3.98.86 \(\int \frac {-24-48 x-33 x^2-29 x^3-25 x^4-2 x^5+(8 x^2+8 x^3-x^4-x^5) \log (x+x^2)}{4 x^4+4 x^5} \, dx\)

Optimal. Leaf size=30 \[ \frac {\left (2-x+\left (4+\frac {x}{4}\right ) x\right ) \left (\frac {1}{x^2}-\log \left (x+x^2\right )\right )}{x} \]

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Rubi [B]  time = 0.45, antiderivative size = 99, normalized size of antiderivative = 3.30, number of steps used = 13, number of rules used = 8, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {1593, 6742, 1620, 2513, 14, 2334, 2414, 894} \begin {gather*} \frac {2}{x^3}+\frac {3}{x^2}+\frac {1}{4 x}-\frac {1}{4} \left (x+\frac {8}{x}\right ) \log (x)-3 \log (x)-\frac {1}{4} x \log (x+1)-\frac {2 \log (x+1)}{x}-3 \log (x+1)+\frac {1}{4} x (\log (x)+\log (x+1)-\log (x (x+1)))+\frac {2 (\log (x)+\log (x+1)-\log (x (x+1)))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24 - 48*x - 33*x^2 - 29*x^3 - 25*x^4 - 2*x^5 + (8*x^2 + 8*x^3 - x^4 - x^5)*Log[x + x^2])/(4*x^4 + 4*x^5)
,x]

[Out]

2/x^3 + 3/x^2 + 1/(4*x) - 3*Log[x] - ((8/x + x)*Log[x])/4 - 3*Log[1 + x] - (2*Log[1 + x])/x - (x*Log[1 + x])/4
+ (2*(Log[x] + Log[1 + x] - Log[x*(1 + x)]))/x + (x*(Log[x] + Log[1 + x] - Log[x*(1 + x)]))/4

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 894

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2334

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

Rule 2414

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*(x_)^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = IntHide[x^m*(f + g*x^r)^q, x]}, Dist[a + b*Log[c*(d + e*x)^n], u, x] - Dist[b*e*n, Int[SimplifyI
ntegrand[u/(d + e*x), x], x], x] /; InverseFunctionFreeQ[u, x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, q, r}, x]
&& IntegerQ[m] && IntegerQ[q] && IntegerQ[r]

Rule 2513

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*(RFx_.), x_Symbol] :> Dist[
p*r, Int[RFx*Log[a + b*x], x], x] + (Dist[q*r, Int[RFx*Log[c + d*x], x], x] - Dist[p*r*Log[a + b*x] + q*r*Log[
c + d*x] - Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r], Int[RFx, x], x]) /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] &&
RationalFunctionQ[RFx, x] && NeQ[b*c - a*d, 0] &&  !MatchQ[RFx, (u_.)*(a + b*x)^(m_.)*(c + d*x)^(n_.) /; Integ
ersQ[m, n]]

