3.88.3 \(\int \frac {40-95 x+30 x^2+e^x (16-38 x+12 x^2)+(15-15 x+15 x^2+e^x (-10+16 x-16 x^2+6 x^3)) \log (-1+x-x^2)}{1-x+x^2} \, dx\)

Optimal. Leaf size=22 \[ \left (5+2 e^x\right ) (-8+3 x) \log \left (-1+x-x^2\right ) \]

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Rubi [B]  time = 1.03, antiderivative size = 57, normalized size of antiderivative = 2.59, number of steps used = 31, number of rules used = 11, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {6688, 6728, 773, 634, 618, 204, 628, 2194, 2178, 2176, 2554} \begin {gather*} -6 e^x \log \left (-x^2+x-1\right )-2 e^x (5-3 x) \log \left (-x^2+x-1\right )+15 x \log \left (-x^2+x-1\right )-40 \log \left (x^2-x+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 - 95*x + 30*x^2 + E^x*(16 - 38*x + 12*x^2) + (15 - 15*x + 15*x^2 + E^x*(-10 + 16*x - 16*x^2 + 6*x^3))*
Log[-1 + x - x^2])/(1 - x + x^2),x]

[Out]

-6*E^x*Log[-1 + x - x^2] - 2*E^x*(5 - 3*x)*Log[-1 + x - x^2] + 15*x*Log[-1 + x - x^2] - 40*Log[1 - x + x^2]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

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 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]]

