3.41.96 $\int \frac{-1650-150x+(968x^3+88x^4+2x^5){\log }^2(3)}{12100x^2+1100x^3+25x^4}dx$

Optimal. Leaf size=24 $$-\frac{3}{(-22-x)x}+\frac{1}{25}{x}^2{\log }^2(3)$$

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Rubi [A]  time = 0.06, antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 4, integrand size = 45, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.089, Rules used = {1594, 27, 12, 1620} $$\begin{array}{c}\frac{1}{25}{x}^2{\log }^2(3)-\frac{3}{22(x+22)}+\frac{3}{22x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-1650 - 150*x + (968*x^3 + 88*x^4 + 2*x^5)*Log[3]^2)/(12100*x^2 + 1100*x^3 + 25*x^4),x]

[Out]

3/(22*x) - 3/(22*(22 + x)) + (x^2*Log[3]^2)/25

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-1650-150x+(968x^3+88x^4+2x^5){\log }^2(3)}{x^2(12100+1100x+25x^2)}dx\\ & =\int \frac{-1650-150x+(968x^3+88x^4+2x^5){\log }^2(3)}{25x^2(22+x{)}^2}dx\\ & =\frac{1}{25}\int \frac{-1650-150x+(968x^3+88x^4+2x^5){\log }^2(3)}{x^2(22+x{)}^2}dx\\ & =\frac{1}{25}\int (-\frac{75}{22x^2}+\frac{75}{22(22+x{)}^2}+2x{\log }^2(3))dx\\ & =\frac{3}{22x}-\frac{3}{22(22+x)}+\frac{1}{25}{x}^2{\log }^2(3)\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.33 $$\begin{array}{c}\frac{2}{25}(\frac{75}{44x}-\frac{75}{44(22+x)}+\frac{1}{2}{x}^2{\log }^2(3))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-1650 - 150*x + (968*x^3 + 88*x^4 + 2*x^5)*Log[3]^2)/(12100*x^2 + 1100*x^3 + 25*x^4),x]

[Out]

(2*(75/(44*x) - 75/(44*(22 + x)) + (x^2*Log[3]^2)/2))/25

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fricas [A]  time = 0.93, size = 27, normalized size = 1.12 $$\begin{array}{c}\frac{(x^4+22x^3)\log (3{)}^2+75}{25(x^2+22x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+88*x^4+968*x^3)*log(3)^2-150*x-1650)/(25*x^4+1100*x^3+12100*x^2),x, algorithm="fricas")

[Out]

1/25*((x^4 + 22*x^3)*log(3)^2 + 75)/(x^2 + 22*x)

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giac [A]  time = 0.14, size = 21, normalized size = 0.88 $$\begin{array}{c}\frac{1}{25}{x}^2\log (3{)}^2+\frac{3}{x^2+22x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+88*x^4+968*x^3)*log(3)^2-150*x-1650)/(25*x^4+1100*x^3+12100*x^2),x, algorithm="giac")

[Out]

1/25*x^2*log(3)^2 + 3/(x^2 + 22*x)

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maple [A]  time = 0.04, size = 21, normalized size = 0.88

method	result	size
risch	$\frac{x^2\ln (3{)}^2}{25}+\frac{3}{x(22+x)}$	$21$
default	$\frac{x^2\ln (3{)}^2}{25}-\frac{3}{22(22+x)}+\frac{3}{22x}$	$23$
gosper	$\frac{x^4\ln (3{)}^2+22x^3\ln (3{)}^2+75}{25x(22+x)}$	$30$
norman	$\frac{3+\frac{22x^3\ln (3{)}^2}{25}+\frac{x^4\ln (3{)}^2}{25}}{x(22+x)}$	$30$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^5+88*x^4+968*x^3)*ln(3)^2-150*x-1650)/(25*x^4+1100*x^3+12100*x^2),x,method=_RETURNVERBOSE)

[Out]

1/25*x^2*ln(3)^2+3/x/(22+x)

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maxima [A]  time = 0.36, size = 21, normalized size = 0.88 $$\begin{array}{c}\frac{1}{25}{x}^2\log (3{)}^2+\frac{3}{x^2+22x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+88*x^4+968*x^3)*log(3)^2-150*x-1650)/(25*x^4+1100*x^3+12100*x^2),x, algorithm="maxima")

[Out]

1/25*x^2*log(3)^2 + 3/(x^2 + 22*x)

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mupad [B]  time = 3.01, size = 20, normalized size = 0.83 $$\begin{array}{c}\frac{x^2{\ln (3)}^2}{25}+\frac{3}{x(x+22)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(150*x - log(3)^2*(968*x^3 + 88*x^4 + 2*x^5) + 1650)/(12100*x^2 + 1100*x^3 + 25*x^4),x)

[Out]

(x^2*log(3)^2)/25 + 3/(x*(x + 22))

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sympy [A]  time = 0.13, size = 17, normalized size = 0.71 $$\begin{array}{c}\frac{x^2\log {(3)}^2}{25}+\frac{3}{x^2+22x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**5+88*x**4+968*x**3)*ln(3)**2-150*x-1650)/(25*x**4+1100*x**3+12100*x**2),x)

[Out]

x**2*log(3)**2/25 + 3/(x**2 + 22*x)

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