\(\int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx\) [6]

Optimal result

Integrand size = 29, antiderivative size = 14 \[ \int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx=-\frac {1}{2 b (a+b x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2083, 32} \[ \int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx=-\frac {1}{2 b (a+b x)^2} \]

[In]

Int[(a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3)^(-1),x]

[Out]

-1/2*1/(b*(a + b*x)^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2083

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P]}, Int[ExpandIntegrand[u^p, x], x] /;  !SumQ[NonfreeFactors[u,
x]]] /; PolyQ[P, x] && ILtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(a+b x)^3} \, dx \\ & = -\frac {1}{2 b (a+b x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx=-\frac {1}{2 b (a+b x)^2} \]

[In]

Integrate[(a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3)^(-1),x]

[Out]

-1/2*1/(b*(a + b*x)^2)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {1}{2 b \left (b x +a \right )^{2}}\)	\(13\)
norman	\(-\frac {1}{2 b \left (b x +a \right )^{2}}\)	\(13\)
gosper	\(-\frac {1}{2 b \left (b^{2} x^{2}+2 a b x +a^{2}\right )}\)	\(24\)
risch	\(-\frac {1}{2 b \left (b^{2} x^{2}+2 a b x +a^{2}\right )}\)	\(24\)
parallelrisch	\(-\frac {1}{2 b \left (b^{2} x^{2}+2 a b x +a^{2}\right )}\)	\(24\)

[In]

int(1/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.71 \[ \int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx=-\frac {1}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

[In]

integrate(1/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3),x, algorithm="fricas")

[Out]

-1/2/(b^3*x^2 + 2*a*b^2*x + a^2*b)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86 \[ \int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx=- \frac {1}{2 a^{2} b + 4 a b^{2} x + 2 b^{3} x^{2}} \]

[In]

integrate(1/(b**3*x**3+3*a*b**2*x**2+3*a**2*b*x+a**3),x)

[Out]

-1/(2*a**2*b + 4*a*b**2*x + 2*b**3*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.71 \[ \int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx=-\frac {1}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

[In]

integrate(1/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3),x, algorithm="maxima")

[Out]

-1/2/(b^3*x^2 + 2*a*b^2*x + a^2*b)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx=-\frac {1}{2 \, {\left (b x + a\right )}^{2} b} \]

[In]

integrate(1/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3),x, algorithm="giac")

[Out]

-1/2/((b*x + a)^2*b)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86 \[ \int \frac {1}{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx=-\frac {1}{2\,a^2\,b+4\,a\,b^2\,x+2\,b^3\,x^2} \]

[In]

int(1/(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x),x)

[Out]

-1/(2*a^2*b + 2*b^3*x^2 + 4*a*b^2*x)