3.30.10 \(\int \frac {4+6 x+2 x^2+(2+2 x) \log ^2(2)+(4+2 x+2 \log ^2(2)) \log (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}})}{x+\log ^2(2)} \, dx\)

Optimal. Leaf size=19 \[ \left (1+x+\log \left (\frac {25 \left (x+\log ^2(2)\right )^2}{e^{10}}\right )\right )^2 \]

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Rubi [A]  time = 0.12, antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6688, 12, 6686} \begin {gather*} \left (-x-\log \left (\left (x+\log ^2(2)\right )^2\right )+9-\log (25)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 6*x + 2*x^2 + (2 + 2*x)*Log[2]^2 + (4 + 2*x + 2*Log[2]^2)*Log[(25*x^2 + 50*x*Log[2]^2 + 25*Log[2]^4)/
E^10])/(x + Log[2]^2),x]

[Out]

(9 - x - Log[25] - Log[(x + Log[2]^2)^2])^2

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (2+x+\log ^2(2)\right ) \left (x-9 \left (1-\frac {2 \log (5)}{9}\right )+\log \left (\left (x+\log ^2(2)\right )^2\right )\right )}{x+\log ^2(2)} \, dx\\ &=2 \int \frac {\left (2+x+\log ^2(2)\right ) \left (x-9 \left (1-\frac {2 \log (5)}{9}\right )+\log \left (\left (x+\log ^2(2)\right )^2\right )\right )}{x+\log ^2(2)} \, dx\\ &=\left (9-x-\log (25)-\log \left (\left (x+\log ^2(2)\right )^2\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.84 \begin {gather*} \left (-9+x+\log (25)+\log \left (\left (x+\log ^2(2)\right )^2\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 6*x + 2*x^2 + (2 + 2*x)*Log[2]^2 + (4 + 2*x + 2*Log[2]^2)*Log[(25*x^2 + 50*x*Log[2]^2 + 25*Log[
2]^4)/E^10])/(x + Log[2]^2),x]

[Out]

(-9 + x + Log[25] + Log[(x + Log[2]^2)^2])^2

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fricas [B]  time = 0.70, size = 54, normalized size = 2.84 \begin {gather*} x^{2} + 2 \, {\left (x + 1\right )} \log \left (25 \, {\left (\log \relax (2)^{4} + 2 \, x \log \relax (2)^{2} + x^{2}\right )} e^{\left (-10\right )}\right ) + \log \left (25 \, {\left (\log \relax (2)^{4} + 2 \, x \log \relax (2)^{2} + x^{2}\right )} e^{\left (-10\right )}\right )^{2} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(2)^2+2*x+4)*log((25*log(2)^4+50*x*log(2)^2+25*x^2)/exp(5)^2)+(2*x+2)*log(2)^2+2*x^2+6*x+4)/(
log(2)^2+x),x, algorithm="fricas")

[Out]

x^2 + 2*(x + 1)*log(25*(log(2)^4 + 2*x*log(2)^2 + x^2)*e^(-10)) + log(25*(log(2)^4 + 2*x*log(2)^2 + x^2)*e^(-1
0))^2 + 2*x

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giac [B]  time = 0.25, size = 61, normalized size = 3.21 \begin {gather*} x^{2} + 2 \, x \log \left (25 \, \log \relax (2)^{4} + 50 \, x \log \relax (2)^{2} + 25 \, x^{2}\right ) + \log \left (25 \, \log \relax (2)^{4} + 50 \, x \log \relax (2)^{2} + 25 \, x^{2}\right )^{2} - 18 \, x - 36 \, \log \left (\log \relax (2)^{2} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(2)^2+2*x+4)*log((25*log(2)^4+50*x*log(2)^2+25*x^2)/exp(5)^2)+(2*x+2)*log(2)^2+2*x^2+6*x+4)/(
log(2)^2+x),x, algorithm="giac")

[Out]

x^2 + 2*x*log(25*log(2)^4 + 50*x*log(2)^2 + 25*x^2) + log(25*log(2)^4 + 50*x*log(2)^2 + 25*x^2)^2 - 18*x - 36*
log(log(2)^2 + x)

