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

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

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Rubi [A]  time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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*}

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.82 \begin {gather*} \frac {c \left ((a+b x)^2\right )^{p+1} (d+e x)^{-2 (p+1)}}{2 (p+1) (b d-a e)} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.22, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 99, normalized size = 1.94 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.27, size = 305, normalized size = 5.98 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.05, size = 59, normalized size = 1.16 \begin {gather*} -\frac {\left (b x +a \right )^{2} c \left (e x +d \right )^{-2 p -2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{2 \left (a e p -b d p +a e -b d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*x+a)^2*(e*x+d)^(-2*p-2)*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 2.25, size = 178, normalized size = 3.49 \begin {gather*} -{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p\,\left (\frac {a^2\,c\,d}{2\,\left (a\,e-b\,d\right )\,\left (p+1\right )\,{\left (d+e\,x\right )}^{2\,p+3}}+\frac {a\,c\,x\,\left (a\,e+2\,b\,d\right )}{2\,\left (a\,e-b\,d\right )\,\left (p+1\right )\,{\left (d+e\,x\right )}^{2\,p+3}}+\frac {b\,c\,x^2\,\left (2\,a\,e+b\,d\right )}{2\,\left (a\,e-b\,d\right )\,\left (p+1\right )\,{\left (d+e\,x\right )}^{2\,p+3}}+\frac {b^2\,c\,e\,x^3}{2\,\left (a\,e-b\,d\right )\,\left (p+1\right )\,{\left (d+e\,x\right )}^{2\,p+3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed

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