3.6.33 $\int \frac{x^2+2x\log (3)+{\log }^2(3)+e^x(-4+4x+4\log (3))\log (\log (4))}{5x^2+x^3+(10x+2x^2)\log (3)+(5+x){\log }^2(3)+(16x^2+32x\log (3)+16{\log }^2(3)+e^x(4x+4\log (3)))\log (\log (4))}dx$

Optimal. Leaf size=25 $$\log (4+\frac{e^x}{x+\log (3)}+\frac{5+x}{4\log (\log (4))})$$

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Rubi [F]  time = 1.90, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.000, Rules used = {} $$\begin{array}{c}\int \frac{x^2+2x\log (3)+{\log }^2(3)+e^x(-4+4x+4\log (3))\log (\log (4))}{5x^2+x^3+(10x+2x^2)\log (3)+(5+x){\log }^2(3)+(16x^2+32x\log (3)+16{\log }^2(3)+e^x(4x+4\log (3)))\log (\log (4))}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(x^2 + 2*x*Log[3] + Log[3]^2 + E^x*(-4 + 4*x + 4*Log[3])*Log[Log[4]])/(5*x^2 + x^3 + (10*x + 2*x^2)*Log[3]
+ (5 + x)*Log[3]^2 + (16*x^2 + 32*x*Log[3] + 16*Log[3]^2 + E^x*(4*x + 4*Log[3]))*Log[Log[4]]),x]

[Out]

x - Log[x + Log[3]] - (5 - Log[81] + 16*(1 - Log[3])*Log[Log[4]])*Defer[Int][(-x^2 - 4*E^x*Log[Log[4]] - 5*Log
[3]*(1 + (16*Log[Log[4]])/5) - 5*x*(1 + (Log[3] + 16*Log[Log[4]])/5))^(-1), x] + Defer[Int][x^2/(-x^2 - 4*E^x*
Log[Log[4]] - 5*Log[3]*(1 + (16*Log[Log[4]])/5) - 5*x*(1 + (Log[3] + 16*Log[Log[4]])/5)), x] - (3 + Log[3] + 1
6*Log[Log[4]])*Defer[Int][x/(x^2 + 4*E^x*Log[Log[4]] + 5*Log[3]*(1 + (16*Log[Log[4]])/5) + 5*x*(1 + (Log[3] +
16*Log[Log[4]])/5)), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{x^2+2x\log (3)+{\log }^2(3)+e^x(-4+4x+4\log (3))\log (\log (4))}{(x+\log (3))(x^2+4e^x\log (\log (4))+5\log (3)(1+\frac{16}{5}\log (\log (4)))+5x(1+\frac{1}{5}(\log (3)+16\log (\log (4)))))}dx\\ & =\int (\frac{-1+x+\log (3)}{x+\log (3)}+\frac{5-x^2-\log (81)+16(1-\log (3))\log (\log (4))-x(3+\log (3)+16\log (\log (4)))}{x^2+4e^x\log (\log (4))+5\log (3)(1+\frac{16}{5}\log (\log (4)))+5x(1+\frac{1}{5}(\log (3)+16\log (\log (4))))})dx\\ & =\int \frac{-1+x+\log (3)}{x+\log (3)}dx+\int \frac{5-x^2-\log (81)+16(1-\log (3))\log (\log (4))-x(3+\log (3)+16\log (\log (4)))}{x^2+4e^x\log (\log (4))+5\log (3)(1+\frac{16}{5}\log (\log (4)))+5x(1+\frac{1}{5}(\log (3)+16\log (\log (4))))}dx\\ & =\int (1+\frac{1}{-x-\log (3)})dx+\int (\frac{x^2}{-x^2-4e^x\log (\log (4))-5\log (3)(1+\frac{16}{5}\log (\log (4)))-5x(1+\frac{1}{5}(\log (3)+16\log (\log (4))))}+\frac{\log (81)(1+\frac{-5+16(-1+\log (3))\log (\log (4))}{\log (81)})}{-x^2-4e^x\log (\log (4))-5\log (3)(1+\frac{16}{5}\log (\log (4)))-5x(1+\frac{1}{5}(\log (3)+16\log (\log (4))))}+\frac{x(-3-\log (3)-16\log (\log (4)))}{x^2+4e^x\log (\log (4))+5\log (3)(1+\frac{16}{5}\log (\log (4)))+5x(1+\frac{1}{5}(\log (3)+16\log (\log (4))))})dx\\ & =x-\log (x+\log (3))+(-3-\log (3)-16\log (\log (4)))\int \frac{x}{x^2+4e^x\log (\log (4))+5\log (3)(1+\frac{16}{5}\log (\log (4)))+5x(1+\frac{1}{5}(\log (3)+16\log (\log (4))))}dx+(-5+\log (81)-16(1-\log (3))\log (\log (4)))\int \frac{1}{-x^2-4e^x\log (\log (4))-5\log (3)(1+\frac{16}{5}\log (\log (4)))-5x(1+\frac{1}{5}(\log (3)+16\log (\log (4))))}dx+\int \frac{x^2}{-x^2-4e^x\log (\log (4))-5\log (3)(1+\frac{16}{5}\log (\log (4)))-5x(1+\frac{1}{5}(\log (3)+16\log (\log (4))))}dx\end{array}\end{array}$$

