\(\int \frac {1}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [1533]

Optimal result

Integrand size = 18, antiderivative size = 14 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {1}{5 b (a+b x)^5} \]

[Out]

-1/5/b/(b*x+a)^5

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {27, 32} \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {1}{5 b (a+b x)^5} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(-3),x]

[Out]

-1/5*1/(b*(a + b*x)^5)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(a+b x)^6} \, dx \\ & = -\frac {1}{5 b (a+b x)^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {1}{5 b (a+b x)^5} \]

[In]

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

[Out]

-1/5*1/(b*(a + b*x)^5)

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {1}{5 b \left (b x +a \right )^{5}}\)	\(13\)
norman	\(-\frac {1}{5 b \left (b x +a \right )^{5}}\)	\(13\)
risch	\(-\frac {1}{5 b \left (b x +a \right )^{5}}\)	\(13\)
gosper	\(-\frac {1}{5 \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2} b}\)	\(31\)
parallelrisch	\(-\frac {1}{5 \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2} b}\)	\(31\)

[In]

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

[Out]

-1/5/b/(b*x+a)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (12) = 24\).

Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 4.07 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {1}{5 \, {\left (b^{6} x^{5} + 5 \, a b^{5} x^{4} + 10 \, a^{2} b^{4} x^{3} + 10 \, a^{3} b^{3} x^{2} + 5 \, a^{4} b^{2} x + a^{5} b\right )}} \]

[In]

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

[Out]

-1/5/(b^6*x^5 + 5*a*b^5*x^4 + 10*a^2*b^4*x^3 + 10*a^3*b^3*x^2 + 5*a^4*b^2*x + a^5*b)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (12) = 24\).

Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 4.36 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=- \frac {1}{5 a^{5} b + 25 a^{4} b^{2} x + 50 a^{3} b^{3} x^{2} + 50 a^{2} b^{4} x^{3} + 25 a b^{5} x^{4} + 5 b^{6} x^{5}} \]

[In]

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

[Out]

-1/(5*a**5*b + 25*a**4*b**2*x + 50*a**3*b**3*x**2 + 50*a**2*b**4*x**3 + 25*a*b**5*x**4 + 5*b**6*x**5)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (12) = 24\).

Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 4.07 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {1}{5 \, {\left (b^{6} x^{5} + 5 \, a b^{5} x^{4} + 10 \, a^{2} b^{4} x^{3} + 10 \, a^{3} b^{3} x^{2} + 5 \, a^{4} b^{2} x + a^{5} b\right )}} \]

[In]

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

[Out]

-1/5/(b^6*x^5 + 5*a*b^5*x^4 + 10*a^2*b^4*x^3 + 10*a^3*b^3*x^2 + 5*a^4*b^2*x + a^5*b)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {1}{5 \, {\left (b x + a\right )}^{5} b} \]

[In]

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

[Out]

-1/5/((b*x + a)^5*b)

Mupad [B] (verification not implemented)

Time = 9.57 (sec) , antiderivative size = 59, normalized size of antiderivative = 4.21 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {1}{5\,a^5\,b+25\,a^4\,b^2\,x+50\,a^3\,b^3\,x^2+50\,a^2\,b^4\,x^3+25\,a\,b^5\,x^4+5\,b^6\,x^5} \]

[In]

int(1/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)

[Out]

-1/(5*a^5*b + 5*b^6*x^5 + 25*a^4*b^2*x + 25*a*b^5*x^4 + 50*a^3*b^3*x^2 + 50*a^2*b^4*x^3)