3.102.49 \(\int \frac {-2 x^3+e^{2 x} (32-50 x+16 x^2)+e^x (-32 x+52 x^2-16 x^3)+(e^{2 x} (-8+12 x-4 x^2)+e^x (8 x-12 x^2+4 x^3)) \log (96-48 x)}{-2 x^3+x^4} \, dx\)

Optimal. Leaf size=30 \[ \frac {2 \left (e^x-x\right )^2 (4-\log (16 (3+3 (1-x))))}{x^2} \]

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Rubi [A]  time = 3.58, antiderivative size = 57, normalized size of antiderivative = 1.90, number of steps used = 37, number of rules used = 8, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {1593, 6688, 12, 6742, 2178, 2177, 2197, 2554} \begin {gather*} \frac {8 e^{2 x}}{x^2}-\frac {2 e^{2 x} \log (96-48 x)}{x^2}-\frac {16 e^x}{x}+\frac {4 e^x \log (96-48 x)}{x}-2 \log (2-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x^3 + E^(2*x)*(32 - 50*x + 16*x^2) + E^x*(-32*x + 52*x^2 - 16*x^3) + (E^(2*x)*(-8 + 12*x - 4*x^2) + E^
x*(8*x - 12*x^2 + 4*x^3))*Log[96 - 48*x])/(-2*x^3 + x^4),x]

[Out]

(8*E^(2*x))/x^2 - (16*E^x)/x - (2*E^(2*x)*Log[96 - 48*x])/x^2 + (4*E^x*Log[96 - 48*x])/x - 2*Log[2 - x]

