3.15.73 \(\int \frac {225-30 x+x^2+(-30 x+2 x^2) \log (5)+(-30 x+2 x^2) \log (x)+5 \log (25-9 \log (5)+\log ^2(5))}{x} \, dx\)

Optimal. Leaf size=26 \[ (\log (5)+\log (x)) \left ((-15+x)^2+5 \log \left ((5-\log (5))^2+\log (5)\right )\right ) \]

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Rubi [B]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 2.31, number of steps used = 6, number of rules used = 3, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {14, 2313, 9} \begin {gather*} -\left (30 x-x^2\right ) \log (x)+\frac {1}{2} x^2 (1+\log (25))-\frac {1}{2} (30-x)^2+5 \left (45+\log \left (25+\log ^2(5)-9 \log (5)\right )\right ) \log (x)-30 x (1+\log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(225 - 30*x + x^2 + (-30*x + 2*x^2)*Log[5] + (-30*x + 2*x^2)*Log[x] + 5*Log[25 - 9*Log[5] + Log[5]^2])/x,x
]

[Out]

-1/2*(30 - x)^2 - 30*x*(1 + Log[5]) + (x^2*(1 + Log[25]))/2 - (30*x - x^2)*Log[x] + 5*Log[x]*(45 + Log[25 - 9*
Log[5] + Log[5]^2])

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, 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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 (-15+x) \log (x)+\frac {-30 x (1+\log (5))+x^2 (1+\log (25))+5 \left (45+\log \left (25-9 \log (5)+\log ^2(5)\right )\right )}{x}\right ) \, dx\\ &=2 \int (-15+x) \log (x) \, dx+\int \frac {-30 x (1+\log (5))+x^2 (1+\log (25))+5 \left (45+\log \left (25-9 \log (5)+\log ^2(5)\right )\right )}{x} \, dx\\ &=-\left (\left (30 x-x^2\right ) \log (x)\right )-2 \int \frac {1}{2} (-30+x) \, dx+\int \left (-30 (1+\log (5))+x (1+\log (25))+\frac {5 \left (45+\log \left (25-9 \log (5)+\log ^2(5)\right )\right )}{x}\right ) \, dx\\ &=-\frac {1}{2} (30-x)^2-30 x (1+\log (5))+\frac {1}{2} x^2 (1+\log (25))-\left (30 x-x^2\right ) \log (x)+5 \log (x) \left (45+\log \left (25-9 \log (5)+\log ^2(5)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 1.73 \begin {gather*} -30 x \log (5)+\frac {1}{2} x^2 \log (25)+225 \log (x)-30 x \log (x)+x^2 \log (x)+5 \log (x) \log \left (25-9 \log (5)+\log ^2(5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(225 - 30*x + x^2 + (-30*x + 2*x^2)*Log[5] + (-30*x + 2*x^2)*Log[x] + 5*Log[25 - 9*Log[5] + Log[5]^2
])/x,x]

[Out]

-30*x*Log[5] + (x^2*Log[25])/2 + 225*Log[x] - 30*x*Log[x] + x^2*Log[x] + 5*Log[x]*Log[25 - 9*Log[5] + Log[5]^2
]

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fricas [A]  time = 0.49, size = 37, normalized size = 1.42 \begin {gather*} {\left (x^{2} - 30 \, x\right )} \log \relax (5) + {\left (x^{2} - 30 \, x + 225\right )} \log \relax (x) + 5 \, \log \left (\log \relax (5)^{2} - 9 \, \log \relax (5) + 25\right ) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(log(5)^2-9*log(5)+25)+(2*x^2-30*x)*log(x)+(2*x^2-30*x)*log(5)+x^2-30*x+225)/x,x, algorithm="f
ricas")

[Out]

(x^2 - 30*x)*log(5) + (x^2 - 30*x + 225)*log(x) + 5*log(log(5)^2 - 9*log(5) + 25)*log(x)

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giac [A]  time = 0.18, size = 39, normalized size = 1.50 \begin {gather*} x^{2} \log \relax (5) - 30 \, x \log \relax (5) + {\left (x^{2} - 30 \, x\right )} \log \relax (x) + 5 \, {\left (\log \left (\log \relax (5)^{2} - 9 \, \log \relax (5) + 25\right ) + 45\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(log(5)^2-9*log(5)+25)+(2*x^2-30*x)*log(x)+(2*x^2-30*x)*log(5)+x^2-30*x+225)/x,x, algorithm="g
iac")

