3.2102 \(\int (d+e x)^{-3-2 p} (a d e+(c d^2+a e^2) x+c d e x^2)^p \, dx\)

Optimal. Leaf size=128 \[ \frac{(d+e x)^{-2 p-3} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{p+1}}{(p+2) \left (c d^2-a e^2\right )}+\frac{c d (d+e x)^{-2 (p+1)} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{p+1}}{(p+1) (p+2) \left (c d^2-a e^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0524983, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.051, Rules used = {658, 650} \[ \frac{(d+e x)^{-2 p-3} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{p+1}}{(p+2) \left (c d^2-a e^2\right )}+\frac{c d (d+e x)^{-2 (p+1)} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{p+1}}{(p+1) (p+2) \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 650

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

Rubi steps

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

Mathematica [A]  time = 0.0554549, size = 76, normalized size = 0.59 \[ \frac{(d+e x)^{-2 p-3} ((d+e x) (a e+c d x))^{p+1} \left (c d (d (p+2)+e x)-a e^2 (p+1)\right )}{(p+1) (p+2) \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 170, normalized size = 1.3 \begin{align*} -{\frac{ \left ( cdx+ae \right ) \left ( ex+d \right ) ^{-2-2\,p} \left ( a{e}^{2}p-c{d}^{2}p-cdex+a{e}^{2}-2\,c{d}^{2} \right ) \left ( cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade \right ) ^{p}}{{a}^{2}{e}^{4}{p}^{2}-2\,ac{d}^{2}{e}^{2}{p}^{2}+{c}^{2}{d}^{4}{p}^{2}+3\,{a}^{2}{e}^{4}p-6\,ac{d}^{2}{e}^{2}p+3\,{c}^{2}{d}^{4}p+2\,{a}^{2}{e}^{4}-4\,ac{d}^{2}{e}^{2}+2\,{c}^{2}{d}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*d*x+a*e)*(e*x+d)^(-2-2*p)*(a*e^2*p-c*d^2*p-c*d*e*x+a*e^2-2*c*d^2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^p/(a^2
*e^4*p^2-2*a*c*d^2*e^2*p^2+c^2*d^4*p^2+3*a^2*e^4*p-6*a*c*d^2*e^2*p+3*c^2*d^4*p+2*a^2*e^4-4*a*c*d^2*e^2+2*c^2*d
^4)

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

[Out]

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

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Fricas [A]  time = 2.30756, size = 505, normalized size = 3.95 \begin{align*} \frac{{\left (c^{2} d^{2} e^{2} x^{3} + 2 \, a c d^{3} e - a^{2} d e^{3} +{\left (3 \, c^{2} d^{3} e +{\left (c^{2} d^{3} e - a c d e^{3}\right )} p\right )} x^{2} +{\left (a c d^{3} e - a^{2} d e^{3}\right )} p +{\left (2 \, c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4} +{\left (c^{2} d^{4} - a^{2} e^{4}\right )} p\right )} x\right )}{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 3}}{2 \, c^{2} d^{4} - 4 \, a c d^{2} e^{2} + 2 \, a^{2} e^{4} +{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} p^{2} + 3 \,{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*d^2*e^2*x^3 + 2*a*c*d^3*e - a^2*d*e^3 + (3*c^2*d^3*e + (c^2*d^3*e - a*c*d*e^3)*p)*x^2 + (a*c*d^3*e - a^2*
d*e^3)*p + (2*c^2*d^4 + 2*a*c*d^2*e^2 - a^2*e^4 + (c^2*d^4 - a^2*e^4)*p)*x)*(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^
2)*x)^p*(e*x + d)^(-2*p - 3)/(2*c^2*d^4 - 4*a*c*d^2*e^2 + 2*a^2*e^4 + (c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*p^2
+ 3*(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*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((e*x+d)**(-3-2*p)*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**p,x)

[Out]

Timed out

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

[Out]

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