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

Optimal. Leaf size=21 \[ x \left (-1+2 \left (4-\frac {1}{25} (-5-2 x)^2+\log (x)\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2295} \begin {gather*} -\frac {8 x^3}{25}-\frac {8 x^2}{5}+5 x+2 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[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 2295

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 1.10 \begin {gather*} 5 x-\frac {8 x^2}{5}-\frac {8 x^3}{25}+2 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.74, size = 19, normalized size = 0.90 \begin {gather*} -\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.25, size = 19, normalized size = 0.90 \begin {gather*} -\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.01, size = 20, normalized size = 0.95

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 19, normalized size = 0.90 \begin {gather*} -\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.08, size = 17, normalized size = 0.81 \begin {gather*} -\frac {x\,\left (40\,x-50\,\ln \relax (x)+8\,x^2-125\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.09, size = 22, normalized size = 1.05 \begin {gather*} - \frac {8 x^{3}}{25} - \frac {8 x^{2}}{5} + 2 x \log {\relax (x )} + 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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