\(\int \frac {6 x^2+10 x^3+(4 x+20 x^2) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx\) [6861]

Optimal result

Integrand size = 57, antiderivative size = 17 \[ \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx=\frac {2 x^2}{10+\frac {4}{x+\log (400)}} \]

[Out]

2*x^2/(2/(1/2*x+ln(20))+10)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(36\) vs. \(2(17)=34\).

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.12, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2011, 27, 1864} \[ \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx=\frac {x^2}{5}-\frac {2 x}{25}-\frac {2 (2+5 \log (400))^2}{125 (5 x+2+5 \log (400))} \]

[In]

Int[(6*x^2 + 10*x^3 + (4*x + 20*x^2)*Log[400] + 10*x*Log[400]^2)/(4 + 20*x + 25*x^2 + (20 + 50*x)*Log[400] + 2
5*Log[400]^2),x]

[Out]

(-2*x)/25 + x^2/5 - (2*(2 + 5*Log[400])^2)/(125*(2 + 5*x + 5*Log[400]))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 2011

Int[(Pq_)*(u_)^(p_.), x_Symbol] :> Int[Pq*ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && PolyQ[Pq, x] && QuadraticQ
[u, x] &&  !QuadraticMatchQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{25 x^2+10 x (2+5 \log (400))+(2+5 \log (400))^2} \, dx \\ & = \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{(2+5 x+5 \log (400))^2} \, dx \\ & = \int \left (-\frac {2}{25}+\frac {2 x}{5}+\frac {2 (2+5 \log (400))^2}{25 (2+5 x+5 \log (400))^2}\right ) \, dx \\ & = -\frac {2 x}{25}+\frac {x^2}{5}-\frac {2 (2+5 \log (400))^2}{125 (2+5 x+5 \log (400))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(17)=34\).

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 3.47 \[ \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx=\frac {125 x^3+125 x^2 \log (400)-(2+5 \log (400))^2 (6+5 \log (400))-5 x \left (12+40 \log (400)+25 \log ^2(400)\right )}{125 (2+5 x+5 \log (400))} \]

[In]

Integrate[(6*x^2 + 10*x^3 + (4*x + 20*x^2)*Log[400] + 10*x*Log[400]^2)/(4 + 20*x + 25*x^2 + (20 + 50*x)*Log[40
0] + 25*Log[400]^2),x]

[Out]

(125*x^3 + 125*x^2*Log[400] - (2 + 5*Log[400])^2*(6 + 5*Log[400]) - 5*x*(12 + 40*Log[400] + 25*Log[400]^2))/(1
25*(2 + 5*x + 5*Log[400]))

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29

method	result	size
gosper	\(\frac {x^{2} \left (x +2 \ln \left (20\right )\right )}{10 \ln \left (20\right )+5 x +2}\)	\(22\)
norman	\(\frac {x^{3}+2 x^{2} \ln \left (20\right )}{10 \ln \left (20\right )+5 x +2}\)	\(24\)
parallelrisch	\(\frac {10 x^{2} \ln \left (20\right )+5 x^{3}}{50 \ln \left (20\right )+25 x +10}\)	\(27\)
default	\(-\frac {2 x}{25}+\frac {x^{2}}{5}-\frac {2 \left (\frac {4 \ln \left (20\right )^{2}}{5}+\frac {8 \ln \left (20\right )}{25}+\frac {4}{125}\right )}{10 \ln \left (20\right )+5 x +2}\)	\(35\)
risch	\(\frac {x^{2}}{5}-\frac {2 x}{25}-\frac {4 \ln \left (5\right )^{2}}{25 \left (2 \ln \left (2\right )+\ln \left (5\right )+\frac {x}{2}+\frac {1}{5}\right )}-\frac {16 \ln \left (2\right ) \ln \left (5\right )}{25 \left (2 \ln \left (2\right )+\ln \left (5\right )+\frac {x}{2}+\frac {1}{5}\right )}-\frac {16 \ln \left (2\right )^{2}}{25 \left (2 \ln \left (2\right )+\ln \left (5\right )+\frac {x}{2}+\frac {1}{5}\right )}-\frac {16 \ln \left (2\right )}{125 \left (2 \ln \left (2\right )+\ln \left (5\right )+\frac {x}{2}+\frac {1}{5}\right )}-\frac {8 \ln \left (5\right )}{125 \left (2 \ln \left (2\right )+\ln \left (5\right )+\frac {x}{2}+\frac {1}{5}\right )}-\frac {4}{625 \left (2 \ln \left (2\right )+\ln \left (5\right )+\frac {x}{2}+\frac {1}{5}\right )}\)	\(116\)
meijerg	\(\frac {4 \left (5 \ln \left (20\right )+1\right )^{2} \left (\frac {8 \ln \left (20\right )}{5}+\frac {6}{25}\right ) \left (\frac {5 x \left (\frac {15 x}{2 \left (5 \ln \left (20\right )+1\right )}+6\right )}{6 \left (5 \ln \left (20\right )+1\right ) \left (1+\frac {5 x}{2 \left (5 \ln \left (20\right )+1\right )}\right )}-2 \ln \left (1+\frac {5 x}{2 \left (5 \ln \left (20\right )+1\right )}\right )\right )}{25 \left (2 \ln \left (20\right )+\frac {2}{5}\right )}+\left (\frac {8 \ln \left (20\right )^{2}}{5}+\frac {8 \ln \left (20\right )}{25}\right ) \left (-\frac {5 x}{2 \left (5 \ln \left (20\right )+1\right ) \left (1+\frac {5 x}{2 \left (5 \ln \left (20\right )+1\right )}\right )}+\ln \left (1+\frac {5 x}{2 \left (5 \ln \left (20\right )+1\right )}\right )\right )+\frac {16 \left (5 \ln \left (20\right )+1\right )^{3} \left (-\frac {5 x \left (-\frac {25 x^{2}}{2 \left (5 \ln \left (20\right )+1\right )^{2}}+\frac {15 x}{5 \ln \left (20\right )+1}+12\right )}{8 \left (5 \ln \left (20\right )+1\right ) \left (1+\frac {5 x}{2 \left (5 \ln \left (20\right )+1\right )}\right )}+3 \ln \left (1+\frac {5 x}{2 \left (5 \ln \left (20\right )+1\right )}\right )\right )}{625 \left (2 \ln \left (20\right )+\frac {2}{5}\right )}\)	\(222\)

