3.10.24 \(\int \frac {(c x^2)^p (a+b x)^{-1-2 p}}{x} \, dx\)

Optimal. Leaf size=26 \[ \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{2 a p} \]

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{2 a p} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.00 \begin {gather*} \frac {\left (c x^2\right )^p (a+b x)^{-2 p}}{2 a p} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^2)^p/(2*a*p*(a + b*x)^(2*p))

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IntegrateAlgebraic [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^2\right )^p (a+b x)^{-1-2 p}}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.18, size = 31, normalized size = 1.19 \begin {gather*} \frac {{\left (b x + a\right )} \left (c x^{2}\right )^{p} {\left (b x + a\right )}^{-2 \, p - 1}}{2 \, a p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 - 1}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 25, normalized size = 0.96 \begin {gather*} \frac {\left (c \,x^{2}\right )^{p} \left (b x +a \right )^{-2 p}}{2 a p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.45, size = 27, normalized size = 1.04 \begin {gather*} \frac {c^{p} e^{\left (-2 \, p \log \left (b x + a\right ) + 2 \, p \log \relax (x)\right )}}{2 \, a p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 26, normalized size = 1.00 \begin {gather*} \frac {{\left (c\,x^2\right )}^p}{2\,a\,p\,{\left (a+b\,x\right )}^{2\,p}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x^2)^p/(2*a*p*(a + b*x)^(2*p))

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sympy [A]  time = 59.69, size = 264, normalized size = 10.15 \begin {gather*} \begin {cases} - \frac {b^{- 2 p} c^{p} x^{- 2 p} \left (x^{2}\right )^{p}}{b x} & \text {for}\: a = 0 \\\frac {0^{- 2 p - 1} c^{p} \left (x^{2}\right )^{p}}{2 p} & \text {for}\: a = - b x \\\frac {c^{p} \left (0^{\frac {1}{p}}\right )^{- 2 p - 1} \left (x^{2}\right )^{p}}{2 p} & \text {for}\: a = 0^{\frac {1}{p}} - b x \\\frac {\log {\relax (x )}}{a} - \frac {\log {\left (\frac {a}{b} + x \right )}}{a} & \text {for}\: p = 0 \\\frac {a^{2} c^{p} \left (x^{2}\right )^{p}}{2 a^{3} p \left (a + b x\right )^{2 p} + 4 a^{2} b p x \left (a + b x\right )^{2 p} + 2 a b^{2} p x^{2} \left (a + b x\right )^{2 p}} + \frac {a b c^{p} x \left (x^{2}\right )^{p}}{2 a^{3} p \left (a + b x\right )^{2 p} + 4 a^{2} b p x \left (a + b x\right )^{2 p} + 2 a b^{2} p x^{2} \left (a + b x\right )^{2 p}} + \frac {b c^{p} x \left (x^{2}\right )^{p}}{2 a^{2} p \left (a + b x\right )^{2 p} + 2 a b p 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*(b*x+a)**(-1-2*p)/x,x)

[Out]

Piecewise((-b**(-2*p)*c**p*x**(-2*p)*(x**2)**p/(b*x), Eq(a, 0)), (0**(-2*p - 1)*c**p*(x**2)**p/(2*p), Eq(a, -b
*x)), (c**p*(0**(1/p))**(-2*p - 1)*(x**2)**p/(2*p), Eq(a, 0**(1/p) - b*x)), (log(x)/a - log(a/b + x)/a, Eq(p,
0)), (a**2*c**p*(x**2)**p/(2*a**3*p*(a + b*x)**(2*p) + 4*a**2*b*p*x*(a + b*x)**(2*p) + 2*a*b**2*p*x**2*(a + b*
x)**(2*p)) + a*b*c**p*x*(x**2)**p/(2*a**3*p*(a + b*x)**(2*p) + 4*a**2*b*p*x*(a + b*x)**(2*p) + 2*a*b**2*p*x**2
*(a + b*x)**(2*p)) + b*c**p*x*(x**2)**p/(2*a**2*p*(a + b*x)**(2*p) + 2*a*b*p*x*(a + b*x)**(2*p)), True))

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