3.4.32 \(\int \frac {50+e^{x+\frac {1}{25} e^x (e^6-10 e^3 \log (4)+25 \log ^2(4))} (-2 e^6+20 e^3 \log (4)-50 \log ^2(4))}{225 e^{\frac {3}{25} e^x (e^6-10 e^3 \log (4)+25 \log ^2(4))}-675 e^{\frac {2}{25} e^x (e^6-10 e^3 \log (4)+25 \log ^2(4))} x+675 e^{\frac {1}{25} e^x (e^6-10 e^3 \log (4)+25 \log ^2(4))} x^2-225 x^3} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{9 \left (-e^{e^x \left (-\frac {e^3}{5}+\log (4)\right )^2}+x\right )^2} \]

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Rubi [A]  time = 1.39, antiderivative size = 29, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 147, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6688, 12, 6686} \begin {gather*} \frac {1}{9 \left (e^{\frac {1}{25} e^x \left (e^3-5 \log (4)\right )^2}-x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50 + E^(x + (E^x*(E^6 - 10*E^3*Log[4] + 25*Log[4]^2))/25)*(-2*E^6 + 20*E^3*Log[4] - 50*Log[4]^2))/(225*E^
((3*E^x*(E^6 - 10*E^3*Log[4] + 25*Log[4]^2))/25) - 675*E^((2*E^x*(E^6 - 10*E^3*Log[4] + 25*Log[4]^2))/25)*x +
675*E^((E^x*(E^6 - 10*E^3*Log[4] + 25*Log[4]^2))/25)*x^2 - 225*x^3),x]

[Out]

1/(9*(E^((E^x*(E^3 - 5*Log[4])^2)/25) - 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 {2 \left (25-e^{6+x+\frac {1}{25} e^x \left (e^3-5 \log (4)\right )^2} \left (1-\frac {5 \left (2 e^3-5 \log (4)\right ) \log (4)}{e^6}\right )\right )}{225 \left (e^{\frac {1}{25} e^x \left (e^3-5 \log (4)\right )^2}-x\right )^3} \, dx\\ &=\frac {2}{225} \int \frac {25-e^{6+x+\frac {1}{25} e^x \left (e^3-5 \log (4)\right )^2} \left (1-\frac {5 \left (2 e^3-5 \log (4)\right ) \log (4)}{e^6}\right )}{\left (e^{\frac {1}{25} e^x \left (e^3-5 \log (4)\right )^2}-x\right )^3} \, dx\\ &=\frac {1}{9 \left (e^{\frac {1}{25} e^x \left (e^3-5 \log (4)\right )^2}-x\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 29, normalized size = 1.04 \begin {gather*} \frac {1}{9 \left (e^{\frac {1}{25} e^x \left (e^3-5 \log (4)\right )^2}-x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50 + E^(x + (E^x*(E^6 - 10*E^3*Log[4] + 25*Log[4]^2))/25)*(-2*E^6 + 20*E^3*Log[4] - 50*Log[4]^2))/(
225*E^((3*E^x*(E^6 - 10*E^3*Log[4] + 25*Log[4]^2))/25) - 675*E^((2*E^x*(E^6 - 10*E^3*Log[4] + 25*Log[4]^2))/25
)*x + 675*E^((E^x*(E^6 - 10*E^3*Log[4] + 25*Log[4]^2))/25)*x^2 - 225*x^3),x]

[Out]

1/(9*(E^((E^x*(E^3 - 5*Log[4])^2)/25) - x)^2)

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fricas [B]  time = 0.75, size = 72, normalized size = 2.57 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{9 \, {\left (x^{2} e^{\left (2 \, x\right )} - 2 \, x e^{\left (-\frac {1}{25} \, {\left (20 \, e^{3} \log \relax (2) - 100 \, \log \relax (2)^{2} - e^{6}\right )} e^{x} + 2 \, x\right )} + e^{\left (-\frac {2}{25} \, {\left (20 \, e^{3} \log \relax (2) - 100 \, \log \relax (2)^{2} - e^{6}\right )} e^{x} + 2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*log(2)^2+40*exp(3)*log(2)-2*exp(3)^2)*exp(x)*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2
)*exp(x))+50)/(225*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2)*exp(x))^3-675*x*exp(1/25*(100*log(2)^2-20
*exp(3)*log(2)+exp(3)^2)*exp(x))^2+675*x^2*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2)*exp(x))-225*x^3),
x, algorithm="fricas")

[Out]

1/9*e^(2*x)/(x^2*e^(2*x) - 2*x*e^(-1/25*(20*e^3*log(2) - 100*log(2)^2 - e^6)*e^x + 2*x) + e^(-2/25*(20*e^3*log
(2) - 100*log(2)^2 - e^6)*e^x + 2*x))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*log(2)^2+40*exp(3)*log(2)-2*exp(3)^2)*exp(x)*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2
)*exp(x))+50)/(225*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2)*exp(x))^3-675*x*exp(1/25*(100*log(2)^2-20
*exp(3)*log(2)+exp(3)^2)*exp(x))^2+675*x^2*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2)*exp(x))-225*x^3),
x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.12, size = 31, normalized size = 1.11

method	result	size
risch	\(\frac {1}{9 \left (x -{\mathrm e}^{-\frac {\left (-{\mathrm e}^{6}+20 \,{\mathrm e}^{3} \ln \relax (2)-100 \ln \relax (2)^{2}\right ) {\mathrm e}^{x}}{25}}\right )^{2}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-200*ln(2)^2+40*exp(3)*ln(2)-2*exp(3)^2)*exp(x)*exp(1/25*(100*ln(2)^2-20*exp(3)*ln(2)+exp(3)^2)*exp(x))+
50)/(225*exp(1/25*(100*ln(2)^2-20*exp(3)*ln(2)+exp(3)^2)*exp(x))^3-675*x*exp(1/25*(100*ln(2)^2-20*exp(3)*ln(2)
+exp(3)^2)*exp(x))^2+675*x^2*exp(1/25*(100*ln(2)^2-20*exp(3)*ln(2)+exp(3)^2)*exp(x))-225*x^3),x,method=_RETURN
VERBOSE)

