3.43.69 \(\int \frac {-40-6 x+9 x^6+4 x^5 \log (x)+(6 x^4+2 x^5) \log ^2(x)}{2 x^5} \, dx\)

Optimal. Leaf size=24 \[ \frac {9 x^2}{4}+\frac {5+x}{x^4}+\log ^2(x) (x+\log (x)) \]

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Rubi [A]  time = 0.06, antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {12, 14, 2295, 2346, 2302, 30, 2296} \begin {gather*} \frac {5}{x^4}+\frac {1}{x^3}+\frac {9 x^2}{4}+\log ^3(x)+x \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40 - 6*x + 9*x^6 + 4*x^5*Log[x] + (6*x^4 + 2*x^5)*Log[x]^2)/(2*x^5),x]

[Out]

5/x^4 + x^(-3) + (9*x^2)/4 + x*Log[x]^2 + Log[x]^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-40-6 x+9 x^6+4 x^5 \log (x)+\left (6 x^4+2 x^5\right ) \log ^2(x)}{x^5} \, dx\\ &=\frac {1}{2} \int \left (\frac {-40-6 x+9 x^6}{x^5}+4 \log (x)+\frac {2 (3+x) \log ^2(x)}{x}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-40-6 x+9 x^6}{x^5} \, dx+2 \int \log (x) \, dx+\int \frac {(3+x) \log ^2(x)}{x} \, dx\\ &=-2 x+2 x \log (x)+\frac {1}{2} \int \left (-\frac {40}{x^5}-\frac {6}{x^4}+9 x\right ) \, dx+3 \int \frac {\log ^2(x)}{x} \, dx+\int \log ^2(x) \, dx\\ &=\frac {5}{x^4}+\frac {1}{x^3}-2 x+\frac {9 x^2}{4}+2 x \log (x)+x \log ^2(x)-2 \int \log (x) \, dx+3 \operatorname {Subst}\left (\int x^2 \, dx,x,\log (x)\right )\\ &=\frac {5}{x^4}+\frac {1}{x^3}+\frac {9 x^2}{4}+x \log ^2(x)+\log ^3(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.08 \begin {gather*} \frac {5}{x^4}+\frac {1}{x^3}+\frac {9 x^2}{4}+x \log ^2(x)+\log ^3(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40 - 6*x + 9*x^6 + 4*x^5*Log[x] + (6*x^4 + 2*x^5)*Log[x]^2)/(2*x^5),x]

[Out]

5/x^4 + x^(-3) + (9*x^2)/4 + x*Log[x]^2 + Log[x]^3

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fricas [A]  time = 0.75, size = 33, normalized size = 1.38 \begin {gather*} \frac {4 \, x^{5} \log \relax (x)^{2} + 4 \, x^{4} \log \relax (x)^{3} + 9 \, x^{6} + 4 \, x + 20}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x^5+6*x^4)*log(x)^2+4*x^5*log(x)+9*x^6-6*x-40)/x^5,x, algorithm="fricas")

[Out]

1/4*(4*x^5*log(x)^2 + 4*x^4*log(x)^3 + 9*x^6 + 4*x + 20)/x^4

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giac [A]  time = 0.23, size = 23, normalized size = 0.96 \begin {gather*} x \log \relax (x)^{2} + \log \relax (x)^{3} + \frac {9}{4} \, x^{2} + \frac {x + 5}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x^5+6*x^4)*log(x)^2+4*x^5*log(x)+9*x^6-6*x-40)/x^5,x, algorithm="giac")

[Out]

x*log(x)^2 + log(x)^3 + 9/4*x^2 + (x + 5)/x^4

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maple [A]  time = 0.02, size = 25, normalized size = 1.04

method	result	size
default	\(x \ln \relax (x )^{2}+\ln \relax (x )^{3}+\frac {9 x^{2}}{4}+\frac {1}{x^{3}}+\frac {5}{x^{4}}\)	\(25\)
risch	\(\ln \relax (x )^{3}+x \ln \relax (x )^{2}+\frac {9 x^{6}+4 x +20}{4 x^{4}}\)	\(27\)
norman	\(\frac {5+x +x^{4} \ln \relax (x )^{3}+x^{5} \ln \relax (x )^{2}+\frac {9 x^{6}}{4}}{x^{4}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((2*x^5+6*x^4)*ln(x)^2+4*x^5*ln(x)+9*x^6-6*x-40)/x^5,x,method=_RETURNVERBOSE)

[Out]

x*ln(x)^2+ln(x)^3+9/4*x^2+1/x^3+5/x^4

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maxima [A]  time = 0.34, size = 38, normalized size = 1.58 \begin {gather*} \log \relax (x)^{3} + {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x + \frac {9}{4} \, x^{2} + 2 \, x \log \relax (x) - 2 \, x + \frac {1}{x^{3}} + \frac {5}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x^5+6*x^4)*log(x)^2+4*x^5*log(x)+9*x^6-6*x-40)/x^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.07, size = 23, normalized size = 0.96 \begin {gather*} \frac {x+5}{x^4}+x\,{\ln \relax (x)}^2+{\ln \relax (x)}^3+\frac {9\,x^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^5*log(x) - 3*x + (log(x)^2*(6*x^4 + 2*x^5))/2 + (9*x^6)/2 - 20)/x^5,x)

[Out]

(x + 5)/x^4 + x*log(x)^2 + log(x)^3 + (9*x^2)/4

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sympy [A]  time = 0.12, size = 27, normalized size = 1.12 \begin {gather*} \frac {9 x^{2}}{4} + x \log {\relax (x )}^{2} + \log {\relax (x )}^{3} + \frac {2 x + 10}{2 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x**5+6*x**4)*ln(x)**2+4*x**5*ln(x)+9*x**6-6*x-40)/x**5,x)

[Out]

9*x**2/4 + x*log(x)**2 + log(x)**3 + (2*x + 10)/(2*x**4)

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