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

Optimal. Leaf size=35 \[ -\frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{2 a (1-p) x^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 35, 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)^{2-2 p}}{2 a (1-p) x^2} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.04, size = 32, normalized size = 0.91 \begin {gather*} \frac {\left (c x^2\right )^p (a+b x)^{2-2 p}}{a (-2+2 p) x^2} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [A]
time = 0.15, size = 32, normalized size = 0.91

method	result	size
gosper	\(\frac {\left (b x +a \right )^{2-2 p} \left (c \,x^{2}\right )^{p}}{2 x^{2} a \left (p -1\right )}\)	\(32\)
risch	\(\frac {\left (b x +a \right )^{1-2 p} \left (b x +a \right ) {\mathrm e}^{\frac {p \left (-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i c \,x^{2}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i c \,x^{2}\right ) \mathrm {csgn}\left (i c \right )-i \pi \mathrm {csgn}\left (i c \,x^{2}\right )^{3}+i \pi \mathrm {csgn}\left (i c \,x^{2}\right )^{2} \mathrm {csgn}\left (i c \right )+2 \ln \left (c \right )+4 \ln \left (x \right )\right )}{2}}}{2 x^{2} a \left (p -1\right )}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.53, size = 37, normalized size = 1.06 \begin {gather*} \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}} \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^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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^{2}\right )^{p} \left (a + b x\right )^{1 - 2 p}}{x^{3}}\, dx \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**3,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.25, size = 50, normalized size = 1.43 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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