3.2160 \(\int (a+b x) (d+e x) (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=83 \[ \frac{(a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^2 (p+1)}+\frac{e (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+3)} \]

[Out]

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

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Rubi [A]  time = 0.054496, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {770, 21, 43} \[ \frac{(a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^2 (p+1)}+\frac{e (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0358717, size = 51, normalized size = 0.61 \[ \frac{\left ((a+b x)^2\right )^{p+1} (-a e+b d (2 p+3)+2 b e (p+1) x)}{2 b^2 (p+1) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 67, normalized size = 0.8 \begin{align*} -{\frac{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p} \left ( -2\,bepx-2\,bdp-2\,bex+ae-3\,bd \right ) \left ( bx+a \right ) ^{2}}{2\,{b}^{2} \left ( 2\,{p}^{2}+5\,p+3 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*x^2+2*a*b*x+a^2)^p*(-2*b*e*p*x-2*b*d*p-2*b*e*x+a*e-3*b*d)*(b*x+a)^2/b^2/(2*p^2+5*p+3)

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Maxima [B]  time = 1.07568, size = 278, normalized size = 3.35 \begin{align*} \frac{{\left (b x + a\right )}{\left (b x + a\right )}^{2 \, p} a d}{b{\left (2 \, p + 1\right )}} + \frac{{\left (b^{2}{\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )}{\left (b x + a\right )}^{2 \, p} d}{2 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b} + \frac{{\left (b^{2}{\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )}{\left (b x + a\right )}^{2 \, p} a e}{2 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac{{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{3} +{\left (2 \, p^{2} + p\right )} a b^{2} x^{2} - 2 \, a^{2} b p x + a^{3}\right )}{\left (b x + a\right )}^{2 \, p} e}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)*(b*x + a)^(2*p)*a*d/(b*(2*p + 1)) + 1/2*(b^2*(2*p + 1)*x^2 + 2*a*b*p*x - a^2)*(b*x + a)^(2*p)*d/((2*
p^2 + 3*p + 1)*b) + 1/2*(b^2*(2*p + 1)*x^2 + 2*a*b*p*x - a^2)*(b*x + a)^(2*p)*a*e/((2*p^2 + 3*p + 1)*b^2) + ((
2*p^2 + 3*p + 1)*b^3*x^3 + (2*p^2 + p)*a*b^2*x^2 - 2*a^2*b*p*x + a^3)*(b*x + a)^(2*p)*e/((4*p^3 + 12*p^2 + 11*
p + 3)*b^2)

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Fricas [A]  time = 1.08743, size = 298, normalized size = 3.59 \begin{align*} \frac{{\left (2 \, a^{2} b d p + 3 \, a^{2} b d - a^{3} e + 2 \,{\left (b^{3} e p + b^{3} e\right )} x^{3} +{\left (3 \, b^{3} d + 3 \, a b^{2} e + 2 \,{\left (b^{3} d + 2 \, a b^{2} e\right )} p\right )} x^{2} + 2 \,{\left (3 \, a b^{2} d +{\left (2 \, a b^{2} d + a^{2} b e\right )} p\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \,{\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*b*d*p + 3*a^2*b*d - a^3*e + 2*(b^3*e*p + b^3*e)*x^3 + (3*b^3*d + 3*a*b^2*e + 2*(b^3*d + 2*a*b^2*e)*
p)*x^2 + 2*(3*a*b^2*d + (2*a*b^2*d + a^2*b*e)*p)*x)*(b^2*x^2 + 2*a*b*x + a^2)^p/(2*b^2*p^2 + 5*b^2*p + 3*b^2)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.15084, size = 477, normalized size = 5.75 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} p x^{3} e + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d p x^{2} + 4 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} p x^{2} e + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} x^{3} e + 4 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d p x + 3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d x^{2} + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b p x e + 3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} x^{2} e + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d p + 6 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d x + 3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d -{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{3} e}{2 \,{\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*p*x^3*e + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*p*x^2 + 4*(b^2*x^2 + 2*a*
b*x + a^2)^p*a*b^2*p*x^2*e + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*x^3*e + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*p
*x + 3*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*x^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*p*x*e + 3*(b^2*x^2 + 2*a*b*
x + a^2)^p*a*b^2*x^2*e + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d*p + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*x + 3
*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d - (b^2*x^2 + 2*a*b*x + a^2)^p*a^3*e)/(2*b^2*p^2 + 5*b^2*p + 3*b^2)