3.675 \(\int \frac{x^5}{(a+c x^4)^3} \, dx\)

Optimal. Leaf size=71 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{3/2} c^{3/2}}+\frac{x^2}{16 a c \left (a+c x^4\right )}-\frac{x^2}{8 c \left (a+c x^4\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0346298, antiderivative size = 71, 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, 288, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{3/2} c^{3/2}}+\frac{x^2}{16 a c \left (a+c x^4\right )}-\frac{x^2}{8 c \left (a+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + c*x^4)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.030823, size = 62, normalized size = 0.87 \[ \frac{\frac{\sqrt{a} \sqrt{c} x^2 \left (c x^4-a\right )}{\left (a+c x^4\right )^2}+\tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{3/2} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + c*x^4)^3,x]

[Out]

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

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Maple [A]  time = 0.012, size = 54, normalized size = 0.8 \begin{align*}{\frac{1}{2\, \left ( c{x}^{4}+a \right ) ^{2}} \left ({\frac{{x}^{6}}{8\,a}}-{\frac{{x}^{2}}{8\,c}} \right ) }+{\frac{1}{16\,ac}\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^5/(c*x^4+a)^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71089, size = 409, normalized size = 5.76 \begin{align*} \left [\frac{2 \, a c^{2} x^{6} - 2 \, a^{2} c x^{2} -{\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^{2} c^{4} x^{8} + 2 \, a^{3} c^{3} x^{4} + a^{4} c^{2}\right )}}, \frac{a c^{2} x^{6} - a^{2} c x^{2} -{\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^{2} c^{4} x^{8} + 2 \, a^{3} c^{3} x^{4} + a^{4} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.2218, size = 116, normalized size = 1.63 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{3} c^{3}}} \log{\left (- a^{2} c \sqrt{- \frac{1}{a^{3} c^{3}}} + x^{2} \right )}}{32} + \frac{\sqrt{- \frac{1}{a^{3} c^{3}}} \log{\left (a^{2} c \sqrt{- \frac{1}{a^{3} c^{3}}} + x^{2} \right )}}{32} + \frac{- a x^{2} + c x^{6}}{16 a^{3} c + 32 a^{2} c^{2} x^{4} + 16 a c^{3} x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**4+a)**3,x)

[Out]

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

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Giac [A]  time = 1.15702, size = 73, normalized size = 1.03 \begin{align*} \frac{\arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} a c} + \frac{c x^{6} - a x^{2}}{16 \,{\left (c x^{4} + a\right )}^{2} a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+a)^3,x, algorithm="giac")

[Out]

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