3.86.99 \(\int \frac {e^x (e^{e^{e^x} x^{5/4}}+x) (4+4 x+e^{e^{e^x} x^{5/4}} (4+e^{e^x} \sqrt [4]{x} (5+4 e^x x)))}{4 e^{e^{e^x} x^{5/4}}+4 x} \, dx\)

Optimal. Leaf size=19 \[ e^x \left (e^{e^{e^x} x^{5/4}}+x\right ) \]

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Rubi [A]  time = 1.96, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {2288} \begin {gather*} e^x \left (e^{e^{e^x} x^{5/4}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(E^(E^E^x*x^(5/4)) + x)*(4 + 4*x + E^(E^E^x*x^(5/4))*(4 + E^E^x*x^(1/4)*(5 + 4*E^x*x))))/(4*E^(E^E^x*
x^(5/4)) + 4*x),x]

[Out]

E^x*(E^(E^E^x*x^(5/4)) + x)

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} &=\operatorname {Subst}\left (\int e^{x^4} x^3 \left (4+4 e^{e^{e^{x^4}} x^5}+5 e^{e^{x^4}+e^{e^{x^4}} x^5} x+4 x^4+4 e^{e^{x^4}+x^4+e^{e^{x^4}} x^5} x^5\right ) \, dx,x,\sqrt [4]{x}\right )\\ &=e^x \left (e^{e^{e^x} x^{5/4}}+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 19, normalized size = 1.00 \begin {gather*} e^x \left (e^{e^{e^x} x^{5/4}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(E^(E^E^x*x^(5/4)) + x)*(4 + 4*x + E^(E^E^x*x^(5/4))*(4 + E^E^x*x^(1/4)*(5 + 4*E^x*x))))/(4*E^(
E^E^x*x^(5/4)) + 4*x),x]

[Out]

E^x*(E^(E^E^x*x^(5/4)) + x)

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fricas [A]  time = 0.69, size = 15, normalized size = 0.79 \begin {gather*} x e^{x} + e^{\left (x^{\frac {5}{4}} e^{\left (e^{x}\right )} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*exp(x)*x+5)*exp(exp(x))*x^(1/4)+4)*exp(x^(5/4)*exp(exp(x)))+4*x+4)*exp(log(exp(x^(5/4)*exp(exp(
x)))+x)+x)/(4*exp(x^(5/4)*exp(exp(x)))+4*x),x, algorithm="fricas")

[Out]

x*e^x + e^(x^(5/4)*e^(e^x) + x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left ({\left (4 \, x e^{x} + 5\right )} x^{\frac {1}{4}} e^{\left (e^{x}\right )} + 4\right )} e^{\left (x^{\frac {5}{4}} e^{\left (e^{x}\right )}\right )} + 4 \, x + 4\right )} e^{\left (x + \log \left (x + e^{\left (x^{\frac {5}{4}} e^{\left (e^{x}\right )}\right )}\right )\right )}}{4 \, {\left (x + e^{\left (x^{\frac {5}{4}} e^{\left (e^{x}\right )}\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*exp(x)*x+5)*exp(exp(x))*x^(1/4)+4)*exp(x^(5/4)*exp(exp(x)))+4*x+4)*exp(log(exp(x^(5/4)*exp(exp(
x)))+x)+x)/(4*exp(x^(5/4)*exp(exp(x)))+4*x),x, algorithm="giac")

[Out]

integrate(1/4*(((4*x*e^x + 5)*x^(1/4)*e^(e^x) + 4)*e^(x^(5/4)*e^(e^x)) + 4*x + 4)*e^(x + log(x + e^(x^(5/4)*e^
(e^x))))/(x + e^(x^(5/4)*e^(e^x))), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (4 \,{\mathrm e}^{x} x +5\right ) {\mathrm e}^{{\mathrm e}^{x}} x^{\frac {1}{4}}+4\right ) {\mathrm e}^{x^{\frac {5}{4}} {\mathrm e}^{{\mathrm e}^{x}}}+4 x +4\right ) {\mathrm e}^{\ln \left ({\mathrm e}^{x^{\frac {5}{4}} {\mathrm e}^{{\mathrm e}^{x}}}+x \right )+x}}{4 \,{\mathrm e}^{x^{\frac {5}{4}} {\mathrm e}^{{\mathrm e}^{x}}}+4 x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*exp(x)*x+5)*exp(exp(x))*x^(1/4)+4)*exp(x^(5/4)*exp(exp(x)))+4*x+4)*exp(ln(exp(x^(5/4)*exp(exp(x)))+x)
+x)/(4*exp(x^(5/4)*exp(exp(x)))+4*x),x)

[Out]

int((((4*exp(x)*x+5)*exp(exp(x))*x^(1/4)+4)*exp(x^(5/4)*exp(exp(x)))+4*x+4)*exp(ln(exp(x^(5/4)*exp(exp(x)))+x)
+x)/(4*exp(x^(5/4)*exp(exp(x)))+4*x),x)

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maxima [A]  time = 0.48, size = 19, normalized size = 1.00 \begin {gather*} {\left (x - 1\right )} e^{x} + e^{\left (x^{\frac {5}{4}} e^{\left (e^{x}\right )} + x\right )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*exp(x)*x+5)*exp(exp(x))*x^(1/4)+4)*exp(x^(5/4)*exp(exp(x)))+4*x+4)*exp(log(exp(x^(5/4)*exp(exp(
x)))+x)+x)/(4*exp(x^(5/4)*exp(exp(x)))+4*x),x, algorithm="maxima")

[Out]

(x - 1)*e^x + e^(x^(5/4)*e^(e^x) + x) + e^x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{x+\ln \left (x+{\mathrm {e}}^{x^{5/4}\,{\mathrm {e}}^{{\mathrm {e}}^x}}\right )}\,\left (4\,x+{\mathrm {e}}^{x^{5/4}\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,\left (x^{1/4}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (4\,x\,{\mathrm {e}}^x+5\right )+4\right )+4\right )}{4\,x+4\,{\mathrm {e}}^{x^{5/4}\,{\mathrm {e}}^{{\mathrm {e}}^x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(x + exp(x^(5/4)*exp(exp(x)))))*(4*x + exp(x^(5/4)*exp(exp(x)))*(x^(1/4)*exp(exp(x))*(4*x*exp(
x) + 5) + 4) + 4))/(4*x + 4*exp(x^(5/4)*exp(exp(x)))),x)

[Out]

int((exp(x + log(x + exp(x^(5/4)*exp(exp(x)))))*(4*x + exp(x^(5/4)*exp(exp(x)))*(x^(1/4)*exp(exp(x))*(4*x*exp(
x) + 5) + 4) + 4))/(4*x + 4*exp(x^(5/4)*exp(exp(x)))), x)

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sympy [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((((4*exp(x)*x+5)*exp(exp(x))*x**(1/4)+4)*exp(x**(5/4)*exp(exp(x)))+4*x+4)*exp(ln(exp(x**(5/4)*exp(ex
p(x)))+x)+x)/(4*exp(x**(5/4)*exp(exp(x)))+4*x),x)

[Out]

Timed out

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