3.13.13 \(\int \frac {-5+10 x-26 x^2+21 x^3-22 x^4+e^{4 x} (-1+x-2 x^2)+e^{2 x} (2 x-2 x^2+4 x^3)+(-5+10 x-28 x^2+22 x^3-22 x^4+e^{4 x} (-1-4 x+6 x^2-8 x^3)+e^{2 x} (4 x+2 x^2-4 x^3+8 x^4)) \log (x)}{5-10 x+25 x^2-20 x^3+20 x^4} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{5} x \left (-5+\frac {\left (-e^{2 x}+x\right )^2}{-1+x-2 x^2}\right ) \log (x) \]

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Rubi [C]  time = 6.67, antiderivative size = 1691, normalized size of antiderivative = 54.55, number of steps used = 157, number of rules used = 23, integrand size = 146, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6742, 614, 618, 204, 638, 722, 738, 773, 634, 628, 800, 2357, 2314, 31, 2317, 2391, 2295, 2270, 2178, 2194, 2177, 2554, 2268}

result too large to display

Antiderivative was successfully verified.

[In]

Int[(-5 + 10*x - 26*x^2 + 21*x^3 - 22*x^4 + E^(4*x)*(-1 + x - 2*x^2) + E^(2*x)*(2*x - 2*x^2 + 4*x^3) + (-5 + 1
0*x - 28*x^2 + 22*x^3 - 22*x^4 + E^(4*x)*(-1 - 4*x + 6*x^2 - 8*x^3) + E^(2*x)*(4*x + 2*x^2 - 4*x^3 + 8*x^4))*L
og[x])/(5 - 10*x + 25*x^2 - 20*x^3 + 20*x^4),x]

[Out]

