\(\int \frac {1}{x^3 (a+c x^4)^2} \, dx\) [665]

Optimal result

Integrand size = 13, antiderivative size = 59 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^2} \, dx=-\frac {3}{4 a^2 x^2}+\frac {1}{4 a x^2 \left (a+c x^4\right )}-\frac {3 \sqrt {c} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 296, 331, 211} \[ \int \frac {1}{x^3 \left (a+c x^4\right )^2} \, dx=-\frac {3 \sqrt {c} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{5/2}}-\frac {3}{4 a^2 x^2}+\frac {1}{4 a x^2 \left (a+c x^4\right )} \]

[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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 281

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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {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 \text {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) \text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.59 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^2} \, dx=\frac {-\frac {\sqrt {a} \left (2 a+3 c x^4\right )}{x^2 \left (a+c x^4\right )}+3 \sqrt {c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+3 \sqrt {c} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{4 a^{5/2}} \]

[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))

Maple [A] (verified)

Time = 3.90 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {1}{2 a^{2} x^{2}}-\frac {c \left (\frac {x^{2}}{2 x^{4} c +2 a}+\frac {3 \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{2 \sqrt {a c}}\right )}{2 a^{2}}\)	\(49\)
risch	\(\frac {-\frac {3 c \,x^{4}}{4 a^{2}}-\frac {1}{2 a}}{x^{2} \left (x^{4} c +a \right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} \textit {\_Z}^{2}+c \right )}{\sum }\textit {\_R} \ln \left (\left (-5 a^{5} \textit {\_R}^{2}-4 c \right ) x^{2}-a^{3} \textit {\_R} \right )\right )}{8}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.58 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^2} \, dx=\left [-\frac {6 \, c x^{4} - 3 \, {\left (c x^{6} + a x^{2}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} - 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) + 4 \, a}{8 \, {\left (a^{2} c x^{6} + a^{3} x^{2}\right )}}, -\frac {3 \, c x^{4} - 3 \, {\left (c x^{6} + a x^{2}\right )} \sqrt {\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{a}}}{c x^{2}}\right ) + 2 \, a}{4 \, {\left (a^{2} c x^{6} + a^{3} x^{2}\right )}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.64 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^2} \, dx=\frac {3 \sqrt {- \frac {c}{a^{5}}} \log {\left (- \frac {a^{3} \sqrt {- \frac {c}{a^{5}}}}{c} + x^{2} \right )}}{8} - \frac {3 \sqrt {- \frac {c}{a^{5}}} \log {\left (\frac {a^{3} \sqrt {- \frac {c}{a^{5}}}}{c} + x^{2} \right )}}{8} + \frac {- 2 a - 3 c x^{4}}{4 a^{3} x^{2} + 4 a^{2} c x^{6}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^2} \, dx=-\frac {3 \, c x^{4} + 2 \, a}{4 \, {\left (a^{2} c x^{6} + a^{3} x^{2}\right )}} - \frac {3 \, c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, \sqrt {a c} a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^2} \, dx=-\frac {3 \, c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, \sqrt {a c} a^{2}} - \frac {3 \, c x^{4} + 2 \, a}{4 \, {\left (c x^{6} + a x^{2}\right )} a^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^2} \, dx=-\frac {\frac {1}{2\,a}+\frac {3\,c\,x^4}{4\,a^2}}{c\,x^6+a\,x^2}-\frac {3\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x^2}{\sqrt {a}}\right )}{4\,a^{5/2}} \]

[In]

int(1/(x^3*(a + c*x^4)^2),x)

[Out]

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