3.54.33 \(\int \frac {1344 x-592 x^2+64 x^3+e^x (48-48 x+12 x^2)+e^x (-252+300 x-111 x^2+12 x^3) \log (\frac {1}{4} (21-4 x))}{-252+300 x-111 x^2+12 x^3} \, dx\)

Optimal. Leaf size=27 \[ -\frac {16 x^2}{3 (2-x)}+e^x \log \left (\frac {21}{4}-x\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6688, 74, 2178, 2194, 2554, 12} \begin {gather*} e^x \log \left (\frac {21}{4}-x\right )-\frac {16 x^2}{3 (2-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1344*x - 592*x^2 + 64*x^3 + E^x*(48 - 48*x + 12*x^2) + E^x*(-252 + 300*x - 111*x^2 + 12*x^3)*Log[(21 - 4*
x)/4])/(-252 + 300*x - 111*x^2 + 12*x^3),x]

[Out]

(-16*x^2)/(3*(2 - x)) + E^x*Log[21/4 - 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 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {16 (-4+x) x}{3 (-2+x)^2}+\frac {4 e^x}{-21+4 x}+e^x \log \left (\frac {21}{4}-x\right )\right ) \, dx\\ &=4 \int \frac {e^x}{-21+4 x} \, dx+\frac {16}{3} \int \frac {(-4+x) x}{(-2+x)^2} \, dx+\int e^x \log \left (\frac {21}{4}-x\right ) \, dx\\ &=-\frac {16 x^2}{3 (2-x)}+e^{21/4} \text {Ei}\left (\frac {1}{4} (-21+4 x)\right )+e^x \log \left (\frac {21}{4}-x\right )-\int \frac {4 e^x}{-21+4 x} \, dx\\ &=-\frac {16 x^2}{3 (2-x)}+e^{21/4} \text {Ei}\left (\frac {1}{4} (-21+4 x)\right )+e^x \log \left (\frac {21}{4}-x\right )-4 \int \frac {e^x}{-21+4 x} \, dx\\ &=-\frac {16 x^2}{3 (2-x)}+e^x \log \left (\frac {21}{4}-x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 27, normalized size = 1.00 \begin {gather*} \frac {64}{3 (-2+x)}+\frac {16 x}{3}+e^x \log \left (\frac {21}{4}-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1344*x - 592*x^2 + 64*x^3 + E^x*(48 - 48*x + 12*x^2) + E^x*(-252 + 300*x - 111*x^2 + 12*x^3)*Log[(2
1 - 4*x)/4])/(-252 + 300*x - 111*x^2 + 12*x^3),x]

[Out]

64/(3*(-2 + x)) + (16*x)/3 + E^x*Log[21/4 - x]

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fricas [A]  time = 0.64, size = 30, normalized size = 1.11 \begin {gather*} \frac {3 \, {\left (x - 2\right )} e^{x} \log \left (-x + \frac {21}{4}\right ) + 16 \, x^{2} - 32 \, x + 64}{3 \, {\left (x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-111*x^2+300*x-252)*exp(x)*log(-x+21/4)+(12*x^2-48*x+48)*exp(x)+64*x^3-592*x^2+1344*x)/(12*x
^3-111*x^2+300*x-252),x, algorithm="fricas")

[Out]

1/3*(3*(x - 2)*e^x*log(-x + 21/4) + 16*x^2 - 32*x + 64)/(x - 2)

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giac [A]  time = 0.13, size = 38, normalized size = 1.41 \begin {gather*} \frac {3 \, x e^{x} \log \left (-x + \frac {21}{4}\right ) + 16 \, x^{2} - 6 \, e^{x} \log \left (-x + \frac {21}{4}\right ) - 32 \, x + 64}{3 \, {\left (x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-111*x^2+300*x-252)*exp(x)*log(-x+21/4)+(12*x^2-48*x+48)*exp(x)+64*x^3-592*x^2+1344*x)/(12*x
^3-111*x^2+300*x-252),x, algorithm="giac")

[Out]

1/3*(3*x*e^x*log(-x + 21/4) + 16*x^2 - 6*e^x*log(-x + 21/4) - 32*x + 64)/(x - 2)

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maple [A]  time = 0.08, size = 21, normalized size = 0.78

method	result	size
default	\({\mathrm e}^{x} \ln \left (-x +\frac {21}{4}\right )+\frac {16 x}{3}+\frac {64}{3 \left (x -2\right )}\)	\(21\)
risch	\({\mathrm e}^{x} \ln \left (-x +\frac {21}{4}\right )+\frac {\frac {16}{3} x^{2}-\frac {32}{3} x +\frac {64}{3}}{x -2}\)	\(26\)
norman	\(\frac {{\mathrm e}^{x} \ln \left (-x +\frac {21}{4}\right ) x +\frac {16 x^{2}}{3}-2 \,{\mathrm e}^{x} \ln \left (-x +\frac {21}{4}\right )}{x -2}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x^3-111*x^2+300*x-252)*exp(x)*ln(-x+21/4)+(12*x^2-48*x+48)*exp(x)+64*x^3-592*x^2+1344*x)/(12*x^3-111*
x^2+300*x-252),x,method=_RETURNVERBOSE)

[Out]

exp(x)*ln(-x+21/4)+16/3*x+64/3/(x-2)

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maxima [A]  time = 0.49, size = 26, normalized size = 0.96 \begin {gather*} -2 \, e^{x} \log \relax (2) + e^{x} \log \left (-4 \, x + 21\right ) + \frac {16}{3} \, x + \frac {64}{3 \, {\left (x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-111*x^2+300*x-252)*exp(x)*log(-x+21/4)+(12*x^2-48*x+48)*exp(x)+64*x^3-592*x^2+1344*x)/(12*x
^3-111*x^2+300*x-252),x, algorithm="maxima")

[Out]

-2*e^x*log(2) + e^x*log(-4*x + 21) + 16/3*x + 64/3/(x - 2)

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mupad [B]  time = 3.71, size = 22, normalized size = 0.81 \begin {gather*} \frac {16\,x}{3}+\frac {64}{3\,\left (x-2\right )}+{\mathrm {e}}^x\,\ln \left (\frac {21}{4}-x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1344*x + exp(x)*(12*x^2 - 48*x + 48) - 592*x^2 + 64*x^3 + exp(x)*log(21/4 - x)*(300*x - 111*x^2 + 12*x^3
- 252))/(300*x - 111*x^2 + 12*x^3 - 252),x)

[Out]

(16*x)/3 + 64/(3*(x - 2)) + exp(x)*log(21/4 - x)

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sympy [A]  time = 0.39, size = 20, normalized size = 0.74 \begin {gather*} \frac {16 x}{3} + e^{x} \log {\left (\frac {21}{4} - x \right )} + \frac {64}{3 x - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x**3-111*x**2+300*x-252)*exp(x)*ln(-x+21/4)+(12*x**2-48*x+48)*exp(x)+64*x**3-592*x**2+1344*x)/(
12*x**3-111*x**2+300*x-252),x)

[Out]

16*x/3 + exp(x)*log(21/4 - x) + 64/(3*x - 6)

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