3.29.2 \(\int \frac {e^{\frac {9 x}{\log (\frac {e^{10}+2 e^5 x+x^2}{e^8})}} (36 x+(-18 e^5-18 x) \log (\frac {e^{10}+2 e^5 x+x^2}{e^8}))}{e^{\frac {18 x}{\log (\frac {e^{10}+2 e^5 x+x^2}{e^8})}} (e^5+x) \log ^2(\frac {e^{10}+2 e^5 x+x^2}{e^8})+e^{\frac {9 x}{\log (\frac {e^{10}+2 e^5 x+x^2}{e^8})}} (2 e^7+2 e^2 x) \log ^2(\frac {e^{10}+2 e^5 x+x^2}{e^8})+(e^9+e^4 x) \log ^2(\frac {e^{10}+2 e^5 x+x^2}{e^8})} \, dx\)

Optimal. Leaf size=27 \[ \frac {2}{e^2+e^{\frac {9 x}{\log \left (\frac {\left (e^5+x\right )^2}{e^8}\right )}}} \]

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Rubi [A]  time = 0.86, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 200, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6688, 12, 6686} \begin {gather*} \frac {2}{e^{-\frac {9 x}{8-\log \left (\left (x+e^5\right )^2\right )}}+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((9*x)/Log[(E^10 + 2*E^5*x + x^2)/E^8])*(36*x + (-18*E^5 - 18*x)*Log[(E^10 + 2*E^5*x + x^2)/E^8]))/(E^(
(18*x)/Log[(E^10 + 2*E^5*x + x^2)/E^8])*(E^5 + x)*Log[(E^10 + 2*E^5*x + x^2)/E^8]^2 + E^((9*x)/Log[(E^10 + 2*E
^5*x + x^2)/E^8])*(2*E^7 + 2*E^2*x)*Log[(E^10 + 2*E^5*x + x^2)/E^8]^2 + (E^9 + E^4*x)*Log[(E^10 + 2*E^5*x + x^
2)/E^8]^2),x]

[Out]

2/(E^2 + E^((-9*x)/(8 - Log[(E^5 + x)^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 {18 e^{\frac {9 x}{-8+\log \left (\left (e^5+x\right )^2\right )}} \left (2 \left (4 e^5+5 x\right )-\left (e^5+x\right ) \log \left (\left (e^5+x\right )^2\right )\right )}{\left (e^2+e^{\frac {9 x}{-8+\log \left (\left (e^5+x\right )^2\right )}}\right )^2 \left (e^5+x\right ) \left (8-\log \left (\left (e^5+x\right )^2\right )\right )^2} \, dx\\ &=18 \int \frac {e^{\frac {9 x}{-8+\log \left (\left (e^5+x\right )^2\right )}} \left (2 \left (4 e^5+5 x\right )-\left (e^5+x\right ) \log \left (\left (e^5+x\right )^2\right )\right )}{\left (e^2+e^{\frac {9 x}{-8+\log \left (\left (e^5+x\right )^2\right )}}\right )^2 \left (e^5+x\right ) \left (8-\log \left (\left (e^5+x\right )^2\right )\right )^2} \, dx\\ &=\frac {2}{e^2+e^{-\frac {9 x}{8-\log \left (\left (e^5+x\right )^2\right )}}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 28, normalized size = 1.04 \begin {gather*} -\frac {2}{e^2 \left (1+e^{2-\frac {9 x}{-8+\log \left (\left (e^5+x\right )^2\right )}}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((9*x)/Log[(E^10 + 2*E^5*x + x^2)/E^8])*(36*x + (-18*E^5 - 18*x)*Log[(E^10 + 2*E^5*x + x^2)/E^8])
)/(E^((18*x)/Log[(E^10 + 2*E^5*x + x^2)/E^8])*(E^5 + x)*Log[(E^10 + 2*E^5*x + x^2)/E^8]^2 + E^((9*x)/Log[(E^10
+ 2*E^5*x + x^2)/E^8])*(2*E^7 + 2*E^2*x)*Log[(E^10 + 2*E^5*x + x^2)/E^8]^2 + (E^9 + E^4*x)*Log[(E^10 + 2*E^5*
x + x^2)/E^8]^2),x]