(-27*x)/35 + (11*x^2)/35 + (1 - 4*x)/(7*(1 - x + 2*x^2)) - (2*(2 - x))/(7*(1 - x + 2*x^2)) + (26*(2 - x)*x)/(3
5*(1 - x + 2*x^2)) - (3*(2 - x)*x^2)/(5*(1 - x + 2*x^2)) + (22*(2 - x)*x^3)/(35*(1 - x + 2*x^2)) - (3*ArcTan[(
1 - 4*x)/Sqrt[7]])/(20*Sqrt[7]) + Log[1 - I*Sqrt[7] - 4*x]/28 - (3*Log[1 - I*Sqrt[7] - 4*x])/(35*(1 - I*Sqrt[7
])) + Log[1 + I*Sqrt[7] - 4*x]/28 - (3*Log[1 + I*Sqrt[7] - 4*x])/(35*(1 + I*Sqrt[7])) + (E^(2*x)*Log[x])/5 - (
4*E^(2*x)*Log[x])/(35*(1 - I*Sqrt[7] - 4*x)) - (3*(1 - I*Sqrt[7])*E^(2*x)*Log[x])/(35*(1 - I*Sqrt[7] - 4*x)) +
(8*E^(4*x)*Log[x])/(35*(1 - I*Sqrt[7] - 4*x)) - ((1 - I*Sqrt[7])*E^(4*x)*Log[x])/(35*(1 - I*Sqrt[7] - 4*x)) -
(4*E^(2*x)*Log[x])/(35*(1 + I*Sqrt[7] - 4*x)) - (3*(1 + I*Sqrt[7])*E^(2*x)*Log[x])/(35*(1 + I*Sqrt[7] - 4*x))
+ (8*E^(4*x)*Log[x])/(35*(1 + I*Sqrt[7] - 4*x)) - ((1 + I*Sqrt[7])*E^(4*x)*Log[x])/(35*(1 + I*Sqrt[7] - 4*x))
- (11*x*Log[x])/10 + (x*Log[x])/(7*(1 - I*Sqrt[7] - 4*x)) - (12*x*Log[x])/(35*(1 - I*Sqrt[7])*(1 - I*Sqrt[7]
- 4*x)) + (x*Log[x])/(7*(1 + I*Sqrt[7] - 4*x)) - (12*x*Log[x])/(35*(1 + I*Sqrt[7])*(1 + I*Sqrt[7] - 4*x)) - (2
*E^(1/2 + (I/2)*Sqrt[7])*ExpIntegralEi[(-1 - I*Sqrt[7] + 4*x)/2]*Log[x])/35 - ((I/5)*E^(1/2 + (I/2)*Sqrt[7])*E
xpIntegralEi[(-1 - I*Sqrt[7] + 4*x)/2]*Log[x])/Sqrt[7] - (3*(1 + I*Sqrt[7])*E^(1/2 + (I/2)*Sqrt[7])*ExpIntegra
lEi[(-1 - I*Sqrt[7] + 4*x)/2]*Log[x])/70 + ((7 + (5*I)*Sqrt[7])*E^(1/2 + (I/2)*Sqrt[7])*ExpIntegralEi[(-1 - I*
Sqrt[7] + 4*x)/2]*Log[x])/70 + (E^(1 + I*Sqrt[7])*ExpIntegralEi[-1 - I*Sqrt[7] + 4*x]*Log[x])/35 + ((I/5)*E^(1