Rule 12

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

Rule 1593

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

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x^3+e^{2 x} \left (32-50 x+16 x^2\right )+e^x \left (-32 x+52 x^2-16 x^3\right )+\left (e^{2 x} \left (-8+12 x-4 x^2\right )+e^x \left (8 x-12 x^2+4 x^3\right )\right ) \log (96-48 x)}{(-2+x) x^3} \, dx\\ &=\int \frac {2 \left (e^x-x\right ) \left (-x^2+e^x \left (-16+25 x-8 x^2\right )+2 e^x \left (2-3 x+x^2\right ) \log (-48 (-2+x))\right )}{(2-x) x^3} \, dx\\ &=2 \int \frac {\left (e^x-x\right ) \left (-x^2+e^x \left (-16+25 x-8 x^2\right )+2 e^x \left (2-3 x+x^2\right ) \log (-48 (-2+x))\right )}{(2-x) x^3} \, dx\\ &=2 \int \left (\frac {1}{2-x}+\frac {2 e^x \left (-8+13 x-4 x^2+2 \log (-48 (-2+x))-3 x \log (-48 (-2+x))+x^2 \log (-48 (-2+x))\right )}{(-2+x) x^2}-\frac {e^{2 x} \left (-16+25 x-8 x^2+4 \log (-48 (-2+x))-6 x \log (-48 (-2+x))+2 x^2 \log (-48 (-2+x))\right )}{(-2+x) x^3}\right ) \, dx\\ &=-2 \log (2-x)-2 \int \frac {e^{2 x} \left (-16+25 x-8 x^2+4 \log (-48 (-2+x))-6 x \log (-48 (-2+x))+2 x^2 \log (-48 (-2+x))\right )}{(-2+x) x^3} \, dx+4 \int \frac {e^x \left (-8+13 x-4 x^2+2 \log (-48 (-2+x))-3 x \log (-48 (-2+x))+x^2 \log (-48 (-2+x))\right )}{(-2+x) x^2} \, dx\\ &=-2 \log (2-x)-2 \int \left (\frac {e^{2 x} \left (-16+25 x-8 x^2\right )}{(-2+x) x^3}+\frac {2 e^{2 x} (-1+x) \log (96-48 x)}{x^3}\right ) \, dx+4 \int \left (\frac {e^x \left (-8+13 x-4 x^2\right )}{(-2+x) x^2}+\frac {e^x (-1+x) \log (96-48 x)}{x^2}\right ) \, dx\\ &=-2 \log (2-x)-2 \int \frac {e^{2 x} \left (-16+25 x-8 x^2\right )}{(-2+x) x^3} \, dx+4 \int \frac {e^x \left (-8+13 x-4 x^2\right )}{(-2+x) x^2} \, dx-4 \int \frac {e^{2 x} (-1+x) \log (96-48 x)}{x^3} \, dx+4 \int \frac {e^x (-1+x) \log (96-48 x)}{x^2} \, dx\\ &=-\frac {2 e^{2 x} \log (96-48 x)}{x^2}+\frac {4 e^x \log (96-48 x)}{x}-2 \log (2-x)-2 \int \left (\frac {e^{2 x}}{4 (-2+x)}+\frac {8 e^{2 x}}{x^3}-\frac {17 e^{2 x}}{2 x^2}-\frac {e^{2 x}}{4 x}\right ) \, dx+4 \int \left (\frac {e^x}{2 (-2+x)}+\frac {4 e^x}{x^2}-\frac {9 e^x}{2 x}\right ) \, dx+4 \int \frac {e^{2 x}}{2 (-2+x) x^2} \, dx-4 \int \frac {e^x}{(-2+x) x} \, dx\\ &=-\frac {2 e^{2 x} \log (96-48 x)}{x^2}+\frac {4 e^x \log (96-48 x)}{x}-2 \log (2-x)-\frac {1}{2} \int \frac {e^{2 x}}{-2+x} \, dx+\frac {1}{2} \int \frac {e^{2 x}}{x} \, dx+2 \int \frac {e^x}{-2+x} \, dx+2 \int \frac {e^{2 x}}{(-2+x) x^2} \, dx-4 \int \left (\frac {e^x}{2 (-2+x)}-\frac {e^x}{2 x}\right ) \, dx-16 \int \frac {e^{2 x}}{x^3} \, dx+16 \int \frac {e^x}{x^2} \, dx+17 \int \frac {e^{2 x}}{x^2} \, dx-18 \int \frac {e^x}{x} \, dx\\ &=\frac {8 e^{2 x}}{x^2}-\frac {16 e^x}{x}-\frac {17 e^{2 x}}{x}-\frac {1}{2} e^4 \text {Ei}(-2 (2-x))+2 e^2 \text {Ei}(-2+x)-18 \text {Ei}(x)+\frac {\text {Ei}(2 x)}{2}-\frac {2 e^{2 x} \log (96-48 x)}{x^2}+\frac {4 e^x \log (96-48 x)}{x}-2 \log (2-x)+2 \int \left (\frac {e^{2 x}}{4 (-2+x)}-\frac {e^{2 x}}{2 x^2}-\frac {e^{2 x}}{4 x}\right ) \, dx-2 \int \frac {e^x}{-2+x} \, dx+2 \int \frac {e^x}{x} \, dx-16 \int \frac {e^{2 x}}{x^2} \, dx+16 \int \frac {e^x}{x} \, dx+34 \int \frac {e^{2 x}}{x} \, dx\\ &=\frac {8 e^{2 x}}{x^2}-\frac {16 e^x}{x}-\frac {e^{2 x}}{x}-\frac {1}{2} e^4 \text {Ei}(-2 (2-x))+\frac {69 \text {Ei}(2 x)}{2}-\frac {2 e^{2 x} \log (96-48 x)}{x^2}+\frac {4 e^x \log (96-48 x)}{x}-2 \log (2-x)+\frac {1}{2} \int \frac {e^{2 x}}{-2+x} \, dx-\frac {1}{2} \int \frac {e^{2 x}}{x} \, dx-32 \int \frac {e^{2 x}}{x} \, dx-\int \frac {e^{2 x}}{x^2} \, dx\\ &=\frac {8 e^{2 x}}{x^2}-\frac {16 e^x}{x}+2 \text {Ei}(2 x)-\frac {2 e^{2 x} \log (96-48 x)}{x^2}+\frac {4 e^x \log (96-48 x)}{x}-2 \log (2-x)-2 \int \frac {e^{2 x}}{x} \, dx\\ &=\frac {8 e^{2 x}}{x^2}-\frac {16 e^x}{x}-\frac {2 e^{2 x} \log (96-48 x)}{x^2}+\frac {4 e^x \log (96-48 x)}{x}-2 \log (2-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.21, size = 34, normalized size = 1.13 \begin {gather*} 2 \left (-\log (2-x)-\frac {e^x \left (e^x-2 x\right ) (-4+\log (-48 (-2+x)))}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^3 + E^(2*x)*(32 - 50*x + 16*x^2) + E^x*(-32*x + 52*x^2 - 16*x^3) + (E^(2*x)*(-8 + 12*x - 4*x^2
) + E^x*(8*x - 12*x^2 + 4*x^3))*Log[96 - 48*x])/(-2*x^3 + x^4),x]

[Out]

2*(-Log[2 - x] - (E^x*(E^x - 2*x)*(-4 + Log[-48*(-2 + x)]))/x^2)

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fricas [A]  time = 0.89, size = 37, normalized size = 1.23 \begin {gather*} -\frac {2 \, {\left (8 \, x e^{x} + {\left (x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} \log \left (-48 \, x + 96\right ) - 4 \, e^{\left (2 \, x\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2+12*x-8)*exp(x)^2+(4*x^3-12*x^2+8*x)*exp(x))*log(-48*x+96)+(16*x^2-50*x+32)*exp(x)^2+(-16*x
^3+52*x^2-32*x)*exp(x)-2*x^3)/(x^4-2*x^3),x, algorithm="fricas")

[Out]

-2*(8*x*e^x + (x^2 - 2*x*e^x + e^(2*x))*log(-48*x + 96) - 4*e^(2*x))/x^2

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giac [B]  time = 0.19, size = 47, normalized size = 1.57 \begin {gather*} -\frac {2 \, {\left (x^{2} \log \left (x - 2\right ) - 2 \, x e^{x} \log \left (-48 \, x + 96\right ) + 8 \, x e^{x} + e^{\left (2 \, x\right )} \log \left (-48 \, x + 96\right ) - 4 \, e^{\left (2 \, x\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2+12*x-8)*exp(x)^2+(4*x^3-12*x^2+8*x)*exp(x))*log(-48*x+96)+(16*x^2-50*x+32)*exp(x)^2+(-16*x
^3+52*x^2-32*x)*exp(x)-2*x^3)/(x^4-2*x^3),x, algorithm="giac")

