3.2.1 \(\int \frac {6 e^4 x+24 e^3 x \log (3)+(64+36 e^2 x) \log ^2(3)+24 e x \log ^3(3)+6 x \log ^4(3)}{e^4+4 e^3 \log (3)+6 e^2 \log ^2(3)+4 e \log ^3(3)+\log ^4(3)} \, dx\)

Optimal. Leaf size=18 \[ x \left (3 x+\frac {64 \log ^2(3)}{(e+\log (3))^4}\right ) \]

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Rubi [B]  time = 0.04, antiderivative size = 60, normalized size of antiderivative = 3.33, number of steps used = 5, number of rules used = 2, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6, 12} \begin {gather*} \frac {3 x^2 \left (e^4+\log ^4(3)+4 e \log ^3(3)+e^3 \log (81)\right )}{(e+\log (3))^4}+\frac {2 \left (9 e^2 x+16\right )^2 \log ^2(3)}{9 e^2 (e+\log (3))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*E^4*x + 24*E^3*x*Log[3] + (64 + 36*E^2*x)*Log[3]^2 + 24*E*x*Log[3]^3 + 6*x*Log[3]^4)/(E^4 + 4*E^3*Log[3
] + 6*E^2*Log[3]^2 + 4*E*Log[3]^3 + Log[3]^4),x]

[Out]

(2*(16 + 9*E^2*x)^2*Log[3]^2)/(9*E^2*(E + Log[3])^4) + (3*x^2*(E^4 + 4*E*Log[3]^3 + Log[3]^4 + E^3*Log[81]))/(
E + Log[3])^4

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (64+36 e^2 x\right ) \log ^2(3)+24 e x \log ^3(3)+6 x \log ^4(3)+x \left (6 e^4+24 e^3 \log (3)\right )}{e^4+4 e^3 \log (3)+6 e^2 \log ^2(3)+4 e \log ^3(3)+\log ^4(3)} \, dx\\ &=\int \frac {\left (64+36 e^2 x\right ) \log ^2(3)+x \left (6 e^4+24 e^3 \log (3)\right )+x \left (24 e \log ^3(3)+6 \log ^4(3)\right )}{e^4+4 e^3 \log (3)+6 e^2 \log ^2(3)+4 e \log ^3(3)+\log ^4(3)} \, dx\\ &=\int \frac {\left (64+36 e^2 x\right ) \log ^2(3)+x \left (6 e^4+24 e^3 \log (3)+24 e \log ^3(3)+6 \log ^4(3)\right )}{e^4+4 e^3 \log (3)+6 e^2 \log ^2(3)+4 e \log ^3(3)+\log ^4(3)} \, dx\\ &=\frac {\int \left (\left (64+36 e^2 x\right ) \log ^2(3)+x \left (6 e^4+24 e^3 \log (3)+24 e \log ^3(3)+6 \log ^4(3)\right )\right ) \, dx}{(e+\log (3))^4}\\ &=\frac {2 \left (16+9 e^2 x\right )^2 \log ^2(3)}{9 e^2 (e+\log (3))^4}+\frac {3 x^2 \left (e^4+4 e \log ^3(3)+\log ^4(3)+e^3 \log (81)\right )}{(e+\log (3))^4}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 64, normalized size = 3.56 \begin {gather*} \frac {3 e^4 x^2+12 e^3 x^2 \log (3)+64 x \log ^2(3)+18 e^2 x^2 \log ^2(3)+12 e x^2 \log ^3(3)+3 x^2 \log ^4(3)}{(e+\log (3))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*E^4*x + 24*E^3*x*Log[3] + (64 + 36*E^2*x)*Log[3]^2 + 24*E*x*Log[3]^3 + 6*x*Log[3]^4)/(E^4 + 4*E^3
*Log[3] + 6*E^2*Log[3]^2 + 4*E*Log[3]^3 + Log[3]^4),x]

[Out]

(3*E^4*x^2 + 12*E^3*x^2*Log[3] + 64*x*Log[3]^2 + 18*E^2*x^2*Log[3]^2 + 12*E*x^2*Log[3]^3 + 3*x^2*Log[3]^4)/(E
+ Log[3])^4

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fricas [B]  time = 1.01, size = 86, normalized size = 4.78 \begin {gather*} \frac {12 \, x^{2} e \log \relax (3)^{3} + 3 \, x^{2} \log \relax (3)^{4} + 12 \, x^{2} e^{3} \log \relax (3) + 3 \, x^{2} e^{4} + 2 \, {\left (9 \, x^{2} e^{2} + 32 \, x\right )} \log \relax (3)^{2}}{4 \, e \log \relax (3)^{3} + \log \relax (3)^{4} + 6 \, e^{2} \log \relax (3)^{2} + 4 \, e^{3} \log \relax (3) + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*log(3)^4+24*x*exp(1)*log(3)^3+(36*x*exp(1)^2+64)*log(3)^2+24*x*exp(1)^3*log(3)+6*x*exp(1)^4)/(l
og(3)^4+4*exp(1)*log(3)^3+6*exp(1)^2*log(3)^2+4*exp(1)^3*log(3)+exp(1)^4),x, algorithm="fricas")

