3.35.13 \(\int e^{3 x+e^{1+e^{e^{5+x}}+e^x \log (5)} x} (1+3 x+e^{1+e^{e^{5+x}}+e^x \log (5)} (x+e^{5+e^{5+x}+x} x^2+e^x x^2 \log (5))) \, dx\)

Optimal. Leaf size=25 \[ e^{\left (3+e^{1+e^{e^{5+x}}+e^x \log (5)}\right ) x} x \]

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Rubi [B]  time = 0.69, antiderivative size = 127, normalized size of antiderivative = 5.08, number of steps used = 1, number of rules used = 1, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2288} \begin {gather*} \frac {e^{5^{e^x} e^{e^{e^{x+5}}+1} x+3 x} \left (5^{e^x} e^{e^{e^{x+5}}+1} \left (e^{x+e^{x+5}+5} x^2+e^x x^2 \log (5)+x\right )+3 x\right )}{5^{e^x} e^{e^{e^{x+5}}+1}+5^{e^x} e^{e^{e^{x+5}}+1} x \left (e^{x+e^{x+5}+5}+e^x \log (5)\right )+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(3*x + E^(1 + E^E^(5 + x) + E^x*Log[5])*x)*(1 + 3*x + E^(1 + E^E^(5 + x) + E^x*Log[5])*(x + E^(5 + E^(5
+ x) + x)*x^2 + E^x*x^2*Log[5])),x]

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.72, size = 25, normalized size = 1.00 \begin {gather*} e^{\left (3+5^{e^x} e^{1+e^{e^{5+x}}}\right ) x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(3*x + E^(1 + E^E^(5 + x) + E^x*Log[5])*x)*(1 + 3*x + E^(1 + E^E^(5 + x) + E^x*Log[5])*(x + E^(5 +
E^(5 + x) + x)*x^2 + E^x*x^2*Log[5])),x]

[Out]

E^((3 + 5^E^x*E^(1 + E^E^(5 + x)))*x)*x

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fricas [A]  time = 0.84, size = 39, normalized size = 1.56 \begin {gather*} x e^{\left (x e^{\left ({\left (e^{\left (2 \, x + 10\right )} \log \relax (5) + e^{\left (x + e^{\left (x + 5\right )} + 10\right )} + e^{\left (x + 10\right )}\right )} e^{\left (-x - 10\right )}\right )} + 3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(5+x)*exp(exp(5+x))+x^2*log(5)*exp(x)+x)*exp(exp(exp(5+x))+exp(x)*log(5)+1)+3*x+1)*exp(x*ex
p(exp(exp(5+x))+exp(x)*log(5)+1)+3*x),x, algorithm="fricas")

[Out]

x*e^(x*e^((e^(2*x + 10)*log(5) + e^(x + e^(x + 5) + 10) + e^(x + 10))*e^(-x - 10)) + 3*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left ({\left (x^{2} e^{x} \log \relax (5) + x^{2} e^{\left (x + e^{\left (x + 5\right )} + 5\right )} + x\right )} e^{\left (e^{x} \log \relax (5) + e^{\left (e^{\left (x + 5\right )}\right )} + 1\right )} + 3 \, x + 1\right )} e^{\left (x e^{\left (e^{x} \log \relax (5) + e^{\left (e^{\left (x + 5\right )}\right )} + 1\right )} + 3 \, x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(5+x)*exp(exp(5+x))+x^2*log(5)*exp(x)+x)*exp(exp(exp(5+x))+exp(x)*log(5)+1)+3*x+1)*exp(x*ex
p(exp(exp(5+x))+exp(x)*log(5)+1)+3*x),x, algorithm="giac")

[Out]

integrate(((x^2*e^x*log(5) + x^2*e^(x + e^(x + 5) + 5) + x)*e^(e^x*log(5) + e^(e^(x + 5)) + 1) + 3*x + 1)*e^(x
*e^(e^x*log(5) + e^(e^(x + 5)) + 1) + 3*x), x)

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maple [A]  time = 0.08, size = 21, normalized size = 0.84

method	result	size
risch	\(x \,{\mathrm e}^{x \left (5^{{\mathrm e}^{x}} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{5+x}}+1}+3\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*exp(5+x)*exp(exp(5+x))+x^2*ln(5)*exp(x)+x)*exp(exp(exp(5+x))+exp(x)*ln(5)+1)+3*x+1)*exp(x*exp(exp(ex
p(5+x))+exp(x)*ln(5)+1)+3*x),x,method=_RETURNVERBOSE)

[Out]

x*exp(x*(5^exp(x)*exp(exp(exp(5+x))+1)+3))

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maxima [A]  time = 1.02, size = 22, normalized size = 0.88 \begin {gather*} x e^{\left (x e^{\left (e^{x} \log \relax (5) + e^{\left (e^{\left (x + 5\right )}\right )} + 1\right )} + 3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(5+x)*exp(exp(5+x))+x^2*log(5)*exp(x)+x)*exp(exp(exp(5+x))+exp(x)*log(5)+1)+3*x+1)*exp(x*ex
p(exp(exp(5+x))+exp(x)*log(5)+1)+3*x),x, algorithm="maxima")

[Out]

x*e^(x*e^(e^x*log(5) + e^(e^(x + 5)) + 1) + 3*x)

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mupad [B]  time = 2.23, size = 22, normalized size = 0.88 \begin {gather*} x\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{5^{{\mathrm {e}}^x}\,x\,\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^5\,{\mathrm {e}}^x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x + x*exp(exp(exp(x + 5)) + exp(x)*log(5) + 1))*(3*x + exp(exp(exp(x + 5)) + exp(x)*log(5) + 1)*(x +
x^2*exp(x + 5)*exp(exp(x + 5)) + x^2*exp(x)*log(5)) + 1),x)

[Out]

x*exp(3*x)*exp(5^exp(x)*x*exp(1)*exp(exp(exp(5)*exp(x))))

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sympy [A]  time = 10.31, size = 26, normalized size = 1.04 \begin {gather*} x e^{x e^{e^{x} \log {\relax (5 )} + e^{e^{5} e^{x}} + 1} + 3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2*exp(5+x)*exp(exp(5+x))+x**2*ln(5)*exp(x)+x)*exp(exp(exp(5+x))+exp(x)*ln(5)+1)+3*x+1)*exp(x*ex
p(exp(exp(5+x))+exp(x)*ln(5)+1)+3*x),x)

[Out]

x*exp(x*exp(exp(x)*log(5) + exp(exp(5)*exp(x)) + 1) + 3*x)

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