[Out]

-2/(E^2*(1 + E^(2 - (9*x)/(-8 + Log[(E^5 + x)^2]))))

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fricas [A]  time = 0.62, size = 28, normalized size = 1.04 \begin {gather*} \frac {2}{e^{2} + e^{\left (\frac {9 \, x}{\log \left ({\left (x^{2} + 2 \, x e^{5} + e^{10}\right )} e^{\left (-8\right )}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*exp(5)-18*x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)+36*x)*exp(9*x/log((exp(5)^2+2*x*exp(5)+x^
2)/exp(4)^2))/((exp(5)+x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2*exp(9*x/log((exp(5)^2+2*x*exp(5)+x^2)/exp(
4)^2))^2+(2*exp(2)*exp(5)+2*exp(2)*x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2*exp(9*x/log((exp(5)^2+2*x*exp(
5)+x^2)/exp(4)^2))+(exp(2)^2*exp(5)+x*exp(2)^2)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2),x, algorithm="frica
s")

[Out]

2/(e^2 + e^(9*x/log((x^2 + 2*x*e^5 + e^10)*e^(-8))))

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giac [B]  time = 2.80, size = 776, normalized size = 28.74 \begin {gather*} \frac {2 \, {\left (x^{2} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} + 2 \, x e^{5} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} - 26 \, x^{2} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} - 50 \, x e^{5} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} + e^{10} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} + 224 \, x^{2} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) + 416 \, x e^{5} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 24 \, e^{10} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} - 640 \, x^{2} - 1152 \, x e^{5} + 192 \, e^{10} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 512 \, e^{10}\right )}}{x^{2} e^{2} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} + x^{2} e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8}\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} - 26 \, x^{2} e^{2} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} - 26 \, x^{2} e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8}\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} + 2 \, x e^{7} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} + 2 \, x e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8} + 5\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} + 224 \, x^{2} e^{2} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) + 224 \, x^{2} e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8}\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 50 \, x e^{7} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} - 50 \, x e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8} + 5\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} + e^{12} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} + e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8} + 10\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{3} - 640 \, x^{2} e^{2} - 640 \, x^{2} e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8}\right )} + 416 \, x e^{7} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) + 416 \, x e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8} + 5\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 24 \, e^{12} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} - 24 \, e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8} + 10\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right )^{2} - 1152 \, x e^{7} - 1152 \, x e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8} + 5\right )} + 192 \, e^{12} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) + 192 \, e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8} + 10\right )} \log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 512 \, e^{12} - 512 \, e^{\left (\frac {9 \, x}{\log \left (x^{2} + 2 \, x e^{5} + e^{10}\right ) - 8} + 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*exp(5)-18*x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)+36*x)*exp(9*x/log((exp(5)^2+2*x*exp(5)+x^
2)/exp(4)^2))/((exp(5)+x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2*exp(9*x/log((exp(5)^2+2*x*exp(5)+x^2)/exp(
4)^2))^2+(2*exp(2)*exp(5)+2*exp(2)*x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2*exp(9*x/log((exp(5)^2+2*x*exp(
5)+x^2)/exp(4)^2))+(exp(2)^2*exp(5)+x*exp(2)^2)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2),x, algorithm="giac"
)

[Out]

