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

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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)} \end {gather*}

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 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]

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]

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 = 0.03, size = 49, normalized size = 0.80 \begin {gather*} -\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} \end {gather*}

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]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.77, size = 49, normalized size = 0.80 \begin {gather*} -\frac {\left ((a+b x)^2\right )^p (c (a+b x))^{-2 p} (a g+b f+2 b g x)}{2 b^2 c^3 (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 52, normalized size = 0.85 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

giac [B]  time = 0.26, size = 221, normalized size = 3.62 \begin {gather*} -\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 \relax (c) - 3 \, \log \left (b x + a\right ) - 3 \, \log \relax (c)\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 \relax (c) - 3 \, \log \left (b x + a\right ) - 3 \, \log \relax (c)\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 \relax (c) - 3 \, \log \left (b x + a\right ) - 3 \, \log \relax (c)\right )} + {\left (b x + a\right )}^{2 \, p} a b f e^{\left (-2 \, p \log \left (b x + a\right ) - 2 \, p \log \relax (c) - 3 \, \log \left (b x + a\right ) - 3 \, \log \relax (c)\right )} + {\left (b x + a\right )}^{2 \, p} a^{2} g e^{\left (-2 \, p \log \left (b x + a\right ) - 2 \, p \log \relax (c) - 3 \, \log \left (b x + a\right ) - 3 \, \log \relax (c)\right )}}{2 \, b^{2}} \end {gather*}

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 55, normalized size = 0.90 \begin {gather*} -\frac {\left (b x +a \right ) \left (2 b g x +a g +b f \right ) \left (b c x +a c \right )^{-2 p -3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{2 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maxima [A]  time = 0.58, size = 101, normalized size = 1.66 \begin {gather*} -\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 {gather*}

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))

________________________________________________________________________________________

mupad [B]  time = 2.14, size = 106, normalized size = 1.74 \begin {gather*} -\left (\frac {g\,a^2+b\,f\,a}{2\,b^2\,{\left (a\,c+b\,c\,x\right )}^{2\,p+3}}+\frac {g\,x^2}{{\left (a\,c+b\,c\,x\right )}^{2\,p+3}}+\frac {x\,\left (f\,b^2+3\,a\,g\,b\right )}{2\,b^2\,{\left (a\,c+b\,c\,x\right )}^{2\,p+3}}\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________