3.48.40 \(\int \frac {e^{\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}} (250 x+e^x (50 x+25 x^2) (i \pi +\log (3))+(-25 x+75 x^2) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x))}{\log (2)} \, dx\) [4740]

Optimal. Leaf size=31 \[ e^{\frac {25 x^2 \left (5+(i \pi +\log (3)) \left (e^x+x-\log (x)\right )\right )}{\log (2)}} \]

[Out]

exp(25*x^2*((x+exp(x)-ln(x))*(ln(3)+I*Pi)+5)/ln(2))

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Rubi [A]
time = 2.04, antiderivative size = 61, normalized size of antiderivative = 1.97, number of steps used = 2, number of rules used = 2, integrand size = 118, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {12, 6838} \begin {gather*} x^{-\frac {25 x^2 (\log (3)+i \pi )}{\log (2)}} \exp \left (\frac {25 \left (x^3 (\log (3)+i \pi )+5 x^2+e^x x^2 (\log (3)+i \pi )\right )}{\log (2)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((125*x^2 + 25*E^x*x^2*(I*Pi + Log[3]) + 25*x^3*(I*Pi + Log[3]) - 25*x^2*(I*Pi + Log[3])*Log[x])/Log[2]
)*(250*x + E^x*(50*x + 25*x^2)*(I*Pi + Log[3]) + (-25*x + 75*x^2)*(I*Pi + Log[3]) - 50*x*(I*Pi + Log[3])*Log[x
]))/Log[2],x]

[Out]

E^((25*(5*x^2 + E^x*x^2*(I*Pi + Log[3]) + x^3*(I*Pi + Log[3])))/Log[2])/x^((25*x^2*(I*Pi + Log[3]))/Log[2])

Rule 12

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

Rule 6838

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} &=\frac {\int \exp \left (\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}\right ) \left (250 x+e^x \left (50 x+25 x^2\right ) (i \pi +\log (3))+\left (-25 x+75 x^2\right ) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x)\right ) \, dx}{\log (2)}\\ &=\exp \left (\frac {25 \left (5 x^2+e^x x^2 (i \pi +\log (3))+x^3 (i \pi +\log (3))\right )}{\log (2)}\right ) x^{-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 2.57, size = 55, normalized size = 1.77 \begin {gather*} e^{\frac {25 x^2 \left (5+i \pi x+x \log (3)+e^x (i \pi +\log (3))\right )}{\log (2)}} x^{-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((125*x^2 + 25*E^x*x^2*(I*Pi + Log[3]) + 25*x^3*(I*Pi + Log[3]) - 25*x^2*(I*Pi + Log[3])*Log[x])/
Log[2])*(250*x + E^x*(50*x + 25*x^2)*(I*Pi + Log[3]) + (-25*x + 75*x^2)*(I*Pi + Log[3]) - 50*x*(I*Pi + Log[3])
*Log[x]))/Log[2],x]

[Out]

E^((25*x^2*(5 + I*Pi*x + x*Log[3] + E^x*(I*Pi + Log[3])))/Log[2])/x^((25*x^2*(I*Pi + Log[3]))/Log[2])

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Maple [A]
time = 1.88, size = 45, normalized size = 1.45

method	result	size
risch	\({\mathrm e}^{\frac {25 x^{2} \left (-i \pi \ln \left (x \right )+i {\mathrm e}^{x} \pi +i x \pi -\ln \left (3\right ) \ln \left (x \right )+\ln \left (3\right ) {\mathrm e}^{x}+x \ln \left (3\right )+5\right )}{\ln \left (2\right )}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-50*x*(ln(3)+I*Pi)*ln(x)+(25*x^2+50*x)*(ln(3)+I*Pi)*exp(x)+(75*x^2-25*x)*(ln(3)+I*Pi)+250*x)*exp((-25*x^2
*(ln(3)+I*Pi)*ln(x)+25*x^2*(ln(3)+I*Pi)*exp(x)+25*x^3*(ln(3)+I*Pi)+125*x^2)/ln(2))/ln(2),x,method=_RETURNVERBO
SE)

[Out]

exp(25*x^2*(-I*Pi*ln(x)+I*exp(x)*Pi+I*x*Pi-ln(3)*ln(x)+ln(3)*exp(x)+x*ln(3)+5)/ln(2))

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Maxima [A]
time = 0.73, size = 82, normalized size = 2.65 \begin {gather*} e^{\left (\frac {25 i \, \pi x^{3}}{\log \left (2\right )} + \frac {25 i \, \pi x^{2} e^{x}}{\log \left (2\right )} + \frac {25 \, x^{3} \log \left (3\right )}{\log \left (2\right )} + \frac {25 \, x^{2} e^{x} \log \left (3\right )}{\log \left (2\right )} - \frac {25 i \, \pi x^{2} \log \left (x\right )}{\log \left (2\right )} - \frac {25 \, x^{2} \log \left (3\right ) \log \left (x\right )}{\log \left (2\right )} + \frac {125 \, x^{2}}{\log \left (2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*x*(log(3)+I*pi)*log(x)+(25*x^2+50*x)*(log(3)+I*pi)*exp(x)+(75*x^2-25*x)*(log(3)+I*pi)+250*x)*ex
p((-25*x^2*(log(3)+I*pi)*log(x)+25*x^2*(log(3)+I*pi)*exp(x)+25*x^3*(log(3)+I*pi)+125*x^2)/log(2))/log(2),x, al
gorithm="maxima")

