3.23.64 \(\int \frac {25 x-25 x^2-44 x^3-12 x^4+(75+35 x+23 x^2+14 x^3+6 x^4) \log (3-x+x^2)}{375 x^2+175 x^3+85 x^4+80 x^5+20 x^6} \, dx\)

Optimal. Leaf size=30 \[ \frac {\log \left (3-x+x^2\right )}{5 \left (-x+\frac {x^2}{5+3 x}\right )} \]

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Rubi [A]  time = 1.07, antiderivative size = 37, normalized size of antiderivative = 1.23, number of steps used = 45, number of rules used = 14, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {6741, 12, 6728, 709, 800, 634, 618, 204, 628, 893, 1628, 2528, 2525, 2074} \begin {gather*} -\frac {\log \left (x^2-x+3\right )}{5 x}-\frac {\log \left (x^2-x+3\right )}{5 (2 x+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25*x - 25*x^2 - 44*x^3 - 12*x^4 + (75 + 35*x + 23*x^2 + 14*x^3 + 6*x^4)*Log[3 - x + x^2])/(375*x^2 + 175*
x^3 + 85*x^4 + 80*x^5 + 20*x^6),x]

[Out]

-1/5*Log[3 - x + x^2]/x - Log[3 - x + x^2]/(5*(5 + 2*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 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 709

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

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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 2528