[Out]

1/9/(x-exp(-1/25*(-exp(6)+20*exp(3)*ln(2)-100*ln(2)^2)*exp(x)))^2

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maxima [B]  time = 1.29, size = 68, normalized size = 2.43 \begin {gather*} \frac {2^{\frac {8}{5} \, e^{\left (x + 3\right )}}}{9 \, {\left (2^{\frac {8}{5} \, e^{\left (x + 3\right )}} x^{2} - 2 \, x e^{\left (4 \, e^{x} \log \relax (2)^{2} + \frac {4}{5} \, e^{\left (x + 3\right )} \log \relax (2) + \frac {1}{25} \, e^{\left (x + 6\right )}\right )} + e^{\left (8 \, e^{x} \log \relax (2)^{2} + \frac {2}{25} \, e^{\left (x + 6\right )}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*log(2)^2+40*exp(3)*log(2)-2*exp(3)^2)*exp(x)*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2
)*exp(x))+50)/(225*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2)*exp(x))^3-675*x*exp(1/25*(100*log(2)^2-20
*exp(3)*log(2)+exp(3)^2)*exp(x))^2+675*x^2*exp(1/25*(100*log(2)^2-20*exp(3)*log(2)+exp(3)^2)*exp(x))-225*x^3),
x, algorithm="maxima")

[Out]

1/9*2^(8/5*e^(x + 3))/(2^(8/5*e^(x + 3))*x^2 - 2*x*e^(4*e^x*log(2)^2 + 4/5*e^(x + 3)*log(2) + 1/25*e^(x + 6))
+ e^(8*e^x*log(2)^2 + 2/25*e^(x + 6)))

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mupad [B]  time = 0.69, size = 65, normalized size = 2.32 \begin {gather*} \frac {1}{9\,\left (\frac {{\mathrm {e}}^{8\,{\mathrm {e}}^x\,{\ln \relax (2)}^2+\frac {2\,{\mathrm {e}}^6\,{\mathrm {e}}^x}{25}}}{2^{\frac {8\,{\mathrm {e}}^{x+3}}{5}}}+x^2-2^{1-\frac {4\,{\mathrm {e}}^{x+3}}{5}}\,x\,{\mathrm {e}}^{4\,{\mathrm {e}}^x\,{\ln \relax (2)}^2+\frac {{\mathrm {e}}^6\,{\mathrm {e}}^x}{25}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(x)*(exp(6) - 20*exp(3)*log(2) + 100*log(2)^2))/25)*exp(x)*(2*exp(6) - 40*exp(3)*log(2) + 200*lo
g(2)^2) - 50)/(225*exp((3*exp(x)*(exp(6) - 20*exp(3)*log(2) + 100*log(2)^2))/25) - 675*x*exp((2*exp(x)*(exp(6)
- 20*exp(3)*log(2) + 100*log(2)^2))/25) - 225*x^3 + 675*x^2*exp((exp(x)*(exp(6) - 20*exp(3)*log(2) + 100*log(
2)^2))/25)),x)

[Out]

1/(9*(exp(8*exp(x)*log(2)^2 + (2*exp(6)*exp(x))/25)/2^((8*exp(x + 3))/5) + x^2 - 2^(1 - (4*exp(x + 3))/5)*x*ex
p(4*exp(x)*log(2)^2 + (exp(6)*exp(x))/25)))

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sympy [A]  time = 0.32, size = 66, normalized size = 2.36 \begin {gather*} \frac {1}{9 x^{2} - 18 x e^{\left (- \frac {4 e^{3} \log {\relax (2 )}}{5} + 4 \log {\relax (2 )}^{2} + \frac {e^{6}}{25}\right ) e^{x}} + 9 e^{2 \left (- \frac {4 e^{3} \log {\relax (2 )}}{5} + 4 \log {\relax (2 )}^{2} + \frac {e^{6}}{25}\right ) e^{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*ln(2)**2+40*exp(3)*ln(2)-2*exp(3)**2)*exp(x)*exp(1/25*(100*ln(2)**2-20*exp(3)*ln(2)+exp(3)**2
)*exp(x))+50)/(225*exp(1/25*(100*ln(2)**2-20*exp(3)*ln(2)+exp(3)**2)*exp(x))**3-675*x*exp(1/25*(100*ln(2)**2-2
0*exp(3)*ln(2)+exp(3)**2)*exp(x))**2+675*x**2*exp(1/25*(100*ln(2)**2-20*exp(3)*ln(2)+exp(3)**2)*exp(x))-225*x*
*3),x)

[Out]

1/(9*x**2 - 18*x*exp((-4*exp(3)*log(2)/5 + 4*log(2)**2 + exp(6)/25)*exp(x)) + 9*exp(2*(-4*exp(3)*log(2)/5 + 4*
log(2)**2 + exp(6)/25)*exp(x)))

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