3.43.16 \(\int \frac {1-256 x+8 x^2+(2-384 x+8 x^2) \log (3)+(1-128 x+2 x^2) \log ^2(3)}{x+2 x \log (3)+x \log ^2(3)} \, dx\)

Optimal. Leaf size=19 \[ \left (64-x-\frac {x}{1+\log (3)}\right )^2+\log (x) \]

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.89, number of steps used = 5, number of rules used = 3, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6, 12, 14} \begin {gather*} \frac {x^2 (2+\log (3))^2}{(1+\log (3))^2}-\frac {128 x \left (2+\log ^2(3)+\log (27)\right )}{(1+\log (3))^2}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 256*x + 8*x^2 + (2 - 384*x + 8*x^2)*Log[3] + (1 - 128*x + 2*x^2)*Log[3]^2)/(x + 2*x*Log[3] + x*Log[3]
^2),x]

[Out]

(x^2*(2 + Log[3])^2)/(1 + Log[3])^2 - (128*x*(2 + Log[3]^2 + Log[27]))/(1 + Log[3])^2 + Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-256 x+8 x^2+\left (2-384 x+8 x^2\right ) \log (3)+\left (1-128 x+2 x^2\right ) \log ^2(3)}{x \log ^2(3)+x (1+2 \log (3))} \, dx\\ &=\int \frac {1-256 x+8 x^2+\left (2-384 x+8 x^2\right ) \log (3)+\left (1-128 x+2 x^2\right ) \log ^2(3)}{x \left (1+2 \log (3)+\log ^2(3)\right )} \, dx\\ &=\frac {\int \frac {1-256 x+8 x^2+\left (2-384 x+8 x^2\right ) \log (3)+\left (1-128 x+2 x^2\right ) \log ^2(3)}{x} \, dx}{(1+\log (3))^2}\\ &=\frac {\int \left (2 x (2+\log (3))^2+\frac {1+2 \log (3)+\log ^2(3)}{x}-128 \left (2+\log ^2(3)+\log (27)\right )\right ) \, dx}{(1+\log (3))^2}\\ &=\frac {x^2 (2+\log (3))^2}{(1+\log (3))^2}-\frac {128 x \left (2+\log ^2(3)+\log (27)\right )}{(1+\log (3))^2}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 40, normalized size = 2.11 \begin {gather*} \frac {x^2 (2+\log (3))^2-128 x \left (2+\log ^2(3)+\log (27)\right )+\left (1+\log ^2(3)+\log (9)\right ) \log (x)}{(1+\log (3))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 256*x + 8*x^2 + (2 - 384*x + 8*x^2)*Log[3] + (1 - 128*x + 2*x^2)*Log[3]^2)/(x + 2*x*Log[3] + x*
Log[3]^2),x]

[Out]

(x^2*(2 + Log[3])^2 - 128*x*(2 + Log[3]^2 + Log[27]) + (1 + Log[3]^2 + Log[9])*Log[x])/(1 + Log[3])^2

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fricas [B]  time = 0.48, size = 58, normalized size = 3.05 \begin {gather*} \frac {{\left (x^{2} - 128 \, x\right )} \log \relax (3)^{2} + 4 \, x^{2} + 4 \, {\left (x^{2} - 96 \, x\right )} \log \relax (3) + {\left (\log \relax (3)^{2} + 2 \, \log \relax (3) + 1\right )} \log \relax (x) - 256 \, x}{\log \relax (3)^{2} + 2 \, \log \relax (3) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-128*x+1)*log(3)^2+(8*x^2-384*x+2)*log(3)+8*x^2-256*x+1)/(x*log(3)^2+2*x*log(3)+x),x, algorit
hm="fricas")

[Out]

((x^2 - 128*x)*log(3)^2 + 4*x^2 + 4*(x^2 - 96*x)*log(3) + (log(3)^2 + 2*log(3) + 1)*log(x) - 256*x)/(log(3)^2
+ 2*log(3) + 1)

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giac [B]  time = 0.20, size = 97, normalized size = 5.11 \begin {gather*} \frac {x^{2} \log \relax (3)^{4} + 6 \, x^{2} \log \relax (3)^{3} - 128 \, x \log \relax (3)^{4} + 13 \, x^{2} \log \relax (3)^{2} - 640 \, x \log \relax (3)^{3} + 12 \, x^{2} \log \relax (3) - 1152 \, x \log \relax (3)^{2} + 4 \, x^{2} - 896 \, x \log \relax (3) - 256 \, x}{\log \relax (3)^{4} + 4 \, \log \relax (3)^{3} + 6 \, \log \relax (3)^{2} + 4 \, \log \relax (3) + 1} + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-128*x+1)*log(3)^2+(8*x^2-384*x+2)*log(3)+8*x^2-256*x+1)/(x*log(3)^2+2*x*log(3)+x),x, algorit
hm="giac")

[Out]

