3.38.66 \(\int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+25 e^8 x^2+e^{2 x} x^2+(-20 x-50 e^4 x^2) \log (3)+25 x^2 \log ^2(3)+(4 x+10 e^4 x^2-10 x^2 \log (3)) \log (5)+x^2 \log ^2(5)+e^x (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5))} \, dx\)

Optimal. Leaf size=23 \[ \frac {9}{2+x \left (5 e^4+e^x-5 \log (3)+\log (5)\right )} \]

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Rubi [A]  time = 0.31, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, integrand size = 135, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {6, 6688, 12, 6686} \begin {gather*} \frac {9}{e^x x+x \left (5 e^4-\log \left (\frac {243}{5}\right )\right )+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45*E^4 + E^x*(-9 - 9*x) + 45*Log[3] - 9*Log[5])/(4 + 20*E^4*x + 25*E^8*x^2 + E^(2*x)*x^2 + (-20*x - 50*E
^4*x^2)*Log[3] + 25*x^2*Log[3]^2 + (4*x + 10*E^4*x^2 - 10*x^2*Log[3])*Log[5] + x^2*Log[5]^2 + E^x*(4*x + 10*E^
4*x^2 - 10*x^2*Log[3] + 2*x^2*Log[5])),x]

[Out]

9/(2 + E^x*x + x*(5*E^4 - Log[243/5]))

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]]

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 {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \left (e^8+\log ^2(3)\right )+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx\\ &=\int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )+x^2 \left (25 \left (e^8+\log ^2(3)\right )+\log ^2(5)\right )} \, dx\\ &=\int \frac {9 \left (-e^x (1+x)-5 e^4 \left (1-\frac {\log (3) \log \left (\frac {243}{5}\right )}{e^4 \log (243)}\right )\right )}{\left (2+e^x x+5 e^4 x \left (1-\frac {\log \left (\frac {243}{5}\right )}{5 e^4}\right )\right )^2} \, dx\\ &=9 \int \frac {-e^x (1+x)-5 e^4 \left (1-\frac {\log (3) \log \left (\frac {243}{5}\right )}{e^4 \log (243)}\right )}{\left (2+e^x x+5 e^4 x \left (1-\frac {\log \left (\frac {243}{5}\right )}{5 e^4}\right )\right )^2} \, dx\\ &=\frac {9}{2+e^x x+x \left (5 e^4-\log \left (\frac {243}{5}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 1.09 \begin {gather*} \frac {9}{2+e^x x+x \left (5 e^4-\log \left (\frac {243}{5}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45*E^4 + E^x*(-9 - 9*x) + 45*Log[3] - 9*Log[5])/(4 + 20*E^4*x + 25*E^8*x^2 + E^(2*x)*x^2 + (-20*x
- 50*E^4*x^2)*Log[3] + 25*x^2*Log[3]^2 + (4*x + 10*E^4*x^2 - 10*x^2*Log[3])*Log[5] + x^2*Log[5]^2 + E^x*(4*x +
10*E^4*x^2 - 10*x^2*Log[3] + 2*x^2*Log[5])),x]

[Out]

9/(2 + E^x*x + x*(5*E^4 - Log[243/5]))

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fricas [A]  time = 0.56, size = 24, normalized size = 1.04 \begin {gather*} \frac {9}{5 \, x e^{4} + x e^{x} + x \log \relax (5) - 5 \, x \log \relax (3) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x-9)*exp(x)-9*log(5)+45*log(3)-45*exp(4))/(exp(x)^2*x^2+(2*x^2*log(5)-10*x^2*log(3)+10*x^2*exp(
4)+4*x)*exp(x)+x^2*log(5)^2+(-10*x^2*log(3)+10*x^2*exp(4)+4*x)*log(5)+25*x^2*log(3)^2+(-50*x^2*exp(4)-20*x)*lo
g(3)+25*x^2*exp(4)^2+20*x*exp(4)+4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.36, size = 24, normalized size = 1.04 \begin {gather*} \frac {9}{5 \, x e^{4} + x e^{x} + x \log \relax (5) - 5 \, x \log \relax (3) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x-9)*exp(x)-9*log(5)+45*log(3)-45*exp(4))/(exp(x)^2*x^2+(2*x^2*log(5)-10*x^2*log(3)+10*x^2*exp(
4)+4*x)*exp(x)+x^2*log(5)^2+(-10*x^2*log(3)+10*x^2*exp(4)+4*x)*log(5)+25*x^2*log(3)^2+(-50*x^2*exp(4)-20*x)*lo
g(3)+25*x^2*exp(4)^2+20*x*exp(4)+4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 27, normalized size = 1.17

