\(\int (d x)^m (c x^2)^p (a+b x)^{-2-m-2 p} \, dx\) [996]

Optimal result

Integrand size = 27, antiderivative size = 39 \[ \int (d x)^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {x (d x)^m \left (c x^2\right )^p (a+b x)^{-1-m-2 p}}{a (1+m+2 p)} \]

[Out]

x*(d*x)^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 = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 20, 37} \[ \int (d x)^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {x \left (c x^2\right )^p (d x)^m (a+b x)^{-m-2 p-1}}{a (m+2 p+1)} \]

[In]

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

[Out]

(x*(d*x)^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 20

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

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 p} (d x)^m (a+b x)^{-2-m-2 p} \, dx \\ & = \left (x^{-m-2 p} (d x)^m \left (c x^2\right )^p\right ) \int x^{m+2 p} (a+b x)^{-2-m-2 p} \, dx \\ & = \frac {x (d x)^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 = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int (d x)^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {x (d x)^m \left (c x^2\right )^p (a+b x)^{-1-m-2 p}}{a (1+m+2 p)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03

method	result	size
gosper	\(\frac {x \left (d x \right )^{m} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-1-m -2 p}}{a \left (1+m +2 p \right )}\)	\(40\)
parallelrisch	\(\frac {x^{2} \left (c \,x^{2}\right )^{p} \left (d x \right )^{m} \left (b x +a \right )^{-2-m -2 p} b +x \left (c \,x^{2}\right )^{p} \left (d x \right )^{m} \left (b x +a \right )^{-2-m -2 p} a}{a \left (1+m +2 p \right )}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.46 \[ \int (d x)^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {{\left (b x^{2} + a x\right )} {\left (b x + a\right )}^{-m - 2 \, p - 2} \left (d x\right )^{m} e^{\left (2 \, p \log \left (d x\right ) + p \log \left (\frac {c}{d^{2}}\right )\right )}}{a m + 2 \, a p + a} \]

[In]

integrate((d*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)*(d*x)^m*e^(2*p*log(d*x) + p*log(c/d^2))/(a*m + 2*a*p + a)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((d*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)*(d*x)^m, x)

Giac [F]

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

[In]

integrate((d*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)*(d*x)^m, x)

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int (d x)^m \left (c x^2\right )^p (a+b x)^{-2-m-2 p} \, dx=\frac {x\,{\left (d\,x\right )}^m\,{\left (c\,x^2\right )}^p}{a\,{\left (a+b\,x\right )}^{m+2\,p+1}\,\left (m+2\,p+1\right )} \]

[In]

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

[Out]

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