\(\int \frac {-20+22 \log ^2(4)+(4-2 \log ^2(4)) \log (4+\log (5))}{-2+\log ^2(4)} \, dx\) [3226]

Optimal result

Integrand size = 31, antiderivative size = 30 \[ \int \frac {-20+22 \log ^2(4)+\left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \, dx=x \left (4+2 \left (3 \left (3-\frac {4}{2-\log ^2(4)}\right )-\log (4+\log (5))\right )\right ) \]

[Out]

x*(22-24/(2-4*ln(2)^2)-2*ln(ln(5)+4))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {8} \[ \int \frac {-20+22 \log ^2(4)+\left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \, dx=\frac {2 x \left (10-11 \log ^2(4)-\left (2-\log ^2(4)\right ) \log (4+\log (5))\right )}{2-\log ^2(4)} \]

[In]

Int[(-20 + 22*Log[4]^2 + (4 - 2*Log[4]^2)*Log[4 + Log[5]])/(-2 + Log[4]^2),x]

[Out]

(2*x*(10 - 11*Log[4]^2 - (2 - Log[4]^2)*Log[4 + Log[5]]))/(2 - Log[4]^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x \left (10-11 \log ^2(4)-\left (2-\log ^2(4)\right ) \log (4+\log (5))\right )}{2-\log ^2(4)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {-20+22 \log ^2(4)+\left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \, dx=-\frac {20 x}{-2+\log ^2(4)}+\frac {22 x \log ^2(4)}{-2+\log ^2(4)}+\frac {x \left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \]

[In]

Integrate[(-20 + 22*Log[4]^2 + (4 - 2*Log[4]^2)*Log[4 + Log[5]])/(-2 + Log[4]^2),x]

[Out]

(-20*x)/(-2 + Log[4]^2) + (22*x*Log[4]^2)/(-2 + Log[4]^2) + (x*(4 - 2*Log[4]^2)*Log[4 + Log[5]])/(-2 + Log[4]^
2)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17

method	result	size
parallelrisch	\(\frac {\left (\left (-8 \ln \left (2\right )^{2}+4\right ) \ln \left (\ln \left (5\right )+4\right )+88 \ln \left (2\right )^{2}-20\right ) x}{4 \ln \left (2\right )^{2}-2}\)	\(35\)
default	\(\frac {\left (\left (-8 \ln \left (2\right )^{2}+4\right ) \ln \left (\ln \left (5\right )+4\right )+88 \ln \left (2\right )^{2}-20\right ) x}{4 \ln \left (2\right )^{2}-2}\)	\(36\)
norman	\(-\frac {2 \left (2 \ln \left (2\right )^{2} \ln \left (\ln \left (5\right )+4\right )-22 \ln \left (2\right )^{2}-\ln \left (\ln \left (5\right )+4\right )+5\right ) x}{2 \ln \left (2\right )^{2}-1}\)	\(40\)
risch	\(-\frac {8 x \ln \left (2\right )^{2} \ln \left (\ln \left (5\right )+4\right )}{4 \ln \left (2\right )^{2}-2}+\frac {88 x \ln \left (2\right )^{2}}{4 \ln \left (2\right )^{2}-2}+\frac {4 x \ln \left (\ln \left (5\right )+4\right )}{4 \ln \left (2\right )^{2}-2}-\frac {20 x}{4 \ln \left (2\right )^{2}-2}\)	\(72\)

[In]

int(((-8*ln(2)^2+4)*ln(ln(5)+4)+88*ln(2)^2-20)/(4*ln(2)^2-2),x,method=_RETURNVERBOSE)

[Out]

((-8*ln(2)^2+4)*ln(ln(5)+4)+88*ln(2)^2-20)/(4*ln(2)^2-2)*x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {-20+22 \log ^2(4)+\left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \, dx=\frac {2 \, {\left (22 \, x \log \left (2\right )^{2} - {\left (2 \, x \log \left (2\right )^{2} - x\right )} \log \left (\log \left (5\right ) + 4\right ) - 5 \, x\right )}}{2 \, \log \left (2\right )^{2} - 1} \]

[In]

integrate(((-8*log(2)^2+4)*log(log(5)+4)+88*log(2)^2-20)/(4*log(2)^2-2),x, algorithm="fricas")

[Out]

2*(22*x*log(2)^2 - (2*x*log(2)^2 - x)*log(log(5) + 4) - 5*x)/(2*log(2)^2 - 1)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-20+22 \log ^2(4)+\left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \, dx=\frac {x \left (-20 + \left (4 - 8 \log {\left (2 \right )}^{2}\right ) \log {\left (\log {\left (5 \right )} + 4 \right )} + 88 \log {\left (2 \right )}^{2}\right )}{-2 + 4 \log {\left (2 \right )}^{2}} \]

[In]

integrate(((-8*ln(2)**2+4)*ln(ln(5)+4)+88*ln(2)**2-20)/(4*ln(2)**2-2),x)

[Out]

x*(-20 + (4 - 8*log(2)**2)*log(log(5) + 4) + 88*log(2)**2)/(-2 + 4*log(2)**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-20+22 \log ^2(4)+\left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \, dx=\frac {2 \, {\left (22 \, \log \left (2\right )^{2} - {\left (2 \, \log \left (2\right )^{2} - 1\right )} \log \left (\log \left (5\right ) + 4\right ) - 5\right )} x}{2 \, \log \left (2\right )^{2} - 1} \]

[In]

integrate(((-8*log(2)^2+4)*log(log(5)+4)+88*log(2)^2-20)/(4*log(2)^2-2),x, algorithm="maxima")

[Out]

2*(22*log(2)^2 - (2*log(2)^2 - 1)*log(log(5) + 4) - 5)*x/(2*log(2)^2 - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-20+22 \log ^2(4)+\left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \, dx=\frac {2 \, {\left (22 \, \log \left (2\right )^{2} - {\left (2 \, \log \left (2\right )^{2} - 1\right )} \log \left (\log \left (5\right ) + 4\right ) - 5\right )} x}{2 \, \log \left (2\right )^{2} - 1} \]

[In]

integrate(((-8*log(2)^2+4)*log(log(5)+4)+88*log(2)^2-20)/(4*log(2)^2-2),x, algorithm="giac")

[Out]

2*(22*log(2)^2 - (2*log(2)^2 - 1)*log(log(5) + 4) - 5)*x/(2*log(2)^2 - 1)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {-20+22 \log ^2(4)+\left (4-2 \log ^2(4)\right ) \log (4+\log (5))}{-2+\log ^2(4)} \, dx=-\frac {x\,\left (\ln \left (\ln \left (5\right )+4\right )\,\left (8\,{\ln \left (2\right )}^2-4\right )-88\,{\ln \left (2\right )}^2+20\right )}{4\,{\ln \left (2\right )}^2-2} \]

[In]

int(-(log(log(5) + 4)*(8*log(2)^2 - 4) - 88*log(2)^2 + 20)/(4*log(2)^2 - 2),x)

[Out]

-(x*(log(log(5) + 4)*(8*log(2)^2 - 4) - 88*log(2)^2 + 20))/(4*log(2)^2 - 2)