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 \left (\frac {\left (5+2 e^x\right ) \left (8-19 x+6 x^2\right )}{1-x+x^2}+\left (15+2 e^x (-5+3 x)\right ) \log \left (-1+x-x^2\right )\right ) \, dx\\ &=\int \frac {\left (5+2 e^x\right ) \left (8-19 x+6 x^2\right )}{1-x+x^2} \, dx+\int \left (15+2 e^x (-5+3 x)\right ) \log \left (-1+x-x^2\right ) \, dx\\ &=-6 e^x \log \left (-1+x-x^2\right )-2 e^x (5-3 x) \log \left (-1+x-x^2\right )+15 x \log \left (-1+x-x^2\right )-\int \frac {(1-2 x) \left (-15 x-2 e^x (-8+3 x)\right )}{1-x+x^2} \, dx+\int \left (\frac {5 (-1+2 x) (-8+3 x)}{1-x+x^2}+\frac {2 e^x (-1+2 x) (-8+3 x)}{1-x+x^2}\right ) \, dx\\ &=-6 e^x \log \left (-1+x-x^2\right )-2 e^x (5-3 x) \log \left (-1+x-x^2\right )+15 x \log \left (-1+x-x^2\right )+2 \int \frac {e^x (-1+2 x) (-8+3 x)}{1-x+x^2} \, dx+5 \int \frac {(-1+2 x) (-8+3 x)}{1-x+x^2} \, dx-\int \left (\frac {15 x (-1+2 x)}{1-x+x^2}+\frac {2 e^x (-1+2 x) (-8+3 x)}{1-x+x^2}\right ) \, dx\\ &=30 x-6 e^x \log \left (-1+x-x^2\right )-2 e^x (5-3 x) \log \left (-1+x-x^2\right )+15 x \log \left (-1+x-x^2\right )-2 \int \frac {e^x (-1+2 x) (-8+3 x)}{1-x+x^2} \, dx+2 \int \left (6 e^x+\frac {e^x (2-13 x)}{1-x+x^2}\right ) \, dx+5 \int \frac {2-13 x}{1-x+x^2} \, dx-15 \int \frac {x (-1+2 x)}{1-x+x^2} \, dx\\ &=-6 e^x \log \left (-1+x-x^2\right )-2 e^x (5-3 x) \log \left (-1+x-x^2\right )+15 x \log \left (-1+x-x^2\right )+2 \int \frac {e^x (2-13 x)}{1-x+x^2} \, dx-2 \int \left (6 e^x+\frac {e^x (2-13 x)}{1-x+x^2}\right ) \, dx+12 \int e^x \, dx-15 \int \frac {-2+x}{1-x+x^2} \, dx-\frac {45}{2} \int \frac {1}{1-x+x^2} \, dx-\frac {65}{2} \int \frac {-1+2 x}{1-x+x^2} \, dx\\ &=12 e^x-6 e^x \log \left (-1+x-x^2\right )-2 e^x (5-3 x) \log \left (-1+x-x^2\right )+15 x \log \left (-1+x-x^2\right )-\frac {65}{2} \log \left (1-x+x^2\right )-2 \int \frac {e^x (2-13 x)}{1-x+x^2} \, dx+2 \int \left (\frac {\left (-13+3 i \sqrt {3}\right ) e^x}{-1-i \sqrt {3}+2 x}+\frac {\left (-13-3 i \sqrt {3}\right ) e^x}{-1+i \sqrt {3}+2 x}\right ) \, dx-\frac {15}{2} \int \frac {-1+2 x}{1-x+x^2} \, dx-12 \int e^x \, dx+\frac {45}{2} \int \frac {1}{1-x+x^2} \, dx+45 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=15 \sqrt {3} \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )-6 e^x \log \left (-1+x-x^2\right )-2 e^x (5-3 x) \log \left (-1+x-x^2\right )+15 x \log \left (-1+x-x^2\right )-40 \log \left (1-x+x^2\right )-2 \int \left (\frac {\left (-13+3 i \sqrt {3}\right ) e^x}{-1-i \sqrt {3}+2 x}+\frac {\left (-13-3 i \sqrt {3}\right ) e^x}{-1+i \sqrt {3}+2 x}\right ) \, dx-45 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )-\left (2 \left (13-3 i \sqrt {3}\right )\right ) \int \frac {e^x}{-1-i \sqrt {3}+2 x} \, dx-\left (2 \left (13+3 i \sqrt {3}\right )\right ) \int \frac {e^x}{-1+i \sqrt {3}+2 x} \, dx\\ &=-\left (\left (13-3 i \sqrt {3}\right ) e^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {3}+2 x\right )\right )\right )-\left (13+3 i \sqrt {3}\right ) e^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {3}+2 x\right )\right )-6 e^x \log \left (-1+x-x^2\right )-2 e^x (5-3 x) \log \left (-1+x-x^2\right )+15 x \log \left (-1+x-x^2\right )-40 \log \left (1-x+x^2\right )+\left (2 \left (13-3 i \sqrt {3}\right )\right ) \int \frac {e^x}{-1-i \sqrt {3}+2 x} \, dx+\left (2 \left (13+3 i \sqrt {3}\right )\right ) \int \frac {e^x}{-1+i \sqrt {3}+2 x} \, dx\\ &=-6 e^x \log \left (-1+x-x^2\right )-2 e^x (5-3 x) \log \left (-1+x-x^2\right )+15 x \log \left (-1+x-x^2\right )-40 \log \left (1-x+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 37, normalized size = 1.68 \begin {gather*} \left (-16 e^x+15 x+6 e^x x\right ) \log \left (-1+x-x^2\right )-40 \log \left (1-x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 - 95*x + 30*x^2 + E^x*(16 - 38*x + 12*x^2) + (15 - 15*x + 15*x^2 + E^x*(-10 + 16*x - 16*x^2 + 6*
x^3))*Log[-1 + x - x^2])/(1 - x + x^2),x]

[Out]

(-16*E^x + 15*x + 6*E^x*x)*Log[-1 + x - x^2] - 40*Log[1 - x + x^2]

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fricas [A]  time = 0.55, size = 24, normalized size = 1.09 \begin {gather*} {\left (2 \, {\left (3 \, x - 8\right )} e^{x} + 15 \, x - 40\right )} \log \left (-x^{2} + x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3-16*x^2+16*x-10)*exp(x)+15*x^2-15*x+15)*log(-x^2+x-1)+(12*x^2-38*x+16)*exp(x)+30*x^2-95*x+40
)/(x^2-x+1),x, algorithm="fricas")

[Out]

(2*(3*x - 8)*e^x + 15*x - 40)*log(-x^2 + x - 1)

