3.38.41 \(\int \frac {-e-x^2+e^{3 x+e^{3 x} x} (-5 x^2-15 x^3)+(-e-x^2) \log (4)}{x^2+x^2 \log (4)} \, dx\)

Optimal. Leaf size=28 \[ \frac {e}{x}-x+\frac {5 e^{e^{3 x} x}}{-1-\log (4)} \]

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Rubi [F]  time = 0.16, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-e-x^2+e^{3 x+e^{3 x} x} \left (-5 x^2-15 x^3\right )+\left (-e-x^2\right ) \log (4)}{x^2+x^2 \log (4)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-E - x^2 + E^(3*x + E^(3*x)*x)*(-5*x^2 - 15*x^3) + (-E - x^2)*Log[4])/(x^2 + x^2*Log[4]),x]

[Out]

E/x - x - (5*Defer[Int][E^(3*x + E^(3*x)*x), x])/(1 + Log[4]) - (15*Defer[Int][E^(3*x + E^(3*x)*x)*x, x])/(1 +
Log[4])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e-x^2+e^{3 x+e^{3 x} x} \left (-5 x^2-15 x^3\right )+\left (-e-x^2\right ) \log (4)}{x^2 (1+\log (4))} \, dx\\ &=\frac {\int \frac {-e-x^2+e^{3 x+e^{3 x} x} \left (-5 x^2-15 x^3\right )+\left (-e-x^2\right ) \log (4)}{x^2} \, dx}{1+\log (4)}\\ &=\frac {\int \left (5 e^{3 x+e^{3 x} x} (-1-3 x)-\frac {\left (e+x^2\right ) (1+\log (4))}{x^2}\right ) \, dx}{1+\log (4)}\\ &=\frac {5 \int e^{3 x+e^{3 x} x} (-1-3 x) \, dx}{1+\log (4)}-\int \frac {e+x^2}{x^2} \, dx\\ &=\frac {5 \int \left (-e^{3 x+e^{3 x} x}-3 e^{3 x+e^{3 x} x} x\right ) \, dx}{1+\log (4)}-\int \left (1+\frac {e}{x^2}\right ) \, dx\\ &=\frac {e}{x}-x-\frac {5 \int e^{3 x+e^{3 x} x} \, dx}{1+\log (4)}-\frac {15 \int e^{3 x+e^{3 x} x} x \, dx}{1+\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 26, normalized size = 0.93 \begin {gather*} \frac {e}{x}-x-\frac {5 e^{e^{3 x} x}}{1+\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E - x^2 + E^(3*x + E^(3*x)*x)*(-5*x^2 - 15*x^3) + (-E - x^2)*Log[4])/(x^2 + x^2*Log[4]),x]

[Out]

E/x - x - (5*E^(E^(3*x)*x))/(1 + Log[4])

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fricas [B]  time = 1.12, size = 55, normalized size = 1.96 \begin {gather*} -\frac {{\left (5 \, x e^{\left (x e^{\left (3 \, x\right )} + 3 \, x\right )} + {\left (x^{2} + 2 \, {\left (x^{2} - e\right )} \log \relax (2) - e\right )} e^{\left (3 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{2 \, x \log \relax (2) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3-5*x^2)*exp(3*x)*exp(x*exp(3*x))+2*(-exp(1)-x^2)*log(2)-exp(1)-x^2)/(2*x^2*log(2)+x^2),x, a
lgorithm="fricas")

[Out]

-(5*x*e^(x*e^(3*x) + 3*x) + (x^2 + 2*(x^2 - e)*log(2) - e)*e^(3*x))*e^(-3*x)/(2*x*log(2) + x)

