3.97.38 \(\int \frac {e^{x+x^2} (9+24 x+13 x^2+2 x^3)+e^{x+x^2} (6+14 x+4 x^2) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x))}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx\)

Optimal. Leaf size=31 \[ -e^{4+e^{1-\frac {16}{3+x+\log (5 x)}} x}+e^{x+x^2} \]

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Rubi [F]  time = 171.34, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{x+x^2} \left (9+24 x+13 x^2+2 x^3\right )+e^{x+x^2} \left (6+14 x+4 x^2\right ) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+\exp \left (4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}\right ) \left (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x)\right )}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x + x^2)*(9 + 24*x + 13*x^2 + 2*x^3) + E^(x + x^2)*(6 + 14*x + 4*x^2)*Log[5*x] + E^(x + x^2)*(1 + 2*x)
*Log[5*x]^2 + E^(4 + E^((-13 + x + Log[5*x])/(3 + x + Log[5*x]))*x + (-13 + x + Log[5*x])/(3 + x + Log[5*x]))*
(-25 - 22*x - x^2 + (-6 - 2*x)*Log[5*x] - Log[5*x]^2))/(9 + 6*x + x^2 + (6 + 2*x)*Log[5*x] + Log[5*x]^2),x]

[Out]

E^(x + x^2) - Defer[Int][5^(5/(3 + x + Log[5*x]))*E^(5^(3 + x + Log[5*x])^(-1)*E^(-13/(3 + x + Log[5*x]) + x/(
3 + x + Log[5*x]))*x^(1 + (3 + x + Log[5*x])^(-1)) - (3 + x + Log[5*x])^(-1) + (5*x)/(3 + x + Log[5*x]))*x^(5/
(3 + x + Log[5*x])), x] - 16*Defer[Int][(5^(5/(3 + x + Log[5*x]))*E^(5^(3 + x + Log[5*x])^(-1)*E^(-13/(3 + x +
Log[5*x]) + x/(3 + x + Log[5*x]))*x^(1 + (3 + x + Log[5*x])^(-1)) - (3 + x + Log[5*x])^(-1) + (5*x)/(3 + x +
Log[5*x]))*x^(5/(3 + x + Log[5*x])))/(3 + x + Log[5*x])^2, x] - 16*Defer[Int][(5^(5/(3 + x + Log[5*x]))*E^(5^(
3 + x + Log[5*x])^(-1)*E^(-13/(3 + x + Log[5*x]) + x/(3 + x + Log[5*x]))*x^(1 + (3 + x + Log[5*x])^(-1)) - (3
+ x + Log[5*x])^(-1) + (5*x)/(3 + x + Log[5*x]))*x^(1 + 5/(3 + x + Log[5*x])))/(3 + x + Log[5*x])^2, x]

Rubi steps

Aborted

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Mathematica [A]  time = 0.67, size = 31, normalized size = 1.00 \begin {gather*} -e^{4+e^{1-\frac {16}{3+x+\log (5 x)}} x}+e^{x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x + x^2)*(9 + 24*x + 13*x^2 + 2*x^3) + E^(x + x^2)*(6 + 14*x + 4*x^2)*Log[5*x] + E^(x + x^2)*(1
+ 2*x)*Log[5*x]^2 + E^(4 + E^((-13 + x + Log[5*x])/(3 + x + Log[5*x]))*x + (-13 + x + Log[5*x])/(3 + x + Log[5
*x]))*(-25 - 22*x - x^2 + (-6 - 2*x)*Log[5*x] - Log[5*x]^2))/(9 + 6*x + x^2 + (6 + 2*x)*Log[5*x] + Log[5*x]^2)
,x]

[Out]

-E^(4 + E^(1 - 16/(3 + x + Log[5*x]))*x) + E^(x + x^2)

