3.2157 \(\int (a c+b c x) (d+e x)^{-3-2 p} (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=51 \[ \frac{c \left (a^2+2 a b x+b^2 x^2\right )^{p+1} (d+e x)^{-2 (p+1)}}{2 (p+1) (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0262107, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.026, Rules used = {767} \[ \frac{c \left (a^2+2 a b x+b^2 x^2\right )^{p+1} (d+e x)^{-2 (p+1)}}{2 (p+1) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 767

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Sim
p[(f*g*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)*(e*f - d*g)), x] /; FreeQ[{a, b, c, d, e, f, g,
m, p}, x] && EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && EqQ[2*c*f - b*g, 0]

Rubi steps

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

Mathematica [A]  time = 0.0276622, size = 42, normalized size = 0.82 \[ \frac{c \left ((a+b x)^2\right )^{p+1} (d+e x)^{-2 (p+1)}}{2 (p+1) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 59, normalized size = 1.2 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{2} \left ( ex+d \right ) ^{-2-2\,p}c \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{2\,aep-2\,bdp+2\,ae-2\,bd}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*c*x+a*c)*(e*x+d)^(-3-2*p)*(b^2*x^2+2*a*b*x+a^2)^p,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b c x + a c\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*c*x + a*c)*(b^2*x^2 + 2*a*b*x + a^2)^p*(e*x + d)^(-2*p - 3), x)

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Fricas [A]  time = 1.10824, size = 216, normalized size = 4.24 \begin{align*} \frac{{\left (b^{2} c e x^{3} + a^{2} c d +{\left (b^{2} c d + 2 \, a b c e\right )} x^{2} +{\left (2 \, a b c d + a^{2} c e\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 3}}{2 \,{\left (b d - a e +{\left (b d - a e\right )} p\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*c*e*x^3 + a^2*c*d + (b^2*c*d + 2*a*b*c*e)*x^2 + (2*a*b*c*d + a^2*c*e)*x)*(b^2*x^2 + 2*a*b*x + a^2)^p*
(e*x + d)^(-2*p - 3)/(b*d - a*e + (b*d - a*e)*p)

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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)*(e*x+d)**(-3-2*p)*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

Timed out

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Giac [B]  time = 1.22283, size = 412, normalized size = 8.08 \begin{align*} \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} c x^{3} e^{\left (-2 \, p \log \left (x e + d\right ) - 3 \, \log \left (x e + d\right ) + 1\right )} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} c d x^{2} e^{\left (-2 \, p \log \left (x e + d\right ) - 3 \, \log \left (x e + d\right )\right )} + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b c x^{2} e^{\left (-2 \, p \log \left (x e + d\right ) - 3 \, \log \left (x e + d\right ) + 1\right )} + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b c d x e^{\left (-2 \, p \log \left (x e + d\right ) - 3 \, \log \left (x e + d\right )\right )} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} c x e^{\left (-2 \, p \log \left (x e + d\right ) - 3 \, \log \left (x e + d\right ) + 1\right )} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} c d e^{\left (-2 \, p \log \left (x e + d\right ) - 3 \, \log \left (x e + d\right )\right )}}{2 \,{\left (b d p - a p e + b d - a e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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