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giac [B]  time = 0.18, size = 72, normalized size = 2.57 \begin {gather*} -\frac {2 \, x^{2} e^{\left (3 \, x\right )} \log \relax (2) + x^{2} e^{\left (3 \, x\right )} + 5 \, x e^{\left (x e^{\left (3 \, x\right )} + 3 \, x\right )} - 2 \, e^{\left (3 \, x + 1\right )} \log \relax (2) - e^{\left (3 \, x + 1\right )}}{2 \, x e^{\left (3 \, x\right )} \log \relax (2) + x e^{\left (3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3-5*x^2)*exp(3*x)*exp(x*exp(3*x))+2*(-exp(1)-x^2)*log(2)-exp(1)-x^2)/(2*x^2*log(2)+x^2),x, a
lgorithm="giac")

[Out]

-(2*x^2*e^(3*x)*log(2) + x^2*e^(3*x) + 5*x*e^(x*e^(3*x) + 3*x) - 2*e^(3*x + 1)*log(2) - e^(3*x + 1))/(2*x*e^(3
*x)*log(2) + x*e^(3*x))

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maple [A]  time = 0.14, size = 28, normalized size = 1.00

method	result	size
risch	\(-x +\frac {{\mathrm e}}{x}-\frac {5 \,{\mathrm e}^{x \,{\mathrm e}^{3 x}}}{1+2 \ln \relax (2)}\)	\(28\)
norman	\(\frac {-x^{2}-\frac {5 x \,{\mathrm e}^{x \,{\mathrm e}^{3 x}}}{1+2 \ln \relax (2)}+{\mathrm e}}{x}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x^3-5*x^2)*exp(3*x)*exp(x*exp(3*x))+2*(-exp(1)-x^2)*ln(2)-exp(1)-x^2)/(2*x^2*ln(2)+x^2),x,method=_RE
TURNVERBOSE)

[Out]

-x+exp(1)/x-5/(1+2*ln(2))*exp(x*exp(3*x))

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maxima [B]  time = 0.49, size = 73, normalized size = 2.61 \begin {gather*} -\frac {2 \, x \log \relax (2)}{2 \, \log \relax (2) + 1} - \frac {x}{2 \, \log \relax (2) + 1} - \frac {5 \, e^{\left (x e^{\left (3 \, x\right )}\right )}}{2 \, \log \relax (2) + 1} + \frac {2 \, e \log \relax (2)}{x {\left (2 \, \log \relax (2) + 1\right )}} + \frac {e}{x {\left (2 \, \log \relax (2) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3-5*x^2)*exp(3*x)*exp(x*exp(3*x))+2*(-exp(1)-x^2)*log(2)-exp(1)-x^2)/(2*x^2*log(2)+x^2),x, a
lgorithm="maxima")

[Out]

-2*x*log(2)/(2*log(2) + 1) - x/(2*log(2) + 1) - 5*e^(x*e^(3*x))/(2*log(2) + 1) + 2*e*log(2)/(x*(2*log(2) + 1))
+ e/(x*(2*log(2) + 1))

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mupad [B]  time = 0.37, size = 25, normalized size = 0.89 \begin {gather*} \frac {\mathrm {e}}{x}-\frac {5\,{\mathrm {e}}^{x\,{\mathrm {e}}^{3\,x}}}{\ln \relax (4)+1}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1) + 2*log(2)*(exp(1) + x^2) + x^2 + exp(3*x)*exp(x*exp(3*x))*(5*x^2 + 15*x^3))/(2*x^2*log(2) + x^2)
,x)

[Out]

exp(1)/x - (5*exp(x*exp(3*x)))/(log(4) + 1) - x

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sympy [A]  time = 0.25, size = 22, normalized size = 0.79 \begin {gather*} - x - \frac {5 e^{x e^{3 x}}}{1 + 2 \log {\relax (2 )}} + \frac {e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x**3-5*x**2)*exp(3*x)*exp(x*exp(3*x))+2*(-exp(1)-x**2)*ln(2)-exp(1)-x**2)/(2*x**2*ln(2)+x**2),
x)

[Out]

-x - 5*exp(x*exp(3*x))/(1 + 2*log(2)) + E/x

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