[In]

int((40*x*ln(20)^2+2*(20*x^2+4*x)*ln(20)+10*x^3+6*x^2)/(100*ln(20)^2+2*(50*x+20)*ln(20)+25*x^2+20*x+4),x,metho
d=_RETURNVERBOSE)

[Out]

x^2*(x+2*ln(20))/(10*ln(20)+5*x+2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (18) = 36\).

Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.53 \[ \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx=\frac {125 \, x^{3} + 10 \, {\left (25 \, x^{2} - 10 \, x - 8\right )} \log \left (20\right ) - 200 \, \log \left (20\right )^{2} - 20 \, x - 8}{125 \, {\left (5 \, x + 10 \, \log \left (20\right ) + 2\right )}} \]

[In]

integrate((40*x*log(20)^2+2*(20*x^2+4*x)*log(20)+10*x^3+6*x^2)/(100*log(20)^2+2*(50*x+20)*log(20)+25*x^2+20*x+
4),x, algorithm="fricas")

[Out]

1/125*(125*x^3 + 10*(25*x^2 - 10*x - 8)*log(20) - 200*log(20)^2 - 20*x - 8)/(5*x + 10*log(20) + 2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (14) = 28\).

Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.00 \[ \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx=\frac {x^{2}}{5} - \frac {2 x}{25} + \frac {- 200 \log {\left (20 \right )}^{2} - 80 \log {\left (20 \right )} - 8}{625 x + 250 + 1250 \log {\left (20 \right )}} \]

[In]

integrate((40*x*ln(20)**2+2*(20*x**2+4*x)*ln(20)+10*x**3+6*x**2)/(100*ln(20)**2+2*(50*x+20)*ln(20)+25*x**2+20*
x+4),x)

[Out]

x**2/5 - 2*x/25 + (-200*log(20)**2 - 80*log(20) - 8)/(625*x + 250 + 1250*log(20))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.00 \[ \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx=\frac {1}{5} \, x^{2} - \frac {2}{25} \, x - \frac {8 \, {\left (25 \, \log \left (20\right )^{2} + 10 \, \log \left (20\right ) + 1\right )}}{125 \, {\left (5 \, x + 10 \, \log \left (20\right ) + 2\right )}} \]

[In]

integrate((40*x*log(20)^2+2*(20*x^2+4*x)*log(20)+10*x^3+6*x^2)/(100*log(20)^2+2*(50*x+20)*log(20)+25*x^2+20*x+
4),x, algorithm="maxima")

[Out]

1/5*x^2 - 2/25*x - 8/125*(25*log(20)^2 + 10*log(20) + 1)/(5*x + 10*log(20) + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.00 \[ \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx=\frac {1}{5} \, x^{2} - \frac {2}{25} \, x - \frac {8 \, {\left (25 \, \log \left (20\right )^{2} + 10 \, \log \left (20\right ) + 1\right )}}{125 \, {\left (5 \, x + 10 \, \log \left (20\right ) + 2\right )}} \]

[In]

integrate((40*x*log(20)^2+2*(20*x^2+4*x)*log(20)+10*x^3+6*x^2)/(100*log(20)^2+2*(50*x+20)*log(20)+25*x^2+20*x+
4),x, algorithm="giac")

[Out]

1/5*x^2 - 2/25*x - 8/125*(25*log(20)^2 + 10*log(20) + 1)/(5*x + 10*log(20) + 2)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.00 \[ \int \frac {6 x^2+10 x^3+\left (4 x+20 x^2\right ) \log (400)+10 x \log ^2(400)}{4+20 x+25 x^2+(20+50 x) \log (400)+25 \log ^2(400)} \, dx=\frac {x^2}{5}-\frac {16\,\ln \left (20\right )+40\,{\ln \left (20\right )}^2+\frac {8}{5}}{125\,x+250\,\ln \left (20\right )+50}-\frac {2\,x}{25} \]

[In]

int((2*log(20)*(4*x + 20*x^2) + 40*x*log(20)^2 + 6*x^2 + 10*x^3)/(20*x + 2*log(20)*(50*x + 20) + 100*log(20)^2
+ 25*x^2 + 4),x)

[Out]

x^2/5 - (16*log(20) + 40*log(20)^2 + 8/5)/(125*x + 250*log(20) + 50) - (2*x)/25