3.82.48 \(\int \frac {3+7 x+11 x^2+15 x^3+7 x^4+x^5+e^8 (-3-5 x-2 x^2)+(-3-5 x-2 x^2) \log (x+x^2)}{9 x^2+15 x^3+7 x^4+x^5} \, dx\)

Optimal. Leaf size=32 \[ x+\frac {e^8}{x (3+x)}-\frac {\log \left (x+x^2\right )}{x-x (4+x)} \]

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Rubi [B]  time = 0.94, antiderivative size = 90, normalized size of antiderivative = 2.81, number of steps used = 33, number of rules used = 16, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {6741, 6742, 44, 88, 72, 77, 74, 2513, 2357, 2304, 2314, 31, 2418, 2395, 36, 29} \begin {gather*} x+\frac {e^8}{(x+3) x}+\frac {x \log (x)}{9 (x+3)}-\frac {\log (x)}{9}-\frac {\log (x+1)}{3 (x+3)}+\frac {\log (x)}{3 x}+\frac {\log (x+1)}{3 x}-\frac {\log (x)+\log (x+1)-\log (x (x+1))}{(x+3) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 7*x + 11*x^2 + 15*x^3 + 7*x^4 + x^5 + E^8*(-3 - 5*x - 2*x^2) + (-3 - 5*x - 2*x^2)*Log[x + x^2])/(9*x^
2 + 15*x^3 + 7*x^4 + x^5),x]

[Out]

x + E^8/(x*(3 + x)) - Log[x]/9 + Log[x]/(3*x) + (x*Log[x])/(9*(3 + x)) + Log[1 + x]/(3*x) - Log[1 + x]/(3*(3 +
x)) - (Log[x] + Log[1 + x] - Log[x*(1 + x)])/(x*(3 + x))

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 74

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2304

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

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 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 2395

