\(\int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log (x^3)}{x} \, dx\) [8275]

Optimal result

Integrand size = 24, antiderivative size = 18 \[ \int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log \left (x^3\right )}{x} \, dx=50625 x^2 \log ^2(2)+\log (x)+x \log \left (x^3\right ) \]

[Out]

x*ln(x^3)+ln(x)+50625*x^2*ln(2)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {14, 2332} \[ \int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log \left (x^3\right )}{x} \, dx=x \log \left (x^3\right )+50625 x^2 \log ^2(2)+\log (x) \]

[In]

Int[(1 + 3*x + 101250*x^2*Log[2]^2 + x*Log[x^3])/x,x]

[Out]

50625*x^2*Log[2]^2 + Log[x] + x*Log[x^3]

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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1+3 x+101250 x^2 \log ^2(2)}{x}+\log \left (x^3\right )\right ) \, dx \\ & = \int \frac {1+3 x+101250 x^2 \log ^2(2)}{x} \, dx+\int \log \left (x^3\right ) \, dx \\ & = -3 x+x \log \left (x^3\right )+\int \left (3+\frac {1}{x}+101250 x \log ^2(2)\right ) \, dx \\ & = 50625 x^2 \log ^2(2)+\log (x)+x \log \left (x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log \left (x^3\right )}{x} \, dx=50625 x^2 \log ^2(2)+\log (x)+x \log \left (x^3\right ) \]

[In]

Integrate[(1 + 3*x + 101250*x^2*Log[2]^2 + x*Log[x^3])/x,x]

[Out]

50625*x^2*Log[2]^2 + Log[x] + x*Log[x^3]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
default	\(x \ln \left (x^{3}\right )+\ln \left (x \right )+50625 x^{2} \ln \left (2\right )^{2}\)	\(19\)
risch	\(x \ln \left (x^{3}\right )+\ln \left (x \right )+50625 x^{2} \ln \left (2\right )^{2}\)	\(19\)
parts	\(x \ln \left (x^{3}\right )+\ln \left (x \right )+50625 x^{2} \ln \left (2\right )^{2}\)	\(19\)
norman	\(x \ln \left (x^{3}\right )+\frac {\ln \left (x^{3}\right )}{3}+50625 x^{2} \ln \left (2\right )^{2}\)	\(23\)
parallelrisch	\(x \ln \left (x^{3}\right )+\frac {\ln \left (x^{3}\right )}{3}+50625 x^{2} \ln \left (2\right )^{2}\)	\(23\)

[In]

int((x*ln(x^3)+101250*x^2*ln(2)^2+3*x+1)/x,x,method=_RETURNVERBOSE)

[Out]

x*ln(x^3)+ln(x)+50625*x^2*ln(2)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log \left (x^3\right )}{x} \, dx=50625 \, x^{2} \log \left (2\right )^{2} + \frac {1}{3} \, {\left (3 \, x + 1\right )} \log \left (x^{3}\right ) \]

[In]

integrate((x*log(x^3)+101250*x^2*log(2)^2+3*x+1)/x,x, algorithm="fricas")

[Out]

50625*x^2*log(2)^2 + 1/3*(3*x + 1)*log(x^3)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log \left (x^3\right )}{x} \, dx=50625 x^{2} \log {\left (2 \right )}^{2} + x \log {\left (x^{3} \right )} + \log {\left (x \right )} \]

[In]

integrate((x*ln(x**3)+101250*x**2*ln(2)**2+3*x+1)/x,x)

[Out]

50625*x**2*log(2)**2 + x*log(x**3) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log \left (x^3\right )}{x} \, dx=50625 \, x^{2} \log \left (2\right )^{2} + x \log \left (x^{3}\right ) + \log \left (x\right ) \]

[In]

integrate((x*log(x^3)+101250*x^2*log(2)^2+3*x+1)/x,x, algorithm="maxima")

[Out]

50625*x^2*log(2)^2 + x*log(x^3) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log \left (x^3\right )}{x} \, dx=50625 \, x^{2} \log \left (2\right )^{2} + x \log \left (x^{3}\right ) + \log \left (x\right ) \]

[In]

integrate((x*log(x^3)+101250*x^2*log(2)^2+3*x+1)/x,x, algorithm="giac")

[Out]

50625*x^2*log(2)^2 + x*log(x^3) + log(x)

Mupad [B] (verification not implemented)

Time = 13.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {1+3 x+101250 x^2 \log ^2(2)+x \log \left (x^3\right )}{x} \, dx=\frac {\ln \left (x^3\right )}{3}+50625\,x^2\,{\ln \left (2\right )}^2+x\,\ln \left (x^3\right ) \]

[In]

int((3*x + 101250*x^2*log(2)^2 + x*log(x^3) + 1)/x,x)

[Out]

log(x^3)/3 + 50625*x^2*log(2)^2 + x*log(x^3)