3.8.85 \(\int \frac {10+17 x-10 x^2}{5 x} \, dx\)

Optimal. Leaf size=20 \[ 7-\frac {3 x}{5}+(4-x) x+\log (4)+\log \left (x^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14} \begin {gather*} -x^2+\frac {17 x}{5}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + 17*x - 10*x^2)/(5*x),x]

[Out]

(17*x)/5 - x^2 + 2*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 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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {10+17 x-10 x^2}{x} \, dx\\ &=\frac {1}{5} \int \left (17+\frac {10}{x}-10 x\right ) \, dx\\ &=\frac {17 x}{5}-x^2+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.75 \begin {gather*} \frac {17 x}{5}-x^2+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + 17*x - 10*x^2)/(5*x),x]

[Out]

(17*x)/5 - x^2 + 2*Log[x]

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fricas [A]  time = 0.67, size = 13, normalized size = 0.65 \begin {gather*} -x^{2} + \frac {17}{5} \, x + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*x^2+17*x+10)/x,x, algorithm="fricas")

[Out]

-x^2 + 17/5*x + 2*log(x)

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giac [A]  time = 0.43, size = 14, normalized size = 0.70 \begin {gather*} -x^{2} + \frac {17}{5} \, x + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*x^2+17*x+10)/x,x, algorithm="giac")

[Out]

-x^2 + 17/5*x + 2*log(abs(x))

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maple [A]  time = 0.02, size = 14, normalized size = 0.70




method result size



default \(-x^{2}+\frac {17 x}{5}+2 \ln \relax (x )\) \(14\)
norman \(-x^{2}+\frac {17 x}{5}+2 \ln \relax (x )\) \(14\)
risch \(-x^{2}+\frac {17 x}{5}+2 \ln \relax (x )\) \(14\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-10*x^2+17*x+10)/x,x,method=_RETURNVERBOSE)

[Out]

-x^2+17/5*x+2*ln(x)

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maxima [A]  time = 0.54, size = 13, normalized size = 0.65 \begin {gather*} -x^{2} + \frac {17}{5} \, x + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*x^2+17*x+10)/x,x, algorithm="maxima")

[Out]

-x^2 + 17/5*x + 2*log(x)

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mupad [B]  time = 0.02, size = 13, normalized size = 0.65 \begin {gather*} \frac {17\,x}{5}+2\,\ln \relax (x)-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((17*x)/5 - 2*x^2 + 2)/x,x)

[Out]

(17*x)/5 + 2*log(x) - x^2

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sympy [A]  time = 0.07, size = 12, normalized size = 0.60 \begin {gather*} - x^{2} + \frac {17 x}{5} + 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*x**2+17*x+10)/x,x)

[Out]

-x**2 + 17*x/5 + 2*log(x)

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