3.74.34 \(\int \frac {e^{1+e^{2 x^2}-x} (-1+4 e^{2 x^2} x)}{e^{1+e^{2 x^2}-x}-15625 e^{9+\log ^2(5)}} \, dx\)

Optimal. Leaf size=26 \[ \log \left (-e^{1+e^{2 x^2}-x}+e^{(3+\log (5))^2}\right ) \]

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Rubi [A]  time = 0.28, antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 1, number of rules used = 1, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6684} \begin {gather*} \log \left (-e^{1-x} \left (e^{e^{2 x^2}}-15625 e^{x+8+\log ^2(5)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 + E^(2*x^2) - x)*(-1 + 4*E^(2*x^2)*x))/(E^(1 + E^(2*x^2) - x) - 15625*E^(9 + Log[5]^2)),x]

[Out]

Log[-(E^(1 - x)*(E^E^(2*x^2) - 15625*E^(8 + x + Log[5]^2)))]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (-e^{1-x} \left (e^{e^{2 x^2}}-15625 e^{8+x+\log ^2(5)}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 26, normalized size = 1.00 \begin {gather*} -x+\log \left (e^{e^{2 x^2}}-15625 e^{8+x+\log ^2(5)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 + E^(2*x^2) - x)*(-1 + 4*E^(2*x^2)*x))/(E^(1 + E^(2*x^2) - x) - 15625*E^(9 + Log[5]^2)),x]

[Out]

-x + Log[E^E^(2*x^2) - 15625*E^(8 + x + Log[5]^2)]

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fricas [A]  time = 0.76, size = 27, normalized size = 1.04 \begin {gather*} \log \left (-e^{\left (\log \relax (5)^{2} + 6 \, \log \relax (5) + 9\right )} + e^{\left (-x + e^{\left (2 \, x^{2}\right )} + 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(2*x^2)-1)*exp(exp(2*x^2)-x+1)/(exp(exp(2*x^2)-x+1)-exp(log(5)^2+6*log(5)+9)),x, algorithm="
fricas")

[Out]

log(-e^(log(5)^2 + 6*log(5) + 9) + e^(-x + e^(2*x^2) + 1))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(2*x^2)-1)*exp(exp(2*x^2)-x+1)/(exp(exp(2*x^2)-x+1)-exp(log(5)^2+6*log(5)+9)),x, algorithm="
giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to isolate sageVARx in 4*sageVARx
*exp(2*sageVARx^2)-1, switching to approx. solutionsSolving by bisection with change of variable x=tan(t) and
t=-1.57..1.57. Try fs

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maple [A]  time = 0.05, size = 26, normalized size = 1.00

method	result	size
norman	\(\ln \left (-{\mathrm e}^{{\mathrm e}^{2 x^{2}}-x +1}+15625 \,{\mathrm e}^{\ln \relax (5)^{2}+9}\right )\)	\(26\)
risch	\(-1+\ln \left ({\mathrm e}^{{\mathrm e}^{2 x^{2}}-x +1}-15625 \,{\mathrm e}^{\ln \relax (5)^{2}+9}\right )\)	\(26\)
derivativedivides	\(\ln \left ({\mathrm e}^{{\mathrm e}^{2 x^{2}}-x +1}-{\mathrm e}^{\ln \relax (5)^{2}+6 \ln \relax (5)+9}\right )\)	\(28\)
default	\(\ln \left ({\mathrm e}^{{\mathrm e}^{2 x^{2}}-x +1}-{\mathrm e}^{\ln \relax (5)^{2}+6 \ln \relax (5)+9}\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*exp(2*x^2)-1)*exp(exp(2*x^2)-x+1)/(exp(exp(2*x^2)-x+1)-exp(ln(5)^2+6*ln(5)+9)),x,method=_RETURNVERBOS
E)

[Out]

ln(-exp(exp(2*x^2)-x+1)+15625*exp(ln(5)^2+9))

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maxima [A]  time = 0.36, size = 25, normalized size = 0.96 \begin {gather*} \log \left (15625 \, e^{\left (\log \relax (5)^{2} + 9\right )} - e^{\left (-x + e^{\left (2 \, x^{2}\right )} + 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(2*x^2)-1)*exp(exp(2*x^2)-x+1)/(exp(exp(2*x^2)-x+1)-exp(log(5)^2+6*log(5)+9)),x, algorithm="
maxima")

[Out]

log(15625*e^(log(5)^2 + 9) - e^(-x + e^(2*x^2) + 1))

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mupad [B]  time = 0.22, size = 23, normalized size = 0.88 \begin {gather*} \ln \left ({\mathrm {e}}^{{\mathrm {e}}^{2\,x^2}-x+1}-15625\,{\mathrm {e}}^{{\ln \relax (5)}^2+9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(2*x^2) - x + 1)*(4*x*exp(2*x^2) - 1))/(exp(exp(2*x^2) - x + 1) - exp(6*log(5) + log(5)^2 + 9)),x)

[Out]

log(exp(exp(2*x^2) - x + 1) - 15625*exp(log(5)^2 + 9))

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sympy [A]  time = 0.35, size = 24, normalized size = 0.92 \begin {gather*} \log {\left (e^{- x + e^{2 x^{2}} + 1} - 15625 e^{9} e^{\log {\relax (5 )}^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(2*x**2)-1)*exp(exp(2*x**2)-x+1)/(exp(exp(2*x**2)-x+1)-exp(ln(5)**2+6*ln(5)+9)),x)

[Out]

log(exp(-x + exp(2*x**2) + 1) - 15625*exp(9)*exp(log(5)**2))

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