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Mathematica [F]  time = 2.13, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{x^2+2x\log (3)+{\log }^2(3)+e^x(-4+4x+4\log (3))\log (\log (4))}{5x^2+x^3+(10x+2x^2)\log (3)+(5+x){\log }^2(3)+(16x^2+32x\log (3)+16{\log }^2(3)+e^x(4x+4\log (3)))\log (\log (4))}dx\end{array}$$

Verification is not applicable to the result.

[In]

Integrate[(x^2 + 2*x*Log[3] + Log[3]^2 + E^x*(-4 + 4*x + 4*Log[3])*Log[Log[4]])/(5*x^2 + x^3 + (10*x + 2*x^2)*
Log[3] + (5 + x)*Log[3]^2 + (16*x^2 + 32*x*Log[3] + 16*Log[3]^2 + E^x*(4*x + 4*Log[3]))*Log[Log[4]]),x]

[Out]

Integrate[(x^2 + 2*x*Log[3] + Log[3]^2 + E^x*(-4 + 4*x + 4*Log[3])*Log[Log[4]])/(5*x^2 + x^3 + (10*x + 2*x^2)*
Log[3] + (5 + x)*Log[3]^2 + (16*x^2 + 32*x*Log[3] + 16*Log[3]^2 + E^x*(4*x + 4*Log[3]))*Log[Log[4]]), x]

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fricas [A]  time = 0.60, size = 39, normalized size = 1.56 $$\begin{array}{c}\log (x^2+(x+5)\log (3)+4(4x+e^x+4\log (3))\log (2\log (2))+5x)-\log (x+\log (3))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)+4*x-4)*exp(x)*log(2*log(2))+log(3)^2+2*x*log(3)+x^2)/(((4*log(3)+4*x)*exp(x)+16*log(3)^2+
32*x*log(3)+16*x^2)*log(2*log(2))+(5+x)*log(3)^2+(2*x^2+10*x)*log(3)+x^3+5*x^2),x, algorithm="fricas")

[Out]

log(x^2 + (x + 5)*log(3) + 4*(4*x + e^x + 4*log(3))*log(2*log(2)) + 5*x) - log(x + log(3))

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giac [B]  time = 0.39, size = 61, normalized size = 2.44 $$\begin{array}{c}\log (x^2+x\log (3)+16x\log (2)+4e^x\log (2)+16\log (3)\log (2)+16x\log (\log (2))+4e^x\log (\log (2))+16\log (3)\log (\log (2))+5x+5\log (3))-\log (x+\log (3))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)+4*x-4)*exp(x)*log(2*log(2))+log(3)^2+2*x*log(3)+x^2)/(((4*log(3)+4*x)*exp(x)+16*log(3)^2+
32*x*log(3)+16*x^2)*log(2*log(2))+(5+x)*log(3)^2+(2*x^2+10*x)*log(3)+x^3+5*x^2),x, algorithm="giac")

[Out]

log(x^2 + x*log(3) + 16*x*log(2) + 4*e^x*log(2) + 16*log(3)*log(2) + 16*x*log(log(2)) + 4*e^x*log(log(2)) + 16
*log(3)*log(log(2)) + 5*x + 5*log(3)) - log(x + log(3))

