3.100.73 \(\int \frac {8+e^{21 x} (-16 x-152 x^2+336 x^3)}{1-4 x+4 x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {2 \left (-1+2 e^{21 x} x^2\right )}{-\frac {1}{2}+x} \]

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Rubi [A]  time = 0.24, antiderivative size = 39, normalized size of antiderivative = 1.86, number of steps used = 11, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {27, 6742, 2199, 2194, 2176, 2177, 2178} \begin {gather*} 4 e^{21 x} x+2 e^{21 x}-\frac {2 e^{21 x}}{1-2 x}+\frac {4}{1-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 + E^(21*x)*(-16*x - 152*x^2 + 336*x^3))/(1 - 4*x + 4*x^2),x]

[Out]

2*E^(21*x) + 4/(1 - 2*x) - (2*E^(21*x))/(1 - 2*x) + 4*E^(21*x)*x

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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 {8+e^{21 x} \left (-16 x-152 x^2+336 x^3\right )}{(-1+2 x)^2} \, dx\\ &=\int \left (\frac {8}{(-1+2 x)^2}+\frac {8 e^{21 x} x \left (-2-19 x+42 x^2\right )}{(-1+2 x)^2}\right ) \, dx\\ &=\frac {4}{1-2 x}+8 \int \frac {e^{21 x} x \left (-2-19 x+42 x^2\right )}{(-1+2 x)^2} \, dx\\ &=\frac {4}{1-2 x}+8 \int \left (\frac {23 e^{21 x}}{4}+\frac {21}{2} e^{21 x} x-\frac {e^{21 x}}{2 (-1+2 x)^2}+\frac {21 e^{21 x}}{4 (-1+2 x)}\right ) \, dx\\ &=\frac {4}{1-2 x}-4 \int \frac {e^{21 x}}{(-1+2 x)^2} \, dx+42 \int \frac {e^{21 x}}{-1+2 x} \, dx+46 \int e^{21 x} \, dx+84 \int e^{21 x} x \, dx\\ &=\frac {46 e^{21 x}}{21}+\frac {4}{1-2 x}-\frac {2 e^{21 x}}{1-2 x}+4 e^{21 x} x+21 e^{21/2} \text {Ei}\left (-\frac {21}{2} (1-2 x)\right )-4 \int e^{21 x} \, dx-42 \int \frac {e^{21 x}}{-1+2 x} \, dx\\ &=2 e^{21 x}+\frac {4}{1-2 x}-\frac {2 e^{21 x}}{1-2 x}+4 e^{21 x} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 20, normalized size = 0.95 \begin {gather*} \frac {4-8 e^{21 x} x^2}{1-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + E^(21*x)*(-16*x - 152*x^2 + 336*x^3))/(1 - 4*x + 4*x^2),x]

[Out]

(4 - 8*E^(21*x)*x^2)/(1 - 2*x)

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fricas [A]  time = 0.74, size = 20, normalized size = 0.95 \begin {gather*} \frac {4 \, {\left (2 \, x^{2} e^{\left (21 \, x\right )} - 1\right )}}{2 \, x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((336*x^3-152*x^2-16*x)*exp(21*x)+8)/(4*x^2-4*x+1),x, algorithm="fricas")

[Out]

4*(2*x^2*e^(21*x) - 1)/(2*x - 1)

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giac [A]  time = 0.16, size = 20, normalized size = 0.95 \begin {gather*} \frac {4 \, {\left (2 \, x^{2} e^{\left (21 \, x\right )} - 1\right )}}{2 \, x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((336*x^3-152*x^2-16*x)*exp(21*x)+8)/(4*x^2-4*x+1),x, algorithm="giac")

[Out]

4*(2*x^2*e^(21*x) - 1)/(2*x - 1)

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maple [A]  time = 0.16, size = 22, normalized size = 1.05

method	result	size
norman	\(\frac {-8 x +8 x^{2} {\mathrm e}^{21 x}}{2 x -1}\)	\(22\)
risch	\(-\frac {2}{x -\frac {1}{2}}+\frac {8 x^{2} {\mathrm e}^{21 x}}{2 x -1}\)	\(25\)
derivativedivides	\(-\frac {84}{42 x -21}+\frac {21 \,{\mathrm e}^{21 x}}{21 x -\frac {21}{2}}+2 \,{\mathrm e}^{21 x}+4 \,{\mathrm e}^{21 x} x\)	\(37\)
default	\(-\frac {84}{42 x -21}+\frac {21 \,{\mathrm e}^{21 x}}{21 x -\frac {21}{2}}+2 \,{\mathrm e}^{21 x}+4 \,{\mathrm e}^{21 x} x\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((336*x^3-152*x^2-16*x)*exp(21*x)+8)/(4*x^2-4*x+1),x,method=_RETURNVERBOSE)

[Out]

(-8*x+8*x^2*exp(21*x))/(2*x-1)

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maxima [A]  time = 0.40, size = 26, normalized size = 1.24 \begin {gather*} \frac {8 \, x^{2} e^{\left (21 \, x\right )}}{2 \, x - 1} - \frac {4}{2 \, x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((336*x^3-152*x^2-16*x)*exp(21*x)+8)/(4*x^2-4*x+1),x, algorithm="maxima")

[Out]

8*x^2*e^(21*x)/(2*x - 1) - 4/(2*x - 1)

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mupad [B]  time = 6.82, size = 18, normalized size = 0.86 \begin {gather*} \frac {8\,x\,\left (x\,{\mathrm {e}}^{21\,x}-1\right )}{2\,x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(21*x)*(16*x + 152*x^2 - 336*x^3) - 8)/(4*x^2 - 4*x + 1),x)

[Out]

(8*x*(x*exp(21*x) - 1))/(2*x - 1)

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sympy [A]  time = 0.13, size = 20, normalized size = 0.95 \begin {gather*} \frac {8 x^{2} e^{21 x}}{2 x - 1} - \frac {8}{4 x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((336*x**3-152*x**2-16*x)*exp(21*x)+8)/(4*x**2-4*x+1),x)

[Out]

8*x**2*exp(21*x)/(2*x - 1) - 8/(4*x - 2)

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