\(\int \frac {1}{(a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3)^3} \, dx\) [8]

Optimal result

Integrand size = 29, antiderivative size = 14 \[ \int \frac {1}{\left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^3} \, dx=-\frac {1}{8 b (a+b x)^8} \]

[Out]

-1/8/b/(b*x+a)^8

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} \[ \int \frac {1}{\left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^3} \, dx=-\frac {1}{8 b (a+b x)^8} \]

[In]

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

[Out]

-1/8*1/(b*(a + b*x)^8)

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)^9} \, dx \\ & = -\frac {1}{8 b (a+b x)^8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^3} \, dx=-\frac {1}{8 b (a+b x)^8} \]

[In]

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

[Out]

-1/8*1/(b*(a + b*x)^8)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {1}{8 b \left (b x +a \right )^{8}}\)	\(13\)
norman	\(-\frac {1}{8 b \left (b x +a \right )^{8}}\)	\(13\)
risch	\(-\frac {1}{8 b \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{3} \left (b x +a \right )^{2}}\)	\(31\)
gosper	\(-\frac {1}{8 \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{2} b}\)	\(53\)
parallelrisch	\(-\frac {1}{8 \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{2} b}\)	\(53\)

[In]

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

[Out]

-1/8/b/(b*x+a)^8

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (12) = 24\).

Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 6.43 \[ \int \frac {1}{\left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^3} \, dx=-\frac {1}{8 \, {\left (b^{9} x^{8} + 8 \, a b^{8} x^{7} + 28 \, a^{2} b^{7} x^{6} + 56 \, a^{3} b^{6} x^{5} + 70 \, a^{4} b^{5} x^{4} + 56 \, a^{5} b^{4} x^{3} + 28 \, a^{6} b^{3} x^{2} + 8 \, a^{7} b^{2} x + a^{8} b\right )}} \]

[In]

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

[Out]

-1/8/(b^9*x^8 + 8*a*b^8*x^7 + 28*a^2*b^7*x^6 + 56*a^3*b^6*x^5 + 70*a^4*b^5*x^4 + 56*a^5*b^4*x^3 + 28*a^6*b^3*x
^2 + 8*a^7*b^2*x + a^8*b)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (12) = 24\).

Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 6.93 \[ \int \frac {1}{\left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^3} \, dx=- \frac {1}{8 a^{8} b + 64 a^{7} b^{2} x + 224 a^{6} b^{3} x^{2} + 448 a^{5} b^{4} x^{3} + 560 a^{4} b^{5} x^{4} + 448 a^{3} b^{6} x^{5} + 224 a^{2} b^{7} x^{6} + 64 a b^{8} x^{7} + 8 b^{9} x^{8}} \]

[In]

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

[Out]

-1/(8*a**8*b + 64*a**7*b**2*x + 224*a**6*b**3*x**2 + 448*a**5*b**4*x**3 + 560*a**4*b**5*x**4 + 448*a**3*b**6*x
**5 + 224*a**2*b**7*x**6 + 64*a*b**8*x**7 + 8*b**9*x**8)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (12) = 24\).

Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 6.43 \[ \int \frac {1}{\left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^3} \, dx=-\frac {1}{8 \, {\left (b^{9} x^{8} + 8 \, a b^{8} x^{7} + 28 \, a^{2} b^{7} x^{6} + 56 \, a^{3} b^{6} x^{5} + 70 \, a^{4} b^{5} x^{4} + 56 \, a^{5} b^{4} x^{3} + 28 \, a^{6} b^{3} x^{2} + 8 \, a^{7} b^{2} x + a^{8} b\right )}} \]

[In]

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

[Out]

-1/8/(b^9*x^8 + 8*a*b^8*x^7 + 28*a^2*b^7*x^6 + 56*a^3*b^6*x^5 + 70*a^4*b^5*x^4 + 56*a^5*b^4*x^3 + 28*a^6*b^3*x
^2 + 8*a^7*b^2*x + a^8*b)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^3} \, dx=-\frac {1}{8 \, {\left (b x + a\right )}^{8} b} \]

[In]

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

[Out]

-1/8/((b*x + a)^8*b)

Mupad [B] (verification not implemented)

Time = 9.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 6.57 \[ \int \frac {1}{\left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^3} \, dx=-\frac {1}{8\,a^8\,b+64\,a^7\,b^2\,x+224\,a^6\,b^3\,x^2+448\,a^5\,b^4\,x^3+560\,a^4\,b^5\,x^4+448\,a^3\,b^6\,x^5+224\,a^2\,b^7\,x^6+64\,a\,b^8\,x^7+8\,b^9\,x^8} \]

[In]

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

[Out]

-1/(8*a^8*b + 8*b^9*x^8 + 64*a^7*b^2*x + 64*a*b^8*x^7 + 224*a^6*b^3*x^2 + 448*a^5*b^4*x^3 + 560*a^4*b^5*x^4 +
448*a^3*b^6*x^5 + 224*a^2*b^7*x^6)