∫ −5−20x−20x2+10x3+44x4+19x5+(−4x4−2x5) log(x)______________________________________

3.96.74 \(\int \frac {-5-20 x-20 x^2+10 x^3+44 x^4+19 x^5+(-4 x^4-2 x^5) \log (x)}{5 x+15 x^2+15 x^3+5 x^4} \, dx\)

Optimal. Leaf size=28 \[ -x+x^2+\left (1+\frac {x^4}{5 (1+x)^2}\right ) (5-\log (x)) \]

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Rubi [B] time = 0.51, antiderivative size = 78, normalized size of antiderivative = 2.79, number of steps used = 21, number of rules used = 11, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {6741, 6742, 44, 37, 43, 2357, 2295, 2304, 2319, 2314, 31} \begin {gather*} -\frac {2 x^2}{(x+1)^2}+2 x^2-\frac {1}{5} x^2 \log (x)-3 x-\frac {8}{x+1}+\frac {3}{(x+1)^2}-\frac {4 x \log (x)}{5 (x+1)}+\frac {2}{5} x \log (x)-\frac {\log (x)}{5 (x+1)^2}-\frac {4 \log (x)}{5} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(-5 - 20*x - 20*x^2 + 10*x^3 + 44*x^4 + 19*x^5 + (-4*x^4 - 2*x^5)*Log[x])/(5*x + 15*x^2 + 15*x^3 + 5*x^4),

[Out]

-3*x + 2*x^2 + 3/(1 + x)^2 - (2*x^2)/(1 + x)^2 - 8/(1 + x) - (4*Log[x])/5 + (2*x*Log[x])/5 - (x^2*Log[x])/5 -

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

Rule 31

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +

1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ

[m, -1]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]

&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 2295

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

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 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1

)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^

(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers

Q[2*p, 2*q] && !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

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 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 {-5-20 x-20 x^2+10 x^3+44 x^4+19 x^5+\left (-4 x^4-2 x^5\right ) \log (x)}{x \left (5+15 x+15 x^2+5 x^3\right )} \, dx\\ &=\int \left (-\frac {4}{(1+x)^3}-\frac {1}{x (1+x)^3}-\frac {4 x}{(1+x)^3}+\frac {2 x^2}{(1+x)^3}+\frac {44 x^3}{5 (1+x)^3}+\frac {19 x^4}{5 (1+x)^3}-\frac {2 x^3 (2+x) \log (x)}{5 (1+x)^3}\right ) \, dx\\ &=\frac {2}{(1+x)^2}-\frac {2}{5} \int \frac {x^3 (2+x) \log (x)}{(1+x)^3} \, dx+2 \int \frac {x^2}{(1+x)^3} \, dx+\frac {19}{5} \int \frac {x^4}{(1+x)^3} \, dx-4 \int \frac {x}{(1+x)^3} \, dx+\frac {44}{5} \int \frac {x^3}{(1+x)^3} \, dx-\int \frac {1}{x (1+x)^3} \, dx\\ &=\frac {2}{(1+x)^2}-\frac {2 x^2}{(1+x)^2}-\frac {2}{5} \int \left (-\log (x)+x \log (x)-\frac {\log (x)}{(1+x)^3}+\frac {2 \log (x)}{(1+x)^2}\right ) \, dx+2 \int \left (\frac {1}{(1+x)^3}-\frac {2}{(1+x)^2}+\frac {1}{1+x}\right ) \, dx+\frac {19}{5} \int \left (-3+x+\frac {1}{(1+x)^3}-\frac {4}{(1+x)^2}+\frac {6}{1+x}\right ) \, dx+\frac {44}{5} \int \left (1-\frac {1}{(1+x)^3}+\frac {3}{(1+x)^2}-\frac {3}{1+x}\right ) \, dx-\int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^3}-\frac {1}{(1+x)^2}\right ) \, dx\\ &=-\frac {13 x}{5}+\frac {19 x^2}{10}+\frac {3}{(1+x)^2}-\frac {2 x^2}{(1+x)^2}-\frac {41}{5 (1+x)}-\log (x)-\frac {3}{5} \log (1+x)+\frac {2}{5} \int \log (x) \, dx-\frac {2}{5} \int x \log (x) \, dx+\frac {2}{5} \int \frac {\log (x)}{(1+x)^3} \, dx-\frac {4}{5} \int \frac {\log (x)}{(1+x)^2} \, dx\\ &=-3 x+2 x^2+\frac {3}{(1+x)^2}-\frac {2 x^2}{(1+x)^2}-\frac {41}{5 (1+x)}-\log (x)+\frac {2}{5} x \log (x)-\frac {1}{5} x^2 \log (x)-\frac {\log (x)}{5 (1+x)^2}-\frac {4 x \log (x)}{5 (1+x)}-\frac {3}{5} \log (1+x)+\frac {1}{5} \int \frac {1}{x (1+x)^2} \, dx+\frac {4}{5} \int \frac {1}{1+x} \, dx\\ &=-3 x+2 x^2+\frac {3}{(1+x)^2}-\frac {2 x^2}{(1+x)^2}-\frac {41}{5 (1+x)}-\log (x)+\frac {2}{5} x \log (x)-\frac {1}{5} x^2 \log (x)-\frac {\log (x)}{5 (1+x)^2}-\frac {4 x \log (x)}{5 (1+x)}+\frac {1}{5} \log (1+x)+\frac {1}{5} \int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^2}\right ) \, dx\\ &=-3 x+2 x^2+\frac {3}{(1+x)^2}-\frac {2 x^2}{(1+x)^2}-\frac {8}{1+x}-\frac {4 \log (x)}{5}+\frac {2}{5} x \log (x)-\frac {1}{5} x^2 \log (x)-\frac {\log (x)}{5 (1+x)^2}-\frac {4 x \log (x)}{5 (1+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.08, size = 47, normalized size = 1.68 \begin {gather*} \frac {5 \left (-3-7 x-4 x^2+x^3+2 x^4\right )-\left (5+10 x+5 x^2+x^4\right ) \log (x)}{5 (1+x)^2} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-5 - 20*x - 20*x^2 + 10*x^3 + 44*x^4 + 19*x^5 + (-4*x^4 - 2*x^5)*Log[x])/(5*x + 15*x^2 + 15*x^3 + 5

