3.78.43 \(\int \frac {-6000+1000 x+(-2200+200 x) \log (x)-200 \log ^2(x)}{3 x^3} \, dx\)

Optimal. Leaf size=16 \[ \frac {100 (6-x+\log (x))^2}{3 x^2} \]

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Rubi [B]  time = 0.07, antiderivative size = 69, normalized size of antiderivative = 4.31, number of steps used = 10, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {12, 14, 37, 2334, 43, 2305, 2304} \begin {gather*} \frac {250 (6-x)^2}{9 x^2}+\frac {200}{x^2}+\frac {100 \log ^2(x)}{3 x^2}+\frac {100 (11-x)^2 \log (x)}{33 x^2}+\frac {100 \log (x)}{3 x^2}-\frac {200}{3 x}-\frac {100 \log (x)}{33} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6000 + 1000*x + (-2200 + 200*x)*Log[x] - 200*Log[x]^2)/(3*x^3),x]

[Out]

200/x^2 + (250*(6 - x)^2)/(9*x^2) - 200/(3*x) - (100*Log[x])/33 + (100*Log[x])/(3*x^2) + (100*(11 - x)^2*Log[x
])/(33*x^2) + (100*Log[x]^2)/(3*x^2)

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 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}{3} \int \frac {-6000+1000 x+(-2200+200 x) \log (x)-200 \log ^2(x)}{x^3} \, dx\\ &=\frac {1}{3} \int \left (\frac {1000 (-6+x)}{x^3}+\frac {200 (-11+x) \log (x)}{x^3}-\frac {200 \log ^2(x)}{x^3}\right ) \, dx\\ &=\frac {200}{3} \int \frac {(-11+x) \log (x)}{x^3} \, dx-\frac {200}{3} \int \frac {\log ^2(x)}{x^3} \, dx+\frac {1000}{3} \int \frac {-6+x}{x^3} \, dx\\ &=\frac {250 (6-x)^2}{9 x^2}+\frac {100 (11-x)^2 \log (x)}{33 x^2}+\frac {100 \log ^2(x)}{3 x^2}-\frac {200}{3} \int \frac {(11-x)^2}{22 x^3} \, dx-\frac {200}{3} \int \frac {\log (x)}{x^3} \, dx\\ &=\frac {50}{3 x^2}+\frac {250 (6-x)^2}{9 x^2}+\frac {100 \log (x)}{3 x^2}+\frac {100 (11-x)^2 \log (x)}{33 x^2}+\frac {100 \log ^2(x)}{3 x^2}-\frac {100}{33} \int \frac {(11-x)^2}{x^3} \, dx\\ &=\frac {50}{3 x^2}+\frac {250 (6-x)^2}{9 x^2}+\frac {100 \log (x)}{3 x^2}+\frac {100 (11-x)^2 \log (x)}{33 x^2}+\frac {100 \log ^2(x)}{3 x^2}-\frac {100}{33} \int \left (\frac {121}{x^3}-\frac {22}{x^2}+\frac {1}{x}\right ) \, dx\\ &=\frac {200}{x^2}+\frac {250 (6-x)^2}{9 x^2}-\frac {200}{3 x}-\frac {100 \log (x)}{33}+\frac {100 \log (x)}{3 x^2}+\frac {100 (11-x)^2 \log (x)}{33 x^2}+\frac {100 \log ^2(x)}{3 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 40, normalized size = 2.50 \begin {gather*} \frac {200}{3} \left (\frac {18}{x^2}-\frac {6}{x}+\frac {6 \log (x)}{x^2}-\frac {\log (x)}{x}+\frac {\log ^2(x)}{2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6000 + 1000*x + (-2200 + 200*x)*Log[x] - 200*Log[x]^2)/(3*x^3),x]

[Out]

