3.64.5 \(\int e^{2+(25-5 x) \log ^2(2 x)} (1+(50-10 x) \log (2 x)-5 x \log ^2(2 x)) \, dx\) [6305]

Optimal. Leaf size=19 \[ e^{2+5 (5-x) \log ^2(2 x)} x \]

[Out]

exp(5*ln(2*x)^2*(5-x)+2)*x

________________________________________________________________________________________

Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(64\) vs. \(2(19)=38\).
time = 0.11, antiderivative size = 64, normalized size of antiderivative = 3.37, number of steps used = 1, number of rules used = 1, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2326} \begin {gather*} \frac {e^{5 (5-x) \log ^2(2 x)+2} \left (2 (5-x) \log (2 x)-x \log ^2(2 x)\right )}{\frac {2 (5-x) \log (2 x)}{x}-\log ^2(2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2 + (25 - 5*x)*Log[2*x]^2)*(1 + (50 - 10*x)*Log[2*x] - 5*x*Log[2*x]^2),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.10, size = 17, normalized size = 0.89 \begin {gather*} e^{2-5 (-5+x) \log ^2(2 x)} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2 + (25 - 5*x)*Log[2*x]^2)*(1 + (50 - 10*x)*Log[2*x] - 5*x*Log[2*x]^2),x]

[Out]

E^(2 - 5*(-5 + x)*Log[2*x]^2)*x

________________________________________________________________________________________

Maple [A]
time = 0.22, size = 18, normalized size = 0.95

method	result	size
norman	\({\mathrm e}^{\left (-5 x +25\right ) \ln \left (2 x \right )^{2}+2} x\)	\(18\)
risch	\({\mathrm e}^{-5 x \ln \left (2 x \right )^{2}+25 \ln \left (2 x \right )^{2}+2} x\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*x*ln(2*x)^2+(-10*x+50)*ln(2*x)+1)*exp((-5*x+25)*ln(2*x)^2+2),x,method=_RETURNVERBOSE)

[Out]

exp((-5*x+25)*ln(2*x)^2+2)*x

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (16) = 32\).
time = 0.71, size = 44, normalized size = 2.32 \begin {gather*} x e^{\left (-5 \, x \log \left (2\right )^{2} - 10 \, x \log \left (2\right ) \log \left (x\right ) - 5 \, x \log \left (x\right )^{2} + 25 \, \log \left (2\right )^{2} + 50 \, \log \left (2\right ) \log \left (x\right ) + 25 \, \log \left (x\right )^{2} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*log(2*x)^2+(-10*x+50)*log(2*x)+1)*exp((-5*x+25)*log(2*x)^2+2),x, algorithm="maxima")

[Out]

x*e^(-5*x*log(2)^2 - 10*x*log(2)*log(x) - 5*x*log(x)^2 + 25*log(2)^2 + 50*log(2)*log(x) + 25*log(x)^2 + 2)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 16, normalized size = 0.84 \begin {gather*} x e^{\left (-5 \, {\left (x - 5\right )} \log \left (2 \, x\right )^{2} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*log(2*x)^2+(-10*x+50)*log(2*x)+1)*exp((-5*x+25)*log(2*x)^2+2),x, algorithm="fricas")

[Out]

x*e^(-5*(x - 5)*log(2*x)^2 + 2)

________________________________________________________________________________________

Sympy [A]
time = 0.14, size = 15, normalized size = 0.79 \begin {gather*} x e^{\left (25 - 5 x\right ) \log {\left (2 x \right )}^{2} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*ln(2*x)**2+(-10*x+50)*ln(2*x)+1)*exp((-5*x+25)*ln(2*x)**2+2),x)

[Out]

x*exp((25 - 5*x)*log(2*x)**2 + 2)

________________________________________________________________________________________

Giac [A]
time = 0.46, size = 22, normalized size = 1.16 \begin {gather*} x e^{\left (-5 \, x \log \left (2 \, x\right )^{2} + 25 \, \log \left (2 \, x\right )^{2} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*log(2*x)^2+(-10*x+50)*log(2*x)+1)*exp((-5*x+25)*log(2*x)^2+2),x, algorithm="giac")

[Out]

x*e^(-5*x*log(2*x)^2 + 25*log(2*x)^2 + 2)

________________________________________________________________________________________

Mupad [B]
time = 4.37, size = 46, normalized size = 2.42 \begin {gather*} x^{50\,\ln \left (2\right )-10\,x\,\ln \left (2\right )+1}\,{\mathrm {e}}^{-5\,x\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{25\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^2\,{\mathrm {e}}^{-5\,x\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{25\,{\ln \left (2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(2 - log(2*x)^2*(5*x - 25))*(5*x*log(2*x)^2 + log(2*x)*(10*x - 50) - 1),x)

[Out]

x^(50*log(2) - 10*x*log(2) + 1)*exp(-5*x*log(2)^2)*exp(25*log(x)^2)*exp(2)*exp(-5*x*log(x)^2)*exp(25*log(2)^2)

________________________________________________________________________________________