3.85.46 \(\int \frac {2+90 x^3}{x+18 x^4} \, dx\)

Optimal. Leaf size=17 \[ \log (4 x)+\log \left (\frac {x}{3}+6 x^4\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1593, 446, 72} \begin {gather*} \log \left (18 x^3+1\right )+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 90*x^3)/(x + 18*x^4),x]

[Out]

2*Log[x] + Log[1 + 18*x^3]

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2+90 x^3}{x \left (1+18 x^3\right )} \, dx\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {2+90 x}{x (1+18 x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {2}{x}+\frac {54}{1+18 x}\right ) \, dx,x,x^3\right )\\ &=2 \log (x)+\log \left (1+18 x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.76 \begin {gather*} 2 \log (x)+\log \left (1+18 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 90*x^3)/(x + 18*x^4),x]

[Out]

2*Log[x] + Log[1 + 18*x^3]

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fricas [A]  time = 0.90, size = 13, normalized size = 0.76 \begin {gather*} \log \left (18 \, x^{3} + 1\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*x^3+2)/(18*x^4+x),x, algorithm="fricas")

[Out]

log(18*x^3 + 1) + 2*log(x)

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giac [A]  time = 0.15, size = 15, normalized size = 0.88 \begin {gather*} \log \left ({\left | 18 \, x^{3} + 1 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*x^3+2)/(18*x^4+x),x, algorithm="giac")

[Out]

log(abs(18*x^3 + 1)) + 2*log(abs(x))

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maple [A]  time = 0.37, size = 14, normalized size = 0.82

method	result	size
default	\(2 \ln \relax (x )+\ln \left (18 x^{3}+1\right )\)	\(14\)
norman	\(2 \ln \relax (x )+\ln \left (18 x^{3}+1\right )\)	\(14\)
risch	\(2 \ln \relax (x )+\ln \left (18 x^{3}+1\right )\)	\(14\)
meijerg	\(\ln \left (18 x^{3}+1\right )+2 \ln \relax (x )+\frac {2 \ln \relax (2)}{3}+\frac {4 \ln \relax (3)}{3}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*x^3+2)/(18*x^4+x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)+ln(18*x^3+1)

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maxima [A]  time = 0.39, size = 13, normalized size = 0.76 \begin {gather*} \log \left (18 \, x^{3} + 1\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*x^3+2)/(18*x^4+x),x, algorithm="maxima")

[Out]

log(18*x^3 + 1) + 2*log(x)

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mupad [B]  time = 5.05, size = 11, normalized size = 0.65 \begin {gather*} \ln \left (x^3+\frac {1}{18}\right )+2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*x^3 + 2)/(x + 18*x^4),x)

[Out]

log(x^3 + 1/18) + 2*log(x)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.71 \begin {gather*} 2 \log {\relax (x )} + \log {\left (18 x^{3} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*x**3+2)/(18*x**4+x),x)

[Out]

2*log(x) + log(18*x**3 + 1)

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