(200*(18/x^2 - 6/x + (6*Log[x])/x^2 - Log[x]/x + Log[x]^2/(2*x^2)))/3

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fricas [A]  time = 0.51, size = 23, normalized size = 1.44 \begin {gather*} -\frac {100 \, {\left (2 \, {\left (x - 6\right )} \log \relax (x) - \log \relax (x)^{2} + 12 \, x - 36\right )}}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-200*log(x)^2+(200*x-2200)*log(x)+1000*x-6000)/x^3,x, algorithm="fricas")

[Out]

-100/3*(2*(x - 6)*log(x) - log(x)^2 + 12*x - 36)/x^2

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giac [A]  time = 0.16, size = 28, normalized size = 1.75 \begin {gather*} -\frac {200 \, {\left (x - 6\right )} \log \relax (x)}{3 \, x^{2}} + \frac {100 \, \log \relax (x)^{2}}{3 \, x^{2}} - \frac {400 \, {\left (x - 3\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-200*log(x)^2+(200*x-2200)*log(x)+1000*x-6000)/x^3,x, algorithm="giac")

[Out]

-200/3*(x - 6)*log(x)/x^2 + 100/3*log(x)^2/x^2 - 400*(x - 3)/x^2

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maple [A]  time = 0.02, size = 25, normalized size = 1.56

method	result	size
norman	\(\frac {1200-400 x +\frac {100 \ln \relax (x )^{2}}{3}-\frac {200 x \ln \relax (x )}{3}+400 \ln \relax (x )}{x^{2}}\)	\(25\)
risch	\(\frac {100 \ln \relax (x )^{2}}{3 x^{2}}-\frac {200 \left (x -6\right ) \ln \relax (x )}{3 x^{2}}-\frac {400 \left (x -3\right )}{x^{2}}\)	\(29\)
default	\(\frac {100 \ln \relax (x )^{2}}{3 x^{2}}+\frac {400 \ln \relax (x )}{x^{2}}+\frac {1200}{x^{2}}-\frac {200 \ln \relax (x )}{3 x}-\frac {400}{x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-200*ln(x)^2+(200*x-2200)*ln(x)+1000*x-6000)/x^3,x,method=_RETURNVERBOSE)

[Out]

(1200-400*x+100/3*ln(x)^2-200/3*x*ln(x)+400*ln(x))/x^2

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maxima [B]  time = 0.35, size = 42, normalized size = 2.62 \begin {gather*} -\frac {200 \, \log \relax (x)}{3 \, x} + \frac {50 \, {\left (2 \, \log \relax (x)^{2} + 2 \, \log \relax (x) + 1\right )}}{3 \, x^{2}} - \frac {400}{x} + \frac {1100 \, \log \relax (x)}{3 \, x^{2}} + \frac {3550}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-200*log(x)^2+(200*x-2200)*log(x)+1000*x-6000)/x^3,x, algorithm="maxima")

[Out]

-200/3*log(x)/x + 50/3*(2*log(x)^2 + 2*log(x) + 1)/x^2 - 400/x + 1100/3*log(x)/x^2 + 3550/3/x^2

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mupad [B]  time = 5.29, size = 16, normalized size = 1.00 \begin {gather*} \frac {100\,\left (\ln \relax (x)+6\right )\,\left (\ln \relax (x)-2\,x+6\right )}{3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1000*x)/3 - (200*log(x)^2)/3 + (log(x)*(200*x - 2200))/3 - 2000)/x^3,x)

[Out]

(100*(log(x) + 6)*(log(x) - 2*x + 6))/(3*x^2)

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sympy [B]  time = 0.17, size = 32, normalized size = 2.00 \begin {gather*} \frac {1200 - 400 x}{x^{2}} + \frac {\left (1200 - 200 x\right ) \log {\relax (x )}}{3 x^{2}} + \frac {100 \log {\relax (x )}^{2}}{3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-200*ln(x)**2+(200*x-2200)*ln(x)+1000*x-6000)/x**3,x)

[Out]

(1200 - 400*x)/x**2 + (1200 - 200*x)*log(x)/(3*x**2) + 100*log(x)**2/(3*x**2)

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