\(\int \frac {1}{(b x^{1-2 m}+a x^m)^3} \, dx\) [337]

Optimal result

Integrand size = 17, antiderivative size = 27 \[ \int \frac {1}{\left (b x^{1-2 m}+a x^m\right )^3} \, dx=-\frac {1}{2 b (1-3 m) \left (a+b x^{1-3 m}\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1607, 267} \[ \int \frac {1}{\left (b x^{1-2 m}+a x^m\right )^3} \, dx=-\frac {1}{2 b (1-3 m) \left (a+b x^{1-3 m}\right )^2} \]

[In]

Int[(b*x^(1 - 2*m) + a*x^m)^(-3),x]

[Out]

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (b x^{1-2 m}+a x^m\right )^3} \, dx=-\frac {1}{2 b (1-3 m) \left (a+b x^{1-3 m}\right )^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.85 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44

method	result	size
risch	\(-\frac {x \left (2 a \,x^{3 m}+b x \right )}{2 \left (3 m -1\right ) a^{2} \left (a \,x^{3 m}+b x \right )^{2}}\)	\(39\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (25) = 50\).

Time = 0.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.04 \[ \int \frac {1}{\left (b x^{1-2 m}+a x^m\right )^3} \, dx=-\frac {2 \, a x x^{3 \, m} + b x^{2}}{2 \, {\left (2 \, {\left (3 \, a^{3} b m - a^{3} b\right )} x x^{3 \, m} + {\left (3 \, a^{2} b^{2} m - a^{2} b^{2}\right )} x^{2} + {\left (3 \, a^{4} m - a^{4}\right )} x^{6 \, m}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {1}{\left (b x^{1-2 m}+a x^m\right )^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (25) = 50\).

Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44 \[ \int \frac {1}{\left (b x^{1-2 m}+a x^m\right )^3} \, dx=-\frac {2 \, a x x^{3 \, m} + b x^{2}}{2 \, {\left (2 \, a^{3} b {\left (3 \, m - 1\right )} x x^{3 \, m} + a^{2} b^{2} {\left (3 \, m - 1\right )} x^{2} + a^{4} {\left (3 \, m - 1\right )} x^{6 \, m}\right )}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate((a*x^m + b*x^(-2*m + 1))^(-3), x)

Mupad [B] (verification not implemented)

Time = 9.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {1}{\left (b x^{1-2 m}+a x^m\right )^3} \, dx=-\frac {x\,\left (b\,x+2\,a\,x^{3\,m}\right )}{2\,a^2\,\left (3\,m-1\right )\,{\left (b\,x+a\,x^{3\,m}\right )}^2} \]

[In]

int(1/(a*x^m + b*x^(1 - 2*m))^3,x)

[Out]

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