3.18.12 $\int \frac{54+84x+25x^2+2x^3+(-15-26x-4x^2)\log (5)+(1+2x){\log }^2(5)}{36+12x+x^2+(-12-2x)\log (5)+{\log }^2(5)}dx$

Optimal. Leaf size=24 $$3+x+x^2+\log (4)+\frac{x}{2+\frac{1}{3}(x-\log (5))}$$

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Rubi [A]  time = 0.08, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, integrand size = 61, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.049, Rules used = {1986, 27, 1850} $$\begin{array}{c}{x}^2+x-\frac{3(6-\log (5))}{x+6-\log (5)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(54 + 84*x + 25*x^2 + 2*x^3 + (-15 - 26*x - 4*x^2)*Log[5] + (1 + 2*x)*Log[5]^2)/(36 + 12*x + x^2 + (-12 -
2*x)*Log[5] + Log[5]^2),x]

[Out]

x + x^2 - (3*(6 - Log[5]))/(6 + x - Log[5])

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{54+84x+25x^2+2x^3+(-15-26x-4x^2)\log (5)+(1+2x){\log }^2(5)}{x^2+2x(6-\log (5))+(-6+\log (5){)}^2}dx\\ & =\int \frac{54+84x+25x^2+2x^3+(-15-26x-4x^2)\log (5)+(1+2x){\log }^2(5)}{(6+x-\log (5){)}^2}dx\\ & =\int (1+2x-\frac{3(-6+\log (5))}{(6+x-\log (5){)}^2})dx\\ & =x+x^2-\frac{3(6-\log (5))}{6+x-\log (5)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.04, size = 39, normalized size = 1.62 $$\begin{array}{c}(6+x-\log (5){)}^2+\frac{3(-6+\log (5))}{6+x-\log (5)}+(6+x-\log (5))(-11+2\log (5))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(54 + 84*x + 25*x^2 + 2*x^3 + (-15 - 26*x - 4*x^2)*Log[5] + (1 + 2*x)*Log[5]^2)/(36 + 12*x + x^2 + (
-12 - 2*x)*Log[5] + Log[5]^2),x]

[Out]

(6 + x - Log[5])^2 + (3*(-6 + Log[5]))/(6 + x - Log[5]) + (6 + x - Log[5])*(-11 + 2*Log[5])

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fricas [A]  time = 0.66, size = 33, normalized size = 1.38 $$\begin{array}{c}\frac{x^3+7x^2-(x^2+x-3)\log (5)+6x-18}{x-\log (5)+6}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*log(5)^2+(-4*x^2-26*x-15)*log(5)+2*x^3+25*x^2+84*x+54)/(log(5)^2+(-2*x-12)*log(5)+x^2+12*x+
36),x, algorithm="fricas")

[Out]

(x^3 + 7*x^2 - (x^2 + x - 3)*log(5) + 6*x - 18)/(x - log(5) + 6)

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giac [A]  time = 0.21, size = 20, normalized size = 0.83 $$\begin{array}{c}{x}^2+x+\frac{3(\log (5)-6)}{x-\log (5)+6}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*log(5)^2+(-4*x^2-26*x-15)*log(5)+2*x^3+25*x^2+84*x+54)/(log(5)^2+(-2*x-12)*log(5)+x^2+12*x+
36),x, algorithm="giac")

[Out]

x^2 + x + 3*(log(5) - 6)/(x - log(5) + 6)

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maple [A]  time = 0.33, size = 23, normalized size = 0.96

