3.26.3 \(\int \frac {e^{-x} (-4 e^{28}+e^2 (-3-x^2)+e^x (-9 x-12 e^{26} x+3 x^2+3 x^3)+(-4 e^{28} x+e^2 (-3 x+x^2+x^3)) \log (\frac {3+4 e^{26}-x-x^2}{x}))}{9 x+12 e^{26} x-3 x^2-3 x^3} \, dx\)

Optimal. Leaf size=34 \[ -x+\frac {1}{3} e^{2-x} \log \left (\frac {3+4 e^{26}-x}{x}-x\right ) \]

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Rubi [B]  time = 1.12, antiderivative size = 115, normalized size of antiderivative = 3.38, number of steps used = 5, number of rules used = 4, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6, 1594, 6728, 2288} \begin {gather*} \frac {e^{2-x} \left (-x^2 \log \left (\frac {-x^2-x+4 e^{26}+3}{x}\right )+\left (3+4 e^{26}\right ) x \log \left (\frac {-x^2-x+4 e^{26}+3}{x}\right )+x^3 \left (-\log \left (\frac {-x^2-x+4 e^{26}+3}{x}\right )\right )\right )}{3 x \left (-x^2-x+4 e^{26}+3\right )}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*E^28 + E^2*(-3 - x^2) + E^x*(-9*x - 12*E^26*x + 3*x^2 + 3*x^3) + (-4*E^28*x + E^2*(-3*x + x^2 + x^3))*
Log[(3 + 4*E^26 - x - x^2)/x])/(E^x*(9*x + 12*E^26*x - 3*x^2 - 3*x^3)),x]

[Out]

-x + (E^(2 - x)*((3 + 4*E^26)*x*Log[(3 + 4*E^26 - x - x^2)/x] - x^2*Log[(3 + 4*E^26 - x - x^2)/x] - x^3*Log[(3
+ 4*E^26 - x - x^2)/x]))/(3*x*(3 + 4*E^26 - x - x^2))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2288

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]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (-4 e^{28}+e^2 \left (-3-x^2\right )+e^x \left (-9 x-12 e^{26} x+3 x^2+3 x^3\right )+\left (-4 e^{28} x+e^2 \left (-3 x+x^2+x^3\right )\right ) \log \left (\frac {3+4 e^{26}-x-x^2}{x}\right )\right )}{\left (9+12 e^{26}\right ) x-3 x^2-3 x^3} \, dx\\ &=\int \frac {e^{-x} \left (-4 e^{28}+e^2 \left (-3-x^2\right )+e^x \left (-9 x-12 e^{26} x+3 x^2+3 x^3\right )+\left (-4 e^{28} x+e^2 \left (-3 x+x^2+x^3\right )\right ) \log \left (\frac {3+4 e^{26}-x-x^2}{x}\right )\right )}{x \left (9+12 e^{26}-3 x-3 x^2\right )} \, dx\\ &=\int \left (-1+\frac {e^{2-x} \left (-3 \left (1+\frac {4 e^{26}}{3}\right )-x^2-3 \left (1+\frac {4 e^{26}}{3}\right ) x \log \left (-\frac {-3-4 e^{26}+x+x^2}{x}\right )+x^2 \log \left (-\frac {-3-4 e^{26}+x+x^2}{x}\right )+x^3 \log \left (-\frac {-3-4 e^{26}+x+x^2}{x}\right )\right )}{3 x \left (3+4 e^{26}-x-x^2\right )}\right ) \, dx\\ &=-x+\frac {1}{3} \int \frac {e^{2-x} \left (-3 \left (1+\frac {4 e^{26}}{3}\right )-x^2-3 \left (1+\frac {4 e^{26}}{3}\right ) x \log \left (-\frac {-3-4 e^{26}+x+x^2}{x}\right )+x^2 \log \left (-\frac {-3-4 e^{26}+x+x^2}{x}\right )+x^3 \log \left (-\frac {-3-4 e^{26}+x+x^2}{x}\right )\right )}{x \left (3+4 e^{26}-x-x^2\right )} \, dx\\ &=-x+\frac {e^{2-x} \left (\left (3+4 e^{26}\right ) x \log \left (\frac {3+4 e^{26}-x-x^2}{x}\right )-x^2 \log \left (\frac {3+4 e^{26}-x-x^2}{x}\right )-x^3 \log \left (\frac {3+4 e^{26}-x-x^2}{x}\right )\right )}{3 x \left (3+4 e^{26}-x-x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 33, normalized size = 0.97 \begin {gather*} \frac {1}{3} \left (-3 x+e^{2-x} \log \left (-\frac {-3-4 e^{26}+x+x^2}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^28 + E^2*(-3 - x^2) + E^x*(-9*x - 12*E^26*x + 3*x^2 + 3*x^3) + (-4*E^28*x + E^2*(-3*x + x^2 +
x^3))*Log[(3 + 4*E^26 - x - x^2)/x])/(E^x*(9*x + 12*E^26*x - 3*x^2 - 3*x^3)),x]