[Out]

e^(25*I*pi*x^3/log(2) + 25*I*pi*x^2*e^x/log(2) + 25*x^3*log(3)/log(2) + 25*x^2*e^x*log(3)/log(2) - 25*I*pi*x^2
*log(x)/log(2) - 25*x^2*log(3)*log(x)/log(2) + 125*x^2/log(2))

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Fricas [A]
time = 0.37, size = 82, normalized size = 2.65 \begin {gather*} e^{\left (\frac {25 i \, \pi x^{3}}{\log \left (2\right )} + \frac {25 i \, \pi x^{2} e^{x}}{\log \left (2\right )} + \frac {25 \, x^{3} \log \left (3\right )}{\log \left (2\right )} + \frac {25 \, x^{2} e^{x} \log \left (3\right )}{\log \left (2\right )} - \frac {25 i \, \pi x^{2} \log \left (x\right )}{\log \left (2\right )} - \frac {25 \, x^{2} \log \left (3\right ) \log \left (x\right )}{\log \left (2\right )} + \frac {125 \, x^{2}}{\log \left (2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*x*(log(3)+I*pi)*log(x)+(25*x^2+50*x)*(log(3)+I*pi)*exp(x)+(75*x^2-25*x)*(log(3)+I*pi)+250*x)*ex
p((-25*x^2*(log(3)+I*pi)*log(x)+25*x^2*(log(3)+I*pi)*exp(x)+25*x^3*(log(3)+I*pi)+125*x^2)/log(2))/log(2),x, al
gorithm="fricas")

[Out]

e^(25*I*pi*x^3/log(2) + 25*I*pi*x^2*e^x/log(2) + 25*x^3*log(3)/log(2) + 25*x^2*e^x*log(3)/log(2) - 25*I*pi*x^2
*log(x)/log(2) - 25*x^2*log(3)*log(x)/log(2) + 125*x^2/log(2))

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Sympy [F(-1)] Timed out
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((-50*x*(ln(3)+I*pi)*ln(x)+(25*x**2+50*x)*(ln(3)+I*pi)*exp(x)+(75*x**2-25*x)*(ln(3)+I*pi)+250*x)*exp(
(-25*x**2*(ln(3)+I*pi)*ln(x)+25*x**2*(ln(3)+I*pi)*exp(x)+25*x**3*(ln(3)+I*pi)+125*x**2)/ln(2))/ln(2),x)

[Out]

Timed out

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Giac [A]
time = 0.50, size = 82, normalized size = 2.65 \begin {gather*} e^{\left (\frac {25 i \, \pi x^{3}}{\log \left (2\right )} + \frac {25 i \, \pi x^{2} e^{x}}{\log \left (2\right )} + \frac {25 \, x^{3} \log \left (3\right )}{\log \left (2\right )} + \frac {25 \, x^{2} e^{x} \log \left (3\right )}{\log \left (2\right )} - \frac {25 i \, \pi x^{2} \log \left (x\right )}{\log \left (2\right )} - \frac {25 \, x^{2} \log \left (3\right ) \log \left (x\right )}{\log \left (2\right )} + \frac {125 \, x^{2}}{\log \left (2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*x*(log(3)+I*pi)*log(x)+(25*x^2+50*x)*(log(3)+I*pi)*exp(x)+(75*x^2-25*x)*(log(3)+I*pi)+250*x)*ex
p((-25*x^2*(log(3)+I*pi)*log(x)+25*x^2*(log(3)+I*pi)*exp(x)+25*x^3*(log(3)+I*pi)+125*x^2)/log(2))/log(2),x, al
gorithm="giac")

[Out]

e^(25*I*pi*x^3/log(2) + 25*I*pi*x^2*e^x/log(2) + 25*x^3*log(3)/log(2) + 25*x^2*e^x*log(3)/log(2) - 25*I*pi*x^2
*log(x)/log(2) - 25*x^2*log(3)*log(x)/log(2) + 125*x^2/log(2))

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Mupad [B]
time = 4.54, size = 78, normalized size = 2.52 \begin {gather*} \frac {{847288609443}^{\frac {x^2\,{\mathrm {e}}^x+x^3}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {125\,x^2}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {\Pi \,x^2\,{\mathrm {e}}^x\,25{}\mathrm {i}}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {\Pi \,x^3\,25{}\mathrm {i}}{\ln \left (2\right )}}}{x^{\frac {25\,x^2\,\ln \left (3\right )+\Pi \,x^2\,25{}\mathrm {i}}{\ln \left (2\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((25*x^3*(Pi*1i + log(3)) + 125*x^2 + 25*x^2*exp(x)*(Pi*1i + log(3)) - 25*x^2*log(x)*(Pi*1i + log(3)))
/log(2))*(250*x - (25*x - 75*x^2)*(Pi*1i + log(3)) + exp(x)*(50*x + 25*x^2)*(Pi*1i + log(3)) - 50*x*log(x)*(Pi
*1i + log(3))))/log(2),x)

[Out]

(847288609443^((x^2*exp(x) + x^3)/log(2))*exp((125*x^2)/log(2))*exp((Pi*x^2*exp(x)*25i)/log(2))*exp((Pi*x^3*25
i)/log(2)))/x^((Pi*x^2*25i + 25*x^2*log(3))/log(2))

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