method	result	size
default	$x^2+x-\frac{-3\ln (5)+18}{-\ln (5)+x+6}$	$23$
risch	$x^2+x-\frac{3\ln (5)}{\ln (5)-x-6}+\frac{18}{\ln (5)-x-6}$	$30$
norman	$\frac{-x^3+(-7+\ln (5))x^2+54+\ln (5{)}^2-15\ln (5)}{\ln (5)-x-6}$	$34$
gosper	$\frac{x^2\ln (5)-x^3+\ln (5{)}^2-7x^2-15\ln (5)+54}{\ln (5)-x-6}$	$37$
meijerg	$(2\ln (5{)}^2-26\ln (5)+84)(-\frac{x}{(-\ln (5)+6)(1+\frac{x}{-\ln (5)+6})}+\ln (1+\frac{x}{-\ln (5)+6}))+(-2\ln (5)+12)(-\ln (5)+6)(-\frac{x(-\frac{2x^2}{{(-\ln (5)+6)}^2}+\frac{6x}{-\ln (5)+6}+12)}{4(-\ln (5)+6)(1+\frac{x}{-\ln (5)+6})}+3\ln (1+\frac{x}{-\ln (5)+6}))+(-4\ln (5)+25)(-\ln (5)+6)(\frac{x(\frac{3x}{-\ln (5)+6}+6)}{3(-\ln (5)+6)(1+\frac{x}{-\ln (5)+6})}-2\ln (1+\frac{x}{-\ln (5)+6}))+\frac{\ln (5{)}^{2}x}{{(-\ln (5)+6)}^2(1+\frac{x}{-\ln (5)+6})}-\frac{15\ln (5)x}{{(-\ln (5)+6)}^2(1+\frac{x}{-\ln (5)+6})}+\frac{54x}{{(-\ln (5)+6)}^2(1+\frac{x}{-\ln (5)+6})}$	$281$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+1)*ln(5)^2+(-4*x^2-26*x-15)*ln(5)+2*x^3+25*x^2+84*x+54)/(ln(5)^2+(-2*x-12)*ln(5)+x^2+12*x+36),x,meth
od=_RETURNVERBOSE)

[Out]

x^2+x-(-3*ln(5)+18)/(-ln(5)+x+6)

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maxima [A]  time = 0.41, size = 20, normalized size = 0.83 $$\begin{array}{c}{x}^2+x+\frac{3(\log (5)-6)}{x-\log (5)+6}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*log(5)^2+(-4*x^2-26*x-15)*log(5)+2*x^3+25*x^2+84*x+54)/(log(5)^2+(-2*x-12)*log(5)+x^2+12*x+
36),x, algorithm="maxima")

[Out]

x^2 + x + 3*(log(5) - 6)/(x - log(5) + 6)

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mupad [B]  time = 1.21, size = 75, normalized size = 3.12 $$\begin{array}{c}x(2\ln \left(25\right)-\ln \left(625\right)+1)+x^2-\frac{\mathrm{atan}\left(\frac{x2\mathrm{i}-\ln \left(25\right)1\mathrm{i}+12\mathrm{i}}{\sqrt{\ln \left(25\right)-2\ln (5)}\sqrt{2\ln (5)+\ln \left(25\right)-24}}\right)(\ln \left(125\right)-18)2\mathrm{i}}{\sqrt{\ln \left(25\right)-2\ln (5)}\sqrt{2\ln (5)+\ln \left(25\right)-24}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((84*x - log(5)*(26*x + 4*x^2 + 15) + log(5)^2*(2*x + 1) + 25*x^2 + 2*x^3 + 54)/(12*x - log(5)*(2*x + 12) +
log(5)^2 + x^2 + 36),x)

[Out]

x*(2*log(25) - log(625) + 1) + x^2 - (atan((x*2i - log(25)*1i + 12i)/((log(25) - 2*log(5))^(1/2)*(2*log(5) + l
og(25) - 24)^(1/2)))*(log(125) - 18)*2i)/((log(25) - 2*log(5))^(1/2)*(2*log(5) + log(25) - 24)^(1/2))

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sympy [A]  time = 0.20, size = 17, normalized size = 0.71 $$\begin{array}{c}{x}^2+x+\frac{-18+3\log (5)}{x-\log (5)+6}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*ln(5)**2+(-4*x**2-26*x-15)*ln(5)+2*x**3+25*x**2+84*x+54)/(ln(5)**2+(-2*x-12)*ln(5)+x**2+12*
x+36),x)

[Out]

x**2 + x + (-18 + 3*log(5))/(x - log(5) + 6)

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