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

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

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 37} \begin {gather*} -\frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{a (1-2 p) x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.97 \begin {gather*} \frac {\left (c x^2\right )^p (a+b x)^{1-2 p}}{a (2 p-1) x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.06, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(c*x^2)^p/(x^2*(a + b*x)^(2*p)),x]

[Out]

Defer[IntegrateAlgebraic][(c*x^2)^p/(x^2*(a + b*x)^(2*p)), x]

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fricas [A]  time = 1.17, size = 37, normalized size = 1.12 \begin {gather*} \frac {{\left (b x + a\right )} \left (c x^{2}\right )^{p}}{{\left (2 \, a p - a\right )} {\left (b x + a\right )}^{2 \, p} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^{2}\right )^{p}}{{\left (b x + a\right )}^{2 \, p} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 38, normalized size = 1.15 \begin {gather*} \frac {\left (b x +a \right ) \left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-2 p}}{\left (2 p -1\right ) a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^{2}\right )^{p}}{{\left (b x + a\right )}^{2 \, p} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.24, size = 32, normalized size = 0.97 \begin {gather*} \frac {{\left (c\,x^2\right )}^p\,{\left (a+b\,x\right )}^{1-2\,p}}{a\,x\,\left (2\,p-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} - \frac {\sqrt {c} \sqrt {x^{2}}}{b x^{2}} & \text {for}\: a = 0 \wedge p = \frac {1}{2} \\- \frac {b^{- 2 p} c^{p} x^{- 2 p} \left (x^{2}\right )^{p}}{x} & \text {for}\: a = 0 \\\int \frac {\sqrt {c x^{2}}}{x^{2} \left (a + b x\right )}\, dx & \text {for}\: p = \frac {1}{2} \\\frac {a c^{p} \left (x^{2}\right )^{p}}{2 a p x \left (a + b x\right )^{2 p} - a x \left (a + b x\right )^{2 p}} + \frac {b c^{p} x \left (x^{2}\right )^{p}}{2 a p x \left (a + b x\right )^{2 p} - a x \left (a + b x\right )^{2 p}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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