3.86.9 \(\int \frac {-30+60 x-32 x^2+4 x^3-2 x^4+e^x (33+27 x-33 x^2+3 x^3)}{6-12 x+6 x^2} \, dx\)

Optimal. Leaf size=30 \[ x \left (-5+\frac {e^x (11-x)}{2 (1-x)}-\frac {x^2}{9}\right ) \]

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Rubi [A]  time = 0.18, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 13, number of rules used = 8, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {27, 12, 6688, 2199, 2194, 2177, 2178, 2176} \begin {gather*} -\frac {x^3}{9}+\frac {e^x x}{2}-5 x-5 e^x+\frac {5 e^x}{1-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30 + 60*x - 32*x^2 + 4*x^3 - 2*x^4 + E^x*(33 + 27*x - 33*x^2 + 3*x^3))/(6 - 12*x + 6*x^2),x]

[Out]

-5*E^x + (5*E^x)/(1 - x) - 5*x + (E^x*x)/2 - x^3/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 \frac {-30+60 x-32 x^2+4 x^3-2 x^4+e^x \left (33+27 x-33 x^2+3 x^3\right )}{6 (-1+x)^2} \, dx\\ &=\frac {1}{6} \int \frac {-30+60 x-32 x^2+4 x^3-2 x^4+e^x \left (33+27 x-33 x^2+3 x^3\right )}{(-1+x)^2} \, dx\\ &=\frac {1}{6} \int \left (-2 \left (15+x^2\right )+\frac {3 e^x \left (11+9 x-11 x^2+x^3\right )}{(-1+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \left (15+x^2\right ) \, dx\right )+\frac {1}{2} \int \frac {e^x \left (11+9 x-11 x^2+x^3\right )}{(-1+x)^2} \, dx\\ &=-5 x-\frac {x^3}{9}+\frac {1}{2} \int \left (-9 e^x+\frac {10 e^x}{(-1+x)^2}-\frac {10 e^x}{-1+x}+e^x x\right ) \, dx\\ &=-5 x-\frac {x^3}{9}+\frac {1}{2} \int e^x x \, dx-\frac {9 \int e^x \, dx}{2}+5 \int \frac {e^x}{(-1+x)^2} \, dx-5 \int \frac {e^x}{-1+x} \, dx\\ &=-\frac {9 e^x}{2}+\frac {5 e^x}{1-x}-5 x+\frac {e^x x}{2}-\frac {x^3}{9}-5 e \text {Ei}(-1+x)-\frac {\int e^x \, dx}{2}+5 \int \frac {e^x}{-1+x} \, dx\\ &=-5 e^x+\frac {5 e^x}{1-x}-5 x+\frac {e^x x}{2}-\frac {x^3}{9}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 28, normalized size = 0.93 \begin {gather*} -5 x-\frac {x^3}{9}+\frac {1}{2} e^x \left (-10-\frac {10}{-1+x}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30 + 60*x - 32*x^2 + 4*x^3 - 2*x^4 + E^x*(33 + 27*x - 33*x^2 + 3*x^3))/(6 - 12*x + 6*x^2),x]

[Out]

-5*x - x^3/9 + (E^x*(-10 - 10/(-1 + x) + x))/2

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fricas [A]  time = 0.63, size = 37, normalized size = 1.23 \begin {gather*} -\frac {2 \, x^{4} - 2 \, x^{3} + 90 \, x^{2} - 9 \, {\left (x^{2} - 11 \, x\right )} e^{x} - 90 \, x}{18 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3-33*x^2+27*x+33)*exp(x)-2*x^4+4*x^3-32*x^2+60*x-30)/(6*x^2-12*x+6),x, algorithm="fricas")

[Out]

-1/18*(2*x^4 - 2*x^3 + 90*x^2 - 9*(x^2 - 11*x)*e^x - 90*x)/(x - 1)

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giac [A]  time = 0.18, size = 38, normalized size = 1.27 \begin {gather*} -\frac {2 \, x^{4} - 2 \, x^{3} - 9 \, x^{2} e^{x} + 90 \, x^{2} + 99 \, x e^{x} - 90 \, x}{18 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3-33*x^2+27*x+33)*exp(x)-2*x^4+4*x^3-32*x^2+60*x-30)/(6*x^2-12*x+6),x, algorithm="giac")

[Out]

-1/18*(2*x^4 - 2*x^3 - 9*x^2*e^x + 90*x^2 + 99*x*e^x - 90*x)/(x - 1)

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maple [A]  time = 0.42, size = 23, normalized size = 0.77

method	result	size
risch	\(-\frac {x^{3}}{9}-5 x +\frac {x \left (x -11\right ) {\mathrm e}^{x}}{2 x -2}\)	\(23\)
default	\(-5 x -\frac {x^{3}}{9}-\frac {5 \,{\mathrm e}^{x}}{x -1}-5 \,{\mathrm e}^{x}+\frac {{\mathrm e}^{x} x}{2}\)	\(28\)
norman	\(\frac {-5 x^{2}+\frac {x^{3}}{9}-\frac {x^{4}}{9}-\frac {11 \,{\mathrm e}^{x} x}{2}+\frac {{\mathrm e}^{x} x^{2}}{2}+5}{x -1}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^3-33*x^2+27*x+33)*exp(x)-2*x^4+4*x^3-32*x^2+60*x-30)/(6*x^2-12*x+6),x,method=_RETURNVERBOSE)

[Out]

-1/9*x^3-5*x+1/2*x*(x-11)/(x-1)*exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{9} \, x^{3} - 5 \, x + \frac {{\left (x^{2} - 11 \, x\right )} e^{x}}{2 \, {\left (x - 1\right )}} - \frac {11 \, e E_{2}\left (-x + 1\right )}{2 \, {\left (x - 1\right )}} - \frac {11}{2} \, \int \frac {e^{x}}{x^{2} - 2 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3-33*x^2+27*x+33)*exp(x)-2*x^4+4*x^3-32*x^2+60*x-30)/(6*x^2-12*x+6),x, algorithm="maxima")

[Out]

-1/9*x^3 - 5*x + 1/2*(x^2 - 11*x)*e^x/(x - 1) - 11/2*e*exp_integral_e(2, -x + 1)/(x - 1) - 11/2*integrate(e^x/
(x^2 - 2*x + 1), x)

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mupad [B]  time = 0.13, size = 27, normalized size = 0.90 \begin {gather*} x\,\left (\frac {{\mathrm {e}}^x}{2}-5\right )-5\,{\mathrm {e}}^x-\frac {5\,{\mathrm {e}}^x}{x-1}-\frac {x^3}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*x - 32*x^2 + 4*x^3 - 2*x^4 + exp(x)*(27*x - 33*x^2 + 3*x^3 + 33) - 30)/(6*x^2 - 12*x + 6),x)

[Out]

x*(exp(x)/2 - 5) - 5*exp(x) - (5*exp(x))/(x - 1) - x^3/9

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sympy [A]  time = 0.14, size = 22, normalized size = 0.73 \begin {gather*} - \frac {x^{3}}{9} - 5 x + \frac {\left (x^{2} - 11 x\right ) e^{x}}{2 x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**3-33*x**2+27*x+33)*exp(x)-2*x**4+4*x**3-32*x**2+60*x-30)/(6*x**2-12*x+6),x)

[Out]

-x**3/9 - 5*x + (x**2 - 11*x)*exp(x)/(2*x - 2)

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