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

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 {gather*} \frac {1}{25} x^2 \log ^2(3)-\frac {3}{22 (x+22)}+\frac {3}{22 x} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-1650-150 x+\left (968 x^3+88 x^4+2 x^5\right ) \log ^2(3)}{x^2 \left (12100+1100 x+25 x^2\right )} \, dx\\ &=\int \frac {-1650-150 x+\left (968 x^3+88 x^4+2 x^5\right ) \log ^2(3)}{25 x^2 (22+x)^2} \, dx\\ &=\frac {1}{25} \int \frac {-1650-150 x+\left (968 x^3+88 x^4+2 x^5\right ) \log ^2(3)}{x^2 (22+x)^2} \, dx\\ &=\frac {1}{25} \int \left (-\frac {75}{22 x^2}+\frac {75}{22 (22+x)^2}+2 x \log ^2(3)\right ) \, dx\\ &=\frac {3}{22 x}-\frac {3}{22 (22+x)}+\frac {1}{25} x^2 \log ^2(3)\\ \end {aligned} \end {gather*}

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

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 {gather*} \frac {{\left (x^{4} + 22 \, x^{3}\right )} \log \relax (3)^{2} + 75}{25 \, {\left (x^{2} + 22 \, x\right )}} \end {gather*}

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 {gather*} \frac {1}{25} \, x^{2} \log \relax (3)^{2} + \frac {3}{x^{2} + 22 \, x} \end {gather*}

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	\(\frac {x^{2} \ln \relax (3)^{2}}{25}+\frac {3}{x \left (22+x \right )}\)	\(21\)
default	\(\frac {x^{2} \ln \relax (3)^{2}}{25}-\frac {3}{22 \left (22+x \right )}+\frac {3}{22 x}\)	\(23\)
gosper	\(\frac {x^{4} \ln \relax (3)^{2}+22 x^{3} \ln \relax (3)^{2}+75}{25 x \left (22+x \right )}\)	\(30\)
norman	\(\frac {3+\frac {22 x^{3} \ln \relax (3)^{2}}{25}+\frac {x^{4} \ln \relax (3)^{2}}{25}}{x \left (22+x \right )}\)	\(30\)

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 {gather*} \frac {1}{25} \, x^{2} \log \relax (3)^{2} + \frac {3}{x^{2} + 22 \, x} \end {gather*}

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 {gather*} \frac {x^2\,{\ln \relax (3)}^2}{25}+\frac {3}{x\,\left (x+22\right )} \end {gather*}

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 {gather*} \frac {x^{2} \log {\relax (3 )}^{2}}{25} + \frac {3}{x^{2} + 22 x} \end {gather*}

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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