\(\int x^m (c x^2)^p (a+b x)^{-2-m-2 p} \, dx\) [995]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 38 \[ \int x^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {x^{1+m} \left (c x^2\right )^p (a+b x)^{-1-m-2 p}}{a (1+m+2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {15, 37} \[ \int x^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {x^{m+1} \left (c x^2\right )^p (a+b x)^{-m-2 p-1}}{a (m+2 p+1)} \]

[In]

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

[Out]

(x^(1 + m)*(c*x^2)^p*(a + b*x)^(-1 - m - 2*p))/(a*(1 + m + 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^{m+2 p} (a+b x)^{-2-m-2 p} \, dx \\ & = \frac {x^{1+m} \left (c x^2\right )^p (a+b x)^{-1-m-2 p}}{a (1+m+2 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int x^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {x^{1+m} \left (c x^2\right )^p (a+b x)^{-1-m-2 p}}{a (1+m+2 p)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03

method result size
gosper \(\frac {x^{1+m} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-1-m -2 p}}{a \left (1+m +2 p \right )}\) \(39\)
parallelrisch \(\frac {x^{2} x^{m} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-2-m -2 p} b +x \,x^{m} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-2-m -2 p} a}{a \left (1+m +2 p \right )}\) \(70\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int x^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {{\left (b x^{2} + a x\right )} {\left (b x + a\right )}^{-m - 2 \, p - 2} x^{m} e^{\left (p \log \left (c\right ) + 2 \, p \log \left (x\right )\right )}}{a m + 2 \, a p + a} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int x^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {x\,x^m\,{\left (c\,x^2\right )}^p}{a\,{\left (a+b\,x\right )}^m\,{\left (a+b\,x\right )}^{2\,p}\,\left (a+b\,x\right )\,\left (m+2\,p+1\right )} \]

[In]

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

[Out]

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