\(\int \frac {x}{a x^2+b x^3} \, dx\) [218]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {x}{a x^2+b x^3} \, dx=\frac {\log (x)}{a}-\frac {\log (a+b x)}{a} \]

[Out]

ln(x)/a-ln(b*x+a)/a

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1598, 36, 29, 31} \[ \int \frac {x}{a x^2+b x^3} \, dx=\frac {\log (x)}{a}-\frac {\log (a+b x)}{a} \]

[In]

Int[x/(a*x^2 + b*x^3),x]

[Out]

Log[x]/a - Log[a + b*x]/a

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x (a+b x)} \, dx \\ & = \frac {\int \frac {1}{x} \, dx}{a}-\frac {b \int \frac {1}{a+b x} \, dx}{a} \\ & = \frac {\log (x)}{a}-\frac {\log (a+b x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x}{a x^2+b x^3} \, dx=\frac {\log (x)}{a}-\frac {\log (a+b x)}{a} \]

[In]

Integrate[x/(a*x^2 + b*x^3),x]

[Out]

Log[x]/a - Log[a + b*x]/a

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(\frac {\ln \left (x \right )-\ln \left (b x +a \right )}{a}\)	\(16\)
default	\(\frac {\ln \left (x \right )}{a}-\frac {\ln \left (b x +a \right )}{a}\)	\(19\)
norman	\(\frac {\ln \left (x \right )}{a}-\frac {\ln \left (b x +a \right )}{a}\)	\(19\)
risch	\(\frac {\ln \left (-x \right )}{a}-\frac {\ln \left (b x +a \right )}{a}\)	\(21\)

[In]

int(x/(b*x^3+a*x^2),x,method=_RETURNVERBOSE)

[Out]

(ln(x)-ln(b*x+a))/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {x}{a x^2+b x^3} \, dx=-\frac {\log \left (b x + a\right ) - \log \left (x\right )}{a} \]

[In]

integrate(x/(b*x^3+a*x^2),x, algorithm="fricas")

[Out]

-(log(b*x + a) - log(x))/a

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {x}{a x^2+b x^3} \, dx=\frac {\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}}{a} \]

[In]

integrate(x/(b*x**3+a*x**2),x)

[Out]

(log(x) - log(a/b + x))/a

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x}{a x^2+b x^3} \, dx=-\frac {\log \left (b x + a\right )}{a} + \frac {\log \left (x\right )}{a} \]

[In]

integrate(x/(b*x^3+a*x^2),x, algorithm="maxima")

[Out]

-log(b*x + a)/a + log(x)/a

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {x}{a x^2+b x^3} \, dx=-\frac {\log \left ({\left | b x + a \right |}\right )}{a} + \frac {\log \left ({\left | x \right |}\right )}{a} \]

[In]

integrate(x/(b*x^3+a*x^2),x, algorithm="giac")

[Out]

-log(abs(b*x + a))/a + log(abs(x))/a

Mupad [B] (verification not implemented)

Time = 8.90 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {x}{a x^2+b x^3} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a} \]

[In]

int(x/(a*x^2 + b*x^3),x)

[Out]

-(2*atanh((2*b*x)/a + 1))/a