+ I*Sqrt[7])*ExpIntegralEi[-1 - I*Sqrt[7] + 4*x]*Log[x])/Sqrt[7] - ((1 + I*Sqrt[7])*E^(1 + I*Sqrt[7])*ExpInte
gralEi[-1 - I*Sqrt[7] + 4*x]*Log[x])/35 - (2*E^(1/2 - (I/2)*Sqrt[7])*ExpIntegralEi[(-1 + I*Sqrt[7] + 4*x)/2]*L
og[x])/35 + ((I/5)*E^(1/2 - (I/2)*Sqrt[7])*ExpIntegralEi[(-1 + I*Sqrt[7] + 4*x)/2]*Log[x])/Sqrt[7] - (3*(1 - I
*Sqrt[7])*E^(1/2 - (I/2)*Sqrt[7])*ExpIntegralEi[(-1 + I*Sqrt[7] + 4*x)/2]*Log[x])/70 + ((7 - (5*I)*Sqrt[7])*E^
(1/2 - (I/2)*Sqrt[7])*ExpIntegralEi[(-1 + I*Sqrt[7] + 4*x)/2]*Log[x])/70 + (E^(1 - I*Sqrt[7])*ExpIntegralEi[-1
+ I*Sqrt[7] + 4*x]*Log[x])/35 - ((I/5)*E^(1 - I*Sqrt[7])*ExpIntegralEi[-1 + I*Sqrt[7] + 4*x]*Log[x])/Sqrt[7]
- ((1 - I*Sqrt[7])*E^(1 - I*Sqrt[7])*ExpIntegralEi[-1 + I*Sqrt[7] + 4*x]*Log[x])/35 - (((11*I)/4)*Log[x]*Log[1
- (4*x)/(1 - I*Sqrt[7])])/Sqrt[7] - (11*(7 - (2*I)*Sqrt[7])*Log[x]*Log[1 - (4*x)/(1 - I*Sqrt[7])])/140 + (11*
(7 + (3*I)*Sqrt[7])*Log[x]*Log[1 - (4*x)/(1 - I*Sqrt[7])])/140 + (((11*I)/4)*Log[x]*Log[1 - (4*x)/(1 + I*Sqrt[
7])])/Sqrt[7] - (11*(7 + (2*I)*Sqrt[7])*Log[x]*Log[1 - (4*x)/(1 + I*Sqrt[7])])/140 + (11*(7 - (3*I)*Sqrt[7])*L
og[x]*Log[1 - (4*x)/(1 + I*Sqrt[7])])/140 - Log[1 - x + 2*x^2]/40 - (((11*I)/4)*PolyLog[2, (4*x)/(1 - I*Sqrt[7
])])/Sqrt[7] - (11*(7 - (2*I)*Sqrt[7])*PolyLog[2, (4*x)/(1 - I*Sqrt[7])])/140 + (11*(7 + (3*I)*Sqrt[7])*PolyLo
g[2, (4*x)/(1 - I*Sqrt[7])])/140 + (((11*I)/4)*PolyLog[2, (4*x)/(1 + I*Sqrt[7])])/Sqrt[7] - (11*(7 + (2*I)*Sqr
t[7])*PolyLog[2, (4*x)/(1 + I*Sqrt[7])])/140 + (11*(7 - (3*I)*Sqrt[7])*PolyLog[2, (4*x)/(1 + I*Sqrt[7])])/140

