3.34.45 \(\int \frac {-256-16 \log (3)}{4096+640 x+25 x^2+(512+40 x) \log (3)+16 \log ^2(3)} \, dx\)

Optimal. Leaf size=15 \[ \frac {x}{-16-\frac {5 x}{4}-\log (3)} \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 1981, 27, 32} \begin {gather*} \frac {16 (16+\log (3))}{5 (5 x+64+\log (81))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-256 - 16*Log[3])/(4096 + 640*x + 25*x^2 + (512 + 40*x)*Log[3] + 16*Log[3]^2),x]

[Out]

(16*(16 + Log[3]))/(5*(64 + 5*x + Log[81]))

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left ((16 (16+\log (3))) \int \frac {1}{4096+640 x+25 x^2+(512+40 x) \log (3)+16 \log ^2(3)} \, dx\right )\\ &=-\left ((16 (16+\log (3))) \int \frac {1}{25 x^2+40 x (16+\log (3))+16 (16+\log (3))^2} \, dx\right )\\ &=-\left ((16 (16+\log (3))) \int \frac {1}{(64+5 x+4 \log (3))^2} \, dx\right )\\ &=\frac {16 (16+\log (3))}{5 (64+5 x+\log (81))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 1.13 \begin {gather*} \frac {16 (16+\log (3))}{5 (64+5 x+\log (81))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-256 - 16*Log[3])/(4096 + 640*x + 25*x^2 + (512 + 40*x)*Log[3] + 16*Log[3]^2),x]

[Out]

(16*(16 + Log[3]))/(5*(64 + 5*x + Log[81]))

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fricas [A]  time = 0.52, size = 17, normalized size = 1.13 \begin {gather*} \frac {16 \, {\left (\log \relax (3) + 16\right )}}{5 \, {\left (5 \, x + 4 \, \log \relax (3) + 64\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(3)-256)/(16*log(3)^2+(40*x+512)*log(3)+25*x^2+640*x+4096),x, algorithm="fricas")

[Out]

16/5*(log(3) + 16)/(5*x + 4*log(3) + 64)

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giac [A]  time = 0.25, size = 17, normalized size = 1.13 \begin {gather*} \frac {16 \, {\left (\log \relax (3) + 16\right )}}{5 \, {\left (5 \, x + 4 \, \log \relax (3) + 64\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(3)-256)/(16*log(3)^2+(40*x+512)*log(3)+25*x^2+640*x+4096),x, algorithm="giac")

[Out]

16/5*(log(3) + 16)/(5*x + 4*log(3) + 64)

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maple [A]  time = 0.48, size = 18, normalized size = 1.20

method	result	size
gosper	\(\frac {\frac {256}{5}+\frac {16 \ln \relax (3)}{5}}{4 \ln \relax (3)+5 x +64}\)	\(18\)
norman	\(\frac {\frac {256}{5}+\frac {16 \ln \relax (3)}{5}}{4 \ln \relax (3)+5 x +64}\)	\(19\)
default	\(-\frac {-16 \ln \relax (3)-256}{5 \left (4 \ln \relax (3)+5 x +64\right )}\)	\(20\)
risch	\(\frac {4 \ln \relax (3)}{5 \left (\ln \relax (3)+\frac {5 x}{4}+16\right )}+\frac {64}{5 \left (\ln \relax (3)+\frac {5 x}{4}+16\right )}\)	\(26\)
meijerg	\(-\frac {64 x}{5 \left (\frac {4 \ln \relax (3)}{5}+\frac {64}{5}\right ) \left (1+\frac {5 x}{4 \left (\ln \relax (3)+16\right )}\right ) \left (\ln \relax (3)+16\right )}-\frac {4 \ln \relax (3) x}{5 \left (\frac {4 \ln \relax (3)}{5}+\frac {64}{5}\right ) \left (1+\frac {5 x}{4 \left (\ln \relax (3)+16\right )}\right ) \left (\ln \relax (3)+16\right )}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*ln(3)-256)/(16*ln(3)^2+(40*x+512)*ln(3)+25*x^2+640*x+4096),x,method=_RETURNVERBOSE)

[Out]

16/5*(ln(3)+16)/(4*ln(3)+5*x+64)

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maxima [A]  time = 0.42, size = 17, normalized size = 1.13 \begin {gather*} \frac {16 \, {\left (\log \relax (3) + 16\right )}}{5 \, {\left (5 \, x + 4 \, \log \relax (3) + 64\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(3)-256)/(16*log(3)^2+(40*x+512)*log(3)+25*x^2+640*x+4096),x, algorithm="maxima")

[Out]

16/5*(log(3) + 16)/(5*x + 4*log(3) + 64)

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mupad [B]  time = 2.01, size = 63, normalized size = 4.20 \begin {gather*} -\frac {\mathrm {atan}\left (\frac {x\,5{}\mathrm {i}+\ln \relax (3)\,4{}\mathrm {i}+64{}\mathrm {i}}{\sqrt {4\,\ln \relax (3)-\ln \left (81\right )}\,\sqrt {4\,\ln \relax (3)+\ln \left (81\right )+128}}\right )\,\left (\ln \relax (3)+16\right )\,16{}\mathrm {i}}{5\,\sqrt {4\,\ln \relax (3)-\ln \left (81\right )}\,\sqrt {4\,\ln \relax (3)+\ln \left (81\right )+128}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*log(3) + 256)/(640*x + log(3)*(40*x + 512) + 16*log(3)^2 + 25*x^2 + 4096),x)

[Out]

-(atan((x*5i + log(3)*4i + 64i)/((4*log(3) - log(81))^(1/2)*(4*log(3) + log(81) + 128)^(1/2)))*(log(3) + 16)*1
6i)/(5*(4*log(3) - log(81))^(1/2)*(4*log(3) + log(81) + 128)^(1/2))

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sympy [A]  time = 0.16, size = 19, normalized size = 1.27 \begin {gather*} - \frac {-256 - 16 \log {\relax (3 )}}{25 x + 20 \log {\relax (3 )} + 320} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*ln(3)-256)/(16*ln(3)**2+(40*x+512)*ln(3)+25*x**2+640*x+4096),x)

[Out]

-(-256 - 16*log(3))/(25*x + 20*log(3) + 320)

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