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 {-24-48 x-33 x^2-29 x^3-25 x^4-2 x^5+\left (8 x^2+8 x^3-x^4-x^5\right ) \log \left (x+x^2\right )}{x^4 (4+4 x)} \, dx\\ &=\int \left (\frac {-24-48 x-33 x^2-29 x^3-25 x^4-2 x^5}{4 x^4 (1+x)}-\frac {\left (-8+x^2\right ) \log (x (1+x))}{4 x^2}\right ) \, dx\\ &=\frac {1}{4} \int \frac {-24-48 x-33 x^2-29 x^3-25 x^4-2 x^5}{x^4 (1+x)} \, dx-\frac {1}{4} \int \frac {\left (-8+x^2\right ) \log (x (1+x))}{x^2} \, dx\\ &=\frac {1}{4} \int \left (-2-\frac {24}{x^4}-\frac {24}{x^3}-\frac {9}{x^2}-\frac {20}{x}-\frac {3}{1+x}\right ) \, dx-\frac {1}{4} \int \frac {\left (-8+x^2\right ) \log (x)}{x^2} \, dx-\frac {1}{4} \int \frac {\left (-8+x^2\right ) \log (1+x)}{x^2} \, dx-\frac {1}{4} (-\log (x)-\log (1+x)+\log (x (1+x))) \int \frac {-8+x^2}{x^2} \, dx\\ &=\frac {2}{x^3}+\frac {3}{x^2}+\frac {9}{4 x}-\frac {x}{2}-5 \log (x)-\frac {1}{4} \left (\frac {8}{x}+x\right ) \log (x)-\frac {3}{4} \log (1+x)-\frac {2 \log (1+x)}{x}-\frac {1}{4} x \log (1+x)+\frac {1}{4} \int \left (1+\frac {8}{x^2}\right ) \, dx+\frac {1}{4} \int \frac {8+x^2}{x (1+x)} \, dx-\frac {1}{4} (-\log (x)-\log (1+x)+\log (x (1+x))) \int \left (1-\frac {8}{x^2}\right ) \, dx\\ &=\frac {2}{x^3}+\frac {3}{x^2}+\frac {1}{4 x}-\frac {x}{4}-5 \log (x)-\frac {1}{4} \left (\frac {8}{x}+x\right ) \log (x)-\frac {3}{4} \log (1+x)-\frac {2 \log (1+x)}{x}-\frac {1}{4} x \log (1+x)+\frac {2 (\log (x)+\log (1+x)-\log (x (1+x)))}{x}+\frac {1}{4} x (\log (x)+\log (1+x)-\log (x (1+x)))+\frac {1}{4} \int \left (1+\frac {8}{x}-\frac {9}{1+x}\right ) \, dx\\ &=\frac {2}{x^3}+\frac {3}{x^2}+\frac {1}{4 x}-3 \log (x)-\frac {1}{4} \left (\frac {8}{x}+x\right ) \log (x)-3 \log (1+x)-\frac {2 \log (1+x)}{x}-\frac {1}{4} x \log (1+x)+\frac {2 (\log (x)+\log (1+x)-\log (x (1+x)))}{x}+\frac {1}{4} x (\log (x)+\log (1+x)-\log (x (1+x)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 48, normalized size = 1.60 \begin {gather*} \frac {1}{4} \left (\frac {8}{x^3}+\frac {12}{x^2}+\frac {1}{x}-12 \log (x)-12 \log (1+x)-\frac {8 \log (x (1+x))}{x}-x \log (x (1+x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24 - 48*x - 33*x^2 - 29*x^3 - 25*x^4 - 2*x^5 + (8*x^2 + 8*x^3 - x^4 - x^5)*Log[x + x^2])/(4*x^4 +
4*x^5),x]

[Out]

(8/x^3 + 12/x^2 + x^(-1) - 12*Log[x] - 12*Log[1 + x] - (8*Log[x*(1 + x)])/x - x*Log[x*(1 + x)])/4

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fricas [A]  time = 0.80, size = 35, normalized size = 1.17 \begin {gather*} \frac {x^{2} - {\left (x^{4} + 12 \, x^{3} + 8 \, x^{2}\right )} \log \left (x^{2} + x\right ) + 12 \, x + 8}{4 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5-x^4+8*x^3+8*x^2)*log(x^2+x)-2*x^5-25*x^4-29*x^3-33*x^2-48*x-24)/(4*x^5+4*x^4),x, algorithm="f
ricas")

[Out]

1/4*(x^2 - (x^4 + 12*x^3 + 8*x^2)*log(x^2 + x) + 12*x + 8)/x^3

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giac [A]  time = 0.23, size = 39, normalized size = 1.30 \begin {gather*} -\frac {1}{4} \, {\left (x + \frac {8}{x}\right )} \log \left (x^{2} + x\right ) + \frac {x^{2} + 12 \, x + 8}{4 \, x^{3}} - 3 \, \log \left (x + 1\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5-x^4+8*x^3+8*x^2)*log(x^2+x)-2*x^5-25*x^4-29*x^3-33*x^2-48*x-24)/(4*x^5+4*x^4),x, algorithm="g
iac")

[Out]