*x^4),x]

[Out]

(5*(-3 - 7*x - 4*x^2 + x^3 + 2*x^4) - (5 + 10*x + 5*x^2 + x^4)*Log[x])/(5*(1 + x)^2)

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fricas [B] time = 0.52, size = 49, normalized size = 1.75 \begin {gather*} \frac {10 \, x^{4} + 5 \, x^{3} - 20 \, x^{2} - {\left (x^{4} + 5 \, x^{2} + 10 \, x + 5\right )} \log \relax (x) - 35 \, x - 15}{5 \, {\left (x^{2} + 2 \, x + 1\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^5-4*x^4)*log(x)+19*x^5+44*x^4+10*x^3-20*x^2-20*x-5)/(5*x^4+15*x^3+15*x^2+5*x),x, algorithm="f

ricas")

[Out]

1/5*(10*x^4 + 5*x^3 - 20*x^2 - (x^4 + 5*x^2 + 10*x + 5)*log(x) - 35*x - 15)/(x^2 + 2*x + 1)

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giac [B] time = 0.14, size = 58, normalized size = 2.07 \begin {gather*} 2 \, x^{2} - \frac {1}{5} \, {\left (x^{2} - 2 \, x - \frac {4 \, x + 3}{x^{2} + 2 \, x + 1}\right )} \log \relax (x) - 3 \, x - \frac {4 \, x + 3}{x^{2} + 2 \, x + 1} - \frac {8}{5} \, \log \relax (x) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^5-4*x^4)*log(x)+19*x^5+44*x^4+10*x^3-20*x^2-20*x-5)/(5*x^4+15*x^3+15*x^2+5*x),x, algorithm="g

iac")

[Out]