(x^2*log(3)^4 + 6*x^2*log(3)^3 - 128*x*log(3)^4 + 13*x^2*log(3)^2 - 640*x*log(3)^3 + 12*x^2*log(3) - 1152*x*lo
g(3)^2 + 4*x^2 - 896*x*log(3) - 256*x)/(log(3)^4 + 4*log(3)^3 + 6*log(3)^2 + 4*log(3) + 1) + log(abs(x))

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maple [B]  time = 0.05, size = 40, normalized size = 2.11

method	result	size
norman	\(\frac {\left (-128 \ln \relax (3)-256\right ) x +\frac {\left (\ln \relax (3)^{2}+4 \ln \relax (3)+4\right ) x^{2}}{\ln \relax (3)+1}}{\ln \relax (3)+1}+\ln \relax (x )\)	\(40\)
default	\(\frac {x^{2} \ln \relax (3)^{2}-128 x \ln \relax (3)^{2}+4 x^{2} \ln \relax (3)-384 x \ln \relax (3)+4 x^{2}-256 x +\left (\ln \relax (3)^{2}+2 \ln \relax (3)+1\right ) \ln \relax (x )}{\ln \relax (3)^{2}+2 \ln \relax (3)+1}\)	\(63\)
risch	\(\frac {\ln \relax (3)^{2} x^{2}}{\ln \relax (3)^{2}+2 \ln \relax (3)+1}-\frac {128 x \ln \relax (3)^{2}}{\ln \relax (3)^{2}+2 \ln \relax (3)+1}+\frac {4 \ln \relax (3) x^{2}}{\ln \relax (3)^{2}+2 \ln \relax (3)+1}-\frac {384 \ln \relax (3) x}{\ln \relax (3)^{2}+2 \ln \relax (3)+1}+\frac {4 x^{2}}{\ln \relax (3)^{2}+2 \ln \relax (3)+1}-\frac {256 x}{\ln \relax (3)^{2}+2 \ln \relax (3)+1}+\ln \relax (x )\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-128*x+1)*ln(3)^2+(8*x^2-384*x+2)*ln(3)+8*x^2-256*x+1)/(x*ln(3)^2+2*x*ln(3)+x),x,method=_RETURNVERB
OSE)

[Out]

((-128*ln(3)-256)*x+(ln(3)^2+4*ln(3)+4)/(ln(3)+1)*x^2)/(ln(3)+1)+ln(x)

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maxima [B]  time = 0.34, size = 44, normalized size = 2.32 \begin {gather*} \frac {{\left (\log \relax (3)^{2} + 4 \, \log \relax (3) + 4\right )} x^{2} - 128 \, {\left (\log \relax (3)^{2} + 3 \, \log \relax (3) + 2\right )} x}{\log \relax (3)^{2} + 2 \, \log \relax (3) + 1} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-128*x+1)*log(3)^2+(8*x^2-384*x+2)*log(3)+8*x^2-256*x+1)/(x*log(3)^2+2*x*log(3)+x),x, algorit
hm="maxima")

[Out]

((log(3)^2 + 4*log(3) + 4)*x^2 - 128*(log(3)^2 + 3*log(3) + 2)*x)/(log(3)^2 + 2*log(3) + 1) + log(x)

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mupad [B]  time = 3.04, size = 58, normalized size = 3.05 \begin {gather*} \ln \relax (x)-\frac {x\,\left (384\,\ln \relax (3)+128\,{\ln \relax (3)}^2+256\right )}{\ln \relax (9)+{\ln \relax (3)}^2+1}+\frac {x^2\,\left (8\,\ln \relax (3)+2\,{\ln \relax (3)}^2+8\right )}{2\,\left (\ln \relax (9)+{\ln \relax (3)}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3)*(8*x^2 - 384*x + 2) - 256*x + log(3)^2*(2*x^2 - 128*x + 1) + 8*x^2 + 1)/(x + 2*x*log(3) + x*log(3)
^2),x)

[Out]

log(x) - (x*(384*log(3) + 128*log(3)^2 + 256))/(log(9) + log(3)^2 + 1) + (x^2*(8*log(3) + 2*log(3)^2 + 8))/(2*
(log(9) + log(3)^2 + 1))

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sympy [B]  time = 0.16, size = 53, normalized size = 2.79 \begin {gather*} \frac {x^{2} \left (\log {\relax (3 )}^{2} + 4 + 4 \log {\relax (3 )}\right ) + x \left (- 384 \log {\relax (3 )} - 256 - 128 \log {\relax (3 )}^{2}\right ) + \left (1 + \log {\relax (3 )}\right )^{2} \log {\relax (x )}}{1 + \log {\relax (3 )}^{2} + 2 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-128*x+1)*ln(3)**2+(8*x**2-384*x+2)*ln(3)+8*x**2-256*x+1)/(x*ln(3)**2+2*x*ln(3)+x),x)

[Out]

(x**2*(log(3)**2 + 4 + 4*log(3)) + x*(-384*log(3) - 256 - 128*log(3)**2) + (1 + log(3))**2*log(x))/(1 + log(3)
**2 + 2*log(3))

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