-1/4*(x + 8/x)*log(x^2 + x) + 1/4*(x^2 + 12*x + 8)/x^3 - 3*log(x + 1) - 3*log(x)

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maple [A]  time = 0.11, size = 44, normalized size = 1.47

method	result	size
risch	\(-\frac {\left (x^{2}+8\right ) \ln \left (x^{2}+x \right )}{4 x}-\frac {12 \ln \left (x^{2}+x \right ) x^{3}-x^{2}-12 x -8}{4 x^{3}}\)	\(44\)
default	\(-\frac {x \ln \left (x^{2}+x \right )}{4}-3 \ln \left (x +1\right )-\frac {2 \ln \left (x^{2}+x \right )}{x}+\frac {1}{4 x}-3 \ln \relax (x )+\frac {2}{x^{3}}+\frac {3}{x^{2}}\)	\(47\)
norman	\(\frac {2-3 \ln \left (x^{2}+x \right ) x^{3}+3 x +\frac {x^{2}}{4}-2 x^{2} \ln \left (x^{2}+x \right )-\frac {\ln \left (x^{2}+x \right ) x^{4}}{4}}{x^{3}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^5-x^4+8*x^3+8*x^2)*ln(x^2+x)-2*x^5-25*x^4-29*x^3-33*x^2-48*x-24)/(4*x^5+4*x^4),x,method=_RETURNVERBOS
E)

[Out]

-1/4*(x^2+8)/x*ln(x^2+x)-1/4*(12*ln(x^2+x)*x^3-x^2-12*x-8)/x^3

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maxima [B]  time = 0.39, size = 81, normalized size = 2.70 \begin {gather*} -\frac {1}{2} \, x + \frac {2 \, x^{2} - {\left (x^{2} + 9 \, x + 8\right )} \log \left (x + 1\right ) - {\left (x^{2} - 8 \, x + 8\right )} \log \relax (x) - 8}{4 \, x} - \frac {6 \, {\left (2 \, x - 1\right )}}{x^{2}} + \frac {33}{4 \, x} + \frac {6 \, x^{2} - 3 \, x + 2}{x^{3}} - \frac {3}{4} \, \log \left (x + 1\right ) - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5-x^4+8*x^3+8*x^2)*log(x^2+x)-2*x^5-25*x^4-29*x^3-33*x^2-48*x-24)/(4*x^5+4*x^4),x, algorithm="m
axima")

[Out]

-1/2*x + 1/4*(2*x^2 - (x^2 + 9*x + 8)*log(x + 1) - (x^2 - 8*x + 8)*log(x) - 8)/x - 6*(2*x - 1)/x^2 + 33/4/x +
(6*x^2 - 3*x + 2)/x^3 - 3/4*log(x + 1) - 5*log(x)

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mupad [B]  time = 5.85, size = 42, normalized size = 1.40 \begin {gather*} \frac {3\,x-x^2\,\left (2\,\ln \left (x^2+x\right )-\frac {1}{4}\right )+2}{x^3}-3\,\ln \left (x\,\left (x+1\right )\right )-\frac {x\,\ln \left (x^2+x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(48*x - log(x + x^2)*(8*x^2 + 8*x^3 - x^4 - x^5) + 33*x^2 + 29*x^3 + 25*x^4 + 2*x^5 + 24)/(4*x^4 + 4*x^5)
,x)

[Out]

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

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sympy [A]  time = 0.22, size = 39, normalized size = 1.30 \begin {gather*} - 3 \log {\left (x^{2} + x \right )} + \frac {\left (- x^{2} - 8\right ) \log {\left (x^{2} + x \right )}}{4 x} - \frac {- x^{2} - 12 x - 8}{4 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**5-x**4+8*x**3+8*x**2)*ln(x**2+x)-2*x**5-25*x**4-29*x**3-33*x**2-48*x-24)/(4*x**5+4*x**4),x)

[Out]

-3*log(x**2 + x) + (-x**2 - 8)*log(x**2 + x)/(4*x) - (-x**2 - 12*x - 8)/(4*x**3)

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