\(\int x^3 (c x^2)^p (a+b x)^{-5-2 p} \, dx\) [987]

Optimal result

Integrand size = 22, antiderivative size = 33 \[ \int x^3 \left (c x^2\right )^p (a+b x)^{-5-2 p} \, dx=\frac {x^4 \left (c x^2\right )^p (a+b x)^{-2 (2+p)}}{2 a (2+p)} \]

[Out]

1/2*x^4*(c*x^2)^p/a/(2+p)/((b*x+a)^(4+2*p))

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {15, 37} \[ \int x^3 \left (c x^2\right )^p (a+b x)^{-5-2 p} \, dx=\frac {x^4 \left (c x^2\right )^p (a+b x)^{-2 (p+2)}}{2 a (p+2)} \]

[In]

Int[x^3*(c*x^2)^p*(a + b*x)^(-5 - 2*p),x]

[Out]

(x^4*(c*x^2)^p)/(2*a*(2 + p)*(a + b*x)^(2*(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^{3+2 p} (a+b x)^{-5-2 p} \, dx \\ & = \frac {x^4 \left (c x^2\right )^p (a+b x)^{-2 (2+p)}}{2 a (2+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int x^3 \left (c x^2\right )^p (a+b x)^{-5-2 p} \, dx=\frac {x^4 \left (c x^2\right )^p (a+b x)^{-4-2 p}}{a (4+2 p)} \]

[In]

Integrate[x^3*(c*x^2)^p*(a + b*x)^(-5 - 2*p),x]

[Out]

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97

method	result	size
gosper	\(\frac {x^{4} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-4-2 p}}{2 a \left (2+p \right )}\)	\(32\)
parallelrisch	\(\frac {x^{5} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-5-2 p} b +x^{4} \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-5-2 p} a}{2 a \left (2+p \right )}\)	\(58\)

[In]

int(x^3*(c*x^2)^p*(b*x+a)^(-5-2*p),x,method=_RETURNVERBOSE)

[Out]

1/2/a*x^4*(c*x^2)^p/(2+p)*(b*x+a)^(-4-2*p)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int x^3 \left (c x^2\right )^p (a+b x)^{-5-2 p} \, dx=\frac {{\left (b x^{5} + a x^{4}\right )} \left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p - 5}}{2 \, {\left (a p + 2 \, a\right )}} \]

[In]

integrate(x^3*(c*x^2)^p*(b*x+a)^(-5-2*p),x, algorithm="fricas")

[Out]

1/2*(b*x^5 + a*x^4)*(c*x^2)^p*(b*x + a)^(-2*p - 5)/(a*p + 2*a)

Sympy [F]

\[ \int x^3 \left (c x^2\right )^p (a+b x)^{-5-2 p} \, dx=\int x^{3} \left (c x^{2}\right )^{p} \left (a + b x\right )^{- 2 p - 5}\, dx \]

[In]

integrate(x**3*(c*x**2)**p*(b*x+a)**(-5-2*p),x)

[Out]

Integral(x**3*(c*x**2)**p*(a + b*x)**(-2*p - 5), x)

Maxima [F]

\[ \int x^3 \left (c x^2\right )^p (a+b x)^{-5-2 p} \, dx=\int { \left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p - 5} x^{3} \,d x } \]

[In]

integrate(x^3*(c*x^2)^p*(b*x+a)^(-5-2*p),x, algorithm="maxima")

[Out]

integrate((c*x^2)^p*(b*x + a)^(-2*p - 5)*x^3, x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (33) = 66\).

Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.24 \[ \int x^3 \left (c x^2\right )^p (a+b x)^{-5-2 p} \, dx=\frac {\left (c x^{2}\right )^{p} b x^{5} e^{\left (-2 \, p \log \left (b x + a\right ) - 5 \, \log \left (b x + a\right )\right )} + \left (c x^{2}\right )^{p} a x^{4} e^{\left (-2 \, p \log \left (b x + a\right ) - 5 \, \log \left (b x + a\right )\right )}}{2 \, {\left (a p + 2 \, a\right )}} \]

[In]

integrate(x^3*(c*x^2)^p*(b*x+a)^(-5-2*p),x, algorithm="giac")

[Out]

1/2*((c*x^2)^p*b*x^5*e^(-2*p*log(b*x + a) - 5*log(b*x + a)) + (c*x^2)^p*a*x^4*e^(-2*p*log(b*x + a) - 5*log(b*x
+ a)))/(a*p + 2*a)

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int x^3 \left (c x^2\right )^p (a+b x)^{-5-2 p} \, dx=\frac {x^4\,{\left (c\,x^2\right )}^p}{2\,a\,\left (p+2\right )\,{\left (a+b\,x\right )}^{2\,p+4}} \]

[In]

int((x^3*(c*x^2)^p)/(a + b*x)^(2*p + 5),x)

[Out]

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