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giac [B]  time = 0.19, size = 51, normalized size = 2.32 \begin {gather*} 6 \, x e^{x} \log \left (-x^{2} + x - 1\right ) + 15 \, x \log \left (-x^{2} + x - 1\right ) - 16 \, e^{x} \log \left (-x^{2} + x - 1\right ) - 40 \, \log \left (x^{2} - x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3-16*x^2+16*x-10)*exp(x)+15*x^2-15*x+15)*log(-x^2+x-1)+(12*x^2-38*x+16)*exp(x)+30*x^2-95*x+40
)/(x^2-x+1),x, algorithm="giac")

[Out]

6*x*e^x*log(-x^2 + x - 1) + 15*x*log(-x^2 + x - 1) - 16*e^x*log(-x^2 + x - 1) - 40*log(x^2 - x + 1)

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maple [A]  time = 0.83, size = 36, normalized size = 1.64

method	result	size
risch	\(\left (6 \,{\mathrm e}^{x} x +15 x -16 \,{\mathrm e}^{x}\right ) \ln \left (-x^{2}+x -1\right )-40 \ln \left (x^{2}-x +1\right )\)	\(36\)
default	\(-16 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right )+6 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right ) x -40 \ln \left (x^{2}-x +1\right )+15 \ln \left (-x^{2}+x -1\right ) x\)	\(52\)
norman	\(-16 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right )+6 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right ) x -40 \ln \left (x^{2}-x +1\right )+15 \ln \left (-x^{2}+x -1\right ) x\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((6*x^3-16*x^2+16*x-10)*exp(x)+15*x^2-15*x+15)*ln(-x^2+x-1)+(12*x^2-38*x+16)*exp(x)+30*x^2-95*x+40)/(x^2-
x+1),x,method=_RETURNVERBOSE)

[Out]

(6*exp(x)*x+15*x-16*exp(x))*ln(-x^2+x-1)-40*ln(x^2-x+1)

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maxima [A]  time = 0.48, size = 37, normalized size = 1.68 \begin {gather*} \frac {1}{2} \, {\left (4 \, {\left (3 \, x - 8\right )} e^{x} + 30 \, x - 15\right )} \log \left (-x^{2} + x - 1\right ) - \frac {65}{2} \, \log \left (x^{2} - x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3-16*x^2+16*x-10)*exp(x)+15*x^2-15*x+15)*log(-x^2+x-1)+(12*x^2-38*x+16)*exp(x)+30*x^2-95*x+40
)/(x^2-x+1),x, algorithm="maxima")

[Out]

1/2*(4*(3*x - 8)*e^x + 30*x - 15)*log(-x^2 + x - 1) - 65/2*log(x^2 - x + 1)

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mupad [B]  time = 5.91, size = 34, normalized size = 1.55 \begin {gather*} \ln \left (-x^2+x-1\right )\,\left (15\,x+{\mathrm {e}}^x\,\left (6\,x-16\right )\right )-40\,\ln \left (x^2-x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - x^2 - 1)*(15*x^2 - 15*x + exp(x)*(16*x - 16*x^2 + 6*x^3 - 10) + 15) - 95*x + exp(x)*(12*x^2 - 38*
x + 16) + 30*x^2 + 40)/(x^2 - x + 1),x)

[Out]

log(x - x^2 - 1)*(15*x + exp(x)*(6*x - 16)) - 40*log(x^2 - x + 1)

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sympy [B]  time = 0.52, size = 46, normalized size = 2.09 \begin {gather*} 15 x \log {\left (- x^{2} + x - 1 \right )} + \left (6 x \log {\left (- x^{2} + x - 1 \right )} - 16 \log {\left (- x^{2} + x - 1 \right )}\right ) e^{x} - 40 \log {\left (x^{2} - x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x**3-16*x**2+16*x-10)*exp(x)+15*x**2-15*x+15)*ln(-x**2+x-1)+(12*x**2-38*x+16)*exp(x)+30*x**2-95
*x+40)/(x**2-x+1),x)

[Out]

15*x*log(-x**2 + x - 1) + (6*x*log(-x**2 + x - 1) - 16*log(-x**2 + x - 1))*exp(x) - 40*log(x**2 - x + 1)

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