method	result	size
norman	\(-\frac {9}{5 x \ln \relax (3)-x \ln \relax (5)-5 x \,{\mathrm e}^{4}-{\mathrm e}^{x} x -2}\)	\(27\)
risch	\(-\frac {9}{5 x \ln \relax (3)-x \ln \relax (5)-5 x \,{\mathrm e}^{4}-{\mathrm e}^{x} x -2}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*x-9)*exp(x)-9*ln(5)+45*ln(3)-45*exp(4))/(exp(x)^2*x^2+(2*x^2*ln(5)-10*x^2*ln(3)+10*x^2*exp(4)+4*x)*ex
p(x)+x^2*ln(5)^2+(-10*x^2*ln(3)+10*x^2*exp(4)+4*x)*ln(5)+25*x^2*ln(3)^2+(-50*x^2*exp(4)-20*x)*ln(3)+25*x^2*exp
(4)^2+20*x*exp(4)+4),x,method=_RETURNVERBOSE)

[Out]

-9/(5*x*ln(3)-x*ln(5)-5*x*exp(4)-exp(x)*x-2)

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maxima [A]  time = 0.50, size = 23, normalized size = 1.00 \begin {gather*} \frac {9}{x {\left (5 \, e^{4} + \log \relax (5) - 5 \, \log \relax (3)\right )} + x e^{x} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x-9)*exp(x)-9*log(5)+45*log(3)-45*exp(4))/(exp(x)^2*x^2+(2*x^2*log(5)-10*x^2*log(3)+10*x^2*exp(
4)+4*x)*exp(x)+x^2*log(5)^2+(-10*x^2*log(3)+10*x^2*exp(4)+4*x)*log(5)+25*x^2*log(3)^2+(-50*x^2*exp(4)-20*x)*lo
g(3)+25*x^2*exp(4)^2+20*x*exp(4)+4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.39, size = 45, normalized size = 1.96 \begin {gather*} -\frac {\frac {9\,x\,{\mathrm {e}}^x}{2}+x\,\left (\frac {45\,{\mathrm {e}}^4}{2}-\ln \left (\frac {31381059609\,\sqrt {15}}{3125}\right )\right )}{5\,x\,{\mathrm {e}}^4-5\,x\,\ln \relax (3)+x\,\ln \relax (5)+x\,{\mathrm {e}}^x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(45*exp(4) - 45*log(3) + 9*log(5) + exp(x)*(9*x + 9))/(25*x^2*log(3)^2 + x^2*log(5)^2 + 20*x*exp(4) + log
(5)*(4*x + 10*x^2*exp(4) - 10*x^2*log(3)) + exp(x)*(4*x + 10*x^2*exp(4) - 10*x^2*log(3) + 2*x^2*log(5)) + x^2*
exp(2*x) + 25*x^2*exp(8) - log(3)*(20*x + 50*x^2*exp(4)) + 4),x)

[Out]

-((9*x*exp(x))/2 + x*((45*exp(4))/2 - log((31381059609*15^(1/2))/3125)))/(5*x*exp(4) - 5*x*log(3) + x*log(5) +
x*exp(x) + 2)

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sympy [A]  time = 0.21, size = 26, normalized size = 1.13 \begin {gather*} \frac {9}{x e^{x} - 5 x \log {\relax (3 )} + x \log {\relax (5 )} + 5 x e^{4} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x-9)*exp(x)-9*ln(5)+45*ln(3)-45*exp(4))/(exp(x)**2*x**2+(2*x**2*ln(5)-10*x**2*ln(3)+10*x**2*exp
(4)+4*x)*exp(x)+x**2*ln(5)**2+(-10*x**2*ln(3)+10*x**2*exp(4)+4*x)*ln(5)+25*x**2*ln(3)**2+(-50*x**2*exp(4)-20*x
)*ln(3)+25*x**2*exp(4)**2+20*x*exp(4)+4),x)

[Out]

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

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