\(\int \frac {(c x^2)^p (a+b x)^{-2 p}}{x^2} \, dx\) [992]

Optimal result

Integrand size = 20, antiderivative size = 33 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{x^2} \, dx=-\frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{a (1-2 p) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 37} \[ \int \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{x^2} \, dx=-\frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{a (1-2 p) x} \]

[In]

Int[(c*x^2)^p/(x^2*(a + b*x)^(2*p)),x]

[Out]

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{x^2} \, dx=\frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{a (-1+2 p) x} \]

[In]

Integrate[(c*x^2)^p/(x^2*(a + b*x)^(2*p)),x]

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15

method	result	size
gosper	\(\frac {\left (b x +a \right ) \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-2 p}}{x a \left (2 p -1\right )}\)	\(38\)
parallelrisch	\(\frac {\left (x \left (c \,x^{2}\right )^{p} b +\left (c \,x^{2}\right )^{p} a \right ) \left (b x +a \right )^{-2 p}}{x a \left (2 p -1\right )}\)	\(46\)
risch	\(\frac {\left (b x +a \right ) \left (b x +a \right )^{-2 p} c^{p} x^{2 p} {\mathrm e}^{\frac {i \pi p \left (-\operatorname {csgn}\left (i x^{2}\right )^{3}+2 \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i c \,x^{2}\right )^{2}-\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i c \,x^{2}\right ) \operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{2}\right )^{3}+\operatorname {csgn}\left (i c \,x^{2}\right )^{2} \operatorname {csgn}\left (i c \right )\right )}{2}}}{\left (2 p -1\right ) a x}\)	\(157\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{x^2} \, dx=\frac {{\left (b x + a\right )} \left (c x^{2}\right )^{p}}{{\left (2 \, a p - a\right )} {\left (b x + a\right )}^{2 \, p} x} \]

[In]

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

[Out]

(b*x + a)*(c*x^2)^p/((2*a*p - a)*(b*x + a)^(2*p)*x)

Sympy [F]

\[ \int \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{x^2} \, dx=\begin {cases} - \frac {\sqrt {c x^{2}}}{b x^{2}} & \text {for}\: a = 0 \wedge p = \frac {1}{2} \\- \frac {\left (b x\right )^{- 2 p} \left (c x^{2}\right )^{p}}{x} & \text {for}\: a = 0 \\\int \frac {\sqrt {c x^{2}}}{x^{2} \left (a + b x\right )}\, dx & \text {for}\: p = \frac {1}{2} \\\frac {a \left (c x^{2}\right )^{p}}{2 a p x \left (a + b x\right )^{2 p} - a x \left (a + b x\right )^{2 p}} + \frac {b x \left (c x^{2}\right )^{p}}{2 a p x \left (a + b x\right )^{2 p} - a x \left (a + b x\right )^{2 p}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**2)**p/x**2/((b*x+a)**(2*p)),x)

[Out]

Piecewise((-sqrt(c*x**2)/(b*x**2), Eq(a, 0) & Eq(p, 1/2)), (-(c*x**2)**p/(x*(b*x)**(2*p)), Eq(a, 0)), (Integra
l(sqrt(c*x**2)/(x**2*(a + b*x)), x), Eq(p, 1/2)), (a*(c*x**2)**p/(2*a*p*x*(a + b*x)**(2*p) - a*x*(a + b*x)**(2
*p)) + b*x*(c*x**2)**p/(2*a*p*x*(a + b*x)**(2*p) - a*x*(a + b*x)**(2*p)), True))

Maxima [F]

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

[In]

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

[Out]

integrate((c*x^2)^p/((b*x + a)^(2*p)*x^2), x)

Giac [F]

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

[In]

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

[Out]

integrate((c*x^2)^p/((b*x + a)^(2*p)*x^2), x)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{x^2} \, dx=\frac {{\left (c\,x^2\right )}^p\,{\left (a+b\,x\right )}^{1-2\,p}}{a\,x\,\left (2\,p-1\right )} \]

[In]

int((c*x^2)^p/(x^2*(a + b*x)^(2*p)),x)

[Out]

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