3.69.61 $\int \frac{6x^2+10x^3+(4x+20x^2)\log (400)+10x{\log }^2(400)}{4+20x+25x^2+(20+50x)\log (400)+25{\log }^2(400)}dx$

Optimal. Leaf size=17 $$\frac{2x^2}{10+\frac{4}{x+\log (400)}}$$

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Rubi [B]  time = 0.09, antiderivative size = 36, normalized size of antiderivative = 2.12, number of steps used = 4, number of rules used = 3, integrand size = 57, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.053, Rules used = {1986, 27, 1850} $$\begin{array}{c}\frac{x^2}{5}-\frac{2x}{25}-\frac{2(2+5\log (400){)}^2}{125(5x+2+5\log (400))}\end{array}$$

Antiderivative was successfully verified.

[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 1850

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 1986

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{6x^2+10x^3+(4x+20x^2)\log (400)+10x{\log }^2(400)}{25x^2+10x(2+5\log (400))+(2+5\log (400){)}^2}dx\\ & =\int \frac{6x^2+10x^3+(4x+20x^2)\log (400)+10x{\log }^2(400)}{(2+5x+5\log (400){)}^2}dx\\ & =\int (-\frac{2}{25}+\frac{2x}{5}+\frac{2(2+5\log (400){)}^2}{25(2+5x+5\log (400){)}^2})dx\\ & =-\frac{2x}{25}+\frac{x^2}{5}-\frac{2(2+5\log (400){)}^2}{125(2+5x+5\log (400))}\end{array}\end{array}$$

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Mathematica [B]  time = 0.03, size = 59, normalized size = 3.47 $$\begin{array}{c}\frac{125x^3+125x^2\log (400)-(2+5\log (400){)}^2(6+5\log (400))-5x(12+40\log (400)+25{\log }^2(400))}{125(2+5x+5\log (400))}\end{array}$$

Antiderivative was successfully verified.

[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]))

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fricas [B]  time = 0.56, size = 43, normalized size = 2.53 $$\begin{array}{c}\frac{125x^3+10(25x^2-10x-8)\log \left(20\right)-200\log {\left(20\right)}^2-20x-8}{125(5x+10\log \left(20\right)+2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.29, size = 34, normalized size = 2.00 $$\begin{array}{c}\frac{1}{5}{x}^2-\frac{2}{25}x-\frac{8(25\log {\left(20\right)}^2+10\log \left(20\right)+1)}{125(5x+10\log \left(20\right)+2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.11, size = 22, normalized size = 1.29

method	result	size
gosper	$\frac{x^2(x+2\ln \left(20\right))}{10\ln \left(20\right)+5x+2}$	$22$
norman	$\frac{x^3+2x^2\ln \left(20\right)}{10\ln \left(20\right)+5x+2}$	$24$
default	$-\frac{2x}{25}+\frac{x^2}{5}-\frac{2(\frac{4\ln {\left(20\right)}^2}{5}+\frac{8\ln \left(20\right)}{25}+\frac{4}{125})}{10\ln \left(20\right)+5x+2}$	$35$
risch	$\frac{x^2}{5}-\frac{2x}{25}-\frac{4\ln (5{)}^2}{25(2\ln (2)+\ln (5)+\frac{x}{2}+\frac{1}{5})}-\frac{16\ln (2)\ln (5)}{25(2\ln (2)+\ln (5)+\frac{x}{2}+\frac{1}{5})}-\frac{16\ln (2{)}^2}{25(2\ln (2)+\ln (5)+\frac{x}{2}+\frac{1}{5})}-\frac{16\ln (2)}{125(2\ln (2)+\ln (5)+\frac{x}{2}+\frac{1}{5})}-\frac{8\ln (5)}{125(2\ln (2)+\ln (5)+\frac{x}{2}+\frac{1}{5})}-\frac{4}{625(2\ln (2)+\ln (5)+\frac{x}{2}+\frac{1}{5})}$	$116$
meijerg	$\frac{4(40\ln \left(20\right)+6){(5\ln \left(20\right)+1)}^2(\frac{5x(\frac{15x}{2(5\ln \left(20\right)+1)}+6)}{6(5\ln \left(20\right)+1)(1+\frac{5x}{2(5\ln \left(20\right)+1)})}-2\ln (1+\frac{5x}{2(5\ln \left(20\right)+1)}))}{625(2\ln \left(20\right)+\frac{2}{5})}+\frac{(40\ln {\left(20\right)}^2+8\ln \left(20\right))(-\frac{5x}{2(5\ln \left(20\right)+1)(1+\frac{5x}{2(5\ln \left(20\right)+1)})}+\ln (1+\frac{5x}{2(5\ln \left(20\right)+1)}))}{25}+\frac{16{(5\ln \left(20\right)+1)}^3(-\frac{5x(-\frac{25x^2}{2{(5\ln \left(20\right)+1)}^2}+\frac{15x}{5\ln \left(20\right)+1}+12)}{8(5\ln \left(20\right)+1)(1+\frac{5x}{2(5\ln \left(20\right)+1)})}+3\ln (1+\frac{5x}{2(5\ln \left(20\right)+1)}))}{625(2\ln \left(20\right)+\frac{2}{5})}$	$223$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.37, size = 34, normalized size = 2.00 $$\begin{array}{c}\frac{1}{5}{x}^2-\frac{2}{25}x-\frac{8(25\log {\left(20\right)}^2+10\log \left(20\right)+1)}{125(5x+10\log \left(20\right)+2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.13, size = 34, normalized size = 2.00 $$\begin{array}{c}\frac{x^2}{5}-\frac{16\ln \left(20\right)+40{\ln \left(20\right)}^2+\frac{8}{5}}{125x+250\ln \left(20\right)+50}-\frac{2x}{25}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [B]  time = 0.28, size = 34, normalized size = 2.00 $$\begin{array}{c}\frac{x^2}{5}-\frac{2x}{25}+\frac{-200\log {\left(20\right)}^2-80\log \left(20\right)-8}{625x+250+1250\log \left(20\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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