2*(x^2*log(x^2 + 2*x*e^5 + e^10)^3 + 2*x*e^5*log(x^2 + 2*x*e^5 + e^10)^3 - 26*x^2*log(x^2 + 2*x*e^5 + e^10)^2
- 50*x*e^5*log(x^2 + 2*x*e^5 + e^10)^2 + e^10*log(x^2 + 2*x*e^5 + e^10)^3 + 224*x^2*log(x^2 + 2*x*e^5 + e^10)
+ 416*x*e^5*log(x^2 + 2*x*e^5 + e^10) - 24*e^10*log(x^2 + 2*x*e^5 + e^10)^2 - 640*x^2 - 1152*x*e^5 + 192*e^10*
log(x^2 + 2*x*e^5 + e^10) - 512*e^10)/(x^2*e^2*log(x^2 + 2*x*e^5 + e^10)^3 + x^2*e^(9*x/(log(x^2 + 2*x*e^5 + e
^10) - 8))*log(x^2 + 2*x*e^5 + e^10)^3 - 26*x^2*e^2*log(x^2 + 2*x*e^5 + e^10)^2 - 26*x^2*e^(9*x/(log(x^2 + 2*x
*e^5 + e^10) - 8))*log(x^2 + 2*x*e^5 + e^10)^2 + 2*x*e^7*log(x^2 + 2*x*e^5 + e^10)^3 + 2*x*e^(9*x/(log(x^2 + 2
*x*e^5 + e^10) - 8) + 5)*log(x^2 + 2*x*e^5 + e^10)^3 + 224*x^2*e^2*log(x^2 + 2*x*e^5 + e^10) + 224*x^2*e^(9*x/
(log(x^2 + 2*x*e^5 + e^10) - 8))*log(x^2 + 2*x*e^5 + e^10) - 50*x*e^7*log(x^2 + 2*x*e^5 + e^10)^2 - 50*x*e^(9*
x/(log(x^2 + 2*x*e^5 + e^10) - 8) + 5)*log(x^2 + 2*x*e^5 + e^10)^2 + e^12*log(x^2 + 2*x*e^5 + e^10)^3 + e^(9*x
/(log(x^2 + 2*x*e^5 + e^10) - 8) + 10)*log(x^2 + 2*x*e^5 + e^10)^3 - 640*x^2*e^2 - 640*x^2*e^(9*x/(log(x^2 + 2
*x*e^5 + e^10) - 8)) + 416*x*e^7*log(x^2 + 2*x*e^5 + e^10) + 416*x*e^(9*x/(log(x^2 + 2*x*e^5 + e^10) - 8) + 5)
*log(x^2 + 2*x*e^5 + e^10) - 24*e^12*log(x^2 + 2*x*e^5 + e^10)^2 - 24*e^(9*x/(log(x^2 + 2*x*e^5 + e^10) - 8) +
10)*log(x^2 + 2*x*e^5 + e^10)^2 - 1152*x*e^7 - 1152*x*e^(9*x/(log(x^2 + 2*x*e^5 + e^10) - 8) + 5) + 192*e^12*
log(x^2 + 2*x*e^5 + e^10) + 192*e^(9*x/(log(x^2 + 2*x*e^5 + e^10) - 8) + 10)*log(x^2 + 2*x*e^5 + e^10) - 512*e
^12 - 512*e^(9*x/(log(x^2 + 2*x*e^5 + e^10) - 8) + 10))

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maple [A]  time = 4.16, size = 29, normalized size = 1.07

method	result	size
risch	\(\frac {2}{{\mathrm e}^{2}+{\mathrm e}^{\frac {9 x}{\ln \left (\left ({\mathrm e}^{10}+2 x \,{\mathrm e}^{5}+x^{2}\right ) {\mathrm e}^{-8}\right )}}}\)	\(29\)
norman	\(\frac {2}{{\mathrm e}^{2}+{\mathrm e}^{\frac {9 x}{\ln \left (\left ({\mathrm e}^{10}+2 x \,{\mathrm e}^{5}+x^{2}\right ) {\mathrm e}^{-8}\right )}}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-18*exp(5)-18*x)*ln((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)+36*x)*exp(9*x/ln((exp(5)^2+2*x*exp(5)+x^2)/exp(4
)^2))/((exp(5)+x)*ln((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2*exp(9*x/ln((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2))^2+(
2*exp(2)*exp(5)+2*exp(2)*x)*ln((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2*exp(9*x/ln((exp(5)^2+2*x*exp(5)+x^2)/exp(
4)^2))+(exp(2)^2*exp(5)+x*exp(2)^2)*ln((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2),x,method=_RETURNVERBOSE)

