32} - \frac{3 a x^{2} + 5 c x^{6}}{16 a^{2} c^{2} + 32 a c^{3} x^{4} + 16 c^{4} x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(c*x**4+a)**3,x)

[Out]

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

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Giac [A]  time = 1.12835, size = 66, normalized size = 0.97 \begin{align*} \frac{3 \, \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} c^{2}} - \frac{5 \, c x^{6} + 3 \, a x^{2}}{16 \,{\left (c x^{4} + a\right )}^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+a)^3,x, algorithm="giac")

[Out]

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