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

Optimal. Leaf size=78 \[ \frac{5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac{15 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{7/2}}-\frac{15}{16 a^3 x^2}+\frac{1}{8 a x^2 \left (a+c x^4\right )^2} \]

[Out]

-15/(16*a^3*x^2) + 1/(8*a*x^2*(a + c*x^4)^2) + 5/(16*a^2*x^2*(a + c*x^4)) - (15*Sqrt[c]*ArcTan[(Sqrt[c]*x^2)/S
qrt[a]])/(16*a^(7/2))

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Rubi [A]  time = 0.0432071, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {275, 290, 325, 205} \[ \frac{5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac{15 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{7/2}}-\frac{15}{16 a^3 x^2}+\frac{1}{8 a x^2 \left (a+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + c*x^4)^3),x]

[Out]

-15/(16*a^3*x^2) + 1/(8*a*x^2*(a + c*x^4)^2) + 5/(16*a^2*x^2*(a + c*x^4)) - (15*Sqrt[c]*ArcTan[(Sqrt[c]*x^2)/S
qrt[a]])/(16*a^(7/2))

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 )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+c x^2\right )^3} \, dx,x,x^2\right )\\ &=\frac{1}{8 a x^2 \left (a+c x^4\right )^2}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+c x^2\right )^2} \, dx,x,x^2\right )}{8 a}\\ &=\frac{1}{8 a x^2 \left (a+c x^4\right )^2}+\frac{5}{16 a^2 x^2 \left (a+c x^4\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+c x^2\right )} \, dx,x,x^2\right )}{16 a^2}\\ &=-\frac{15}{16 a^3 x^2}+\frac{1}{8 a x^2 \left (a+c x^4\right )^2}+\frac{5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac{(15 c) \operatorname{Subst}\left (\int \frac{1}{a+c x^2} \, dx,x,x^2\right )}{16 a^3}\\ &=-\frac{15}{16 a^3 x^2}+\frac{1}{8 a x^2 \left (a+c x^4\right )^2}+\frac{5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac{15 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{16 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.080893, size = 105, normalized size = 1.35 \[ \frac{-\frac{\sqrt{a} \left (8 a^2+25 a c x^4+15 c^2 x^8\right )}{x^2 \left (a+c x^4\right )^2}+15 \sqrt{c} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+15 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{16 a^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + c*x^4)^3),x]

[Out]

(-((Sqrt[a]*(8*a^2 + 25*a*c*x^4 + 15*c^2*x^8))/(x^2*(a + c*x^4)^2)) + 15*Sqrt[c]*ArcTan[1 - (Sqrt[2]*c^(1/4)*x
)/a^(1/4)] + 15*Sqrt[c]*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(16*a^(7/2))

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Maple [A]  time = 0.015, size = 70, normalized size = 0.9 \begin{align*} -{\frac{7\,{c}^{2}{x}^{6}}{16\,{a}^{3} \left ( c{x}^{4}+a \right ) ^{2}}}-{\frac{9\,c{x}^{2}}{16\,{a}^{2} \left ( c{x}^{4}+a \right ) ^{2}}}-{\frac{15\,c}{16\,{a}^{3}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{1}{2\,{x}^{2}{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/16*c^2/a^3/(c*x^4+a)^2*x^6-9/16*c/a^2/(c*x^4+a)^2*x^2-15/16*c/a^3/(a*c)^(1/2)*arctan(x^2*c/(a*c)^(1/2))-1/2
/x^2/a^3

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83311, size = 459, normalized size = 5.88 \begin{align*} \left [-\frac{30 \, c^{2} x^{8} + 50 \, a c x^{4} - 15 \,{\left (c^{2} x^{10} + 2 \, a c x^{6} + a^{2} 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 ) + 16 \, a^{2}}{32 \,{\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}}, -\frac{15 \, c^{2} x^{8} + 25 \, a c x^{4} - 15 \,{\left (c^{2} x^{10} + 2 \, a c x^{6} + a^{2} x^{2}\right )} \sqrt{\frac{c}{a}} \arctan \left (\frac{a \sqrt{\frac{c}{a}}}{c x^{2}}\right ) + 8 \, a^{2}}{16 \,{\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} 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)^3,x, algorithm="fricas")

[Out]

[-1/32*(30*c^2*x^8 + 50*a*c*x^4 - 15*(c^2*x^10 + 2*a*c*x^6 + a^2*x^2)*sqrt(-c/a)*log((c*x^4 - 2*a*x^2*sqrt(-c/
a) - a)/(c*x^4 + a)) + 16*a^2)/(a^3*c^2*x^10 + 2*a^4*c*x^6 + a^5*x^2), -1/16*(15*c^2*x^8 + 25*a*c*x^4 - 15*(c^
2*x^10 + 2*a*c*x^6 + a^2*x^2)*sqrt(c/a)*arctan(a*sqrt(c/a)/(c*x^2)) + 8*a^2)/(a^3*c^2*x^10 + 2*a^4*c*x^6 + a^5
*x^2)]

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Sympy [A]  time = 9.4252, size = 119, normalized size = 1.53 \begin{align*} \frac{15 \sqrt{- \frac{c}{a^{7}}} \log{\left (- \frac{a^{4} \sqrt{- \frac{c}{a^{7}}}}{c} + x^{2} \right )}}{32} - \frac{15 \sqrt{- \frac{c}{a^{7}}} \log{\left (\frac{a^{4} \sqrt{- \frac{c}{a^{7}}}}{c} + x^{2} \right )}}{32} - \frac{8 a^{2} + 25 a c x^{4} + 15 c^{2} x^{8}}{16 a^{5} x^{2} + 32 a^{4} c x^{6} + 16 a^{3} c^{2} x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*sqrt(-c/a**7)*log(-a**4*sqrt(-c/a**7)/c + x**2)/32 - 15*sqrt(-c/a**7)*log(a**4*sqrt(-c/a**7)/c + x**2)/32 -
(8*a**2 + 25*a*c*x**4 + 15*c**2*x**8)/(16*a**5*x**2 + 32*a**4*c*x**6 + 16*a**3*c**2*x**10)

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Giac [A]  time = 1.15483, size = 82, normalized size = 1.05 \begin{align*} -\frac{15 \, c \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} a^{3}} - \frac{7 \, c^{2} x^{6} + 9 \, a c x^{2}}{16 \,{\left (c x^{4} + a\right )}^{2} a^{3}} - \frac{1}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/16*c*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*a^3) - 1/16*(7*c^2*x^6 + 9*a*c*x^2)/((c*x^4 + a)^2*a^3) - 1/2/(a^3
*x^2)