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

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maple [A] time = 0.06, size = 34, normalized size = 1.21


method	result	size

norman	\(\frac {x^{3}-\frac {x^{2}}{2}+2 x^{4}-\frac {x^{4} \ln \relax (x )}{5}+\frac {1}{2}}{\left (x +1\right )^{2}}-\ln \relax (x )\)	\(34\)
default	\(2 x^{2}-3 x -\ln \relax (x )+\frac {1}{\left (x +1\right )^{2}}-\frac {4}{x +1}-\frac {x^{2} \ln \relax (x )}{5}+\frac {2 x \ln \relax (x )}{5}-\frac {4 \ln \relax (x ) x}{5 \left (x +1\right )}+\frac {\ln \relax (x ) x \left (2+x \right )}{5 \left (x +1\right )^{2}}\)	\(61\)
risch	\(-\frac {\left (x^{4}-3 x^{2}-6 x -3\right ) \ln \relax (x )}{5 \left (x^{2}+2 x +1\right )}-\frac {-10 x^{4}+8 x^{2} \ln \relax (x )-5 x^{3}+16 x \ln \relax (x )+20 x^{2}+8 \ln \relax (x )+35 x +15}{5 \left (x^{2}+2 x +1\right )}\)	\(77\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^5-4*x^4)*ln(x)+19*x^5+44*x^4+10*x^3-20*x^2-20*x-5)/(5*x^4+15*x^3+15*x^2+5*x),x,method=_RETURNVERBOS

[Out]

(x^3-1/2*x^2+2*x^4-1/5*x^4*ln(x)+1/2)/(x+1)^2-ln(x)

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maxima [B] time = 0.42, size = 148, normalized size = 5.29 \begin {gather*} \frac {19}{10} \, x^{2} - \frac {13}{5} \, x - \frac {2 \, x^{4} \log \relax (x) - x^{4} + 2 \, x^{3} + 7 \, x^{2} + 2 \, x - 2}{10 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {19 \, {\left (8 \, x + 7\right )}}{10 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {22 \, {\left (6 \, x + 5\right )}}{5 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {4 \, x + 3}{x^{2} + 2 \, x + 1} - \frac {2 \, x + 3}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {2 \, {\left (2 \, x + 1\right )}}{x^{2} + 2 \, x + 1} + \frac {2}{x^{2} + 2 \, x + 1} - \log \relax (x) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^5-4*x^4)*log(x)+19*x^5+44*x^4+10*x^3-20*x^2-20*x-5)/(5*x^4+15*x^3+15*x^2+5*x),x, algorithm="m

axima")

[Out]

19/10*x^2 - 13/5*x - 1/10*(2*x^4*log(x) - x^4 + 2*x^3 + 7*x^2 + 2*x - 2)/(x^2 + 2*x + 1) + 19/10*(8*x + 7)/(x^

2 + 2*x + 1) - 22/5*(6*x + 5)/(x^2 + 2*x + 1) + (4*x + 3)/(x^2 + 2*x + 1) - 1/2*(2*x + 3)/(x^2 + 2*x + 1) + 2*

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

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mupad [B] time = 9.42, size = 42, normalized size = 1.50 \begin {gather*} -\frac {x+\ln \relax (x)+x^2\,\ln \relax (x)+\frac {x^4\,\ln \relax (x)}{5}+2\,x\,\ln \relax (x)+x^2-x^3-2\,x^4}{{\left (x+1\right )}^2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x + log(x)*(4*x^4 + 2*x^5) + 20*x^2 - 10*x^3 - 44*x^4 - 19*x^5 + 5)/(5*x + 15*x^2 + 15*x^3 + 5*x^4),x

)

[Out]

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

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sympy [B] time = 0.23, size = 56, normalized size = 2.00 \begin {gather*} 2 x^{2} - 3 x + \frac {- 4 x - 3}{x^{2} + 2 x + 1} - \frac {8 \log {\relax (x )}}{5} + \frac {\left (- x^{4} + 3 x^{2} + 6 x + 3\right ) \log {\relax (x )}}{5 x^{2} + 10 x + 5} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**5-4*x**4)*ln(x)+19*x**5+44*x**4+10*x**3-20*x**2-20*x-5)/(5*x**4+15*x**3+15*x**2+5*x),x)

[Out]

2*x**2 - 3*x + (-4*x - 3)/(x**2 + 2*x + 1) - 8*log(x)/5 + (-x**4 + 3*x**2 + 6*x + 3)*log(x)/(5*x**2 + 10*x + 5

)

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