3.994 \(\int \frac{(c x^2)^p (a+b x)^{2-2 p}}{x^4} \, dx\)

Optimal. Leaf size=33 \[ -\frac{\left (c x^2\right )^p (a+b x)^{3-2 p}}{a (3-2 p) x^3} \]

[Out]

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

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Rubi [A]  time = 0.0104211, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {15, 37} \[ -\frac{\left (c x^2\right )^p (a+b x)^{3-2 p}}{a (3-2 p) x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0098643, size = 32, normalized size = 0.97 \[ \frac{\left (c x^2\right )^p (a+b x)^{3-2 p}}{a (2 p-3) x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.002, size = 33, normalized size = 1. \begin{align*}{\frac{ \left ( bx+a \right ) ^{3-2\,p} \left ( c{x}^{2} \right ) ^{p}}{{x}^{3}a \left ( 2\,p-3 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x^{2}\right )^{p}{\left (b x + a\right )}^{-2 \, p + 2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.6147, size = 84, normalized size = 2.55 \begin{align*} \frac{{\left (b x + a\right )} \left (c x^{2}\right )^{p}{\left (b x + a\right )}^{-2 \, p + 2}}{{\left (2 \, a p - 3 \, a\right )} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x^{2}\right )^{p}{\left (b x + a\right )}^{-2 \, p + 2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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