[Out]

x^2*log(5) - 30*x*log(5) + (x^2 - 30*x)*log(x) + 5*(log(log(5)^2 - 9*log(5) + 25) + 45)*log(x)

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maple [A]  time = 0.04, size = 42, normalized size = 1.62

method	result	size
norman	\(x^{2} \ln \relax (5)+x^{2} \ln \relax (x )+\left (5 \ln \left (\ln \relax (5)^{2}-9 \ln \relax (5)+25\right )+225\right ) \ln \relax (x )-30 x \ln \relax (5)-30 x \ln \relax (x )\)	\(42\)
risch	\(\left (x^{2}-30 x \right ) \ln \relax (x )+x^{2} \ln \relax (5)+5 \ln \left (\ln \relax (5)^{2}-9 \ln \relax (5)+25\right ) \ln \relax (x )-30 x \ln \relax (5)+225 \ln \relax (x )\)	\(42\)
default	\(x^{2} \ln \relax (x )+x^{2} \ln \relax (5)-30 x \ln \relax (x )-30 x \ln \relax (5)+5 \ln \left (\ln \relax (5)^{2}-9 \ln \relax (5)+25\right ) \ln \relax (x )+225 \ln \relax (x )\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*ln(ln(5)^2-9*ln(5)+25)+(2*x^2-30*x)*ln(x)+(2*x^2-30*x)*ln(5)+x^2-30*x+225)/x,x,method=_RETURNVERBOSE)

[Out]

x^2*ln(5)+x^2*ln(x)+(5*ln(ln(5)^2-9*ln(5)+25)+225)*ln(x)-30*x*ln(5)-30*x*ln(x)

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maxima [A]  time = 0.40, size = 42, normalized size = 1.62 \begin {gather*} x^{2} \log \relax (5) + x^{2} \log \relax (x) - 30 \, x \log \relax (5) - 30 \, x \log \relax (x) + 5 \, \log \left (\log \relax (5)^{2} - 9 \, \log \relax (5) + 25\right ) \log \relax (x) + 225 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(log(5)^2-9*log(5)+25)+(2*x^2-30*x)*log(x)+(2*x^2-30*x)*log(5)+x^2-30*x+225)/x,x, algorithm="m
axima")

[Out]

x^2*log(5) + x^2*log(x) - 30*x*log(5) - 30*x*log(x) + 5*log(log(5)^2 - 9*log(5) + 25)*log(x) + 225*log(x)

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mupad [B]  time = 1.00, size = 42, normalized size = 1.62 \begin {gather*} 225\,\ln \relax (x)+x^2\,\ln \relax (x)-30\,x\,\ln \relax (5)+x^2\,\ln \relax (5)+5\,\ln \left ({\ln \relax (5)}^2-9\,\ln \relax (5)+25\right )\,\ln \relax (x)-30\,x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*x - 5*log(log(5)^2 - 9*log(5) + 25) + log(5)*(30*x - 2*x^2) + log(x)*(30*x - 2*x^2) - x^2 - 225)/x,x)

[Out]

225*log(x) + x^2*log(x) - 30*x*log(5) + x^2*log(5) + 5*log(log(5)^2 - 9*log(5) + 25)*log(x) - 30*x*log(x)

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sympy [A]  time = 0.25, size = 42, normalized size = 1.62 \begin {gather*} x^{2} \log {\relax (5 )} - 30 x \log {\relax (5 )} + \left (x^{2} - 30 x\right ) \log {\relax (x )} + 5 \left (\log {\left (- 9 \log {\relax (5 )} + \log {\relax (5 )}^{2} + 25 \right )} + 45\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*ln(ln(5)**2-9*ln(5)+25)+(2*x**2-30*x)*ln(x)+(2*x**2-30*x)*ln(5)+x**2-30*x+225)/x,x)

[Out]

x**2*log(5) - 30*x*log(5) + (x**2 - 30*x)*log(x) + 5*(log(-9*log(5) + log(5)**2 + 25) + 45)*log(x)

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