3.103.13 \(\int \frac {12+1246 x-622 x^3+(-8-623 x) \log (x)}{x^3} \, dx\)

Optimal. Leaf size=28 \[ -5+x+\frac {(2 (-2+x)-625 x) \left (x+\frac {1-\log (x)}{x}\right )}{x} \]

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Rubi [A]  time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 8, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {14, 37, 2334, 12, 43} \begin {gather*} -\frac {4}{x^2}+\frac {(623 x+8)^2 \log (x)}{16 x^2}-622 x-\frac {623}{x}-\frac {388129 \log (x)}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + 1246*x - 622*x^3 + (-8 - 623*x)*Log[x])/x^3,x]

[Out]

-4/x^2 - 623/x - 622*x - (388129*Log[x])/16 + ((8 + 623*x)^2*Log[x])/(16*x^2)

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 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 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])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 \left (-6-623 x+311 x^3\right )}{x^3}-\frac {(8+623 x) \log (x)}{x^3}\right ) \, dx\\ &=-\left (2 \int \frac {-6-623 x+311 x^3}{x^3} \, dx\right )-\int \frac {(8+623 x) \log (x)}{x^3} \, dx\\ &=\frac {(8+623 x)^2 \log (x)}{16 x^2}-2 \int \left (311-\frac {6}{x^3}-\frac {623}{x^2}\right ) \, dx+\int -\frac {(8+623 x)^2}{16 x^3} \, dx\\ &=-\frac {6}{x^2}-\frac {1246}{x}-622 x+\frac {(8+623 x)^2 \log (x)}{16 x^2}-\frac {1}{16} \int \frac {(8+623 x)^2}{x^3} \, dx\\ &=-\frac {6}{x^2}-\frac {1246}{x}-622 x+\frac {(8+623 x)^2 \log (x)}{16 x^2}-\frac {1}{16} \int \left (\frac {64}{x^3}+\frac {9968}{x^2}+\frac {388129}{x}\right ) \, dx\\ &=-\frac {4}{x^2}-\frac {623}{x}-622 x-\frac {388129 \log (x)}{16}+\frac {(8+623 x)^2 \log (x)}{16 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \begin {gather*} -\frac {4}{x^2}-\frac {623}{x}-622 x+\frac {4 \log (x)}{x^2}+\frac {623 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + 1246*x - 622*x^3 + (-8 - 623*x)*Log[x])/x^3,x]

[Out]

-4/x^2 - 623/x - 622*x + (4*Log[x])/x^2 + (623*Log[x])/x

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fricas [A]  time = 0.56, size = 24, normalized size = 0.86 \begin {gather*} -\frac {622 \, x^{3} - {\left (623 \, x + 4\right )} \log \relax (x) + 623 \, x + 4}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-623*x-8)*log(x)-622*x^3+1246*x+12)/x^3,x, algorithm="fricas")

[Out]

-(622*x^3 - (623*x + 4)*log(x) + 623*x + 4)/x^2

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giac [A]  time = 0.22, size = 25, normalized size = 0.89 \begin {gather*} -622 \, x + \frac {{\left (623 \, x + 4\right )} \log \relax (x)}{x^{2}} - \frac {623 \, x + 4}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-623*x-8)*log(x)-622*x^3+1246*x+12)/x^3,x, algorithm="giac")

[Out]

-622*x + (623*x + 4)*log(x)/x^2 - (623*x + 4)/x^2

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maple [A]  time = 0.02, size = 24, normalized size = 0.86

method	result	size
norman	\(\frac {-4-623 x -622 x^{3}+623 x \ln \relax (x )+4 \ln \relax (x )}{x^{2}}\)	\(24\)
risch	\(\frac {\left (623 x +4\right ) \ln \relax (x )}{x^{2}}-\frac {622 x^{3}+623 x +4}{x^{2}}\)	\(28\)
default	\(-622 x +\frac {623 \ln \relax (x )}{x}-\frac {623}{x}+\frac {4 \ln \relax (x )}{x^{2}}-\frac {4}{x^{2}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-623*x-8)*ln(x)-622*x^3+1246*x+12)/x^3,x,method=_RETURNVERBOSE)

[Out]

(-4-623*x-622*x^3+623*x*ln(x)+4*ln(x))/x^2

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maxima [A]  time = 0.34, size = 28, normalized size = 1.00 \begin {gather*} -622 \, x + \frac {623 \, \log \relax (x)}{x} - \frac {623}{x} + \frac {4 \, \log \relax (x)}{x^{2}} - \frac {4}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-623*x-8)*log(x)-622*x^3+1246*x+12)/x^3,x, algorithm="maxima")

[Out]

-622*x + 623*log(x)/x - 623/x + 4*log(x)/x^2 - 4/x^2

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mupad [B]  time = 7.06, size = 27, normalized size = 0.96 \begin {gather*} \frac {x\,\left (4\,\ln \relax (x)-4\right )+x^2\,\left (623\,\ln \relax (x)-623\right )}{x^3}-622\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1246*x - log(x)*(623*x + 8) - 622*x^3 + 12)/x^3,x)

[Out]

(x*(4*log(x) - 4) + x^2*(623*log(x) - 623))/x^3 - 622*x

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sympy [A]  time = 0.13, size = 22, normalized size = 0.79 \begin {gather*} - 622 x + \frac {\left (623 x + 4\right ) \log {\relax (x )}}{x^{2}} - \frac {623 x + 4}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-623*x-8)*ln(x)-622*x**3+1246*x+12)/x**3,x)

[Out]

-622*x + (623*x + 4)*log(x)/x**2 - (623*x + 4)/x**2

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