3.86.28 \(\int \frac {-3+x-3 x^2+2 x^3+e^3 (-1+2 x)+e^{2 x} (x+x^2)+e^x (-2 x+2 x^2+x^3+e^3 (1+x))}{10+2 e^6-12 x+2 x^2+2 e^{2 x} x^2-4 x^3+2 x^4+e^3 (-4 x+4 x^2)+e^x (4 e^3 x-4 x^2+4 x^3)} \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{4} \log \left (-5 (1-x)+x-\left (-e^3+x-x \left (e^x+x\right )\right )^2\right ) \]

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Rubi [B]  time = 0.73, antiderivative size = 70, normalized size of antiderivative = 2.12, number of steps used = 3, number of rules used = 3, integrand size = 128, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6741, 12, 6684} \begin {gather*} \frac {1}{4} \log \left (x^4+2 e^x x^3-2 x^3-2 e^x x^2+e^{2 x} x^2+\left (1+2 e^3\right ) x^2+2 e^{x+3} x-2 \left (3+e^3\right ) x+e^6+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + x - 3*x^2 + 2*x^3 + E^3*(-1 + 2*x) + E^(2*x)*(x + x^2) + E^x*(-2*x + 2*x^2 + x^3 + E^3*(1 + x)))/(10
+ 2*E^6 - 12*x + 2*x^2 + 2*E^(2*x)*x^2 - 4*x^3 + 2*x^4 + E^3*(-4*x + 4*x^2) + E^x*(4*E^3*x - 4*x^2 + 4*x^3)),
x]

[Out]

Log[5 + E^6 + 2*E^(3 + x)*x - 2*(3 + E^3)*x - 2*E^x*x^2 + E^(2*x)*x^2 + (1 + 2*E^3)*x^2 - 2*x^3 + 2*E^x*x^3 +
x^4]/4

Rule 12

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

Rule 6684

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

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3+x-3 x^2+2 x^3+e^3 (-1+2 x)+e^{2 x} \left (x+x^2\right )+e^x \left (-2 x+2 x^2+x^3+e^3 (1+x)\right )}{2 \left (5 \left (1+\frac {e^6}{5}\right )+2 e^{3+x} x-6 \left (1+\frac {e^3}{3}\right ) x-2 e^x x^2+e^{2 x} x^2+\left (1+2 e^3\right ) x^2-2 x^3+2 e^x x^3+x^4\right )} \, dx\\ &=\frac {1}{2} \int \frac {-3+x-3 x^2+2 x^3+e^3 (-1+2 x)+e^{2 x} \left (x+x^2\right )+e^x \left (-2 x+2 x^2+x^3+e^3 (1+x)\right )}{5 \left (1+\frac {e^6}{5}\right )+2 e^{3+x} x-6 \left (1+\frac {e^3}{3}\right ) x-2 e^x x^2+e^{2 x} x^2+\left (1+2 e^3\right ) x^2-2 x^3+2 e^x x^3+x^4} \, dx\\ &=\frac {1}{4} \log \left (5+e^6+2 e^{3+x} x-2 \left (3+e^3\right ) x-2 e^x x^2+e^{2 x} x^2+\left (1+2 e^3\right ) x^2-2 x^3+2 e^x x^3+x^4\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.50, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-3+x-3 x^2+2 x^3+e^3 (-1+2 x)+e^{2 x} \left (x+x^2\right )+e^x \left (-2 x+2 x^2+x^3+e^3 (1+x)\right )}{10+2 e^6-12 x+2 x^2+2 e^{2 x} x^2-4 x^3+2 x^4+e^3 \left (-4 x+4 x^2\right )+e^x \left (4 e^3 x-4 x^2+4 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(-3 + x - 3*x^2 + 2*x^3 + E^3*(-1 + 2*x) + E^(2*x)*(x + x^2) + E^x*(-2*x + 2*x^2 + x^3 + E^3*(1 + x)
))/(10 + 2*E^6 - 12*x + 2*x^2 + 2*E^(2*x)*x^2 - 4*x^3 + 2*x^4 + E^3*(-4*x + 4*x^2) + E^x*(4*E^3*x - 4*x^2 + 4*
x^3)),x]

[Out]

