\(\int \frac {x}{a x^3+b x^4} \, dx\) [308]

Optimal result

Integrand size = 15, antiderivative size = 28 \[ \int \frac {x}{a x^3+b x^4} \, dx=-\frac {1}{a x}-\frac {b \log (x)}{a^2}+\frac {b \log (a+b x)}{a^2} \]

[Out]

-1/a/x-b*ln(x)/a^2+b*ln(b*x+a)/a^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1598, 46} \[ \int \frac {x}{a x^3+b x^4} \, dx=-\frac {b \log (x)}{a^2}+\frac {b \log (a+b x)}{a^2}-\frac {1}{a x} \]

[In]

Int[x/(a*x^3 + b*x^4),x]

[Out]

-(1/(a*x)) - (b*Log[x])/a^2 + (b*Log[a + b*x])/a^2

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 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^2 (a+b x)} \, dx \\ & = \int \left (\frac {1}{a x^2}-\frac {b}{a^2 x}+\frac {b^2}{a^2 (a+b x)}\right ) \, dx \\ & = -\frac {1}{a x}-\frac {b \log (x)}{a^2}+\frac {b \log (a+b x)}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {x}{a x^3+b x^4} \, dx=-\frac {1}{a x}-\frac {b \log (x)}{a^2}+\frac {b \log (a+b x)}{a^2} \]

[In]

Integrate[x/(a*x^3 + b*x^4),x]

[Out]

-(1/(a*x)) - (b*Log[x])/a^2 + (b*Log[a + b*x])/a^2

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

method	result	size
parallelrisch	\(-\frac {b \ln \left (x \right ) x -b \ln \left (b x +a \right ) x +a}{a^{2} x}\)	\(26\)
default	\(-\frac {1}{a x}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {b \ln \left (b x +a \right )}{a^{2}}\)	\(29\)
norman	\(-\frac {1}{a x}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {b \ln \left (b x +a \right )}{a^{2}}\)	\(29\)
risch	\(-\frac {1}{a x}+\frac {b \ln \left (-b x -a \right )}{a^{2}}-\frac {b \ln \left (x \right )}{a^{2}}\)	\(32\)

[In]

int(x/(b*x^4+a*x^3),x,method=_RETURNVERBOSE)

[Out]

-(b*ln(x)*x-b*ln(b*x+a)*x+a)/a^2/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {x}{a x^3+b x^4} \, dx=\frac {b x \log \left (b x + a\right ) - b x \log \left (x\right ) - a}{a^{2} x} \]

[In]

integrate(x/(b*x^4+a*x^3),x, algorithm="fricas")

[Out]

(b*x*log(b*x + a) - b*x*log(x) - a)/(a^2*x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {x}{a x^3+b x^4} \, dx=- \frac {1}{a x} + \frac {b \left (- \log {\left (x \right )} + \log {\left (\frac {a}{b} + x \right )}\right )}{a^{2}} \]

[In]

integrate(x/(b*x**4+a*x**3),x)

[Out]

-1/(a*x) + b*(-log(x) + log(a/b + x))/a**2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {x}{a x^3+b x^4} \, dx=\frac {b \log \left (b x + a\right )}{a^{2}} - \frac {b \log \left (x\right )}{a^{2}} - \frac {1}{a x} \]

[In]

integrate(x/(b*x^4+a*x^3),x, algorithm="maxima")

[Out]

b*log(b*x + a)/a^2 - b*log(x)/a^2 - 1/(a*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {x}{a x^3+b x^4} \, dx=\frac {b \log \left ({\left | b x + a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {1}{a x} \]

[In]

integrate(x/(b*x^4+a*x^3),x, algorithm="giac")

[Out]

b*log(abs(b*x + a))/a^2 - b*log(abs(x))/a^2 - 1/(a*x)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {x}{a x^3+b x^4} \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^2}-\frac {1}{a\,x} \]

[In]

int(x/(a*x^3 + b*x^4),x)

[Out]

(2*b*atanh((2*b*x)/a + 1))/a^2 - 1/(a*x)