\(\int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx\) [366]

Optimal result

Integrand size = 20, antiderivative size = 16 \[ \int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx=\frac {\frac {15 x^2}{2}+\log ^2(x)}{x^4} \]

[Out]

(ln(x)^2+15/2*x^2)/x^4

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 2341, 2342} \[ \int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx=\frac {\log ^2(x)}{x^4}+\frac {15}{2 x^2} \]

[In]

Int[(-15*x^2 + 2*Log[x] - 4*Log[x]^2)/x^5,x]

[Out]

15/(2*x^2) + Log[x]^2/x^4

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {15}{x^3}+\frac {2 \log (x)}{x^5}-\frac {4 \log ^2(x)}{x^5}\right ) \, dx \\ & = \frac {15}{2 x^2}+2 \int \frac {\log (x)}{x^5} \, dx-4 \int \frac {\log ^2(x)}{x^5} \, dx \\ & = -\frac {1}{8 x^4}+\frac {15}{2 x^2}-\frac {\log (x)}{2 x^4}+\frac {\log ^2(x)}{x^4}-2 \int \frac {\log (x)}{x^5} \, dx \\ & = \frac {15}{2 x^2}+\frac {\log ^2(x)}{x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx=\frac {15}{2 x^2}+\frac {\log ^2(x)}{x^4} \]

[In]

Integrate[(-15*x^2 + 2*Log[x] - 4*Log[x]^2)/x^5,x]

[Out]

15/(2*x^2) + Log[x]^2/x^4

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\ln \left (x \right )^{2}}{x^{4}}+\frac {15}{2 x^{2}}\)	\(15\)
norman	\(\frac {\ln \left (x \right )^{2}+\frac {15 x^{2}}{2}}{x^{4}}\)	\(15\)
risch	\(\frac {\ln \left (x \right )^{2}}{x^{4}}+\frac {15}{2 x^{2}}\)	\(15\)
parts	\(\frac {\ln \left (x \right )^{2}}{x^{4}}+\frac {15}{2 x^{2}}\)	\(15\)
parallelrisch	\(\frac {15 x^{2}+2 \ln \left (x \right )^{2}}{2 x^{4}}\)	\(18\)

[In]

int((-4*ln(x)^2+2*ln(x)-15*x^2)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/x^4*ln(x)^2+15/2/x^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx=\frac {15 \, x^{2} + 2 \, \log \left (x\right )^{2}}{2 \, x^{4}} \]

[In]

integrate((-4*log(x)^2+2*log(x)-15*x^2)/x^5,x, algorithm="fricas")

[Out]

1/2*(15*x^2 + 2*log(x)^2)/x^4

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx=\frac {15}{2 x^{2}} + \frac {\log {\left (x \right )}^{2}}{x^{4}} \]

[In]

integrate((-4*ln(x)**2+2*ln(x)-15*x**2)/x**5,x)

[Out]

15/(2*x**2) + log(x)**2/x**4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (17) = 34\).

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.19 \[ \int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx=\frac {15}{2 \, x^{2}} + \frac {8 \, \log \left (x\right )^{2} + 4 \, \log \left (x\right ) + 1}{8 \, x^{4}} - \frac {\log \left (x\right )}{2 \, x^{4}} - \frac {1}{8 \, x^{4}} \]

[In]

integrate((-4*log(x)^2+2*log(x)-15*x^2)/x^5,x, algorithm="maxima")

[Out]

15/2/x^2 + 1/8*(8*log(x)^2 + 4*log(x) + 1)/x^4 - 1/2*log(x)/x^4 - 1/8/x^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx=\frac {15}{2 \, x^{2}} + \frac {\log \left (x\right )^{2}}{x^{4}} \]

[In]

integrate((-4*log(x)^2+2*log(x)-15*x^2)/x^5,x, algorithm="giac")

[Out]

15/2/x^2 + log(x)^2/x^4

Mupad [B] (verification not implemented)

Time = 7.75 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-15 x^2+2 \log (x)-4 \log ^2(x)}{x^5} \, dx=\frac {\frac {15\,x^2}{2}+{\ln \left (x\right )}^2}{x^4} \]

[In]

int(-(4*log(x)^2 - 2*log(x) + 15*x^2)/x^5,x)

[Out]

(log(x)^2 + (15*x^2)/2)/x^4