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

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

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 6741

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

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 {3+7 x+11 x^2+15 x^3+7 x^4+x^5+e^8 \left (-3-5 x-2 x^2\right )+\left (-3-5 x-2 x^2\right ) \log \left (x+x^2\right )}{x^2 (1+x) (3+x)^2} \, dx\\ &=\int \left (\frac {11}{(1+x) (3+x)^2}+\frac {3}{x^2 (1+x) (3+x)^2}+\frac {7}{x (1+x) (3+x)^2}+\frac {15 x}{(1+x) (3+x)^2}+\frac {7 x^2}{(1+x) (3+x)^2}+\frac {x^3}{(1+x) (3+x)^2}-\frac {e^8 (3+2 x)}{x^2 (3+x)^2}-\frac {(3+2 x) \log (x (1+x))}{x^2 (3+x)^2}\right ) \, dx\\ &=3 \int \frac {1}{x^2 (1+x) (3+x)^2} \, dx+7 \int \frac {1}{x (1+x) (3+x)^2} \, dx+7 \int \frac {x^2}{(1+x) (3+x)^2} \, dx+11 \int \frac {1}{(1+x) (3+x)^2} \, dx+15 \int \frac {x}{(1+x) (3+x)^2} \, dx-e^8 \int \frac {3+2 x}{x^2 (3+x)^2} \, dx+\int \frac {x^3}{(1+x) (3+x)^2} \, dx-\int \frac {(3+2 x) \log (x (1+x))}{x^2 (3+x)^2} \, dx\\ &=\frac {e^8}{x (3+x)}+3 \int \left (\frac {1}{9 x^2}-\frac {5}{27 x}+\frac {1}{4 (1+x)}-\frac {1}{18 (3+x)^2}-\frac {7}{108 (3+x)}\right ) \, dx+7 \int \left (\frac {1}{9 x}-\frac {1}{4 (1+x)}+\frac {1}{6 (3+x)^2}+\frac {5}{36 (3+x)}\right ) \, dx+7 \int \left (\frac {1}{4 (1+x)}-\frac {9}{2 (3+x)^2}+\frac {3}{4 (3+x)}\right ) \, dx+11 \int \left (\frac {1}{4 (1+x)}-\frac {1}{2 (3+x)^2}-\frac {1}{4 (3+x)}\right ) \, dx+15 \int \left (-\frac {1}{4 (1+x)}+\frac {3}{2 (3+x)^2}+\frac {1}{4 (3+x)}\right ) \, dx+(\log (x)+\log (1+x)-\log (x (1+x))) \int \frac {3+2 x}{x^2 (3+x)^2} \, dx+\int \left (1-\frac {1}{4 (1+x)}+\frac {27}{2 (3+x)^2}-\frac {27}{4 (3+x)}\right ) \, dx-\int \frac {(3+2 x) \log (x)}{x^2 (3+x)^2} \, dx-\int \frac {(3+2 x) \log (1+x)}{x^2 (3+x)^2} \, dx\\ &=-\frac {1}{3 x}+x+\frac {e^8}{x (3+x)}+\frac {2 \log (x)}{9}-\frac {1}{2} \log (1+x)-\frac {\log (x)+\log (1+x)-\log (x (1+x))}{x (3+x)}+\frac {5}{18} \log (3+x)-\int \left (\frac {\log (x)}{3 x^2}-\frac {\log (x)}{3 (3+x)^2}\right ) \, dx-\int \left (\frac {\log (1+x)}{3 x^2}-\frac {\log (1+x)}{3 (3+x)^2}\right ) \, dx\\ &=-\frac {1}{3 x}+x+\frac {e^8}{x (3+x)}+\frac {2 \log (x)}{9}-\frac {1}{2} \log (1+x)-\frac {\log (x)+\log (1+x)-\log (x (1+x))}{x (3+x)}+\frac {5}{18} \log (3+x)-\frac {1}{3} \int \frac {\log (x)}{x^2} \, dx+\frac {1}{3} \int \frac {\log (x)}{(3+x)^2} \, dx-\frac {1}{3} \int \frac {\log (1+x)}{x^2} \, dx+\frac {1}{3} \int \frac {\log (1+x)}{(3+x)^2} \, dx\\ &=x+\frac {e^8}{x (3+x)}+\frac {2 \log (x)}{9}+\frac {\log (x)}{3 x}+\frac {x \log (x)}{9 (3+x)}-\frac {1}{2} \log (1+x)+\frac {\log (1+x)}{3 x}-\frac {\log (1+x)}{3 (3+x)}-\frac {\log (x)+\log (1+x)-\log (x (1+x))}{x (3+x)}+\frac {5}{18} \log (3+x)-\frac {1}{9} \int \frac {1}{3+x} \, dx-\frac {1}{3} \int \frac {1}{x (1+x)} \, dx+\frac {1}{3} \int \frac {1}{(1+x) (3+x)} \, dx\\ &=x+\frac {e^8}{x (3+x)}+\frac {2 \log (x)}{9}+\frac {\log (x)}{3 x}+\frac {x \log (x)}{9 (3+x)}-\frac {1}{2} \log (1+x)+\frac {\log (1+x)}{3 x}-\frac {\log (1+x)}{3 (3+x)}-\frac {\log (x)+\log (1+x)-\log (x (1+x))}{x (3+x)}+\frac {1}{6} \log (3+x)+\frac {1}{6} \int \frac {1}{1+x} \, dx-\frac {1}{6} \int \frac {1}{3+x} \, dx-\frac {1}{3} \int \frac {1}{x} \, dx+\frac {1}{3} \int \frac {1}{1+x} \, dx\\ &=x+\frac {e^8}{x (3+x)}-\frac {\log (x)}{9}+\frac {\log (x)}{3 x}+\frac {x \log (x)}{9 (3+x)}+\frac {\log (1+x)}{3 x}-\frac {\log (1+x)}{3 (3+x)}-\frac {\log (x)+\log (1+x)-\log (x (1+x))}{x (3+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 26, normalized size = 0.81 \begin {gather*} \frac {e^8+x^2 (3+x)+\log (x (1+x))}{x (3+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 7*x + 11*x^2 + 15*x^3 + 7*x^4 + x^5 + E^8*(-3 - 5*x - 2*x^2) + (-3 - 5*x - 2*x^2)*Log[x + x^2])
/(9*x^2 + 15*x^3 + 7*x^4 + x^5),x]

[Out]

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

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fricas [A]  time = 0.76, size = 27, normalized size = 0.84 \begin {gather*} \frac {x^{3} + 3 \, x^{2} + e^{8} + \log \left (x^{2} + x\right )}{x^{2} + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-5*x-3)*log(x^2+x)+(-2*x^2-5*x-3)*exp(4)^2+x^5+7*x^4+15*x^3+11*x^2+7*x+3)/(x^5+7*x^4+15*x^3+
9*x^2),x, algorithm="fricas")

[Out]

(x^3 + 3*x^2 + e^8 + log(x^2 + x))/(x^2 + 3*x)

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giac [A]  time = 0.15, size = 27, normalized size = 0.84 \begin {gather*} \frac {x^{3} + 3 \, x^{2} + e^{8} + \log \left (x^{2} + x\right )}{x^{2} + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-5*x-3)*log(x^2+x)+(-2*x^2-5*x-3)*exp(4)^2+x^5+7*x^4+15*x^3+11*x^2+7*x+3)/(x^5+7*x^4+15*x^3+
9*x^2),x, algorithm="giac")

[Out]

(x^3 + 3*x^2 + e^8 + log(x^2 + x))/(x^2 + 3*x)