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

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]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 x-25 x^2-44 x^3-12 x^4+\left (75+35 x+23 x^2+14 x^3+6 x^4\right ) \log \left (3-x+x^2\right )}{5 x^2 (5+2 x)^2 \left (3-x+x^2\right )} \, dx\\ &=\frac {1}{5} \int \frac {25 x-25 x^2-44 x^3-12 x^4+\left (75+35 x+23 x^2+14 x^3+6 x^4\right ) \log \left (3-x+x^2\right )}{x^2 (5+2 x)^2 \left (3-x+x^2\right )} \, dx\\ &=\frac {1}{5} \int \left (-\frac {25}{(5+2 x)^2 \left (3-x+x^2\right )}+\frac {25}{x (5+2 x)^2 \left (3-x+x^2\right )}-\frac {44 x}{(5+2 x)^2 \left (3-x+x^2\right )}-\frac {12 x^2}{(5+2 x)^2 \left (3-x+x^2\right )}+\frac {\left (25+20 x+6 x^2\right ) \log \left (3-x+x^2\right )}{x^2 (5+2 x)^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {\left (25+20 x+6 x^2\right ) \log \left (3-x+x^2\right )}{x^2 (5+2 x)^2} \, dx-\frac {12}{5} \int \frac {x^2}{(5+2 x)^2 \left (3-x+x^2\right )} \, dx-5 \int \frac {1}{(5+2 x)^2 \left (3-x+x^2\right )} \, dx+5 \int \frac {1}{x (5+2 x)^2 \left (3-x+x^2\right )} \, dx-\frac {44}{5} \int \frac {x}{(5+2 x)^2 \left (3-x+x^2\right )} \, dx\\ &=\frac {10}{47 (5+2 x)}-\frac {5}{47} \int \frac {7-2 x}{(5+2 x) \left (3-x+x^2\right )} \, dx+\frac {1}{5} \int \left (\frac {\log \left (3-x+x^2\right )}{x^2}+\frac {2 \log \left (3-x+x^2\right )}{(5+2 x)^2}\right ) \, dx-\frac {12}{5} \int \left (\frac {25}{47 (5+2 x)^2}-\frac {170}{2209 (5+2 x)}+\frac {-39+85 x}{2209 \left (3-x+x^2\right )}\right ) \, dx+5 \int \left (\frac {1}{75 x}-\frac {8}{235 (5+2 x)^2}-\frac {856}{55225 (5+2 x)}+\frac {-35-37 x}{6627 \left (3-x+x^2\right )}\right ) \, dx-\frac {44}{5} \int \left (-\frac {10}{47 (5+2 x)^2}-\frac {26}{2209 (5+2 x)}+\frac {72+13 x}{2209 \left (3-x+x^2\right )}\right ) \, dx\\ &=\frac {\log (x)}{15}+\frac {1164 \log (5+2 x)}{11045}+\frac {5 \int \frac {-35-37 x}{3-x+x^2} \, dx}{6627}-\frac {12 \int \frac {-39+85 x}{3-x+x^2} \, dx}{11045}-\frac {44 \int \frac {72+13 x}{3-x+x^2} \, dx}{11045}-\frac {5}{47} \int \left (\frac {48}{47 (5+2 x)}+\frac {37-24 x}{47 \left (3-x+x^2\right )}\right ) \, dx+\frac {1}{5} \int \frac {\log \left (3-x+x^2\right )}{x^2} \, dx+\frac {2}{5} \int \frac {\log \left (3-x+x^2\right )}{(5+2 x)^2} \, dx\\ &=\frac {\log (x)}{15}+\frac {12}{235} \log (5+2 x)-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}-\frac {5 \int \frac {37-24 x}{3-x+x^2} \, dx}{2209}-\frac {42 \int \frac {1}{3-x+x^2} \, dx}{11045}-\frac {185 \int \frac {-1+2 x}{3-x+x^2} \, dx}{13254}-\frac {286 \int \frac {-1+2 x}{3-x+x^2} \, dx}{11045}-\frac {535 \int \frac {1}{3-x+x^2} \, dx}{13254}-\frac {102 \int \frac {-1+2 x}{3-x+x^2} \, dx}{2209}+\frac {1}{5} \int \frac {-1+2 x}{x \left (3-x+x^2\right )} \, dx+\frac {1}{5} \int \frac {-1+2 x}{15+x+3 x^2+2 x^3} \, dx-\frac {3454 \int \frac {1}{3-x+x^2} \, dx}{11045}\\ &=\frac {\log (x)}{15}+\frac {12}{235} \log (5+2 x)-\frac {5701 \log \left (3-x+x^2\right )}{66270}-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}+\frac {84 \operatorname {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )}{11045}+\frac {60 \int \frac {-1+2 x}{3-x+x^2} \, dx}{2209}-\frac {125 \int \frac {1}{3-x+x^2} \, dx}{2209}+\frac {535 \operatorname {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )}{6627}+\frac {1}{5} \int \left (-\frac {1}{3 x}+\frac {5+x}{3 \left (3-x+x^2\right )}\right ) \, dx+\frac {1}{5} \int \left (-\frac {24}{47 (5+2 x)}+\frac {5+12 x}{47 \left (3-x+x^2\right )}\right ) \, dx+\frac {6908 \operatorname {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )}{11045}\\ &=\frac {2927 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {11}}\right )}{33135 \sqrt {11}}+\frac {628 \sqrt {11} \tan ^{-1}\left (\frac {1-2 x}{\sqrt {11}}\right )}{11045}-\frac {83 \log \left (3-x+x^2\right )}{1410}-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}+\frac {1}{235} \int \frac {5+12 x}{3-x+x^2} \, dx+\frac {1}{15} \int \frac {5+x}{3-x+x^2} \, dx+\frac {250 \operatorname {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )}{2209}\\ &=\frac {53}{705} \sqrt {11} \tan ^{-1}\left (\frac {1-2 x}{\sqrt {11}}\right )-\frac {83 \log \left (3-x+x^2\right )}{1410}-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}+\frac {6}{235} \int \frac {-1+2 x}{3-x+x^2} \, dx+\frac {1}{30} \int \frac {-1+2 x}{3-x+x^2} \, dx+\frac {11}{235} \int \frac {1}{3-x+x^2} \, dx+\frac {11}{30} \int \frac {1}{3-x+x^2} \, dx\\ &=\frac {53}{705} \sqrt {11} \tan ^{-1}\left (\frac {1-2 x}{\sqrt {11}}\right )-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}-\frac {22}{235} \operatorname {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )-\frac {11}{15} \operatorname {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,-1+2 x\right )\\ &=-\frac {\log \left (3-x+x^2\right )}{5 x}-\frac {\log \left (3-x+x^2\right )}{5 (5+2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 28, normalized size = 0.93 \begin {gather*} -\frac {(5+3 x) \log \left (3-x+x^2\right )}{5 x (5+2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x - 25*x^2 - 44*x^3 - 12*x^4 + (75 + 35*x + 23*x^2 + 14*x^3 + 6*x^4)*Log[3 - x + x^2])/(375*x^2
+ 175*x^3 + 85*x^4 + 80*x^5 + 20*x^6),x]

