\(\int \frac {2+56 x^2+6 x^3+(-4+6 x^3) \log (x)}{x^3} \, dx\) [7947]

Optimal result

Integrand size = 26, antiderivative size = 14 \[ \int \frac {2+56 x^2+6 x^3+\left (-4+6 x^3\right ) \log (x)}{x^3} \, dx=2 \left (4+\frac {1}{x^2}+3 (8+x)\right ) \log (x) \]

[Out]

2*(1/x^2+28+3*x)*ln(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21, number of steps used = 6, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14, 2372} \[ \int \frac {2+56 x^2+6 x^3+\left (-4+6 x^3\right ) \log (x)}{x^3} \, dx=\frac {2 \log (x)}{x^2}+6 x \log (x)+56 \log (x) \]

[In]

Int[(2 + 56*x^2 + 6*x^3 + (-4 + 6*x^3)*Log[x])/x^3,x]

[Out]

56*Log[x] + (2*Log[x])/x^2 + 6*x*Log[x]

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 2372

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]]
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 \left (1+28 x^2+3 x^3\right )}{x^3}+\frac {2 \left (-2+3 x^3\right ) \log (x)}{x^3}\right ) \, dx \\ & = 2 \int \frac {1+28 x^2+3 x^3}{x^3} \, dx+2 \int \frac {\left (-2+3 x^3\right ) \log (x)}{x^3} \, dx \\ & = \frac {2 \log (x)}{x^2}+6 x \log (x)-2 \int \left (3+\frac {1}{x^3}\right ) \, dx+2 \int \left (3+\frac {1}{x^3}+\frac {28}{x}\right ) \, dx \\ & = 56 \log (x)+\frac {2 \log (x)}{x^2}+6 x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {2+56 x^2+6 x^3+\left (-4+6 x^3\right ) \log (x)}{x^3} \, dx=56 \log (x)+\frac {2 \log (x)}{x^2}+6 x \log (x) \]

[In]

Integrate[(2 + 56*x^2 + 6*x^3 + (-4 + 6*x^3)*Log[x])/x^3,x]

[Out]

56*Log[x] + (2*Log[x])/x^2 + 6*x*Log[x]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29

method	result	size
default	\(6 x \ln \left (x \right )+56 \ln \left (x \right )+\frac {2 \ln \left (x \right )}{x^{2}}\)	\(18\)
parts	\(6 x \ln \left (x \right )+56 \ln \left (x \right )+\frac {2 \ln \left (x \right )}{x^{2}}\)	\(18\)
risch	\(\frac {2 \left (3 x^{3}+1\right ) \ln \left (x \right )}{x^{2}}+56 \ln \left (x \right )\)	\(20\)
norman	\(\frac {56 x^{2} \ln \left (x \right )+6 x^{3} \ln \left (x \right )+2 \ln \left (x \right )}{x^{2}}\)	\(24\)
parallelrisch	\(\frac {56 x^{2} \ln \left (x \right )+6 x^{3} \ln \left (x \right )+2 \ln \left (x \right )}{x^{2}}\)	\(24\)

[In]

int(((6*x^3-4)*ln(x)+6*x^3+56*x^2+2)/x^3,x,method=_RETURNVERBOSE)

[Out]

6*x*ln(x)+56*ln(x)+2*ln(x)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {2+56 x^2+6 x^3+\left (-4+6 x^3\right ) \log (x)}{x^3} \, dx=\frac {2 \, {\left (3 \, x^{3} + 28 \, x^{2} + 1\right )} \log \left (x\right )}{x^{2}} \]

[In]

integrate(((6*x^3-4)*log(x)+6*x^3+56*x^2+2)/x^3,x, algorithm="fricas")

[Out]

2*(3*x^3 + 28*x^2 + 1)*log(x)/x^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {2+56 x^2+6 x^3+\left (-4+6 x^3\right ) \log (x)}{x^3} \, dx=56 \log {\left (x \right )} + \frac {\left (6 x^{3} + 2\right ) \log {\left (x \right )}}{x^{2}} \]

[In]

integrate(((6*x**3-4)*ln(x)+6*x**3+56*x**2+2)/x**3,x)

[Out]

56*log(x) + (6*x**3 + 2)*log(x)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {2+56 x^2+6 x^3+\left (-4+6 x^3\right ) \log (x)}{x^3} \, dx=6 \, x \log \left (x\right ) + \frac {2 \, \log \left (x\right )}{x^{2}} + 56 \, \log \left (x\right ) \]

[In]

integrate(((6*x^3-4)*log(x)+6*x^3+56*x^2+2)/x^3,x, algorithm="maxima")

[Out]

6*x*log(x) + 2*log(x)/x^2 + 56*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {2+56 x^2+6 x^3+\left (-4+6 x^3\right ) \log (x)}{x^3} \, dx=2 \, {\left (3 \, x + \frac {1}{x^{2}}\right )} \log \left (x\right ) + 56 \, \log \left (x\right ) \]

[In]

integrate(((6*x^3-4)*log(x)+6*x^3+56*x^2+2)/x^3,x, algorithm="giac")

[Out]

2*(3*x + 1/x^2)*log(x) + 56*log(x)

Mupad [B] (verification not implemented)

Time = 12.91 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {2+56 x^2+6 x^3+\left (-4+6 x^3\right ) \log (x)}{x^3} \, dx=\frac {2\,\ln \left (x\right )\,\left (3\,x^3+28\,x^2+1\right )}{x^2} \]

[In]

int((56*x^2 + 6*x^3 + log(x)*(6*x^3 - 4) + 2)/x^3,x)

[Out]

(2*log(x)*(28*x^2 + 3*x^3 + 1))/x^2