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maple [A]  time = 0.10, size = 27, normalized size = 0.84

method	result	size
norman	\(\frac {x^{3}+{\mathrm e}^{8}-9 x +\ln \left (x^{2}+x \right )}{\left (3+x \right ) x}\)	\(27\)
risch	\(\frac {\ln \left (x^{2}+x \right )}{\left (3+x \right ) x}+\frac {x^{3}+{\mathrm e}^{8}+3 x^{2}}{\left (3+x \right ) x}\)	\(37\)
default	\(x -\frac {{\mathrm e}^{8}}{3 \left (3+x \right )}-\frac {\ln \relax (x )}{2}+\frac {{\mathrm e}^{8}}{3 x}-\frac {1}{3 x}-\frac {\ln \left (x +1\right )}{2}+\frac {1+\frac {x}{3}+\frac {3 \ln \left (x^{2}+x \right ) x}{2}+\frac {x^{2} \ln \left (x^{2}+x \right )}{2}+\ln \left (x^{2}+x \right )}{\left (3+x \right ) x}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-5*x-3)*ln(x^2+x)+(-2*x^2-5*x-3)*exp(4)^2+x^5+7*x^4+15*x^3+11*x^2+7*x+3)/(x^5+7*x^4+15*x^3+9*x^2),
x,method=_RETURNVERBOSE)

[Out]

(x^3+exp(4)^2-9*x+ln(x^2+x))/(3+x)/x

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maxima [B]  time = 0.44, size = 162, normalized size = 5.06 \begin {gather*} \frac {1}{36} \, {\left (\frac {6 \, {\left (x + 6\right )}}{x^{2} + 3 \, x} + 7 \, \log \left (x + 3\right ) - 27 \, \log \left (x + 1\right ) + 20 \, \log \relax (x)\right )} e^{8} + \frac {5}{36} \, {\left (\frac {6}{x + 3} - 5 \, \log \left (x + 3\right ) + 9 \, \log \left (x + 1\right ) - 4 \, \log \relax (x)\right )} e^{8} - \frac {1}{2} \, {\left (\frac {2}{x + 3} - \log \left (x + 3\right ) + \log \left (x + 1\right )\right )} e^{8} + x + \frac {9 \, {\left (x^{2} + 3 \, x + 2\right )} \log \left (x + 1\right ) - 2 \, {\left (2 \, x^{2} + 6 \, x - 9\right )} \log \relax (x) + 6 \, x + 18}{18 \, {\left (x^{2} + 3 \, x\right )}} - \frac {x + 6}{6 \, {\left (x^{2} + 3 \, x\right )}} - \frac {1}{6 \, {\left (x + 3\right )}} - \frac {1}{2} \, \log \left (x + 1\right ) + \frac {2}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-5*x-3)*log(x^2+x)+(-2*x^2-5*x-3)*exp(4)^2+x^5+7*x^4+15*x^3+11*x^2+7*x+3)/(x^5+7*x^4+15*x^3+
9*x^2),x, algorithm="maxima")

[Out]

1/36*(6*(x + 6)/(x^2 + 3*x) + 7*log(x + 3) - 27*log(x + 1) + 20*log(x))*e^8 + 5/36*(6/(x + 3) - 5*log(x + 3) +
9*log(x + 1) - 4*log(x))*e^8 - 1/2*(2/(x + 3) - log(x + 3) + log(x + 1))*e^8 + x + 1/18*(9*(x^2 + 3*x + 2)*lo
g(x + 1) - 2*(2*x^2 + 6*x - 9)*log(x) + 6*x + 18)/(x^2 + 3*x) - 1/6*(x + 6)/(x^2 + 3*x) - 1/6/(x + 3) - 1/2*lo
g(x + 1) + 2/9*log(x)

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mupad [B]  time = 5.34, size = 20, normalized size = 0.62 \begin {gather*} x+\frac {\ln \left (x^2+x\right )+{\mathrm {e}}^8}{x\,\left (x+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((7*x - exp(8)*(5*x + 2*x^2 + 3) + 11*x^2 + 15*x^3 + 7*x^4 + x^5 - log(x + x^2)*(5*x + 2*x^2 + 3) + 3)/(9*x
^2 + 15*x^3 + 7*x^4 + x^5),x)

[Out]

x + (log(x + x^2) + exp(8))/(x*(x + 3))

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sympy [A]  time = 0.25, size = 24, normalized size = 0.75 \begin {gather*} x + \frac {\log {\left (x^{2} + x \right )}}{x^{2} + 3 x} + \frac {e^{8}}{x^{2} + 3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-5*x-3)*ln(x**2+x)+(-2*x**2-5*x-3)*exp(4)**2+x**5+7*x**4+15*x**3+11*x**2+7*x+3)/(x**5+7*x**
4+15*x**3+9*x**2),x)

[Out]

x + log(x**2 + x)/(x**2 + 3*x) + exp(8)/(x**2 + 3*x)

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