[Out]

-1/5*((5 + 3*x)*Log[3 - x + x^2])/(x*(5 + 2*x))

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fricas [A]  time = 0.77, size = 27, normalized size = 0.90 \begin {gather*} -\frac {{\left (3 \, x + 5\right )} \log \left (x^{2} - x + 3\right )}{5 \, {\left (2 \, x^{2} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^4+14*x^3+23*x^2+35*x+75)*log(x^2-x+3)-12*x^4-44*x^3-25*x^2+25*x)/(20*x^6+80*x^5+85*x^4+175*x^3
+375*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 22, normalized size = 0.73 \begin {gather*} -\frac {1}{5} \, {\left (\frac {1}{2 \, x + 5} + \frac {1}{x}\right )} \log \left (x^{2} - x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^4+14*x^3+23*x^2+35*x+75)*log(x^2-x+3)-12*x^4-44*x^3-25*x^2+25*x)/(20*x^6+80*x^5+85*x^4+175*x^3
+375*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 27, normalized size = 0.90

method	result	size
risch	\(-\frac {\left (3 x +5\right ) \ln \left (x^{2}-x +3\right )}{5 \left (5+2 x \right ) x}\)	\(27\)
norman	\(\frac {-\frac {3 \ln \left (x^{2}-x +3\right ) x}{5}-\ln \left (x^{2}-x +3\right )}{\left (5+2 x \right ) x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^4+14*x^3+23*x^2+35*x+75)*ln(x^2-x+3)-12*x^4-44*x^3-25*x^2+25*x)/(20*x^6+80*x^5+85*x^4+175*x^3+375*x^
2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.97, size = 27, normalized size = 0.90 \begin {gather*} -\frac {{\left (3 \, x + 5\right )} \log \left (x^{2} - x + 3\right )}{5 \, {\left (2 \, x^{2} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^4+14*x^3+23*x^2+35*x+75)*log(x^2-x+3)-12*x^4-44*x^3-25*x^2+25*x)/(20*x^6+80*x^5+85*x^4+175*x^3
+375*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.27, size = 26, normalized size = 0.87 \begin {gather*} -\frac {\ln \left (x^2-x+3\right )\,\left (3\,x+5\right )}{5\,x\,\left (2\,x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x^2 - log(x^2 - x + 3)*(35*x + 23*x^2 + 14*x^3 + 6*x^4 + 75) - 25*x + 44*x^3 + 12*x^4)/(375*x^2 + 175
*x^3 + 85*x^4 + 80*x^5 + 20*x^6),x)

[Out]

-(log(x^2 - x + 3)*(3*x + 5))/(5*x*(2*x + 5))

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sympy [A]  time = 0.23, size = 22, normalized size = 0.73 \begin {gather*} \frac {\left (- 3 x - 5\right ) \log {\left (x^{2} - x + 3 \right )}}{10 x^{2} + 25 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**4+14*x**3+23*x**2+35*x+75)*ln(x**2-x+3)-12*x**4-44*x**3-25*x**2+25*x)/(20*x**6+80*x**5+85*x**
4+175*x**3+375*x**2),x)

[Out]

(-3*x - 5)*log(x**2 - x + 3)/(10*x**2 + 25*x)

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