3.85.46 $\int \frac{2+90x^3}{x+18x^4}dx$

Optimal. Leaf size=17 $$\log (4x)+\log (\frac{x}{3}+6x^4)$$

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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} $$\begin{array}{c}\log (18x^3+1)+2\log (x)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2+90x^3}{x(1+18x^3)}dx\\ & =\frac{1}{3}\mathrm{Subst}(\int \frac{2+90x}{x(1+18x)}dx,x,x^3)\\ & =\frac{1}{3}\mathrm{Subst}(\int (\frac{2}{x}+\frac{54}{1+18x})dx,x,x^3)\\ & =2\log (x)+\log (1+18x^3)\end{array}\end{array}$$

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.76 $$\begin{array}{c}2\log (x)+\log (1+18x^3)\end{array}$$

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 $$\begin{array}{c}\log (18x^3+1)+2\log (x)\end{array}$$

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 $$\begin{array}{c}\log \left(|18x^3+1|\right)+2\log \left(\left|x\right|\right)\end{array}$$

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	$2\ln (x)+\ln (18x^3+1)$	$14$
norman	$2\ln (x)+\ln (18x^3+1)$	$14$
risch	$2\ln (x)+\ln (18x^3+1)$	$14$
meijerg	$\ln (18x^3+1)+2\ln (x)+\frac{2\ln (2)}{3}+\frac{4\ln (3)}{3}$	$22$

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 $$\begin{array}{c}\log (18x^3+1)+2\log (x)\end{array}$$

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 $$\begin{array}{c}\ln (x^3+\frac{1}{18})+2\ln (x)\end{array}$$

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 $$\begin{array}{c}2\log (x)+\log (18x^3+1)\end{array}$$

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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