3.18.57 \(\int (d+e x)^{-3-2 p} (a^2+2 a b x+b^2 x^2)^p \, dx\) [1757]

Optimal. Leaf size=115 \[ \frac {b (a+b x) (d+e x)^{-1-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p}{2 (b d-a e)^2 (1+p) (1+2 p)}+\frac {(a+b x) (d+e x)^{-2 (1+p)} \left (a^2+2 a b x+b^2 x^2\right )^p}{2 (b d-a e) (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {660, 47, 37} \begin {gather*} \frac {b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{-2 p-1}}{2 (p+1) (2 p+1) (b d-a e)^2}+\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{-2 (p+1)}}{2 (p+1) (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 47

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] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 660

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

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 72, normalized size = 0.63 \begin {gather*} \frac {(a+b x) \left ((a+b x)^2\right )^p (d+e x)^{-2 (1+p)} (2 b d (1+p)-a e (1+2 p)+b e x)}{2 (b d-a e)^2 (1+p) (1+2 p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.72, size = 139, normalized size = 1.21

method	result	size
gosper	\(-\frac {\left (e x +d \right )^{-2-2 p} \left (b x +a \right ) \left (2 a e p -2 b d p -b e x +a e -2 b d \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{2 \left (2 a^{2} e^{2} p^{2}-4 a b d e \,p^{2}+2 b^{2} d^{2} p^{2}+3 a^{2} e^{2} p -6 a b d e p +3 b^{2} d^{2} p +a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(e*x+d)^(-2-2*p)*(b*x+a)*(2*a*e*p-2*b*d*p-b*e*x+a*e-2*b*d)*(b^2*x^2+2*a*b*x+a^2)^p/(2*a^2*e^2*p^2-4*a*b*d
*e*p^2+2*b^2*d^2*p^2+3*a^2*e^2*p-6*a*b*d*e*p+3*b^2*d^2*p+a^2*e^2-2*a*b*d*e+b^2*d^2)

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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)*(b^2*x^2+2*a*b*x+a^2)^p,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.88, size = 216, normalized size = 1.88 \begin {gather*} \frac {{\left (2 \, a b d^{2} p + 2 \, a b d^{2} + 2 \, {\left (b^{2} d^{2} p + b^{2} d^{2}\right )} x - {\left (2 \, a b p x^{2} - b^{2} x^{3} + {\left (2 \, a^{2} p + a^{2}\right )} x\right )} e^{2} - {\left (2 \, a^{2} d p - 2 \, a b d x + a^{2} d - {\left (2 \, b^{2} d p + 3 \, b^{2} d\right )} x^{2}\right )} e\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} {\left (x e + d\right )}^{-2 \, p - 3}}{2 \, {\left (2 \, b^{2} d^{2} p^{2} + 3 \, b^{2} d^{2} p + b^{2} d^{2} + {\left (2 \, a^{2} p^{2} + 3 \, a^{2} p + a^{2}\right )} e^{2} - 2 \, {\left (2 \, a b d p^{2} + 3 \, a b d p + a b d\right )} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*(b**2*x**2+2*a*b*x+a**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)*(b^2*x^2+2*a*b*x+a^2)^p,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.83, size = 259, normalized size = 2.25 \begin {gather*} {\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p\,\left (\frac {x\,\left (2\,b^2\,d^2-a^2\,e^2-2\,a^2\,e^2\,p+2\,b^2\,d^2\,p+2\,a\,b\,d\,e\right )}{2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (2\,p^2+3\,p+1\right )}+\frac {b^2\,e^2\,x^3}{2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (2\,p^2+3\,p+1\right )}-\frac {a\,d\,\left (a\,e-2\,b\,d+2\,a\,e\,p-2\,b\,d\,p\right )}{2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (2\,p^2+3\,p+1\right )}+\frac {b\,e\,x^2\,\left (3\,b\,d-2\,a\,e\,p+2\,b\,d\,p\right )}{2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (2\,p^2+3\,p+1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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