3.88.42 \(\int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log (-2+e^2+e^5)}{x}} (e^{200+40 x+2 x^2} (40 x^2+4 x^3)+\log (-2+e^2+e^5))}{x^2} \, dx\)

Optimal. Leaf size=29 \[ 1+e^{3+e^{2 (10+x)^2}-\frac {\log \left (-2+e^2+e^5\right )}{x}} \]

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Rubi [A]  time = 0.48, antiderivative size = 39, normalized size of antiderivative = 1.34, number of steps used = 1, number of rules used = 1, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6706} \begin {gather*} e^{\frac {e^{2 x^2+40 x+200} x+3 x}{x}} \left (-2+e^2+e^5\right )^{-1/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((3*x + E^(200 + 40*x + 2*x^2)*x - Log[-2 + E^2 + E^5])/x)*(E^(200 + 40*x + 2*x^2)*(40*x^2 + 4*x^3) + L
og[-2 + E^2 + E^5]))/x^2,x]

[Out]

E^((3*x + E^(200 + 40*x + 2*x^2)*x)/x)/(-2 + E^2 + E^5)^x^(-1)

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{\frac {3 x+e^{200+40 x+2 x^2} x}{x}} \left (-2+e^2+e^5\right )^{-1/x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 28, normalized size = 0.97 \begin {gather*} e^{3+e^{2 (10+x)^2}} \left (-2+e^2+e^5\right )^{-1/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3*x + E^(200 + 40*x + 2*x^2)*x - Log[-2 + E^2 + E^5])/x)*(E^(200 + 40*x + 2*x^2)*(40*x^2 + 4*x^
3) + Log[-2 + E^2 + E^5]))/x^2,x]

[Out]

E^(3 + E^(2*(10 + x)^2))/(-2 + E^2 + E^5)^x^(-1)

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fricas [A]  time = 0.73, size = 31, normalized size = 1.07 \begin {gather*} e^{\left (\frac {x e^{\left (2 \, x^{2} + 40 \, x + 200\right )} + 3 \, x - \log \left (e^{5} + e^{2} - 2\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(5)+exp(2)-2)+(4*x^3+40*x^2)*exp(x^2+20*x+100)^2)*exp((-log(exp(5)+exp(2)-2)+x*exp(x^2+20*x+
100)^2+3*x)/x)/x^2,x, algorithm="fricas")

[Out]

e^((x*e^(2*x^2 + 40*x + 200) + 3*x - log(e^5 + e^2 - 2))/x)

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giac [A]  time = 0.29, size = 26, normalized size = 0.90 \begin {gather*} e^{\left (-\frac {\log \left (e^{5} + e^{2} - 2\right )}{x} + e^{\left (2 \, x^{2} + 40 \, x + 200\right )} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(5)+exp(2)-2)+(4*x^3+40*x^2)*exp(x^2+20*x+100)^2)*exp((-log(exp(5)+exp(2)-2)+x*exp(x^2+20*x+
100)^2+3*x)/x)/x^2,x, algorithm="giac")

[Out]

e^(-log(e^5 + e^2 - 2)/x + e^(2*x^2 + 40*x + 200) + 3)

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maple [A]  time = 0.14, size = 32, normalized size = 1.10

method	result	size
norman	\({\mathrm e}^{\frac {-\ln \left ({\mathrm e}^{5}+{\mathrm e}^{2}-2\right )+x \,{\mathrm e}^{2 x^{2}+40 x +200}+3 x}{x}}\)	\(32\)
risch	\(\left ({\mathrm e}-1\right )^{-\frac {1}{x}} \left ({\mathrm e}^{4}+{\mathrm e}^{3}+{\mathrm e}^{2}+2 \,{\mathrm e}+2\right )^{-\frac {1}{x}} {\mathrm e}^{{\mathrm e}^{2 \left (x +10\right )^{2}}+3}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(exp(5)+exp(2)-2)+(4*x^3+40*x^2)*exp(x^2+20*x+100)^2)*exp((-ln(exp(5)+exp(2)-2)+x*exp(x^2+20*x+100)^2+3
*x)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp((-ln(exp(5)+exp(2)-2)+x*exp(x^2+20*x+100)^2+3*x)/x)

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maxima [A]  time = 0.47, size = 42, normalized size = 1.45 \begin {gather*} e^{\left (-\frac {\log \left (e^{4} + e^{3} + e^{2} + 2 \, e + 2\right )}{x} - \frac {\log \left (e - 1\right )}{x} + e^{\left (2 \, x^{2} + 40 \, x + 200\right )} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(5)+exp(2)-2)+(4*x^3+40*x^2)*exp(x^2+20*x+100)^2)*exp((-log(exp(5)+exp(2)-2)+x*exp(x^2+20*x+
100)^2+3*x)/x)/x^2,x, algorithm="maxima")

[Out]

e^(-log(e^4 + e^3 + e^2 + 2*e + 2)/x - log(e - 1)/x + e^(2*x^2 + 40*x + 200) + 3)

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mupad [B]  time = 5.59, size = 29, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{40\,x}\,{\mathrm {e}}^{200}\,{\mathrm {e}}^{2\,x^2}}\,{\mathrm {e}}^3}{{\left ({\mathrm {e}}^2+{\mathrm {e}}^5-2\right )}^{1/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((3*x - log(exp(2) + exp(5) - 2) + x*exp(40*x + 2*x^2 + 200))/x)*(log(exp(2) + exp(5) - 2) + exp(40*x
+ 2*x^2 + 200)*(40*x^2 + 4*x^3)))/x^2,x)

[Out]

(exp(exp(40*x)*exp(200)*exp(2*x^2))*exp(3))/(exp(2) + exp(5) - 2)^(1/x)

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sympy [A]  time = 0.43, size = 29, normalized size = 1.00 \begin {gather*} e^{\frac {x e^{2 x^{2} + 40 x + 200} + 3 x - \log {\left (-2 + e^{2} + e^{5} \right )}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(exp(5)+exp(2)-2)+(4*x**3+40*x**2)*exp(x**2+20*x+100)**2)*exp((-ln(exp(5)+exp(2)-2)+x*exp(x**2+20
*x+100)**2+3*x)/x)/x**2,x)

[Out]

exp((x*exp(2*x**2 + 40*x + 200) + 3*x - log(-2 + exp(2) + exp(5)))/x)

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