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maple [B]  time = 0.46, size = 88, normalized size = 4.63

method	result	size
default	\(-18 x +x^{2}-36 \ln \left (\ln \relax (2)^{2}+x \right )+8 \ln \relax (5) \ln \left (\ln \relax (2)^{2}+x \right )+4 x \ln \relax (5)+2 x \ln \left (\ln \relax (2)^{4}+2 x \ln \relax (2)^{2}+x^{2}\right )+4 \ln \left (\ln \relax (2)^{2}+x \right ) \ln \left (\ln \relax (2)^{4}+2 x \ln \relax (2)^{2}+x^{2}\right )-4 \ln \left (\ln \relax (2)^{2}+x \right )^{2}\)	\(88\)
norman	\(x^{2}+\ln \left (\left (25 \ln \relax (2)^{4}+50 x \ln \relax (2)^{2}+25 x^{2}\right ) {\mathrm e}^{-10}\right )^{2}+2 \ln \left (\left (25 \ln \relax (2)^{4}+50 x \ln \relax (2)^{2}+25 x^{2}\right ) {\mathrm e}^{-10}\right )+2 x +2 x \ln \left (\left (25 \ln \relax (2)^{4}+50 x \ln \relax (2)^{2}+25 x^{2}\right ) {\mathrm e}^{-10}\right )\)	\(90\)
risch	\(x^{2}+2 x +4 \ln \left (\ln \relax (2)^{2}+x \right )+2 \ln \left (\left (\ln \relax (2)^{4}+2 x \ln \relax (2)^{2}+x^{2}\right ) {\mathrm e}^{-10}\right ) x +4 \ln \left (\ln \relax (2)^{2}+x \right ) \ln \left (\left (\ln \relax (2)^{4}+2 x \ln \relax (2)^{2}+x^{2}\right ) {\mathrm e}^{-10}\right )-4 \ln \left (\ln \relax (2)^{2}+x \right )^{2}+8 \ln \relax (5) \ln \left (\ln \relax (2)^{2}+x \right )+4 x \ln \relax (5)\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*ln(2)^2+2*x+4)*ln((25*ln(2)^4+50*x*ln(2)^2+25*x^2)/exp(5)^2)+(2*x+2)*ln(2)^2+2*x^2+6*x+4)/(ln(2)^2+x),
x,method=_RETURNVERBOSE)

[Out]

-18*x+x^2-36*ln(ln(2)^2+x)+8*ln(5)*ln(ln(2)^2+x)+4*x*ln(5)+2*x*ln(ln(2)^4+2*x*ln(2)^2+x^2)+4*ln(ln(2)^2+x)*ln(
ln(2)^4+2*x*ln(2)^2+x^2)-4*ln(ln(2)^2+x)^2

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maxima [B]  time = 0.55, size = 244, normalized size = 12.84 \begin {gather*} 2 \, \log \relax (2)^{4} \log \left (\log \relax (2)^{2} + x\right ) + 2 \, \log \relax (2)^{2} \log \left (25 \, e^{\left (-10\right )} \log \relax (2)^{4} + 50 \, x e^{\left (-10\right )} \log \relax (2)^{2} + 25 \, x^{2} e^{\left (-10\right )}\right ) \log \left (\log \relax (2)^{2} + x\right ) + 2 \, \log \relax (2)^{2} \log \left (\log \relax (2)^{2} + x\right )^{2} - 2 \, {\left (\log \relax (2)^{2} \log \left (\log \relax (2)^{2} + x\right ) - x\right )} \log \relax (2)^{2} + 2 \, {\left (2 \, {\left (\log \relax (5) - 5\right )} \log \left (\log \relax (2)^{2} + x\right ) - \log \left (25 \, e^{\left (-10\right )} \log \relax (2)^{4} + 50 \, x e^{\left (-10\right )} \log \relax (2)^{2} + 25 \, x^{2} e^{\left (-10\right )}\right ) \log \left (\log \relax (2)^{2} + x\right ) + \log \left (\log \relax (2)^{2} + x\right )^{2}\right )} \log \relax (2)^{2} - 2 \, x \log \relax (2)^{2} + x^{2} - 2 \, {\left (\log \relax (2)^{2} \log \left (\log \relax (2)^{2} + x\right ) - x\right )} \log \left (25 \, e^{\left (-10\right )} \log \relax (2)^{4} + 50 \, x e^{\left (-10\right )} \log \relax (2)^{2} + 25 \, x^{2} e^{\left (-10\right )}\right ) + 8 \, {\left (\log \relax (5) - 5\right )} \log \left (\log \relax (2)^{2} + x\right ) + 4 \, \log \left (\log \relax (2)^{2} + x\right )^{2} + 2 \, x + 4 \, \log \left (\log \relax (2)^{2} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(2)^2+2*x+4)*log((25*log(2)^4+50*x*log(2)^2+25*x^2)/exp(5)^2)+(2*x+2)*log(2)^2+2*x^2+6*x+4)/(
log(2)^2+x),x, algorithm="maxima")

