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

Optimal. Leaf size=33 \[ -\frac {\left (c x^2\right )^p (a+b x)^{3-2 p}}{a (3-2 p) x^3} \]

[Out]

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

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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 = 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)^{3-2 p}}{a (3-2 p) x^3} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.03, size = 32, normalized size = 0.97 \begin {gather*} \frac {\left (c x^2\right )^p (a+b x)^{3-2 p}}{a (-3+2 p) x^3} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [A]
time = 0.19, size = 33, normalized size = 1.00

method	result	size
gosper	\(\frac {\left (b x +a \right )^{3-2 p} \left (c \,x^{2}\right )^{p}}{x^{3} a \left (2 p -3\right )}\)	\(33\)
risch	\(\frac {\left (b x +a \right )^{2-2 p} \left (b x +a \right ) {\mathrm e}^{\frac {p \left (-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i c \,x^{2}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i c \,x^{2}\right ) \mathrm {csgn}\left (i c \right )-i \pi \mathrm {csgn}\left (i c \,x^{2}\right )^{3}+i \pi \mathrm {csgn}\left (i c \,x^{2}\right )^{2} \mathrm {csgn}\left (i c \right )+2 \ln \left (c \right )+4 \ln \left (x \right )\right )}{2}}}{x^{3} a \left (2 p -3\right )}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x^3*(b*x+a)^(3-2*p)/a/(2*p-3)*(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)^(2-2*p)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.90, size = 37, normalized size = 1.12 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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 )^{2 - 2 p}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*x**2)**p*(a + b*x)**(2 - 2*p)/x**4, 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)^(2-2*p)/x^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.25, size = 51, normalized size = 1.55 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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