Integrate[(-3 + x - 3*x^2 + 2*x^3 + E^3*(-1 + 2*x) + E^(2*x)*(x + x^2) + E^x*(-2*x + 2*x^2 + x^3 + E^3*(1 + x)
))/(10 + 2*E^6 - 12*x + 2*x^2 + 2*E^(2*x)*x^2 - 4*x^3 + 2*x^4 + E^3*(-4*x + 4*x^2) + E^x*(4*E^3*x - 4*x^2 + 4*
x^3)), x]

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fricas [B]  time = 0.63, size = 66, normalized size = 2.00 \begin {gather*} \frac {1}{2} \, \log \relax (x) + \frac {1}{4} \, \log \left (\frac {x^{4} - 2 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + x^{2} + 2 \, {\left (x^{2} - x\right )} e^{3} + 2 \, {\left (x^{3} - x^{2} + x e^{3}\right )} e^{x} - 6 \, x + e^{6} + 5}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+x)*exp(x)^2+((x+1)*exp(3)+x^3+2*x^2-2*x)*exp(x)+(2*x-1)*exp(3)+2*x^3-3*x^2+x-3)/(2*exp(x)^2*x^
2+(4*x*exp(3)+4*x^3-4*x^2)*exp(x)+2*exp(3)^2+(4*x^2-4*x)*exp(3)+2*x^4-4*x^3+2*x^2-12*x+10),x, algorithm="frica
s")

[Out]

1/2*log(x) + 1/4*log((x^4 - 2*x^3 + x^2*e^(2*x) + x^2 + 2*(x^2 - x)*e^3 + 2*(x^3 - x^2 + x*e^3)*e^x - 6*x + e^
6 + 5)/x^2)

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giac [B]  time = 0.27, size = 62, normalized size = 1.88 \begin {gather*} \frac {1}{4} \, \log \left (x^{4} + 2 \, x^{3} e^{x} - 2 \, x^{3} + 2 \, x^{2} e^{3} + x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} + x^{2} - 2 \, x e^{3} + 2 \, x e^{\left (x + 3\right )} - 6 \, x + e^{6} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+x)*exp(x)^2+((x+1)*exp(3)+x^3+2*x^2-2*x)*exp(x)+(2*x-1)*exp(3)+2*x^3-3*x^2+x-3)/(2*exp(x)^2*x^
2+(4*x*exp(3)+4*x^3-4*x^2)*exp(x)+2*exp(3)^2+(4*x^2-4*x)*exp(3)+2*x^4-4*x^3+2*x^2-12*x+10),x, algorithm="giac"
)

[Out]

1/4*log(x^4 + 2*x^3*e^x - 2*x^3 + 2*x^2*e^3 + x^2*e^(2*x) - 2*x^2*e^x + x^2 - 2*x*e^3 + 2*x*e^(x + 3) - 6*x +
e^6 + 5)

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maple [B]  time = 0.16, size = 64, normalized size = 1.94

method	result	size
risch	\(\frac {\ln \relax (x )}{2}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 \left (x^{2}+{\mathrm e}^{3}-x \right ) {\mathrm e}^{x}}{x}+\frac {x^{4}+2 x^{2} {\mathrm e}^{3}-2 x^{3}+{\mathrm e}^{6}-2 x \,{\mathrm e}^{3}+x^{2}-6 x +5}{x^{2}}\right )}{4}\)	\(64\)
norman	\(\frac {\ln \left (2 \,{\mathrm e}^{2 x} x^{2}+4 \,{\mathrm e}^{x} x^{3}+2 x^{4}+4 x \,{\mathrm e}^{3} {\mathrm e}^{x}-4 \,{\mathrm e}^{x} x^{2}+4 x^{2} {\mathrm e}^{3}-4 x^{3}+2 \,{\mathrm e}^{6}-4 x \,{\mathrm e}^{3}+2 x^{2}-12 x +10\right )}{4}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+x)*exp(x)^2+((x+1)*exp(3)+x^3+2*x^2-2*x)*exp(x)+(2*x-1)*exp(3)+2*x^3-3*x^2+x-3)/(2*exp(x)^2*x^2+(4*x
*exp(3)+4*x^3-4*x^2)*exp(x)+2*exp(3)^2+(4*x^2-4*x)*exp(3)+2*x^4-4*x^3+2*x^2-12*x+10),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(x)+1/4*ln(exp(2*x)+2/x*(x^2+exp(3)-x)*exp(x)+(x^4+2*x^2*exp(3)-2*x^3+exp(6)-2*x*exp(3)+x^2-6*x+5)/x^2)