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 722

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*(2*p + 3)*(c*d
^2 - b*d*e + a*e^2))/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ
[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && LtQ[p, -1]

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + 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] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 2268

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[F^(g*(d + e*x)^n), 1/(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x]

Rule 2270

Int[((F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))*(u_)^(m_.))/((a_.) + (b_.)*(x_) + (c_)*(x_)^2), x_Symbol] :> Int[
ExpandIntegrand[F^(g*(d + e*x)^n), u^m/(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x] && Poly
nomialQ[u, x] && IntegerQ[m]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2317

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

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{\left (1-x+2 x^2\right )^2}+\frac {2 x}{\left (1-x+2 x^2\right )^2}-\frac {26 x^2}{5 \left (1-x+2 x^2\right )^2}+\frac {21 x^3}{5 \left (1-x+2 x^2\right )^2}-\frac {22 x^4}{5 \left (1-x+2 x^2\right )^2}-\frac {\log (x)}{\left (1-x+2 x^2\right )^2}+\frac {2 x \log (x)}{\left (1-x+2 x^2\right )^2}-\frac {28 x^2 \log (x)}{5 \left (1-x+2 x^2\right )^2}+\frac {22 x^3 \log (x)}{5 \left (1-x+2 x^2\right )^2}-\frac {22 x^4 \log (x)}{5 \left (1-x+2 x^2\right )^2}+\frac {2 e^{2 x} x \left (1-x+2 x^2+2 \log (x)+x \log (x)-2 x^2 \log (x)+4 x^3 \log (x)\right )}{5 \left (1-x+2 x^2\right )^2}-\frac {e^{4 x} \left (1-x+2 x^2+\log (x)+4 x \log (x)-6 x^2 \log (x)+8 x^3 \log (x)\right )}{5 \left (1-x+2 x^2\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{4 x} \left (1-x+2 x^2+\log (x)+4 x \log (x)-6 x^2 \log (x)+8 x^3 \log (x)\right )}{\left (1-x+2 x^2\right )^2} \, dx\right )+\frac {2}{5} \int \frac {e^{2 x} x \left (1-x+2 x^2+2 \log (x)+x \log (x)-2 x^2 \log (x)+4 x^3 \log (x)\right )}{\left (1-x+2 x^2\right )^2} \, dx+2 \int \frac {x}{\left (1-x+2 x^2\right )^2} \, dx+2 \int \frac {x \log (x)}{\left (1-x+2 x^2\right )^2} \, dx+\frac {21}{5} \int \frac {x^3}{\left (1-x+2 x^2\right )^2} \, dx-\frac {22}{5} \int \frac {x^4}{\left (1-x+2 x^2\right )^2} \, dx+\frac {22}{5} \int \frac {x^3 \log (x)}{\left (1-x+2 x^2\right )^2} \, dx-\frac {22}{5} \int \frac {x^4 \log (x)}{\left (1-x+2 x^2\right )^2} \, dx-\frac {26}{5} \int \frac {x^2}{\left (1-x+2 x^2\right )^2} \, dx-\frac {28}{5} \int \frac {x^2 \log (x)}{\left (1-x+2 x^2\right )^2} \, dx-\int \frac {1}{\left (1-x+2 x^2\right )^2} \, dx-\int \frac {\log (x)}{\left (1-x+2 x^2\right )^2} \, dx\\ &=\frac {1-4 x}{7 \left (1-x+2 x^2\right )}-\frac {2 (2-x)}{7 \left (1-x+2 x^2\right )}+\frac {26 (2-x) x}{35 \left (1-x+2 x^2\right )}-\frac {3 (2-x) x^2}{5 \left (1-x+2 x^2\right )}+\frac {22 (2-x) x^3}{35 \left (1-x+2 x^2\right )}-\frac {1}{5} \int \left (\frac {e^{4 x}}{1-x+2 x^2}+\frac {e^{4 x} \left (1+4 x-6 x^2+8 x^3\right ) \log (x)}{\left (1-x+2 x^2\right )^2}\right ) \, dx+\frac {2}{7} \int \frac {1}{1-x+2 x^2} \, dx+\frac {2}{5} \int \left (\frac {e^{2 x} x}{1-x+2 x^2}+\frac {e^{2 x} x \left (2+x-2 x^2+4 x^3\right ) \log (x)}{\left (1-x+2 x^2\right )^2}\right ) \, dx-\frac {4}{7} \int \frac {1}{1-x+2 x^2} \, dx+\frac {3}{5} \int \frac {(4-x) x}{1-x+2 x^2} \, dx-\frac {22}{35} \int \frac {(6-2 x) x^2}{1-x+2 x^2} \, dx-\frac {52}{35} \int \frac {1}{1-x+2 x^2} \, dx+2 \int \left (-\frac {4 \left (1+i \sqrt {7}\right ) \log (x)}{7 \left (1+i \sqrt {7}-4 x\right )^2}+\frac {4 i \log (x)}{7 \sqrt {7} \left (1+i \sqrt {7}-4 x\right )}-\frac {4 \left (1-i \sqrt {7}\right ) \log (x)}{7 \left (-1+i \sqrt {7}+4 x\right )^2}+\frac {4 i \log (x)}{7 \sqrt {7} \left (-1+i \sqrt {7}+4 x\right )}\right ) \, dx+\frac {22}{5} \int \left (\frac {(-1-x) \log (x)}{4 \left (1-x+2 x^2\right )^2}+\frac {(1+2 x) \log (x)}{4 \left (1-x+2 x^2\right )}\right ) \, dx-\frac {22}{5} \int \left (\frac {\log (x)}{4}+\frac {(1-3 x) \log (x)}{8 \left (1-x+2 x^2\right )^2}+\frac {(-3+4 x) \log (x)}{8 \left (1-x+2 x^2\right )}\right ) \, dx-\frac {28}{5} \int \left (\frac {(-1+x) \log (x)}{2 \left (1-x+2 x^2\right )^2}+\frac {\log (x)}{2 \left (1-x+2 x^2\right )}\right ) \, dx-\int \left (-\frac {16 \log (x)}{7 \left (1+i \sqrt {7}-4 x\right )^2}+\frac {16 i \log (x)}{7 \sqrt {7} \left (1+i \sqrt {7}-4 x\right )}-\frac {16 \log (x)}{7 \left (-1+i \sqrt {7}+4 x\right )^2}+\frac {16 i \log (x)}{7 \sqrt {7} \left (-1+i \sqrt {7}+4 x\right )}\right ) \, dx\\ &=-\frac {3 x}{10}+\frac {1-4 x}{7 \left (1-x+2 x^2\right )}-\frac {2 (2-x)}{7 \left (1-x+2 x^2\right )}+\frac {26 (2-x) x}{35 \left (1-x+2 x^2\right )}-\frac {3 (2-x) x^2}{5 \left (1-x+2 x^2\right )}+\frac {22 (2-x) x^3}{35 \left (1-x+2 x^2\right )}-\frac {1}{5} \int \frac {e^{4 x}}{1-x+2 x^2} \, dx-\frac {1}{5} \int \frac {e^{4 x} \left (1+4 x-6 x^2+8 x^3\right ) \log (x)}{\left (1-x+2 x^2\right )^2} \, dx+\frac {3}{10} \int \frac {1+7 x}{1-x+2 x^2} \, dx+\frac {2}{5} \int \frac {e^{2 x} x}{1-x+2 x^2} \, dx+\frac {2}{5} \int \frac {e^{2 x} x \left (2+x-2 x^2+4 x^3\right ) \log (x)}{\left (1-x+2 x^2\right )^2} \, dx-\frac {11}{20} \int \frac {(1-3 x) \log (x)}{\left (1-x+2 x^2\right )^2} \, dx-\frac {11}{20} \int \frac {(-3+4 x) \log (x)}{1-x+2 x^2} \, dx-\frac {4}{7} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,-1+4 x\right )-\frac {22}{35} \int \left (\frac {5}{2}-x-\frac {5-7 x}{2 \left (1-x+2 x^2\right )}\right ) \, dx-\frac {11}{10} \int \log (x) \, dx+\frac {11}{10} \int \frac {(-1-x) \log (x)}{\left (1-x+2 x^2\right )^2} \, dx+\frac {11}{10} \int \frac {(1+2 x) \log (x)}{1-x+2 x^2} \, dx+\frac {8}{7} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,-1+4 x\right )+\frac {16}{7} \int \frac {\log (x)}{\left (1+i \sqrt {7}-4 x\right )^2} \, dx+\frac {16}{7} \int \frac {\log (x)}{\left (-1+i \sqrt {7}+4 x\right )^2} \, dx-\frac {14}{5} \int \frac {(-1+x) \log (x)}{\left (1-x+2 x^2\right )^2} \, dx-\frac {14}{5} \int \frac {\log (x)}{1-x+2 x^2} \, dx+\frac {104}{35} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,-1+4 x\right )+\frac {(8 i) \int \frac {\log (x)}{1+i \sqrt {7}-4 x} \, dx}{7 \sqrt {7}}+\frac {(8 i) \int \frac {\log (x)}{-1+i \sqrt {7}+4 x} \, dx}{7 \sqrt {7}}-\frac {(16 i) \int \frac {\log (x)}{1+i \sqrt {7}-4 x} \, dx}{7 \sqrt {7}}-\frac {(16 i) \int \frac {\log (x)}{-1+i \sqrt {7}+4 x} \, dx}{7 \sqrt {7}}-\frac {1}{7} \left (8 \left (1-i \sqrt {7}\right )\right ) \int \frac {\log (x)}{\left (-1+i \sqrt {7}+4 x\right )^2} \, dx-\frac {1}{7} \left (8 \left (1+i \sqrt {7}\right )\right ) \int \frac {\log (x)}{\left (1+i \sqrt {7}-4 x\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 42, normalized size = 1.35 \begin {gather*} -\frac {x \left (5+e^{4 x}-5 x-2 e^{2 x} x+11 x^2\right ) \log (x)}{5 \left (1-x+2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 10*x - 26*x^2 + 21*x^3 - 22*x^4 + E^(4*x)*(-1 + x - 2*x^2) + E^(2*x)*(2*x - 2*x^2 + 4*x^3) + (
-5 + 10*x - 28*x^2 + 22*x^3 - 22*x^4 + E^(4*x)*(-1 - 4*x + 6*x^2 - 8*x^3) + E^(2*x)*(4*x + 2*x^2 - 4*x^3 + 8*x
^4))*Log[x])/(5 - 10*x + 25*x^2 - 20*x^3 + 20*x^4),x]

