\(\int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx\) [6147]

Optimal result

Integrand size = 27, antiderivative size = 22 \[ \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx=3+x-2 \left (5+\frac {3 x}{5}\right ) \left (4 x^2-\log (x)\right ) \]

[Out]

3+x-2*(4*x^2-ln(x))*(5+3/5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 14, 2332} \[ \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx=-\frac {24 x^3}{5}-40 x^2+x+\frac {6}{5} x \log (x)+10 \log (x) \]

[In]

Int[(50 + 11*x - 400*x^2 - 72*x^3 + 6*x*Log[x])/(5*x),x]

[Out]

x - 40*x^2 - (24*x^3)/5 + 10*Log[x] + (6*x*Log[x])/5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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}& = \frac {1}{5} \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{x} \, dx \\ & = \frac {1}{5} \int \left (\frac {50+11 x-400 x^2-72 x^3}{x}+6 \log (x)\right ) \, dx \\ & = \frac {1}{5} \int \frac {50+11 x-400 x^2-72 x^3}{x} \, dx+\frac {6}{5} \int \log (x) \, dx \\ & = -\frac {6 x}{5}+\frac {6}{5} x \log (x)+\frac {1}{5} \int \left (11+\frac {50}{x}-400 x-72 x^2\right ) \, dx \\ & = x-40 x^2-\frac {24 x^3}{5}+10 \log (x)+\frac {6}{5} x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx=x-40 x^2-\frac {24 x^3}{5}+10 \log (x)+\frac {6}{5} x \log (x) \]

[In]

Integrate[(50 + 11*x - 400*x^2 - 72*x^3 + 6*x*Log[x])/(5*x),x]

[Out]

x - 40*x^2 - (24*x^3)/5 + 10*Log[x] + (6*x*Log[x])/5

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {6 x \ln \left (x \right )}{5}+x -\frac {24 x^{3}}{5}-40 x^{2}+10 \ln \left (x \right )\)	\(22\)
norman	\(\frac {6 x \ln \left (x \right )}{5}+x -\frac {24 x^{3}}{5}-40 x^{2}+10 \ln \left (x \right )\)	\(22\)
risch	\(\frac {6 x \ln \left (x \right )}{5}+x -\frac {24 x^{3}}{5}-40 x^{2}+10 \ln \left (x \right )\)	\(22\)
parallelrisch	\(\frac {6 x \ln \left (x \right )}{5}+x -\frac {24 x^{3}}{5}-40 x^{2}+10 \ln \left (x \right )\)	\(22\)
parts	\(\frac {6 x \ln \left (x \right )}{5}+x -\frac {24 x^{3}}{5}-40 x^{2}+10 \ln \left (x \right )\)	\(22\)

[In]

int(1/5*(6*x*ln(x)-72*x^3-400*x^2+11*x+50)/x,x,method=_RETURNVERBOSE)

[Out]

6/5*x*ln(x)+x-24/5*x^3-40*x^2+10*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx=-\frac {24}{5} \, x^{3} - 40 \, x^{2} + \frac {2}{5} \, {\left (3 \, x + 25\right )} \log \left (x\right ) + x \]

[In]

integrate(1/5*(6*x*log(x)-72*x^3-400*x^2+11*x+50)/x,x, algorithm="fricas")

[Out]

-24/5*x^3 - 40*x^2 + 2/5*(3*x + 25)*log(x) + x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx=- \frac {24 x^{3}}{5} - 40 x^{2} + \frac {6 x \log {\left (x \right )}}{5} + x + 10 \log {\left (x \right )} \]

[In]

integrate(1/5*(6*x*ln(x)-72*x**3-400*x**2+11*x+50)/x,x)

[Out]

-24*x**3/5 - 40*x**2 + 6*x*log(x)/5 + x + 10*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx=-\frac {24}{5} \, x^{3} - 40 \, x^{2} + \frac {6}{5} \, x \log \left (x\right ) + x + 10 \, \log \left (x\right ) \]

[In]

integrate(1/5*(6*x*log(x)-72*x^3-400*x^2+11*x+50)/x,x, algorithm="maxima")

[Out]

-24/5*x^3 - 40*x^2 + 6/5*x*log(x) + x + 10*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx=-\frac {24}{5} \, x^{3} - 40 \, x^{2} + \frac {6}{5} \, x \log \left (x\right ) + x + 10 \, \log \left (x\right ) \]

[In]

integrate(1/5*(6*x*log(x)-72*x^3-400*x^2+11*x+50)/x,x, algorithm="giac")

[Out]

-24/5*x^3 - 40*x^2 + 6/5*x*log(x) + x + 10*log(x)

Mupad [B] (verification not implemented)

Time = 11.69 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {50+11 x-400 x^2-72 x^3+6 x \log (x)}{5 x} \, dx=x+10\,\ln \left (x\right )+\frac {6\,x\,\ln \left (x\right )}{5}-40\,x^2-\frac {24\,x^3}{5} \]

[In]

int(((11*x)/5 + (6*x*log(x))/5 - 80*x^2 - (72*x^3)/5 + 10)/x,x)

[Out]

x + 10*log(x) + (6*x*log(x))/5 - 40*x^2 - (24*x^3)/5