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maple [B]  time = 0.36, size = 51, normalized size = 2.04

method	result	size
norman	$-\ln (\ln (3)+x)+\ln (16\ln (3)\ln (2\ln (2))+x\ln (3)+16x\ln (2\ln (2))+4{\mathrm{e}}^x\ln (2\ln (2))+x^2+5\ln (3)+5x)$	$51$
risch	$-\ln (\ln (3)+x)+\ln ({\mathrm{e}}^x+\frac{16\ln (2)\ln (3)+16x\ln (2)+16\ln (3)\ln (\ln (2))+x\ln (3)+16x\ln (\ln (2))+x^2+5\ln (3)+5x}{4\ln (2)+4\ln (\ln (2))})$	$62$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*ln(3)+4*x-4)*exp(x)*ln(2*ln(2))+ln(3)^2+2*x*ln(3)+x^2)/(((4*ln(3)+4*x)*exp(x)+16*ln(3)^2+32*x*ln(3)+16
*x^2)*ln(2*ln(2))+(5+x)*ln(3)^2+(2*x^2+10*x)*ln(3)+x^3+5*x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(3)+x)+ln(16*ln(3)*ln(2*ln(2))+x*ln(3)+16*x*ln(2*ln(2))+4*exp(x)*ln(2*ln(2))+x^2+5*ln(3)+5*x)

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maxima [B]  time = 0.50, size = 64, normalized size = 2.56 $$\begin{array}{c}-\log (x+\log (3))+\log \left(\frac{x^2+x(\log (3)+16\log (2)+16\log (\log (2))+5)+4(\log (2)+\log (\log (2)))e^x+(16\log (\log (2))+5)\log (3)+16\log (3)\log (2)}{4(\log (2)+\log (\log (2)))}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)+4*x-4)*exp(x)*log(2*log(2))+log(3)^2+2*x*log(3)+x^2)/(((4*log(3)+4*x)*exp(x)+16*log(3)^2+
32*x*log(3)+16*x^2)*log(2*log(2))+(5+x)*log(3)^2+(2*x^2+10*x)*log(3)+x^3+5*x^2),x, algorithm="maxima")

[Out]

-log(x + log(3)) + log(1/4*(x^2 + x*(log(3) + 16*log(2) + 16*log(log(2)) + 5) + 4*(log(2) + log(log(2)))*e^x +
(16*log(log(2)) + 5)*log(3) + 16*log(3)*log(2))/(log(2) + log(log(2))))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\int \frac{2x\ln (3)+{\ln (3)}^2+x^2+\ln (2\ln (2)){\mathrm{e}}^x(4x+4\ln (3)-4)}{{\ln (3)}^2(x+5)+\ln (3)(2x^2+10x)+5x^2+x^3+\ln (2\ln (2))({\mathrm{e}}^x(4x+4\ln (3))+32x\ln (3)+16{\ln (3)}^2+16x^2)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*log(3) + log(3)^2 + x^2 + log(2*log(2))*exp(x)*(4*x + 4*log(3) - 4))/(log(3)^2*(x + 5) + log(3)*(10*x
+ 2*x^2) + 5*x^2 + x^3 + log(2*log(2))*(exp(x)*(4*x + 4*log(3)) + 32*x*log(3) + 16*log(3)^2 + 16*x^2)),x)

[Out]

int((2*x*log(3) + log(3)^2 + x^2 + log(2*log(2))*exp(x)*(4*x + 4*log(3) - 4))/(log(3)^2*(x + 5) + log(3)*(10*x
+ 2*x^2) + 5*x^2 + x^3 + log(2*log(2))*(exp(x)*(4*x + 4*log(3)) + 32*x*log(3) + 16*log(3)^2 + 16*x^2)), x)

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sympy [B]  time = 0.60, size = 73, normalized size = 2.92 $$\begin{array}{c}-\log (x+\log (3))+\log (\frac{x^2+16x\log (\log (2))+x\log (3)+5x+16x\log (2)+16\log (3)\log (\log (2))+5\log (3)+16\log (2)\log (3)}{4\log (\log (2))+4\log (2)}+e^x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*ln(3)+4*x-4)*exp(x)*ln(2*ln(2))+ln(3)**2+2*x*ln(3)+x**2)/(((4*ln(3)+4*x)*exp(x)+16*ln(3)**2+32*x
*ln(3)+16*x**2)*ln(2*ln(2))+(5+x)*ln(3)**2+(2*x**2+10*x)*ln(3)+x**3+5*x**2),x)

[Out]

-log(x + log(3)) + log((x**2 + 16*x*log(log(2)) + x*log(3) + 5*x + 16*x*log(2) + 16*log(3)*log(log(2)) + 5*log
(3) + 16*log(2)*log(3))/(4*log(log(2)) + 4*log(2)) + exp(x))

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