[Out]

2/(exp(2)+exp(9*x/ln((exp(10)+2*x*exp(5)+x^2)*exp(-8))))

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maxima [A]  time = 0.52, size = 20, normalized size = 0.74 \begin {gather*} \frac {2}{e^{2} + e^{\left (\frac {9 \, x}{2 \, {\left (\log \left (x + e^{5}\right ) - 4\right )}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*exp(5)-18*x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)+36*x)*exp(9*x/log((exp(5)^2+2*x*exp(5)+x^
2)/exp(4)^2))/((exp(5)+x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2*exp(9*x/log((exp(5)^2+2*x*exp(5)+x^2)/exp(
4)^2))^2+(2*exp(2)*exp(5)+2*exp(2)*x)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2*exp(9*x/log((exp(5)^2+2*x*exp(
5)+x^2)/exp(4)^2))+(exp(2)^2*exp(5)+x*exp(2)^2)*log((exp(5)^2+2*x*exp(5)+x^2)/exp(4)^2)^2),x, algorithm="maxim
a")

[Out]

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

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mupad [B]  time = 5.13, size = 28, normalized size = 1.04 \begin {gather*} \frac {2}{{\mathrm {e}}^2+{\mathrm {e}}^{\frac {9\,x}{\ln \left ({\mathrm {e}}^{-8}\,x^2+2\,{\mathrm {e}}^{-3}\,x+{\mathrm {e}}^2\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((9*x)/log(exp(-8)*(exp(10) + 2*x*exp(5) + x^2)))*(36*x - log(exp(-8)*(exp(10) + 2*x*exp(5) + x^2))*(1
8*x + 18*exp(5))))/(log(exp(-8)*(exp(10) + 2*x*exp(5) + x^2))^2*(exp(9) + x*exp(4)) + log(exp(-8)*(exp(10) + 2
*x*exp(5) + x^2))^2*exp((18*x)/log(exp(-8)*(exp(10) + 2*x*exp(5) + x^2)))*(x + exp(5)) + log(exp(-8)*(exp(10)
+ 2*x*exp(5) + x^2))^2*exp((9*x)/log(exp(-8)*(exp(10) + 2*x*exp(5) + x^2)))*(2*exp(7) + 2*x*exp(2))),x)

[Out]

2/(exp(2) + exp((9*x)/log(exp(2) + 2*x*exp(-3) + x^2*exp(-8))))

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sympy [A]  time = 0.45, size = 27, normalized size = 1.00 \begin {gather*} \frac {2}{e^{\frac {9 x}{\log {\left (\frac {x^{2} + 2 x e^{5} + e^{10}}{e^{8}} \right )}}} + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*exp(5)-18*x)*ln((exp(5)**2+2*x*exp(5)+x**2)/exp(4)**2)+36*x)*exp(9*x/ln((exp(5)**2+2*x*exp(5)+
x**2)/exp(4)**2))/((exp(5)+x)*ln((exp(5)**2+2*x*exp(5)+x**2)/exp(4)**2)**2*exp(9*x/ln((exp(5)**2+2*x*exp(5)+x*
*2)/exp(4)**2))**2+(2*exp(2)*exp(5)+2*exp(2)*x)*ln((exp(5)**2+2*x*exp(5)+x**2)/exp(4)**2)**2*exp(9*x/ln((exp(5
)**2+2*x*exp(5)+x**2)/exp(4)**2))+(exp(2)**2*exp(5)+x*exp(2)**2)*ln((exp(5)**2+2*x*exp(5)+x**2)/exp(4)**2)**2)
,x)

[Out]

2/(exp(9*x/log((x**2 + 2*x*exp(5) + exp(10))*exp(-8))) + exp(2))

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