3.989 \(\int x (c x^2)^p (a+b x)^{-3-2 p} \, dx\)

Optimal. Leaf size=33 \[ \frac{x^2 \left (c x^2\right )^p (a+b x)^{-2 (p+1)}}{2 a (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0097638, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 37} \[ \frac{x^2 \left (c x^2\right )^p (a+b x)^{-2 (p+1)}}{2 a (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(c*x^2)^p)/(2*a*(1 + p)*(a + b*x)^(2*(1 + 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*} \int x \left (c x^2\right )^p (a+b x)^{-3-2 p} \, dx &=\left (x^{-2 p} \left (c x^2\right )^p\right ) \int x^{1+2 p} (a+b x)^{-3-2 p} \, dx\\ &=\frac{x^2 \left (c x^2\right )^p (a+b x)^{-2 (1+p)}}{2 a (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0157256, size = 32, normalized size = 0.97 \[ \frac{x^2 \left (c x^2\right )^p (a+b x)^{-2 p-2}}{a (2 p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 32, normalized size = 1. \begin{align*}{\frac{{x}^{2} \left ( bx+a \right ) ^{-2-2\,p} \left ( c{x}^{2} \right ) ^{p}}{2\,a \left ( 1+p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c x^{2}\right )^{p}{\left (b x + a\right )}^{-2 \, p - 3} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.63409, size = 84, normalized size = 2.55 \begin{align*} \frac{{\left (b x^{3} + a x^{2}\right )} \left (c x^{2}\right )^{p}{\left (b x + a\right )}^{-2 \, p - 3}}{2 \,{\left (a p + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.09472, size = 97, normalized size = 2.94 \begin{align*} \frac{\left (c x^{2}\right )^{p} b x^{3} e^{\left (-2 \, p \log \left (b x + a\right ) - 3 \, \log \left (b x + a\right )\right )} + \left (c x^{2}\right )^{p} a x^{2} e^{\left (-2 \, p \log \left (b x + a\right ) - 3 \, \log \left (b x + a\right )\right )}}{2 \,{\left (a p + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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