3.22.2 \(\int (d+e x)^{-3-2 p} (a d e+(c d^2+a e^2) x+c d e x^2)^p \, dx\) [2102]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 128, normalized size of antiderivative = 1.00, 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 = {672, 664} \begin {gather*} \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} \end {gather*}

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 664

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]

Rule 672

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]

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*}

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Mathematica [A]
time = 0.07, size = 76, normalized size = 0.59 \begin {gather*} \frac {(d+e x)^{-3-2 p} ((a e+c d x) (d+e x))^{1+p} \left (-a e^2 (1+p)+c d (d (2+p)+e x)\right )}{\left (c d^2-a e^2\right )^2 (1+p) (2+p)} \end {gather*}

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 = 0.78, size = 170, normalized size = 1.33

method	result	size
gosper	\(-\frac {\left (c d x +a e \right ) \left (e x +d \right )^{-2-2 p} \left (a \,e^{2} p -c \,d^{2} p -c d e x +e^{2} a -2 c \,d^{2}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{p}}{a^{2} e^{4} p^{2}-2 a c \,d^{2} e^{2} p^{2}+c^{2} d^{4} p^{2}+3 a^{2} e^{4} p -6 a c \,d^{2} e^{2} p +3 c^{2} d^{4} p +2 a^{2} e^{4}-4 a c \,d^{2} e^{2}+2 c^{2} d^{4}}\)	\(170\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 3.41, size = 242, normalized size = 1.89 \begin {gather*} -\frac {{\left ({\left (a^{2} p + a^{2}\right )} x e^{4} - {\left (c^{2} d^{4} p + 2 \, c^{2} d^{4}\right )} x + {\left (a c d p x^{2} + a^{2} d p + a^{2} d\right )} e^{3} - {\left (c^{2} d^{2} x^{3} + 2 \, a c d^{2} x\right )} e^{2} - {\left (a c d^{3} p + 2 \, a c d^{3} + {\left (c^{2} d^{3} p + 3 \, c^{2} d^{3}\right )} x^{2}\right )} e\right )} {\left (c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e\right )}^{p} {\left (x e + d\right )}^{-2 \, p - 3}}{c^{2} d^{4} p^{2} + 3 \, c^{2} d^{4} p + 2 \, c^{2} d^{4} + {\left (a^{2} p^{2} + 3 \, a^{2} p + 2 \, a^{2}\right )} e^{4} - 2 \, {\left (a c d^{2} p^{2} + 3 \, a c d^{2} p + 2 \, a c d^{2}\right )} e^{2}} \end {gather*}

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]

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

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [B]
time = 0.98, size = 293, normalized size = 2.29 \begin {gather*} {\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^p\,\left (\frac {x\,\left (2\,c^2\,d^4-a^2\,e^4-a^2\,e^4\,p+c^2\,d^4\,p+2\,a\,c\,d^2\,e^2\right )}{{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (p^2+3\,p+2\right )}+\frac {c^2\,d^2\,e^2\,x^3}{{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (p^2+3\,p+2\right )}-\frac {a\,d\,e\,\left (a\,e^2-2\,c\,d^2+a\,e^2\,p-c\,d^2\,p\right )}{{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (p^2+3\,p+2\right )}+\frac {c\,d\,e\,x^2\,\left (3\,c\,d^2-a\,e^2\,p+c\,d^2\,p\right )}{{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (p^2+3\,p+2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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