\(\int \frac {1}{25} (175-80 x-24 x^2+50 \log (x)) \, dx\) [1746]

Optimal result

Integrand size = 18, antiderivative size = 21 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=x \left (-1+2 \left (4-\frac {1}{25} (-5-2 x)^2+\log (x)\right )\right ) \]

[Out]

x*(2*ln(x)+7-2/5*(-2*x-5)*(-2/5*x-1))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2332} \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {8 x^3}{25}-\frac {8 x^2}{5}+5 x+2 x \log (x) \]

[In]

Int[(175 - 80*x - 24*x^2 + 50*Log[x])/25,x]

[Out]

5*x - (8*x^2)/5 - (8*x^3)/25 + 2*x*Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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}{25} \int \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx \\ & = 7 x-\frac {8 x^2}{5}-\frac {8 x^3}{25}+2 \int \log (x) \, dx \\ & = 5 x-\frac {8 x^2}{5}-\frac {8 x^3}{25}+2 x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=5 x-\frac {8 x^2}{5}-\frac {8 x^3}{25}+2 x \log (x) \]

[In]

Integrate[(175 - 80*x - 24*x^2 + 50*Log[x])/25,x]

[Out]

5*x - (8*x^2)/5 - (8*x^3)/25 + 2*x*Log[x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\)	\(20\)
norman	\(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\)	\(20\)
risch	\(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\)	\(20\)
parallelrisch	\(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\)	\(20\)
parts	\(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\)	\(20\)

[In]

int(2*ln(x)-24/25*x^2-16/5*x+7,x,method=_RETURNVERBOSE)

[Out]

-8/25*x^3-8/5*x^2+5*x+2*x*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \left (x\right ) + 5 \, x \]

[In]

integrate(2*log(x)-24/25*x^2-16/5*x+7,x, algorithm="fricas")

[Out]

-8/25*x^3 - 8/5*x^2 + 2*x*log(x) + 5*x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=- \frac {8 x^{3}}{25} - \frac {8 x^{2}}{5} + 2 x \log {\left (x \right )} + 5 x \]

[In]

integrate(2*ln(x)-24/25*x**2-16/5*x+7,x)

[Out]

-8*x**3/25 - 8*x**2/5 + 2*x*log(x) + 5*x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \left (x\right ) + 5 \, x \]

[In]

integrate(2*log(x)-24/25*x^2-16/5*x+7,x, algorithm="maxima")

[Out]

-8/25*x^3 - 8/5*x^2 + 2*x*log(x) + 5*x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \left (x\right ) + 5 \, x \]

[In]

integrate(2*log(x)-24/25*x^2-16/5*x+7,x, algorithm="giac")

[Out]

-8/25*x^3 - 8/5*x^2 + 2*x*log(x) + 5*x

Mupad [B] (verification not implemented)

Time = 9.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {x\,\left (40\,x-50\,\ln \left (x\right )+8\,x^2-125\right )}{25} \]

[In]

int(2*log(x) - (16*x)/5 - (24*x^2)/25 + 7,x)

[Out]

-(x*(40*x - 50*log(x) + 8*x^2 - 125))/25