[Out]

(-3*x + E^(2 - x)*Log[-((-3 - 4*E^26 + x + x^2)/x)])/3

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fricas [A]  time = 0.68, size = 32, normalized size = 0.94 \begin {gather*} -\frac {1}{3} \, {\left (3 \, x e^{x} - e^{2} \log \left (-\frac {x^{2} + x - 4 \, e^{26} - 3}{x}\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(2)*exp(26)+(x^3+x^2-3*x)*exp(2))*log((4*exp(26)-x^2-x+3)/x)+(-12*x*exp(26)+3*x^3+3*x^2-9*
x)*exp(x)-4*exp(2)*exp(26)+(-x^2-3)*exp(2))/(12*x*exp(26)-3*x^3-3*x^2+9*x)/exp(x),x, algorithm="fricas")

[Out]

-1/3*(3*x*e^x - e^2*log(-(x^2 + x - 4*e^26 - 3)/x))*e^(-x)

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giac [A]  time = 0.19, size = 28, normalized size = 0.82 \begin {gather*} \frac {1}{3} \, e^{\left (-x + 2\right )} \log \left (-\frac {x^{2} + x - 4 \, e^{26} - 3}{x}\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(2)*exp(26)+(x^3+x^2-3*x)*exp(2))*log((4*exp(26)-x^2-x+3)/x)+(-12*x*exp(26)+3*x^3+3*x^2-9*
x)*exp(x)-4*exp(2)*exp(26)+(-x^2-3)*exp(2))/(12*x*exp(26)-3*x^3-3*x^2+9*x)/exp(x),x, algorithm="giac")

[Out]

1/3*e^(-x + 2)*log(-(x^2 + x - 4*e^26 - 3)/x) - x

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maple [A]  time = 0.26, size = 32, normalized size = 0.94

method	result	size
default	\(-x +\frac {{\mathrm e}^{2} \ln \left (\frac {4 \,{\mathrm e}^{26}-x^{2}-x +3}{x}\right ) {\mathrm e}^{-x}}{3}\)	\(32\)
norman	\(\left (\frac {{\mathrm e}^{2} \ln \left (\frac {4 \,{\mathrm e}^{26}-x^{2}-x +3}{x}\right )}{3}-{\mathrm e}^{x} x \right ) {\mathrm e}^{-x}\)	\(35\)
risch	\(\frac {\ln \left (-\frac {x^{2}}{4}+{\mathrm e}^{26}-\frac {x}{4}+\frac {3}{4}\right ) {\mathrm e}^{2-x}}{3}-\frac {\left (i \pi \,{\mathrm e}^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (-\frac {x^{2}}{4}+{\mathrm e}^{26}-\frac {x}{4}+\frac {3}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{4}+{\mathrm e}^{26}-\frac {x}{4}+\frac {3}{4}\right )}{x}\right )-i \pi \,{\mathrm e}^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{4}+{\mathrm e}^{26}-\frac {x}{4}+\frac {3}{4}\right )}{x}\right )^{2}-i \pi \,{\mathrm e}^{2} \mathrm {csgn}\left (i \left (-\frac {x^{2}}{4}+{\mathrm e}^{26}-\frac {x}{4}+\frac {3}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{4}+{\mathrm e}^{26}-\frac {x}{4}+\frac {3}{4}\right )}{x}\right )^{2}+i \pi \,{\mathrm e}^{2} \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{4}+{\mathrm e}^{26}-\frac {x}{4}+\frac {3}{4}\right )}{x}\right )^{3}-4 \,{\mathrm e}^{2} \ln \relax (2)+2 \,{\mathrm e}^{2} \ln \relax (x )+6 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{-x}}{6}\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*exp(2)*exp(26)+(x^3+x^2-3*x)*exp(2))*ln((4*exp(26)-x^2-x+3)/x)+(-12*x*exp(26)+3*x^3+3*x^2-9*x)*exp(
x)-4*exp(2)*exp(26)+(-x^2-3)*exp(2))/(12*x*exp(26)-3*x^3-3*x^2+9*x)/exp(x),x,method=_RETURNVERBOSE)

