\(\int \frac {225-30 x+x^2+(-30 x+2 x^2) \log (5)+(-30 x+2 x^2) \log (x)+5 \log (25-9 \log (5)+\log ^2(5))}{x} \, dx\) [1473]

Optimal result

Integrand size = 49, antiderivative size = 26 \[ \int \frac {225-30 x+x^2+\left (-30 x+2 x^2\right ) \log (5)+\left (-30 x+2 x^2\right ) \log (x)+5 \log \left (25-9 \log (5)+\log ^2(5)\right )}{x} \, dx=(\log (5)+\log (x)) \left ((-15+x)^2+5 \log \left ((5-\log (5))^2+\log (5)\right )\right ) \]

[Out]

((x-15)^2+5*ln(ln(5)+(-ln(5)+5)^2))*(ln(5)+ln(x))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(26)=52\).

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.23, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {14, 2350, 9} \[ \int \frac {225-30 x+x^2+\left (-30 x+2 x^2\right ) \log (5)+\left (-30 x+2 x^2\right ) \log (x)+5 \log \left (25-9 \log (5)+\log ^2(5)\right )}{x} \, dx=x^2 \log (x)+\frac {1}{2} x^2 (1+\log (25))-\frac {1}{2} (30-x)^2+5 \left (45+\log \left (25+\log ^2(5)-9 \log (5)\right )\right ) \log (x)-30 x \log (x)-30 x (1+\log (5)) \]

[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*Log[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 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, 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{align*} \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 \\ & = -30 x \log (x)+x^2 \log (x)-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))-30 x \log (x)+x^2 \log (x)+5 \log (x) \left (45+\log \left (25-9 \log (5)+\log ^2(5)\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {225-30 x+x^2+\left (-30 x+2 x^2\right ) \log (5)+\left (-30 x+2 x^2\right ) \log (x)+5 \log \left (25-9 \log (5)+\log ^2(5)\right )}{x} \, dx=-30 x \log (5)+\frac {1}{2} x^2 \log (25)+225 \log (x)-30 x \log (x)+x^2 \log (x)+5 \log (x) \log \left (25-9 \log (5)+\log ^2(5)\right ) \]

[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
]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {225-30 x+x^2+\left (-30 x+2 x^2\right ) \log (5)+\left (-30 x+2 x^2\right ) \log (x)+5 \log \left (25-9 \log (5)+\log ^2(5)\right )}{x} \, dx={\left (x^{2} - 30 \, x\right )} \log \left (5\right ) + {\left (x^{2} - 30 \, x + 225\right )} \log \left (x\right ) + 5 \, \log \left (\log \left (5\right )^{2} - 9 \, \log \left (5\right ) + 25\right ) \log \left (x\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {225-30 x+x^2+\left (-30 x+2 x^2\right ) \log (5)+\left (-30 x+2 x^2\right ) \log (x)+5 \log \left (25-9 \log (5)+\log ^2(5)\right )}{x} \, dx=x^{2} \log {\left (5 \right )} - 30 x \log {\left (5 \right )} + \left (x^{2} - 30 x\right ) \log {\left (x \right )} + 5 \left (\log {\left (- 9 \log {\left (5 \right )} + \log {\left (5 \right )}^{2} + 25 \right )} + 45\right ) \log {\left (x \right )} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {225-30 x+x^2+\left (-30 x+2 x^2\right ) \log (5)+\left (-30 x+2 x^2\right ) \log (x)+5 \log \left (25-9 \log (5)+\log ^2(5)\right )}{x} \, dx=x^{2} \log \left (5\right ) + x^{2} \log \left (x\right ) - 30 \, x \log \left (5\right ) - 30 \, x \log \left (x\right ) + 5 \, \log \left (\log \left (5\right )^{2} - 9 \, \log \left (5\right ) + 25\right ) \log \left (x\right ) + 225 \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {225-30 x+x^2+\left (-30 x+2 x^2\right ) \log (5)+\left (-30 x+2 x^2\right ) \log (x)+5 \log \left (25-9 \log (5)+\log ^2(5)\right )}{x} \, dx=x^{2} \log \left (5\right ) - 30 \, x \log \left (5\right ) + {\left (x^{2} - 30 \, x\right )} \log \left (x\right ) + 5 \, {\left (\log \left (\log \left (5\right )^{2} - 9 \, \log \left (5\right ) + 25\right ) + 45\right )} \log \left (x\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 8.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {225-30 x+x^2+\left (-30 x+2 x^2\right ) \log (5)+\left (-30 x+2 x^2\right ) \log (x)+5 \log \left (25-9 \log (5)+\log ^2(5)\right )}{x} \, dx=225\,\ln \left (x\right )+x^2\,\ln \left (x\right )-30\,x\,\ln \left (5\right )+x^2\,\ln \left (5\right )+5\,\ln \left ({\ln \left (5\right )}^2-9\,\ln \left (5\right )+25\right )\,\ln \left (x\right )-30\,x\,\ln \left (x\right ) \]

[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)