\(\int \frac {e^{e^x} (96+32 x+e^x (96 x+16 x^2))}{i \pi +\log (25)} \, dx\) [7492]

Optimal result

Integrand size = 34, antiderivative size = 23 \[ \int \frac {e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right )}{i \pi +\log (25)} \, dx=-3+\frac {16 e^{e^x} x (6+x)}{i \pi +\log (25)} \]

[Out]

16*exp(exp(x))*x*(6+x)/(2*ln(5)+I*Pi)-3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 2326} \[ \int \frac {e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right )}{i \pi +\log (25)} \, dx=\frac {16 e^{e^x} \left (x^2+6 x\right )}{\log (25)+i \pi } \]

[In]

Int[(E^E^x*(96 + 32*x + E^x*(96*x + 16*x^2)))/(I*Pi + Log[25]),x]

[Out]

(16*E^E^x*(6*x + x^2))/(I*Pi + Log[25])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2326

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{align*} \text {integral}& = \frac {\int e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right ) \, dx}{i \pi +\log (25)} \\ & = \frac {16 e^{e^x} \left (6 x+x^2\right )}{i \pi +\log (25)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right )}{i \pi +\log (25)} \, dx=\frac {16 e^{e^x} x (6+x)}{i \pi +\log (25)} \]

[In]

Integrate[(E^E^x*(96 + 32*x + E^x*(96*x + 16*x^2)))/(I*Pi + Log[25]),x]

[Out]

(16*E^E^x*x*(6 + x))/(I*Pi + Log[25])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
risch	\(\frac {\left (16 x^{2}+96 x \right ) {\mathrm e}^{{\mathrm e}^{x}}}{2 \ln \left (5\right )+i \pi }\)	\(25\)
parallelrisch	\(\frac {16 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}+96 x \,{\mathrm e}^{{\mathrm e}^{x}}}{2 \ln \left (5\right )+i \pi }\)	\(28\)
norman	\(-\frac {96 \left (i \pi -2 \ln \left (5\right )\right ) x \,{\mathrm e}^{{\mathrm e}^{x}}}{\pi ^{2}+4 \ln \left (5\right )^{2}}-\frac {16 \left (i \pi -2 \ln \left (5\right )\right ) x^{2} {\mathrm e}^{{\mathrm e}^{x}}}{\pi ^{2}+4 \ln \left (5\right )^{2}}\)	\(58\)

[In]

int(((16*x^2+96*x)*exp(x)+32*x+96)*exp(exp(x))/(2*ln(5)+I*Pi),x,method=_RETURNVERBOSE)

[Out]

1/(2*ln(5)+I*Pi)*(16*x^2+96*x)*exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right )}{i \pi +\log (25)} \, dx=\frac {16 \, {\left (x^{2} + 6 \, x\right )} e^{\left (e^{x}\right )}}{i \, \pi + 2 \, \log \left (5\right )} \]

[In]

integrate(((16*x^2+96*x)*exp(x)+32*x+96)*exp(exp(x))/(2*log(5)+I*pi),x, algorithm="fricas")

[Out]

16*(x^2 + 6*x)*e^(e^x)/(I*pi + 2*log(5))

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right )}{i \pi +\log (25)} \, dx=\frac {\left (16 x^{2} + 96 x\right ) e^{e^{x}}}{2 \log {\left (5 \right )} + i \pi } \]

[In]

integrate(((16*x**2+96*x)*exp(x)+32*x+96)*exp(exp(x))/(2*ln(5)+I*pi),x)

[Out]

(16*x**2 + 96*x)*exp(exp(x))/(2*log(5) + I*pi)

Maxima [F]

\[ \int \frac {e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right )}{i \pi +\log (25)} \, dx=\int { \frac {16 \, {\left ({\left (x^{2} + 6 \, x\right )} e^{x} + 2 \, x + 6\right )} e^{\left (e^{x}\right )}}{i \, \pi + 2 \, \log \left (5\right )} \,d x } \]

[In]

integrate(((16*x^2+96*x)*exp(x)+32*x+96)*exp(exp(x))/(2*log(5)+I*pi),x, algorithm="maxima")

[Out]

16*((x^2 + 6*x)*e^(e^x) + 6*Ei(e^x) - 6*integrate(e^(e^x), x))/(I*pi + 2*log(5))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right )}{i \pi +\log (25)} \, dx=\frac {16 \, {\left (x^{2} e^{\left (x + e^{x}\right )} + 6 \, x e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}}{i \, \pi + 2 \, \log \left (5\right )} \]

[In]

integrate(((16*x^2+96*x)*exp(x)+32*x+96)*exp(exp(x))/(2*log(5)+I*pi),x, algorithm="giac")

[Out]

16*(x^2*e^(x + e^x) + 6*x*e^(x + e^x))*e^(-x)/(I*pi + 2*log(5))

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^{e^x} \left (96+32 x+e^x \left (96 x+16 x^2\right )\right )}{i \pi +\log (25)} \, dx=-\frac {16\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x+6\right )\,\left (-\ln \left (25\right )+\Pi \,1{}\mathrm {i}\right )}{\Pi ^2+4\,{\ln \left (5\right )}^2} \]

[In]

int((exp(exp(x))*(32*x + exp(x)*(96*x + 16*x^2) + 96))/(Pi*1i + 2*log(5)),x)

[Out]

-(16*x*exp(exp(x))*(x + 6)*(Pi*1i - log(25)))/(Pi^2 + 4*log(5)^2)