[Out]

-x+1/3*exp(2)*ln((4*exp(26)-x^2-x+3)/x)/exp(x)

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maxima [A]  time = 0.49, size = 36, normalized size = 1.06 \begin {gather*} -\frac {1}{3} \, {\left (3 \, x e^{x} - e^{2} \log \left (-x^{2} - x + 4 \, e^{26} + 3\right ) + e^{2} \log \relax (x)\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(2)*exp(26)+(x^3+x^2-3*x)*exp(2))*log((4*exp(26)-x^2-x+3)/x)+(-12*x*exp(26)+3*x^3+3*x^2-9*
x)*exp(x)-4*exp(2)*exp(26)+(-x^2-3)*exp(2))/(12*x*exp(26)-3*x^3-3*x^2+9*x)/exp(x),x, algorithm="maxima")

[Out]

-1/3*(3*x*e^x - e^2*log(-x^2 - x + 4*e^26 + 3) + e^2*log(x))*e^(-x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{-x}\,\left (4\,{\mathrm {e}}^{28}+{\mathrm {e}}^x\,\left (9\,x+12\,x\,{\mathrm {e}}^{26}-3\,x^2-3\,x^3\right )+\ln \left (-\frac {x^2+x-4\,{\mathrm {e}}^{26}-3}{x}\right )\,\left (4\,x\,{\mathrm {e}}^{28}-{\mathrm {e}}^2\,\left (x^3+x^2-3\,x\right )\right )+{\mathrm {e}}^2\,\left (x^2+3\right )\right )}{9\,x+12\,x\,{\mathrm {e}}^{26}-3\,x^2-3\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(4*exp(28) + exp(x)*(9*x + 12*x*exp(26) - 3*x^2 - 3*x^3) + log(-(x - 4*exp(26) + x^2 - 3)/x)*(4*
x*exp(28) - exp(2)*(x^2 - 3*x + x^3)) + exp(2)*(x^2 + 3)))/(9*x + 12*x*exp(26) - 3*x^2 - 3*x^3),x)

[Out]

int(-(exp(-x)*(4*exp(28) + exp(x)*(9*x + 12*x*exp(26) - 3*x^2 - 3*x^3) + log(-(x - 4*exp(26) + x^2 - 3)/x)*(4*
x*exp(28) - exp(2)*(x^2 - 3*x + x^3)) + exp(2)*(x^2 + 3)))/(9*x + 12*x*exp(26) - 3*x^2 - 3*x^3), x)

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sympy [A]  time = 5.36, size = 24, normalized size = 0.71 \begin {gather*} - x + \frac {e^{2} e^{- x} \log {\left (\frac {- x^{2} - x + 3 + 4 e^{26}}{x} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(2)*exp(26)+(x**3+x**2-3*x)*exp(2))*ln((4*exp(26)-x**2-x+3)/x)+(-12*x*exp(26)+3*x**3+3*x**
2-9*x)*exp(x)-4*exp(2)*exp(26)+(-x**2-3)*exp(2))/(12*x*exp(26)-3*x**3-3*x**2+9*x)/exp(x),x)

[Out]

-x + exp(2)*exp(-x)*log((-x**2 - x + 3 + 4*exp(26))/x)/3

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