\(\int \frac {(c x^2)^p (a+b x)^{1-2 p}}{x^3} \, dx\) [993]

Optimal result

Integrand size = 22, antiderivative size = 35 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{x^3} \, dx=-\frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{2 a (1-p) x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, 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} \[ \int \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{x^3} \, dx=-\frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{2 a (1-p) x^2} \]

[In]

Int[((c*x^2)^p*(a + b*x)^(1 - 2*p))/x^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{x^3} \, dx=\frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{a (-2+2 p) x^2} \]

[In]

Integrate[((c*x^2)^p*(a + b*x)^(1 - 2*p))/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91

method	result	size
gosper	\(\frac {\left (c \,x^{2}\right )^{p} \left (b x +a \right )^{2-2 p}}{2 a \,x^{2} \left (p -1\right )}\)	\(32\)
parallelrisch	\(\frac {x \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{1-2 p} b^{2}+\left (c \,x^{2}\right )^{p} \left (b x +a \right )^{1-2 p} a b}{2 x^{2} \left (p -1\right ) a b}\)	\(62\)
risch	\(\frac {\left (b x +a \right )^{1-2 p} \left (b x +a \right ) c^{p} x^{2 p} {\mathrm e}^{\frac {i \pi p \left (-\operatorname {csgn}\left (i x^{2}\right )^{3}+2 \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i c \,x^{2}\right )^{2}-\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i c \,x^{2}\right ) \operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{2}\right )^{3}+\operatorname {csgn}\left (i c \,x^{2}\right )^{2} \operatorname {csgn}\left (i c \right )\right )}{2}}}{2 x^{2} a \left (p -1\right )}\)	\(156\)

[In]

int((c*x^2)^p*(b*x+a)^(1-2*p)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{x^3} \, dx=\frac {{\left (b x + a\right )} \left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p + 1}}{2 \, {\left (a p - a\right )} x^{2}} \]

[In]

integrate((c*x^2)^p*(b*x+a)^(1-2*p)/x^3,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (27) = 54\).

Time = 2.51 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.11 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{x^3} \, dx=\begin {cases} - \frac {c}{b x} & \text {for}\: a = 0 \wedge p = 1 \\- \frac {\left (b x\right )^{1 - 2 p} \left (c x^{2}\right )^{p}}{x^{2}} & \text {for}\: a = 0 \\c \left (\frac {\log {\left (x \right )}}{a} - \frac {\log {\left (\frac {a}{b} + x \right )}}{a}\right ) & \text {for}\: p = 1 \\\frac {a \left (c x^{2}\right )^{p} \left (a + b x\right )^{1 - 2 p}}{2 a p x^{2} - 2 a x^{2}} + \frac {b x \left (c x^{2}\right )^{p} \left (a + b x\right )^{1 - 2 p}}{2 a p x^{2} - 2 a x^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-c/(b*x), Eq(a, 0) & Eq(p, 1)), (-(b*x)**(1 - 2*p)*(c*x**2)**p/x**2, Eq(a, 0)), (c*(log(x)/a - log(
a/b + x)/a), Eq(p, 1)), (a*(c*x**2)**p*(a + b*x)**(1 - 2*p)/(2*a*p*x**2 - 2*a*x**2) + b*x*(c*x**2)**p*(a + b*x
)**(1 - 2*p)/(2*a*p*x**2 - 2*a*x**2), True))

Maxima [F]

\[ \int \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{x^3} \, dx=\int { \frac {\left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p + 1}}{x^{3}} \,d x } \]

[In]

integrate((c*x^2)^p*(b*x+a)^(1-2*p)/x^3,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{x^3} \, dx=\int { \frac {\left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p + 1}}{x^{3}} \,d x } \]

[In]

integrate((c*x^2)^p*(b*x+a)^(1-2*p)/x^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.43 \[ \int \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{x^3} \, dx=\frac {\left (\frac {{\left (c\,x^2\right )}^p}{2\,\left (p-1\right )}+\frac {b\,x\,{\left (c\,x^2\right )}^p}{2\,a\,\left (p-1\right )}\right )\,{\left (a+b\,x\right )}^{1-2\,p}}{x^2} \]

[In]

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

[Out]

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