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fricas [B]  time = 0.64, size = 100, normalized size = 3.23 \begin {gather*} {\left (e^{\left (x^{2} + x + \frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3}\right )} - e^{\left (\frac {{\left (x^{2} + x \log \left (5 \, x\right ) + 3 \, x\right )} e^{\left (\frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3}\right )} + 5 \, x + 5 \, \log \left (5 \, x\right ) - 1}{x + \log \left (5 \, x\right ) + 3}\right )}\right )} e^{\left (-\frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(5*x)^2+(-2*x-6)*log(5*x)-x^2-22*x-25)*exp((log(5*x)+x-13)/(log(5*x)+3+x))*exp(x*exp((log(5*x)
+x-13)/(log(5*x)+3+x))+4)+(2*x+1)*exp(x^2+x)*log(5*x)^2+(4*x^2+14*x+6)*exp(x^2+x)*log(5*x)+(2*x^3+13*x^2+24*x+
9)*exp(x^2+x))/(log(5*x)^2+(2*x+6)*log(5*x)+x^2+6*x+9),x, algorithm="fricas")

[Out]

(e^(x^2 + x + (x + log(5*x) - 13)/(x + log(5*x) + 3)) - e^(((x^2 + x*log(5*x) + 3*x)*e^((x + log(5*x) - 13)/(x
+ log(5*x) + 3)) + 5*x + 5*log(5*x) - 1)/(x + log(5*x) + 3)))*e^(-(x + log(5*x) - 13)/(x + log(5*x) + 3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x + 1\right )} e^{\left (x^{2} + x\right )} \log \left (5 \, x\right )^{2} + 2 \, {\left (2 \, x^{2} + 7 \, x + 3\right )} e^{\left (x^{2} + x\right )} \log \left (5 \, x\right ) + {\left (2 \, x^{3} + 13 \, x^{2} + 24 \, x + 9\right )} e^{\left (x^{2} + x\right )} - {\left (x^{2} + 2 \, {\left (x + 3\right )} \log \left (5 \, x\right ) + \log \left (5 \, x\right )^{2} + 22 \, x + 25\right )} e^{\left (x e^{\left (\frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3}\right )} + \frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3} + 4\right )}}{x^{2} + 2 \, {\left (x + 3\right )} \log \left (5 \, x\right ) + \log \left (5 \, x\right )^{2} + 6 \, x + 9}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(5*x)^2+(-2*x-6)*log(5*x)-x^2-22*x-25)*exp((log(5*x)+x-13)/(log(5*x)+3+x))*exp(x*exp((log(5*x)
+x-13)/(log(5*x)+3+x))+4)+(2*x+1)*exp(x^2+x)*log(5*x)^2+(4*x^2+14*x+6)*exp(x^2+x)*log(5*x)+(2*x^3+13*x^2+24*x+
9)*exp(x^2+x))/(log(5*x)^2+(2*x+6)*log(5*x)+x^2+6*x+9),x, algorithm="giac")

[Out]

integrate(((2*x + 1)*e^(x^2 + x)*log(5*x)^2 + 2*(2*x^2 + 7*x + 3)*e^(x^2 + x)*log(5*x) + (2*x^3 + 13*x^2 + 24*
x + 9)*e^(x^2 + x) - (x^2 + 2*(x + 3)*log(5*x) + log(5*x)^2 + 22*x + 25)*e^(x*e^((x + log(5*x) - 13)/(x + log(
5*x) + 3)) + (x + log(5*x) - 13)/(x + log(5*x) + 3) + 4))/(x^2 + 2*(x + 3)*log(5*x) + log(5*x)^2 + 6*x + 9), x
)

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maple [A]  time = 0.20, size = 33, normalized size = 1.06

method	result	size
risch	\({\mathrm e}^{\left (x +1\right ) x}-{\mathrm e}^{x \,{\mathrm e}^{\frac {\ln \left (5 x \right )+x -13}{\ln \left (5 x \right )+3+x}}+4}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(5*x)^2+(-2*x-6)*ln(5*x)-x^2-22*x-25)*exp((ln(5*x)+x-13)/(ln(5*x)+3+x))*exp(x*exp((ln(5*x)+x-13)/(ln(
5*x)+3+x))+4)+(2*x+1)*exp(x^2+x)*ln(5*x)^2+(4*x^2+14*x+6)*exp(x^2+x)*ln(5*x)+(2*x^3+13*x^2+24*x+9)*exp(x^2+x))
/(ln(5*x)^2+(2*x+6)*ln(5*x)+x^2+6*x+9),x,method=_RETURNVERBOSE)

