\(\int x^2 (c x^2)^p (a+b x)^{-4-2 p} \, dx\) [988]

Optimal result

Integrand size = 22, antiderivative size = 32 \[ \int x^2 \left (c x^2\right )^p (a+b x)^{-4-2 p} \, dx=\frac {x^3 \left (c x^2\right )^p (a+b x)^{-3-2 p}}{a (3+2 p)} \]

[Out]

x^3*(c*x^2)^p*(b*x+a)^(-3-2*p)/a/(3+2*p)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {15, 37} \[ \int x^2 \left (c x^2\right )^p (a+b x)^{-4-2 p} \, dx=\frac {x^3 \left (c x^2\right )^p (a+b x)^{-2 p-3}}{a (2 p+3)} \]

[In]

Int[x^2*(c*x^2)^p*(a + b*x)^(-4 - 2*p),x]

[Out]

(x^3*(c*x^2)^p*(a + b*x)^(-3 - 2*p))/(a*(3 + 2*p))

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)^{-4-2 p} \, dx \\ & = \frac {x^3 \left (c x^2\right )^p (a+b x)^{-3-2 p}}{a (3+2 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int x^2 \left (c x^2\right )^p (a+b x)^{-4-2 p} \, dx=\frac {x^3 \left (c x^2\right )^p (a+b x)^{1-2 (2+p)}}{a (3+2 p)} \]

[In]

Integrate[x^2*(c*x^2)^p*(a + b*x)^(-4 - 2*p),x]

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03

method	result	size
gosper	\(\frac {x^{3} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-3-2 p}}{a \left (3+2 p \right )}\)	\(33\)
parallelrisch	\(\frac {x^{4} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-4-2 p} b +x^{3} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-4-2 p} a}{a \left (3+2 p \right )}\)	\(59\)

[In]

int(x^2*(c*x^2)^p*(b*x+a)^(-4-2*p),x,method=_RETURNVERBOSE)

[Out]

x^3*(c*x^2)^p*(b*x+a)^(-3-2*p)/a/(3+2*p)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int x^2 \left (c x^2\right )^p (a+b x)^{-4-2 p} \, dx=\frac {{\left (b x^{4} + a x^{3}\right )} \left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p - 4}}{2 \, a p + 3 \, a} \]

[In]

integrate(x^2*(c*x^2)^p*(b*x+a)^(-4-2*p),x, algorithm="fricas")

[Out]

(b*x^4 + a*x^3)*(c*x^2)^p*(b*x + a)^(-2*p - 4)/(2*a*p + 3*a)

Sympy [F]

\[ \int x^2 \left (c x^2\right )^p (a+b x)^{-4-2 p} \, dx=\int x^{2} \left (c x^{2}\right )^{p} \left (a + b x\right )^{- 2 p - 4}\, dx \]

[In]

integrate(x**2*(c*x**2)**p*(b*x+a)**(-4-2*p),x)

[Out]

Integral(x**2*(c*x**2)**p*(a + b*x)**(-2*p - 4), x)

Maxima [F]

\[ \int x^2 \left (c x^2\right )^p (a+b x)^{-4-2 p} \, dx=\int { \left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p - 4} x^{2} \,d x } \]

[In]

integrate(x^2*(c*x^2)^p*(b*x+a)^(-4-2*p),x, algorithm="maxima")

[Out]

integrate((c*x^2)^p*(b*x + a)^(-2*p - 4)*x^2, x)

Giac [F]

\[ \int x^2 \left (c x^2\right )^p (a+b x)^{-4-2 p} \, dx=\int { \left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p - 4} x^{2} \,d x } \]

[In]

integrate(x^2*(c*x^2)^p*(b*x+a)^(-4-2*p),x, algorithm="giac")

[Out]

integrate((c*x^2)^p*(b*x + a)^(-2*p - 4)*x^2, x)

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int x^2 \left (c x^2\right )^p (a+b x)^{-4-2 p} \, dx=\frac {x^3\,{\left (c\,x^2\right )}^p}{a\,\left (2\,p+3\right )\,{\left (a+b\,x\right )}^{2\,p+3}} \]

[In]

int((x^2*(c*x^2)^p)/(a + b*x)^(2*p + 4),x)

[Out]

(x^3*(c*x^2)^p)/(a*(2*p + 3)*(a + b*x)^(2*p + 3))