3.96.46 \(\int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} (144 x^3+18 e^3 x^4)}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3)} \, dx\)

Optimal. Leaf size=21 \[ \log \left (-5+9 e^{\frac {x^4}{\left (4+e^3 x\right )^2}}+x\right ) \]

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Rubi [A]  time = 0.50, antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, integrand size = 149, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6688, 6684} \begin {gather*} \log \left (-9 e^{\frac {x^4}{\left (e^3 x+4\right )^2}}-x+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(64 + 48*E^3*x + 12*E^6*x^2 + E^9*x^3 + E^(x^4/(16 + 8*E^3*x + E^6*x^2))*(144*x^3 + 18*E^3*x^4))/(-320 + 6
4*x + E^9*(-5 + x)*x^3 + E^6*x^2*(-60 + 12*x) + E^3*x*(-240 + 48*x) + E^(x^4/(16 + 8*E^3*x + E^6*x^2))*(576 +
432*E^3*x + 108*E^6*x^2 + 9*E^9*x^3)),x]

[Out]

Log[5 - 9*E^(x^4/(4 + E^3*x)^2) - x]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {-64-48 e^3 x-12 e^6 x^2-e^9 x^3-144 e^{\frac {x^4}{\left (4+e^3 x\right )^2}} x^3-18 e^{3+\frac {x^4}{\left (4+e^3 x\right )^2}} x^4}{\left (5-9 e^{\frac {x^4}{\left (4+e^3 x\right )^2}}-x\right ) \left (4+e^3 x\right )^3} \, dx\\ &=\log \left (5-9 e^{\frac {x^4}{\left (4+e^3 x\right )^2}}-x\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 57, normalized size = 2.71 \begin {gather*} \log \left (5-9 e^{\frac {48}{e^{12}}-\frac {8 x}{e^9}+\frac {x^2}{e^6}+\frac {256}{e^{12} \left (4+e^3 x\right )^2}-\frac {256}{e^{12} \left (4+e^3 x\right )}}-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64 + 48*E^3*x + 12*E^6*x^2 + E^9*x^3 + E^(x^4/(16 + 8*E^3*x + E^6*x^2))*(144*x^3 + 18*E^3*x^4))/(-3
20 + 64*x + E^9*(-5 + x)*x^3 + E^6*x^2*(-60 + 12*x) + E^3*x*(-240 + 48*x) + E^(x^4/(16 + 8*E^3*x + E^6*x^2))*(
576 + 432*E^3*x + 108*E^6*x^2 + 9*E^9*x^3)),x]

[Out]

Log[5 - 9*E^(48/E^12 - (8*x)/E^9 + x^2/E^6 + 256/(E^12*(4 + E^3*x)^2) - 256/(E^12*(4 + E^3*x))) - x]

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fricas [A]  time = 0.70, size = 26, normalized size = 1.24 \begin {gather*} \log \left (x + 9 \, e^{\left (\frac {x^{4}}{x^{2} e^{6} + 8 \, x e^{3} + 16}\right )} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x^3*exp(3+log(x))+144*x^3)*exp(x^4/(exp(3+log(x))^2+8*exp(3+log(x))+16))+exp(3+log(x))^3+12*exp
(3+log(x))^2+48*exp(3+log(x))+64)/((9*exp(3+log(x))^3+108*exp(3+log(x))^2+432*exp(3+log(x))+576)*exp(x^4/(exp(
3+log(x))^2+8*exp(3+log(x))+16))+(x-5)*exp(3+log(x))^3+(12*x-60)*exp(3+log(x))^2+(48*x-240)*exp(3+log(x))+64*x
-320),x, algorithm="fricas")

[Out]

log(x + 9*e^(x^4/(x^2*e^6 + 8*x*e^3 + 16)) - 5)

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giac [A]  time = 0.27, size = 26, normalized size = 1.24 \begin {gather*} \log \left (x + 9 \, e^{\left (\frac {x^{4}}{x^{2} e^{6} + 8 \, x e^{3} + 16}\right )} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x^3*exp(3+log(x))+144*x^3)*exp(x^4/(exp(3+log(x))^2+8*exp(3+log(x))+16))+exp(3+log(x))^3+12*exp
(3+log(x))^2+48*exp(3+log(x))+64)/((9*exp(3+log(x))^3+108*exp(3+log(x))^2+432*exp(3+log(x))+576)*exp(x^4/(exp(
3+log(x))^2+8*exp(3+log(x))+16))+(x-5)*exp(3+log(x))^3+(12*x-60)*exp(3+log(x))^2+(48*x-240)*exp(3+log(x))+64*x
-320),x, algorithm="giac")

[Out]

log(x + 9*e^(x^4/(x^2*e^6 + 8*x*e^3 + 16)) - 5)

