\(\int \frac {(c x^2)^p (a+b x)^{2-2 p}}{x^4} \, dx\) [994]

Optimal result

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

[Out]

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

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

[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*} \text {integral}& = \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] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{x^4} \, dx=\frac {\left (c x^2\right )^p (a+b x)^{3-2 p}}{a (-3+2 p) x^3} \]

[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)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00

method	result	size
gosper	\(\frac {\left (c \,x^{2}\right )^{p} \left (b x +a \right )^{3-2 p}}{a \,x^{3} \left (2 p -3\right )}\)	\(33\)
parallelrisch	\(\frac {x \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{2-2 p} b +\left (c \,x^{2}\right )^{p} \left (b x +a \right )^{2-2 p} a}{x^{3} a \left (2 p -3\right )}\)	\(57\)
risch	\(\frac {\left (b x +a \right )^{2-2 p} \left (b x +a \right ) c^{p} x^{2 p} {\mathrm e}^{\frac {i \pi p \left (-\operatorname {csgn}\left (i x^{2}\right )^{3}+2 \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i c \,x^{2}\right )^{2}-\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i c \,x^{2}\right ) \operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{2}\right )^{3}+\operatorname {csgn}\left (i c \,x^{2}\right )^{2} \operatorname {csgn}\left (i c \right )\right )}{2}}}{x^{3} a \left (2 p -3\right )}\)	\(157\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{x^4} \, dx=\frac {{\left (b x + a\right )} \left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p + 2}}{{\left (2 \, a p - 3 \, a\right )} x^{3}} \]

[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)

Sympy [F]

\[ \int \frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{x^4} \, dx=\begin {cases} - \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{b x^{4}} & \text {for}\: a = 0 \wedge p = \frac {3}{2} \\- \frac {\left (b x\right )^{2 - 2 p} \left (c x^{2}\right )^{p}}{x^{3}} & \text {for}\: a = 0 \\\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{4} \left (a + b x\right )}\, dx & \text {for}\: p = \frac {3}{2} \\\frac {a \left (c x^{2}\right )^{p} \left (a + b x\right )^{2 - 2 p}}{2 a p x^{3} - 3 a x^{3}} + \frac {b x \left (c x^{2}\right )^{p} \left (a + b x\right )^{2 - 2 p}}{2 a p x^{3} - 3 a x^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-(c*x**2)**(3/2)/(b*x**4), Eq(a, 0) & Eq(p, 3/2)), (-(b*x)**(2 - 2*p)*(c*x**2)**p/x**3, Eq(a, 0)),
(Integral((c*x**2)**(3/2)/(x**4*(a + b*x)), x), Eq(p, 3/2)), (a*(c*x**2)**p*(a + b*x)**(2 - 2*p)/(2*a*p*x**3 -
3*a*x**3) + b*x*(c*x**2)**p*(a + b*x)**(2 - 2*p)/(2*a*p*x**3 - 3*a*x**3), True))

Maxima [F]

\[ \int \frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{x^4} \, dx=\int { \frac {\left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p + 2}}{x^{4}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{x^4} \, dx=\int { \frac {\left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p + 2}}{x^{4}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{x^4} \, dx=\frac {\left (\frac {{\left (c\,x^2\right )}^p}{2\,p-3}+\frac {b\,x\,{\left (c\,x^2\right )}^p}{a\,\left (2\,p-3\right )}\right )\,{\left (a+b\,x\right )}^{2-2\,p}}{x^3} \]

[In]

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

[Out]

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