[Out]

(12*x^2*e*log(3)^3 + 3*x^2*log(3)^4 + 12*x^2*e^3*log(3) + 3*x^2*e^4 + 2*(9*x^2*e^2 + 32*x)*log(3)^2)/(4*e*log(
3)^3 + log(3)^4 + 6*e^2*log(3)^2 + 4*e^3*log(3) + e^4)

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giac [B]  time = 0.45, size = 86, normalized size = 4.78 \begin {gather*} \frac {12 \, x^{2} e \log \relax (3)^{3} + 3 \, x^{2} \log \relax (3)^{4} + 12 \, x^{2} e^{3} \log \relax (3) + 3 \, x^{2} e^{4} + 2 \, {\left (9 \, x^{2} e^{2} + 32 \, x\right )} \log \relax (3)^{2}}{4 \, e \log \relax (3)^{3} + \log \relax (3)^{4} + 6 \, e^{2} \log \relax (3)^{2} + 4 \, e^{3} \log \relax (3) + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*log(3)^4+24*x*exp(1)*log(3)^3+(36*x*exp(1)^2+64)*log(3)^2+24*x*exp(1)^3*log(3)+6*x*exp(1)^4)/(l
og(3)^4+4*exp(1)*log(3)^3+6*exp(1)^2*log(3)^2+4*exp(1)^3*log(3)+exp(1)^4),x, algorithm="giac")

[Out]

(12*x^2*e*log(3)^3 + 3*x^2*log(3)^4 + 12*x^2*e^3*log(3) + 3*x^2*e^4 + 2*(9*x^2*e^2 + 32*x)*log(3)^2)/(4*e*log(
3)^3 + log(3)^4 + 6*e^2*log(3)^2 + 4*e^3*log(3) + e^4)

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maple [B]  time = 0.04, size = 57, normalized size = 3.17

method	result	size
norman	\(\frac {\left (3 \,{\mathrm e}^{3}+9 \,{\mathrm e}^{2} \ln \relax (3)+9 \,{\mathrm e} \ln \relax (3)^{2}+3 \ln \relax (3)^{3}\right ) x^{2}+\frac {64 \ln \relax (3)^{2} x}{\ln \relax (3)+{\mathrm e}}}{\left (\ln \relax (3)+{\mathrm e}\right )^{3}}\)	\(57\)
gosper	\(\frac {x \left (3 x \,{\mathrm e}^{4}+12 x \,{\mathrm e}^{3} \ln \relax (3)+18 \,{\mathrm e}^{2} \ln \relax (3)^{2} x +12 x \,{\mathrm e} \ln \relax (3)^{3}+3 x \ln \relax (3)^{4}+64 \ln \relax (3)^{2}\right )}{\ln \relax (3)^{4}+4 \,{\mathrm e} \ln \relax (3)^{3}+6 \,{\mathrm e}^{2} \ln \relax (3)^{2}+4 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{4}}\)	\(90\)
default	\(\frac {3 x^{2} \ln \relax (3)^{4}+12 x^{2} {\mathrm e} \ln \relax (3)^{3}+\ln \relax (3)^{2} \left (18 x^{2} {\mathrm e}^{2}+64 x \right )+12 x^{2} {\mathrm e}^{3} \ln \relax (3)+3 x^{2} {\mathrm e}^{4}}{\ln \relax (3)^{4}+4 \,{\mathrm e} \ln \relax (3)^{3}+6 \,{\mathrm e}^{2} \ln \relax (3)^{2}+4 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{4}}\)	\(98\)
risch	\(\frac {3 x^{2} {\mathrm e}^{4}}{\ln \relax (3)^{4}+4 \,{\mathrm e} \ln \relax (3)^{3}+6 \,{\mathrm e}^{2} \ln \relax (3)^{2}+4 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{4}}+\frac {12 x^{2} {\mathrm e}^{3} \ln \relax (3)}{\ln \relax (3)^{4}+4 \,{\mathrm e} \ln \relax (3)^{3}+6 \,{\mathrm e}^{2} \ln \relax (3)^{2}+4 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{4}}+\frac {18 \,{\mathrm e}^{2} \ln \relax (3)^{2} x^{2}}{\ln \relax (3)^{4}+4 \,{\mathrm e} \ln \relax (3)^{3}+6 \,{\mathrm e}^{2} \ln \relax (3)^{2}+4 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{4}}+\frac {12 x^{2} {\mathrm e} \ln \relax (3)^{3}}{\ln \relax (3)^{4}+4 \,{\mathrm e} \ln \relax (3)^{3}+6 \,{\mathrm e}^{2} \ln \relax (3)^{2}+4 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{4}}+\frac {3 x^{2} \ln \relax (3)^{4}}{\ln \relax (3)^{4}+4 \,{\mathrm e} \ln \relax (3)^{3}+6 \,{\mathrm e}^{2} \ln \relax (3)^{2}+4 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{4}}+\frac {64 x \ln \relax (3)^{2}}{\ln \relax (3)^{4}+4 \,{\mathrm e} \ln \relax (3)^{3}+6 \,{\mathrm e}^{2} \ln \relax (3)^{2}+4 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{4}}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x*ln(3)^4+24*x*exp(1)*ln(3)^3+(36*x*exp(1)^2+64)*ln(3)^2+24*x*exp(1)^3*ln(3)+6*x*exp(1)^4)/(ln(3)^4+4*e
xp(1)*ln(3)^3+6*exp(1)^2*ln(3)^2+4*exp(1)^3*ln(3)+exp(1)^4),x,method=_RETURNVERBOSE)