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maple [A]  time = 0.57, size = 29, normalized size = 1.38

method	result	size
norman	\(\ln \left (x +9 \,{\mathrm e}^{\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}}-5\right )\)	\(29\)
risch	\({\mathrm e}^{3} {\mathrm e}^{-9} x^{2}-8 x \,{\mathrm e}^{-9}+\frac {\left (-256 x -768 \,{\mathrm e}^{-3}\right ) {\mathrm e}^{-9}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}-\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}+\ln \left ({\mathrm e}^{\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}}+\frac {x}{9}-\frac {5}{9}\right )\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((18*x^3*exp(3+ln(x))+144*x^3)*exp(x^4/(exp(3+ln(x))^2+8*exp(3+ln(x))+16))+exp(3+ln(x))^3+12*exp(3+ln(x))^
2+48*exp(3+ln(x))+64)/((9*exp(3+ln(x))^3+108*exp(3+ln(x))^2+432*exp(3+ln(x))+576)*exp(x^4/(exp(3+ln(x))^2+8*ex
p(3+ln(x))+16))+(x-5)*exp(3+ln(x))^3+(12*x-60)*exp(3+ln(x))^2+(48*x-240)*exp(3+ln(x))+64*x-320),x,method=_RETU
RNVERBOSE)

[Out]

ln(x+9*exp(x^4/(x^2*exp(3)^2+8*x*exp(3)+16))-5)

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maxima [B]  time = 0.40, size = 108, normalized size = 5.14 \begin {gather*} \frac {x^{3} e^{9} - 4 \, x^{2} e^{6} - 32 \, x e^{3} - 256}{x e^{15} + 4 \, e^{12}} + \log \left (\frac {1}{9} \, {\left ({\left (x - 5\right )} e^{\left (8 \, x e^{\left (-9\right )} + \frac {256}{x e^{15} + 4 \, e^{12}}\right )} + 9 \, e^{\left (x^{2} e^{\left (-6\right )} + \frac {256}{x^{2} e^{18} + 8 \, x e^{15} + 16 \, e^{12}} + 48 \, e^{\left (-12\right )}\right )}\right )} e^{\left (-x^{2} e^{\left (-6\right )} - 48 \, e^{\left (-12\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x^3*exp(3+log(x))+144*x^3)*exp(x^4/(exp(3+log(x))^2+8*exp(3+log(x))+16))+exp(3+log(x))^3+12*exp
(3+log(x))^2+48*exp(3+log(x))+64)/((9*exp(3+log(x))^3+108*exp(3+log(x))^2+432*exp(3+log(x))+576)*exp(x^4/(exp(
3+log(x))^2+8*exp(3+log(x))+16))+(x-5)*exp(3+log(x))^3+(12*x-60)*exp(3+log(x))^2+(48*x-240)*exp(3+log(x))+64*x
-320),x, algorithm="maxima")

[Out]

(x^3*e^9 - 4*x^2*e^6 - 32*x*e^3 - 256)/(x*e^15 + 4*e^12) + log(1/9*((x - 5)*e^(8*x*e^(-9) + 256/(x*e^15 + 4*e^
12)) + 9*e^(x^2*e^(-6) + 256/(x^2*e^18 + 8*x*e^15 + 16*e^12) + 48*e^(-12)))*e^(-x^2*e^(-6) - 48*e^(-12)))

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mupad [B]  time = 8.57, size = 19, normalized size = 0.90 \begin {gather*} \ln \left (x+9\,{\mathrm {e}}^{\frac {x^4}{{\left (x\,{\mathrm {e}}^3+4\right )}^2}}-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*exp(2*log(x) + 6) + exp(3*log(x) + 9) + 48*exp(log(x) + 3) + exp(x^4/(exp(2*log(x) + 6) + 8*exp(log(x)
+ 3) + 16))*(144*x^3 + 18*x^3*exp(log(x) + 3)) + 64)/(64*x + exp(x^4/(exp(2*log(x) + 6) + 8*exp(log(x) + 3) +
16))*(108*exp(2*log(x) + 6) + 9*exp(3*log(x) + 9) + 432*exp(log(x) + 3) + 576) + exp(3*log(x) + 9)*(x - 5) +
exp(log(x) + 3)*(48*x - 240) + exp(2*log(x) + 6)*(12*x - 60) - 320),x)

[Out]

log(x + 9*exp(x^4/(x*exp(3) + 4)^2) - 5)

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sympy [A]  time = 0.64, size = 27, normalized size = 1.29 \begin {gather*} \log {\left (\frac {x}{9} + e^{\frac {x^{4}}{x^{2} e^{6} + 8 x e^{3} + 16}} - \frac {5}{9} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x**3*exp(3+ln(x))+144*x**3)*exp(x**4/(exp(3+ln(x))**2+8*exp(3+ln(x))+16))+exp(3+ln(x))**3+12*ex
p(3+ln(x))**2+48*exp(3+ln(x))+64)/((9*exp(3+ln(x))**3+108*exp(3+ln(x))**2+432*exp(3+ln(x))+576)*exp(x**4/(exp(
3+ln(x))**2+8*exp(3+ln(x))+16))+(x-5)*exp(3+ln(x))**3+(12*x-60)*exp(3+ln(x))**2+(48*x-240)*exp(3+ln(x))+64*x-3
20),x)

[Out]

log(x/9 + exp(x**4/(x**2*exp(6) + 8*x*exp(3) + 16)) - 5/9)

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