[Out]

2*log(2)^4*log(log(2)^2 + x) + 2*log(2)^2*log(25*e^(-10)*log(2)^4 + 50*x*e^(-10)*log(2)^2 + 25*x^2*e^(-10))*lo
g(log(2)^2 + x) + 2*log(2)^2*log(log(2)^2 + x)^2 - 2*(log(2)^2*log(log(2)^2 + x) - x)*log(2)^2 + 2*(2*(log(5)
- 5)*log(log(2)^2 + x) - log(25*e^(-10)*log(2)^4 + 50*x*e^(-10)*log(2)^2 + 25*x^2*e^(-10))*log(log(2)^2 + x) +
log(log(2)^2 + x)^2)*log(2)^2 - 2*x*log(2)^2 + x^2 - 2*(log(2)^2*log(log(2)^2 + x) - x)*log(25*e^(-10)*log(2)
^4 + 50*x*e^(-10)*log(2)^2 + 25*x^2*e^(-10)) + 8*(log(5) - 5)*log(log(2)^2 + x) + 4*log(log(2)^2 + x)^2 + 2*x
+ 4*log(log(2)^2 + x)

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mupad [B]  time = 2.34, size = 63, normalized size = 3.32 \begin {gather*} 2\,x+2\,\ln \left ({\left (x+{\ln \relax (2)}^2\right )}^2\right )+{\ln \left (25\,{\mathrm {e}}^{-10}\,\left (x^2+2\,{\ln \relax (2)}^2\,x+{\ln \relax (2)}^4\right )\right )}^2+2\,x\,\ln \left (25\,{\mathrm {e}}^{-10}\,\left (x^2+2\,{\ln \relax (2)}^2\,x+{\ln \relax (2)}^4\right )\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + log(exp(-10)*(50*x*log(2)^2 + 25*log(2)^4 + 25*x^2))*(2*x + 2*log(2)^2 + 4) + log(2)^2*(2*x + 2) +
2*x^2 + 4)/(x + log(2)^2),x)

[Out]

2*x + 2*log((x + log(2)^2)^2) + log(25*exp(-10)*(2*x*log(2)^2 + log(2)^4 + x^2))^2 + 2*x*log(25*exp(-10)*(2*x*
log(2)^2 + log(2)^4 + x^2)) + x^2

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sympy [B]  time = 0.19, size = 71, normalized size = 3.74 \begin {gather*} x^{2} + 2 x \log {\left (\frac {25 x^{2} + 50 x \log {\relax (2 )}^{2} + 25 \log {\relax (2 )}^{4}}{e^{10}} \right )} + 2 x + \log {\left (\frac {25 x^{2} + 50 x \log {\relax (2 )}^{2} + 25 \log {\relax (2 )}^{4}}{e^{10}} \right )}^{2} + 4 \log {\left (x + \log {\relax (2 )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*ln(2)**2+2*x+4)*ln((25*ln(2)**4+50*x*ln(2)**2+25*x**2)/exp(5)**2)+(2*x+2)*ln(2)**2+2*x**2+6*x+4)
/(ln(2)**2+x),x)

[Out]

x**2 + 2*x*log((25*x**2 + 50*x*log(2)**2 + 25*log(2)**4)*exp(-10)) + 2*x + log((25*x**2 + 50*x*log(2)**2 + 25*
log(2)**4)*exp(-10))**2 + 4*log(x + log(2)**2)

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