[Out]

-1/5*(x*(5 + E^(4*x) - 5*x - 2*E^(2*x)*x + 11*x^2)*Log[x])/(1 - x + 2*x^2)

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fricas [A]  time = 0.68, size = 45, normalized size = 1.45 \begin {gather*} -\frac {{\left (11 \, x^{3} - 2 \, x^{2} e^{\left (2 \, x\right )} - 5 \, x^{2} + x e^{\left (4 \, x\right )} + 5 \, x\right )} \log \relax (x)}{5 \, {\left (2 \, x^{2} - x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^3+6*x^2-4*x-1)*exp(x)^4+(8*x^4-4*x^3+2*x^2+4*x)*exp(x)^2-22*x^4+22*x^3-28*x^2+10*x-5)*log(x)
+(-2*x^2+x-1)*exp(x)^4+(4*x^3-2*x^2+2*x)*exp(x)^2-22*x^4+21*x^3-26*x^2+10*x-5)/(20*x^4-20*x^3+25*x^2-10*x+5),x
, algorithm="fricas")

[Out]

-1/5*(11*x^3 - 2*x^2*e^(2*x) - 5*x^2 + x*e^(4*x) + 5*x)*log(x)/(2*x^2 - x + 1)

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giac [A]  time = 0.59, size = 53, normalized size = 1.71 \begin {gather*} -\frac {11 \, x^{3} \log \relax (x) - 2 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x) - 5 \, x^{2} \log \relax (x) + x e^{\left (4 \, x\right )} \log \relax (x) + 5 \, x \log \relax (x)}{5 \, {\left (2 \, x^{2} - x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^3+6*x^2-4*x-1)*exp(x)^4+(8*x^4-4*x^3+2*x^2+4*x)*exp(x)^2-22*x^4+22*x^3-28*x^2+10*x-5)*log(x)
+(-2*x^2+x-1)*exp(x)^4+(4*x^3-2*x^2+2*x)*exp(x)^2-22*x^4+21*x^3-26*x^2+10*x-5)/(20*x^4-20*x^3+25*x^2-10*x+5),x
, algorithm="giac")

[Out]

-1/5*(11*x^3*log(x) - 2*x^2*e^(2*x)*log(x) - 5*x^2*log(x) + x*e^(4*x)*log(x) + 5*x*log(x))/(2*x^2 - x + 1)

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maple [A]  time = 0.09, size = 53, normalized size = 1.71

method	result	size
risch	\(-\frac {\left (4 x \,{\mathrm e}^{4 x}-8 \,{\mathrm e}^{2 x} x^{2}+44 x^{3}-22 x^{2}+21 x -1\right ) \ln \relax (x )}{20 \left (2 x^{2}-x +1\right )}-\frac {\ln \relax (x )}{20}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x^3+6*x^2-4*x-1)*exp(x)^4+(8*x^4-4*x^3+2*x^2+4*x)*exp(x)^2-22*x^4+22*x^3-28*x^2+10*x-5)*ln(x)+(-2*x^
2+x-1)*exp(x)^4+(4*x^3-2*x^2+2*x)*exp(x)^2-22*x^4+21*x^3-26*x^2+10*x-5)/(20*x^4-20*x^3+25*x^2-10*x+5),x,method
=_RETURNVERBOSE)

[Out]

-1/20*(4*x*exp(4*x)-8*exp(2*x)*x^2+44*x^3-22*x^2+21*x-1)/(2*x^2-x+1)*ln(x)-1/20*ln(x)

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maxima [B]  time = 0.58, size = 161, normalized size = 5.19 \begin {gather*} -\frac {11}{10} \, x + \frac {4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x) + 22 \, x^{3} - 2 \, x e^{\left (4 \, x\right )} \log \relax (x) - 11 \, x^{2} - 2 \, {\left (11 \, x^{3} - 5 \, x^{2} + 5 \, x\right )} \log \relax (x) + 11 \, x}{10 \, {\left (2 \, x^{2} - x + 1\right )}} - \frac {3 \, {\left (5 \, x - 3\right )}}{20 \, {\left (2 \, x^{2} - x + 1\right )}} - \frac {4 \, x - 1}{7 \, {\left (2 \, x^{2} - x + 1\right )}} + \frac {13 \, {\left (3 \, x + 1\right )}}{35 \, {\left (2 \, x^{2} - x + 1\right )}} - \frac {11 \, {\left (x + 5\right )}}{140 \, {\left (2 \, x^{2} - x + 1\right )}} + \frac {2 \, {\left (x - 2\right )}}{7 \, {\left (2 \, x^{2} - x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^3+6*x^2-4*x-1)*exp(x)^4+(8*x^4-4*x^3+2*x^2+4*x)*exp(x)^2-22*x^4+22*x^3-28*x^2+10*x-5)*log(x)
+(-2*x^2+x-1)*exp(x)^4+(4*x^3-2*x^2+2*x)*exp(x)^2-22*x^4+21*x^3-26*x^2+10*x-5)/(20*x^4-20*x^3+25*x^2-10*x+5),x
, algorithm="maxima")

[Out]

