\(\int \frac {1}{x (a+c x^4)^2} \, dx\) [664]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int \frac {1}{x \left (a+c x^4\right )^2} \, dx=\frac {1}{4 a \left (a+c x^4\right )}+\frac {\log (x)}{a^2}-\frac {\log \left (a+c x^4\right )}{4 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 46} \[ \int \frac {1}{x \left (a+c x^4\right )^2} \, dx=-\frac {\log \left (a+c x^4\right )}{4 a^2}+\frac {\log (x)}{a^2}+\frac {1}{4 a \left (a+c x^4\right )} \]

[In]

Int[1/(x*(a + c*x^4)^2),x]

[Out]

1/(4*a*(a + c*x^4)) + Log[x]/a^2 - Log[a + c*x^4]/(4*a^2)

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {1}{x (a+c x)^2} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{a^2 x}-\frac {c}{a (a+c x)^2}-\frac {c}{a^2 (a+c x)}\right ) \, dx,x,x^4\right ) \\ & = \frac {1}{4 a \left (a+c x^4\right )}+\frac {\log (x)}{a^2}-\frac {\log \left (a+c x^4\right )}{4 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x \left (a+c x^4\right )^2} \, dx=\frac {\frac {a}{a+c x^4}+4 \log (x)-\log \left (a+c x^4\right )}{4 a^2} \]

[In]

Integrate[1/(x*(a + c*x^4)^2),x]

[Out]

(a/(a + c*x^4) + 4*Log[x] - Log[a + c*x^4])/(4*a^2)

Maple [A] (verified)

Time = 3.88 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {1}{4 a \left (x^{4} c +a \right )}+\frac {\ln \left (x \right )}{a^{2}}-\frac {\ln \left (x^{4} c +a \right )}{4 a^{2}}\)	\(35\)
norman	\(-\frac {c \,x^{4}}{4 a^{2} \left (x^{4} c +a \right )}+\frac {\ln \left (x \right )}{a^{2}}-\frac {\ln \left (x^{4} c +a \right )}{4 a^{2}}\)	\(39\)
default	\(\frac {\ln \left (x \right )}{a^{2}}-\frac {c \left (-\frac {a}{2 c \left (x^{4} c +a \right )}+\frac {\ln \left (x^{4} c +a \right )}{2 c}\right )}{2 a^{2}}\)	\(43\)
parallelrisch	\(\frac {4 c \ln \left (x \right ) x^{4}-c \ln \left (x^{4} c +a \right ) x^{4}-x^{4} c +4 a \ln \left (x \right )-a \ln \left (x^{4} c +a \right )}{4 a^{2} \left (x^{4} c +a \right )}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x \left (a+c x^4\right )^2} \, dx=-\frac {{\left (c x^{4} + a\right )} \log \left (c x^{4} + a\right ) - 4 \, {\left (c x^{4} + a\right )} \log \left (x\right ) - a}{4 \, {\left (a^{2} c x^{4} + a^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (a+c x^4\right )^2} \, dx=\frac {1}{4 a^{2} + 4 a c x^{4}} + \frac {\log {\left (x \right )}}{a^{2}} - \frac {\log {\left (\frac {a}{c} + x^{4} \right )}}{4 a^{2}} \]

[In]

integrate(1/x/(c*x**4+a)**2,x)

[Out]

1/(4*a**2 + 4*a*c*x**4) + log(x)/a**2 - log(a/c + x**4)/(4*a**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x \left (a+c x^4\right )^2} \, dx=\frac {1}{4 \, {\left (a c x^{4} + a^{2}\right )}} - \frac {\log \left (c x^{4} + a\right )}{4 \, a^{2}} + \frac {\log \left (x^{4}\right )}{4 \, a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x \left (a+c x^4\right )^2} \, dx=\frac {\log \left (x^{4}\right )}{4 \, a^{2}} - \frac {\log \left ({\left | c x^{4} + a \right |}\right )}{4 \, a^{2}} + \frac {c x^{4} + 2 \, a}{4 \, {\left (c x^{4} + a\right )} a^{2}} \]

[In]

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

[Out]

1/4*log(x^4)/a^2 - 1/4*log(abs(c*x^4 + a))/a^2 + 1/4*(c*x^4 + 2*a)/((c*x^4 + a)*a^2)

Mupad [B] (verification not implemented)

Time = 5.86 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (a+c x^4\right )^2} \, dx=\frac {\ln \left (x\right )}{a^2}+\frac {1}{4\,a\,\left (c\,x^4+a\right )}-\frac {\ln \left (c\,x^4+a\right )}{4\,a^2} \]

[In]

int(1/(x*(a + c*x^4)^2),x)

[Out]

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