[Out]

((3*exp(1)^3+9*ln(3)*exp(1)^2+9*exp(1)*ln(3)^2+3*ln(3)^3)*x^2+64*ln(3)^2/(ln(3)+exp(1))*x)/(ln(3)+exp(1))^3

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maxima [B]  time = 0.37, size = 86, normalized size = 4.78 \begin {gather*} \frac {12 \, x^{2} e \log \relax (3)^{3} + 3 \, x^{2} \log \relax (3)^{4} + 12 \, x^{2} e^{3} \log \relax (3) + 3 \, x^{2} e^{4} + 2 \, {\left (9 \, x^{2} e^{2} + 32 \, x\right )} \log \relax (3)^{2}}{4 \, e \log \relax (3)^{3} + \log \relax (3)^{4} + 6 \, e^{2} \log \relax (3)^{2} + 4 \, e^{3} \log \relax (3) + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*log(3)^4+24*x*exp(1)*log(3)^3+(36*x*exp(1)^2+64)*log(3)^2+24*x*exp(1)^3*log(3)+6*x*exp(1)^4)/(l
og(3)^4+4*exp(1)*log(3)^3+6*exp(1)^2*log(3)^2+4*exp(1)^3*log(3)+exp(1)^4),x, algorithm="maxima")

[Out]

(12*x^2*e*log(3)^3 + 3*x^2*log(3)^4 + 12*x^2*e^3*log(3) + 3*x^2*e^4 + 2*(9*x^2*e^2 + 32*x)*log(3)^2)/(4*e*log(
3)^3 + log(3)^4 + 6*e^2*log(3)^2 + 4*e^3*log(3) + e^4)

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mupad [B]  time = 0.12, size = 53, normalized size = 2.94 \begin {gather*} \frac {x\,\left (3\,x\,{\mathrm {e}}^4+3\,x\,{\ln \relax (3)}^4+64\,{\ln \relax (3)}^2+12\,x\,\mathrm {e}\,{\ln \relax (3)}^3+18\,x\,{\mathrm {e}}^2\,{\ln \relax (3)}^2+12\,x\,{\mathrm {e}}^3\,\ln \relax (3)\right )}{{\left (\mathrm {e}+\ln \relax (3)\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x*exp(4) + 6*x*log(3)^4 + log(3)^2*(36*x*exp(2) + 64) + 24*x*exp(1)*log(3)^3 + 24*x*exp(3)*log(3))/(exp
(4) + 4*exp(3)*log(3) + 4*exp(1)*log(3)^3 + 6*exp(2)*log(3)^2 + log(3)^4),x)

[Out]

(x*(3*x*exp(4) + 3*x*log(3)^4 + 64*log(3)^2 + 12*x*exp(1)*log(3)^3 + 18*x*exp(2)*log(3)^2 + 12*x*exp(3)*log(3)
))/(exp(1) + log(3))^4

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sympy [B]  time = 0.08, size = 49, normalized size = 2.72 \begin {gather*} 3 x^{2} + \frac {64 x \log {\relax (3 )}^{2}}{\log {\relax (3 )}^{4} + 4 e \log {\relax (3 )}^{3} + 6 e^{2} \log {\relax (3 )}^{2} + e^{4} + 4 e^{3} \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*ln(3)**4+24*x*exp(1)*ln(3)**3+(36*x*exp(1)**2+64)*ln(3)**2+24*x*exp(1)**3*ln(3)+6*x*exp(1)**4)/
(ln(3)**4+4*exp(1)*ln(3)**3+6*exp(1)**2*ln(3)**2+4*exp(1)**3*ln(3)+exp(1)**4),x)

[Out]

3*x**2 + 64*x*log(3)**2/(log(3)**4 + 4*E*log(3)**3 + 6*exp(2)*log(3)**2 + exp(4) + 4*exp(3)*log(3))

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