[Out]

-2*(x^2*log(x - 2) - 2*x*e^x*log(-48*x + 96) + 8*x*e^x + e^(2*x)*log(-48*x + 96) - 4*e^(2*x))/x^2

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maple [B]  time = 0.22, size = 48, normalized size = 1.60

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{x} \left (2 x -{\mathrm e}^{x}\right ) \ln \left (-48 x +96\right )}{x^{2}}-\frac {2 \left (x^{2} \ln \left (x -2\right )+8 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{2 x}\right )}{x^{2}}\)	\(48\)
default	\(\frac {\left (4 \ln \left (48\right )-16\right ) {\mathrm e}^{x}+4 \ln \left (2-x \right ) {\mathrm e}^{x}}{x}+\frac {\left (-2 \ln \left (48\right )+8\right ) {\mathrm e}^{2 x}-2 \,{\mathrm e}^{2 x} \ln \left (2-x \right )}{x^{2}}-2 \ln \left (x -2\right )\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^2+12*x-8)*exp(x)^2+(4*x^3-12*x^2+8*x)*exp(x))*ln(-48*x+96)+(16*x^2-50*x+32)*exp(x)^2+(-16*x^3+52*x
^2-32*x)*exp(x)-2*x^3)/(x^4-2*x^3),x,method=_RETURNVERBOSE)

[Out]

2*exp(x)*(2*x-exp(x))/x^2*ln(-48*x+96)-2*(x^2*ln(x-2)+8*exp(x)*x-4*exp(2*x))/x^2

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maxima [C]  time = 0.48, size = 63, normalized size = 2.10 \begin {gather*} \frac {2 \, {\left (2 \, {\left (i \, \pi + \log \relax (3) + 4 \, \log \relax (2) - 4\right )} x e^{x} - {\left (i \, \pi + \log \relax (3) + 4 \, \log \relax (2) - 4\right )} e^{\left (2 \, x\right )} + {\left (2 \, x e^{x} - e^{\left (2 \, x\right )}\right )} \log \left (x - 2\right )\right )}}{x^{2}} - 2 \, \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2+12*x-8)*exp(x)^2+(4*x^3-12*x^2+8*x)*exp(x))*log(-48*x+96)+(16*x^2-50*x+32)*exp(x)^2+(-16*x
^3+52*x^2-32*x)*exp(x)-2*x^3)/(x^4-2*x^3),x, algorithm="maxima")

[Out]

2*(2*(I*pi + log(3) + 4*log(2) - 4)*x*e^x - (I*pi + log(3) + 4*log(2) - 4)*e^(2*x) + (2*x*e^x - e^(2*x))*log(x
- 2))/x^2 - 2*log(x - 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\ln \left (96-48\,x\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (4\,x^2-12\,x+8\right )-{\mathrm {e}}^x\,\left (4\,x^3-12\,x^2+8\,x\right )\right )-{\mathrm {e}}^{2\,x}\,\left (16\,x^2-50\,x+32\right )+2\,x^3+{\mathrm {e}}^x\,\left (16\,x^3-52\,x^2+32\,x\right )}{2\,x^3-x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(96 - 48*x)*(exp(2*x)*(4*x^2 - 12*x + 8) - exp(x)*(8*x - 12*x^2 + 4*x^3)) - exp(2*x)*(16*x^2 - 50*x +
32) + 2*x^3 + exp(x)*(32*x - 52*x^2 + 16*x^3))/(2*x^3 - x^4),x)

[Out]

int((log(96 - 48*x)*(exp(2*x)*(4*x^2 - 12*x + 8) - exp(x)*(8*x - 12*x^2 + 4*x^3)) - exp(2*x)*(16*x^2 - 50*x +
32) + 2*x^3 + exp(x)*(32*x - 52*x^2 + 16*x^3))/(2*x^3 - x^4), x)

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sympy [B]  time = 0.44, size = 48, normalized size = 1.60 \begin {gather*} - 2 \log {\left (x - 2 \right )} + \frac {\left (- 2 x \log {\left (96 - 48 x \right )} + 8 x\right ) e^{2 x} + \left (4 x^{2} \log {\left (96 - 48 x \right )} - 16 x^{2}\right ) e^{x}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**2+12*x-8)*exp(x)**2+(4*x**3-12*x**2+8*x)*exp(x))*ln(-48*x+96)+(16*x**2-50*x+32)*exp(x)**2+(
-16*x**3+52*x**2-32*x)*exp(x)-2*x**3)/(x**4-2*x**3),x)

[Out]

-2*log(x - 2) + ((-2*x*log(96 - 48*x) + 8*x)*exp(2*x) + (4*x**2*log(96 - 48*x) - 16*x**2)*exp(x))/x**3

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