3.2168 \(\int (a c+b c x)^{-3-2 p} (f+g x) (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=61 \[ -\frac{(f+g x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p (a c+b c x)^{-2 p}}{2 c^3 (a+b x)^2 (b f-a g)} \]

[Out]

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

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Rubi [A]  time = 0.0426444, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {770, 23, 37} \[ -\frac{(f+g x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p (a c+b c x)^{-2 p}}{2 c^3 (a+b x)^2 (b f-a g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

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

Mathematica [A]  time = 0.031913, size = 49, normalized size = 0.8 \[ -\frac{\left ((a+b x)^2\right )^p (c (a+b x))^{-2 p} (a g+b (f+2 g x))}{2 b^2 c^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 55, normalized size = 0.9 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 2\,bgx+ag+bf \right ) \left ( bcx+ac \right ) ^{-3-2\,p} \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*c*x+a*c)^(-3-2*p)*(g*x+f)*(b^2*x^2+2*a*b*x+a^2)^p,x)

[Out]

-1/2*(b*x+a)*(2*b*g*x+a*g+b*f)*(b*c*x+a*c)^(-3-2*p)*(b^2*x^2+2*a*b*x+a^2)^p/b^2

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Maxima [A]  time = 1.20349, size = 136, normalized size = 2.23 \begin{align*} -\frac{{\left (2 \, b x + a\right )} g}{2 \,{\left (b^{4} c^{2 \, p + 3} x^{2} + 2 \, a b^{3} c^{2 \, p + 3} x + a^{2} b^{2} c^{2 \, p + 3}\right )}} - \frac{f}{2 \,{\left (b^{3} c^{2 \, p + 3} x^{2} + 2 \, a b^{2} c^{2 \, p + 3} x + a^{2} b c^{2 \, p + 3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x + a)*g/(b^4*c^(2*p + 3)*x^2 + 2*a*b^3*c^(2*p + 3)*x + a^2*b^2*c^(2*p + 3)) - 1/2*f/(b^3*c^(2*p + 3
)*x^2 + 2*a*b^2*c^(2*p + 3)*x + a^2*b*c^(2*p + 3))

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Fricas [A]  time = 1.54678, size = 112, normalized size = 1.84 \begin{align*} -\frac{{\left (2 \, b g x + b f + a g\right )} \frac{1}{c^{2}}^{p}}{2 \,{\left (b^{4} c^{3} x^{2} + 2 \, a b^{3} c^{3} x + a^{2} b^{2} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*g*x + b*f + a*g)*(c^(-2))^p/(b^4*c^3*x^2 + 2*a*b^3*c^3*x + a^2*b^2*c^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*c*x+a*c)**(-3-2*p)*(g*x+f)*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

Timed out

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Giac [B]  time = 1.16449, size = 298, normalized size = 4.89 \begin{align*} -\frac{2 \,{\left (b x + a\right )}^{2 \, p} b^{2} g x^{2} e^{\left (-2 \, p \log \left (b x + a\right ) - 2 \, p \log \left (c\right ) - 3 \, \log \left (b x + a\right ) - 3 \, \log \left (c\right )\right )} +{\left (b x + a\right )}^{2 \, p} b^{2} f x e^{\left (-2 \, p \log \left (b x + a\right ) - 2 \, p \log \left (c\right ) - 3 \, \log \left (b x + a\right ) - 3 \, \log \left (c\right )\right )} + 3 \,{\left (b x + a\right )}^{2 \, p} a b g x e^{\left (-2 \, p \log \left (b x + a\right ) - 2 \, p \log \left (c\right ) - 3 \, \log \left (b x + a\right ) - 3 \, \log \left (c\right )\right )} +{\left (b x + a\right )}^{2 \, p} a b f e^{\left (-2 \, p \log \left (b x + a\right ) - 2 \, p \log \left (c\right ) - 3 \, \log \left (b x + a\right ) - 3 \, \log \left (c\right )\right )} +{\left (b x + a\right )}^{2 \, p} a^{2} g e^{\left (-2 \, p \log \left (b x + a\right ) - 2 \, p \log \left (c\right ) - 3 \, \log \left (b x + a\right ) - 3 \, \log \left (c\right )\right )}}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(b*x + a)^(2*p)*b^2*g*x^2*e^(-2*p*log(b*x + a) - 2*p*log(c) - 3*log(b*x + a) - 3*log(c)) + (b*x + a)^(
2*p)*b^2*f*x*e^(-2*p*log(b*x + a) - 2*p*log(c) - 3*log(b*x + a) - 3*log(c)) + 3*(b*x + a)^(2*p)*a*b*g*x*e^(-2*
p*log(b*x + a) - 2*p*log(c) - 3*log(b*x + a) - 3*log(c)) + (b*x + a)^(2*p)*a*b*f*e^(-2*p*log(b*x + a) - 2*p*lo
g(c) - 3*log(b*x + a) - 3*log(c)) + (b*x + a)^(2*p)*a^2*g*e^(-2*p*log(b*x + a) - 2*p*log(c) - 3*log(b*x + a) -
3*log(c)))/b^2