3.3.89 \(\int \frac {100+20 x+(-50-6 x^2) \log (x)}{5 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {2 \left (3+\frac {5}{x}\right ) (-5+x) (x-x \log (x))}{5 x} \]

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Rubi [A]  time = 0.06, antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 14, 43, 2334} \begin {gather*} \frac {6 x}{5}-\frac {10}{x}+\frac {2}{5} \left (\frac {25}{x}-3 x\right ) \log (x)+4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(100 + 20*x + (-50 - 6*x^2)*Log[x])/(5*x^2),x]

[Out]

-10/x + (6*x)/5 + 4*Log[x] + (2*(25/x - 3*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {20 (5+x)}{x^2}-\frac {2 \left (25+3 x^2\right ) \log (x)}{x^2}\right ) \, dx\\ &=-\left (\frac {2}{5} \int \frac {\left (25+3 x^2\right ) \log (x)}{x^2} \, dx\right )+4 \int \frac {5+x}{x^2} \, dx\\ &=\frac {2}{5} \left (\frac {25}{x}-3 x\right ) \log (x)+\frac {2}{5} \int \left (3-\frac {25}{x^2}\right ) \, dx+4 \int \left (\frac {5}{x^2}+\frac {1}{x}\right ) \, dx\\ &=-\frac {10}{x}+\frac {6 x}{5}+4 \log (x)+\frac {2}{5} \left (\frac {25}{x}-3 x\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 29, normalized size = 1.21 \begin {gather*} -\frac {10}{x}+\frac {6 x}{5}+4 \log (x)+\frac {10 \log (x)}{x}-\frac {6}{5} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 + 20*x + (-50 - 6*x^2)*Log[x])/(5*x^2),x]

[Out]

-10/x + (6*x)/5 + 4*Log[x] + (10*Log[x])/x - (6*x*Log[x])/5

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fricas [A]  time = 0.55, size = 26, normalized size = 1.08 \begin {gather*} \frac {2 \, {\left (3 \, x^{2} - {\left (3 \, x^{2} - 10 \, x - 25\right )} \log \relax (x) - 25\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-6*x^2-50)*log(x)+20*x+100)/x^2,x, algorithm="fricas")

[Out]

2/5*(3*x^2 - (3*x^2 - 10*x - 25)*log(x) - 25)/x

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giac [A]  time = 0.28, size = 26, normalized size = 1.08 \begin {gather*} -\frac {2}{5} \, {\left (3 \, x - \frac {25}{x}\right )} \log \relax (x) + \frac {6}{5} \, x - \frac {10}{x} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-6*x^2-50)*log(x)+20*x+100)/x^2,x, algorithm="giac")

[Out]

-2/5*(3*x - 25/x)*log(x) + 6/5*x - 10/x + 4*log(x)

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maple [A]  time = 0.02, size = 26, normalized size = 1.08

method	result	size
default	\(-\frac {6 x \ln \relax (x )}{5}+\frac {6 x}{5}+\frac {10 \ln \relax (x )}{x}-\frac {10}{x}+4 \ln \relax (x )\)	\(26\)
norman	\(\frac {-10+4 x \ln \relax (x )+\frac {6 x^{2}}{5}-\frac {6 x^{2} \ln \relax (x )}{5}+10 \ln \relax (x )}{x}\)	\(28\)
risch	\(-\frac {2 \left (3 x^{2}-25\right ) \ln \relax (x )}{5 x}+\frac {4 x \ln \relax (x )+\frac {6 x^{2}}{5}-10}{x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-6*x^2-50)*ln(x)+20*x+100)/x^2,x,method=_RETURNVERBOSE)

[Out]

-6/5*x*ln(x)+6/5*x+10*ln(x)/x-10/x+4*ln(x)

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maxima [A]  time = 0.47, size = 25, normalized size = 1.04 \begin {gather*} -\frac {6}{5} \, x \log \relax (x) + \frac {6}{5} \, x + \frac {10 \, \log \relax (x)}{x} - \frac {10}{x} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-6*x^2-50)*log(x)+20*x+100)/x^2,x, algorithm="maxima")

[Out]

-6/5*x*log(x) + 6/5*x + 10*log(x)/x - 10/x + 4*log(x)

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mupad [B]  time = 0.33, size = 24, normalized size = 1.00 \begin {gather*} 4\,\ln \relax (x)-x\,\left (\frac {6\,\ln \relax (x)}{5}-\frac {6}{5}\right )+\frac {10\,\ln \relax (x)-10}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - (log(x)*(6*x^2 + 50))/5 + 20)/x^2,x)

[Out]

4*log(x) - x*((6*log(x))/5 - 6/5) + (10*log(x) - 10)/x

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sympy [A]  time = 0.14, size = 26, normalized size = 1.08 \begin {gather*} \frac {6 x}{5} + 4 \log {\relax (x )} + \frac {\left (50 - 6 x^{2}\right ) \log {\relax (x )}}{5 x} - \frac {10}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-6*x**2-50)*ln(x)+20*x+100)/x**2,x)

[Out]

6*x/5 + 4*log(x) + (50 - 6*x**2)*log(x)/(5*x) - 10/x

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