[Out]

exp((x+1)*x)-exp(x*exp((ln(5*x)+x-13)/(ln(5*x)+3+x))+4)

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maxima [A]  time = 0.52, size = 28, normalized size = 0.90 \begin {gather*} e^{\left (x^{2} + x\right )} - e^{\left (x e^{\left (-\frac {16}{x + \log \relax (5) + \log \relax (x) + 3} + 1\right )} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(5*x)^2+(-2*x-6)*log(5*x)-x^2-22*x-25)*exp((log(5*x)+x-13)/(log(5*x)+3+x))*exp(x*exp((log(5*x)
+x-13)/(log(5*x)+3+x))+4)+(2*x+1)*exp(x^2+x)*log(5*x)^2+(4*x^2+14*x+6)*exp(x^2+x)*log(5*x)+(2*x^3+13*x^2+24*x+
9)*exp(x^2+x))/(log(5*x)^2+(2*x+6)*log(5*x)+x^2+6*x+9),x, algorithm="maxima")

[Out]

e^(x^2 + x) - e^(x*e^(-16/(x + log(5) + log(x) + 3) + 1) + 4)

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mupad [B]  time = 5.96, size = 60, normalized size = 1.94 \begin {gather*} {\mathrm {e}}^{x^2+x}-{\mathrm {e}}^{5^{\frac {1}{x+\ln \left (5\,x\right )+3}}\,x\,x^{\frac {1}{x+\ln \left (5\,x\right )+3}}\,{\mathrm {e}}^{-\frac {13}{x+\ln \left (5\,x\right )+3}}\,{\mathrm {e}}^{\frac {x}{x+\ln \left (5\,x\right )+3}}+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + x^2)*(24*x + 13*x^2 + 2*x^3 + 9) + log(5*x)*exp(x + x^2)*(14*x + 4*x^2 + 6) - exp((x + log(5*x) -
13)/(x + log(5*x) + 3))*exp(x*exp((x + log(5*x) - 13)/(x + log(5*x) + 3)) + 4)*(22*x + log(5*x)^2 + x^2 + log
(5*x)*(2*x + 6) + 25) + log(5*x)^2*exp(x + x^2)*(2*x + 1))/(6*x + log(5*x)^2 + x^2 + log(5*x)*(2*x + 6) + 9),x
)

[Out]

exp(x + x^2) - exp(5^(1/(x + log(5*x) + 3))*x*x^(1/(x + log(5*x) + 3))*exp(-13/(x + log(5*x) + 3))*exp(x/(x +
log(5*x) + 3)) + 4)

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sympy [A]  time = 21.00, size = 29, normalized size = 0.94 \begin {gather*} e^{x^{2} + x} - e^{x e^{\frac {x + \log {\left (5 x \right )} - 13}{x + \log {\left (5 x \right )} + 3}} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-ln(5*x)**2+(-2*x-6)*ln(5*x)-x**2-22*x-25)*exp((ln(5*x)+x-13)/(ln(5*x)+3+x))*exp(x*exp((ln(5*x)+x-
13)/(ln(5*x)+3+x))+4)+(2*x+1)*exp(x**2+x)*ln(5*x)**2+(4*x**2+14*x+6)*exp(x**2+x)*ln(5*x)+(2*x**3+13*x**2+24*x+
9)*exp(x**2+x))/(ln(5*x)**2+(2*x+6)*ln(5*x)+x**2+6*x+9),x)

[Out]

exp(x**2 + x) - exp(x*exp((x + log(5*x) - 13)/(x + log(5*x) + 3)) + 4)

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