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maxima [B]  time = 0.40, size = 66, normalized size = 2.00 \begin {gather*} \frac {1}{2} \, \log \relax (x) + \frac {1}{4} \, \log \left (\frac {x^{4} - 2 \, x^{3} + x^{2} {\left (2 \, e^{3} + 1\right )} + x^{2} e^{\left (2 \, x\right )} - 2 \, x {\left (e^{3} + 3\right )} + 2 \, {\left (x^{3} - x^{2} + x e^{3}\right )} e^{x} + e^{6} + 5}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+x)*exp(x)^2+((x+1)*exp(3)+x^3+2*x^2-2*x)*exp(x)+(2*x-1)*exp(3)+2*x^3-3*x^2+x-3)/(2*exp(x)^2*x^
2+(4*x*exp(3)+4*x^3-4*x^2)*exp(x)+2*exp(3)^2+(4*x^2-4*x)*exp(3)+2*x^4-4*x^3+2*x^2-12*x+10),x, algorithm="maxim
a")

[Out]

1/2*log(x) + 1/4*log((x^4 - 2*x^3 + x^2*(2*e^3 + 1) + x^2*e^(2*x) - 2*x*(e^3 + 3) + 2*(x^3 - x^2 + x*e^3)*e^x
+ e^6 + 5)/x^2)

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mupad [B]  time = 5.52, size = 62, normalized size = 1.88 \begin {gather*} \frac {\ln \left ({\mathrm {e}}^6-6\,x+2\,x\,{\mathrm {e}}^{x+3}-2\,x^2\,{\mathrm {e}}^x+2\,x^3\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^3+x^2\,{\mathrm {e}}^{2\,x}+2\,x^2\,{\mathrm {e}}^3+x^2-2\,x^3+x^4+5\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + exp(x)*(exp(3)*(x + 1) - 2*x + 2*x^2 + x^3) + exp(2*x)*(x + x^2) - 3*x^2 + 2*x^3 + exp(3)*(2*x - 1) -
3)/(2*exp(6) - 12*x - exp(3)*(4*x - 4*x^2) + exp(x)*(4*x*exp(3) - 4*x^2 + 4*x^3) + 2*x^2*exp(2*x) + 2*x^2 - 4
*x^3 + 2*x^4 + 10),x)

[Out]

log(exp(6) - 6*x + 2*x*exp(x + 3) - 2*x^2*exp(x) + 2*x^3*exp(x) - 2*x*exp(3) + x^2*exp(2*x) + 2*x^2*exp(3) + x
^2 - 2*x^3 + x^4 + 5)/4

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sympy [B]  time = 0.55, size = 70, normalized size = 2.12 \begin {gather*} \frac {\log {\relax (x )}}{2} + \frac {\log {\left (e^{2 x} + \frac {\left (2 x^{2} - 2 x + 2 e^{3}\right ) e^{x}}{x} + \frac {x^{4} - 2 x^{3} + x^{2} + 2 x^{2} e^{3} - 2 x e^{3} - 6 x + 5 + e^{6}}{x^{2}} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+x)*exp(x)**2+((x+1)*exp(3)+x**3+2*x**2-2*x)*exp(x)+(2*x-1)*exp(3)+2*x**3-3*x**2+x-3)/(2*exp(x
)**2*x**2+(4*x*exp(3)+4*x**3-4*x**2)*exp(x)+2*exp(3)**2+(4*x**2-4*x)*exp(3)+2*x**4-4*x**3+2*x**2-12*x+10),x)

[Out]

log(x)/2 + log(exp(2*x) + (2*x**2 - 2*x + 2*exp(3))*exp(x)/x + (x**4 - 2*x**3 + x**2 + 2*x**2*exp(3) - 2*x*exp
(3) - 6*x + 5 + exp(6))/x**2)/4

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