\(\int \frac {x^3}{a x^2+b x^3} \, dx\) [216]

Optimal result

Integrand size = 17, antiderivative size = 18 \[ \int \frac {x^3}{a x^2+b x^3} \, dx=\frac {x}{b}-\frac {a \log (a+b x)}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1598, 45} \[ \int \frac {x^3}{a x^2+b x^3} \, dx=\frac {x}{b}-\frac {a \log (a+b x)}{b^2} \]

[In]

Int[x^3/(a*x^2 + b*x^3),x]

[Out]

x/b - (a*Log[a + b*x])/b^2

Rule 45

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 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 {x}{a+b x} \, dx \\ & = \int \left (\frac {1}{b}-\frac {a}{b (a+b x)}\right ) \, dx \\ & = \frac {x}{b}-\frac {a \log (a+b x)}{b^2} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[x^3/(a*x^2 + b*x^3),x]

[Out]

x/b - (a*Log[a + b*x])/b^2

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {x}{b}-\frac {a \ln \left (b x +a \right )}{b^{2}}\)	\(19\)
norman	\(\frac {x}{b}-\frac {a \ln \left (b x +a \right )}{b^{2}}\)	\(19\)
risch	\(\frac {x}{b}-\frac {a \ln \left (b x +a \right )}{b^{2}}\)	\(19\)
parallelrisch	\(-\frac {a \ln \left (b x +a \right )-b x}{b^{2}}\)	\(19\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {x^3}{a x^2+b x^3} \, dx=\frac {b x - a \log \left (b x + a\right )}{b^{2}} \]

[In]

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

[Out]

(b*x - a*log(b*x + a))/b^2

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{a x^2+b x^3} \, dx=- \frac {a \log {\left (a + b x \right )}}{b^{2}} + \frac {x}{b} \]

[In]

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

[Out]

-a*log(a + b*x)/b**2 + x/b

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{a x^2+b x^3} \, dx=\frac {x}{b} - \frac {a \log \left (b x + a\right )}{b^{2}} \]

[In]

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

[Out]

x/b - a*log(b*x + a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{a x^2+b x^3} \, dx=\frac {x}{b} - \frac {a \log \left ({\left | b x + a \right |}\right )}{b^{2}} \]

[In]

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

[Out]

x/b - a*log(abs(b*x + a))/b^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{a x^2+b x^3} \, dx=-\frac {a\,\ln \left (a+b\,x\right )-b\,x}{b^2} \]

[In]

int(x^3/(a*x^2 + b*x^3),x)

[Out]

-(a*log(a + b*x) - b*x)/b^2