-11/10*x + 1/10*(4*x^2*e^(2*x)*log(x) + 22*x^3 - 2*x*e^(4*x)*log(x) - 11*x^2 - 2*(11*x^3 - 5*x^2 + 5*x)*log(x)
+ 11*x)/(2*x^2 - x + 1) - 3/20*(5*x - 3)/(2*x^2 - x + 1) - 1/7*(4*x - 1)/(2*x^2 - x + 1) + 13/35*(3*x + 1)/(2
*x^2 - x + 1) - 11/140*(x + 5)/(2*x^2 - x + 1) + 2/7*(x - 2)/(2*x^2 - x + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{4\,x}\,\left (2\,x^2-x+1\right )-10\,x-{\mathrm {e}}^{2\,x}\,\left (4\,x^3-2\,x^2+2\,x\right )+26\,x^2-21\,x^3+22\,x^4+\ln \relax (x)\,\left ({\mathrm {e}}^{4\,x}\,\left (8\,x^3-6\,x^2+4\,x+1\right )-10\,x-{\mathrm {e}}^{2\,x}\,\left (8\,x^4-4\,x^3+2\,x^2+4\,x\right )+28\,x^2-22\,x^3+22\,x^4+5\right )+5}{20\,x^4-20\,x^3+25\,x^2-10\,x+5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4*x)*(2*x^2 - x + 1) - 10*x - exp(2*x)*(2*x - 2*x^2 + 4*x^3) + 26*x^2 - 21*x^3 + 22*x^4 + log(x)*(ex
p(4*x)*(4*x - 6*x^2 + 8*x^3 + 1) - 10*x - exp(2*x)*(4*x + 2*x^2 - 4*x^3 + 8*x^4) + 28*x^2 - 22*x^3 + 22*x^4 +
5) + 5)/(25*x^2 - 10*x - 20*x^3 + 20*x^4 + 5),x)

[Out]

int(-(exp(4*x)*(2*x^2 - x + 1) - 10*x - exp(2*x)*(2*x - 2*x^2 + 4*x^3) + 26*x^2 - 21*x^3 + 22*x^4 + log(x)*(ex
p(4*x)*(4*x - 6*x^2 + 8*x^3 + 1) - 10*x - exp(2*x)*(4*x + 2*x^2 - 4*x^3 + 8*x^4) + 28*x^2 - 22*x^3 + 22*x^4 +
5) + 5)/(25*x^2 - 10*x - 20*x^3 + 20*x^4 + 5), x)

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sympy [B]  time = 0.58, size = 112, normalized size = 3.61 \begin {gather*} \frac {\left (- 10 x^{3} \log {\relax (x )} + 5 x^{2} \log {\relax (x )} - 5 x \log {\relax (x )}\right ) e^{4 x} + \left (20 x^{4} \log {\relax (x )} - 10 x^{3} \log {\relax (x )} + 10 x^{2} \log {\relax (x )}\right ) e^{2 x}}{100 x^{4} - 100 x^{3} + 125 x^{2} - 50 x + 25} - \frac {\log {\relax (x )}}{20} + \frac {\left (- 44 x^{3} + 22 x^{2} - 21 x + 1\right ) \log {\relax (x )}}{40 x^{2} - 20 x + 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x**3+6*x**2-4*x-1)*exp(x)**4+(8*x**4-4*x**3+2*x**2+4*x)*exp(x)**2-22*x**4+22*x**3-28*x**2+10*x
-5)*ln(x)+(-2*x**2+x-1)*exp(x)**4+(4*x**3-2*x**2+2*x)*exp(x)**2-22*x**4+21*x**3-26*x**2+10*x-5)/(20*x**4-20*x*
*3+25*x**2-10*x+5),x)

[Out]

((-10*x**3*log(x) + 5*x**2*log(x) - 5*x*log(x))*exp(4*x) + (20*x**4*log(x) - 10*x**3*log(x) + 10*x**2*log(x))*
exp(2*x))/(100*x**4 - 100*x**3 + 125*x**2 - 50*x + 25) - log(x)/20 + (-44*x**3 + 22*x**2 - 21*x + 1)*log(x)/(4
0*x**2 - 20*x + 20)

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