3.2159 \(\int (a+b x) (d+e x)^2 (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=134 \[ \frac{2 e (a+b x)^3 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+3)}+\frac{(a+b x)^2 (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^3 (p+1)}+\frac{e^2 (a+b x)^4 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^3 (p+2)} \]

[Out]

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

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Rubi [A]  time = 0.0858001, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {770, 21, 43} \[ \frac{2 e (a+b x)^3 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+3)}+\frac{(a+b x)^2 (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^3 (p+1)}+\frac{e^2 (a+b x)^4 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^3 (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)^2*(a + b*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^p)/(2*b^3*(1 + p)) + (2*e*(b*d - a*e)*(a + b*x)^3*(a^2 +
2*a*b*x + b^2*x^2)^p)/(b^3*(3 + 2*p)) + (e^2*(a + b*x)^4*(a^2 + 2*a*b*x + b^2*x^2)^p)/(2*b^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)^2 \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)^2 \, 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)^2 \, 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)^2 \left (a b+b^2 x\right )^{1+2 p}}{b^2}+\frac{2 e (b d-a e) \left (a b+b^2 x\right )^{2+2 p}}{b^3}+\frac{e^2 \left (a b+b^2 x\right )^{3+2 p}}{b^4}\right ) \, dx}{b}\\ &=\frac{(b d-a e)^2 (a+b x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^3 (1+p)}+\frac{2 e (b d-a e) (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (3+2 p)}+\frac{e^2 (a+b x)^4 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^3 (2+p)}\\ \end{align*}

Mathematica [A]  time = 0.0722754, size = 109, normalized size = 0.81 \[ \frac{\left ((a+b x)^2\right )^{p+1} \left (a^2 e^2-2 a b e (d (p+2)+e (p+1) x)+b^2 \left (d^2 \left (2 p^2+7 p+6\right )+4 d e \left (p^2+3 p+2\right ) x+e^2 \left (2 p^2+5 p+3\right ) x^2\right )\right )}{2 b^3 (p+1) (p+2) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((a + b*x)^2)^(1 + p)*(a^2*e^2 - 2*a*b*e*(d*(2 + p) + e*(1 + p)*x) + b^2*(d^2*(6 + 7*p + 2*p^2) + 4*d*e*(2 +
3*p + p^2)*x + e^2*(3 + 5*p + 2*p^2)*x^2)))/(2*b^3*(1 + p)*(2 + p)*(3 + 2*p))

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Maple [A]  time = 0.006, size = 179, normalized size = 1.3 \begin{align*}{\frac{ \left ( 2\,{b}^{2}{e}^{2}{p}^{2}{x}^{2}+4\,{b}^{2}de{p}^{2}x+5\,{b}^{2}{e}^{2}p{x}^{2}-2\,ab{e}^{2}px+2\,{b}^{2}{d}^{2}{p}^{2}+12\,{b}^{2}depx+3\,{x}^{2}{b}^{2}{e}^{2}-2\,abdep-2\,xab{e}^{2}+7\,{b}^{2}{d}^{2}p+8\,x{b}^{2}de+{a}^{2}{e}^{2}-4\,abde+6\,{b}^{2}{d}^{2} \right ) \left ( bx+a \right ) ^{2} \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{2\,{b}^{3} \left ( 2\,{p}^{3}+9\,{p}^{2}+13\,p+6 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x+a)^2*(2*b^2*e^2*p^2*x^2+4*b^2*d*e*p^2*x+5*b^2*e^2*p*x^2-2*a*b*e^2*p*x+2*b^2*d^2*p^2+12*b^2*d*e*p*x+3*
b^2*e^2*x^2-2*a*b*d*e*p-2*a*b*e^2*x+7*b^2*d^2*p+8*b^2*d*e*x+a^2*e^2-4*a*b*d*e+6*b^2*d^2)*(b^2*x^2+2*a*b*x+a^2)
^p/b^3/(2*p^3+9*p^2+13*p+6)

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Maxima [B]  time = 1.10984, size = 545, normalized size = 4.07 \begin{align*} \frac{{\left (b x + a\right )}{\left (b x + a\right )}^{2 \, p} a d^{2}}{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}}{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 d e}{{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac{2 \,{\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} d e}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\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} a e^{2}}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} + \frac{{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{4} + 2 \,{\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{3} - 3 \,{\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b p x - 3 \, a^{4}\right )}{\left (b x + a\right )}^{2 \, p} e^{2}}{2 \,{\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^2*(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^2/(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/
((2*p^2 + 3*p + 1)*b) + (b^2*(2*p + 1)*x^2 + 2*a*b*p*x - a^2)*(b*x + a)^(2*p)*a*d*e/((2*p^2 + 3*p + 1)*b^2) +
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)*d*e/((4*p^3 + 12*p^2
+ 11*p + 3)*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)*a*
e^2/((4*p^3 + 12*p^2 + 11*p + 3)*b^3) + 1/2*((4*p^3 + 12*p^2 + 11*p + 3)*b^4*x^4 + 2*(2*p^3 + 3*p^2 + p)*a*b^3
*x^3 - 3*(2*p^2 + p)*a^2*b^2*x^2 + 6*a^3*b*p*x - 3*a^4)*(b*x + a)^(2*p)*e^2/((4*p^4 + 20*p^3 + 35*p^2 + 25*p +
6)*b^3)

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Fricas [B]  time = 1.08294, size = 718, normalized size = 5.36 \begin{align*} \frac{{\left (2 \, a^{2} b^{2} d^{2} p^{2} + 6 \, a^{2} b^{2} d^{2} - 4 \, a^{3} b d e + a^{4} e^{2} +{\left (2 \, b^{4} e^{2} p^{2} + 5 \, b^{4} e^{2} p + 3 \, b^{4} e^{2}\right )} x^{4} + 4 \,{\left (2 \, b^{4} d e + a b^{3} e^{2} +{\left (b^{4} d e + a b^{3} e^{2}\right )} p^{2} +{\left (3 \, b^{4} d e + 2 \, a b^{3} e^{2}\right )} p\right )} x^{3} +{\left (6 \, b^{4} d^{2} + 12 \, a b^{3} d e + 2 \,{\left (b^{4} d^{2} + 4 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} p^{2} +{\left (7 \, b^{4} d^{2} + 22 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} p\right )} x^{2} +{\left (7 \, a^{2} b^{2} d^{2} - 2 \, a^{3} b d e\right )} p + 2 \,{\left (6 \, a b^{3} d^{2} + 2 \,{\left (a b^{3} d^{2} + a^{2} b^{2} d e\right )} p^{2} +{\left (7 \, a b^{3} d^{2} + 4 \, a^{2} b^{2} d e - a^{3} b e^{2}\right )} p\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \,{\left (2 \, b^{3} p^{3} + 9 \, b^{3} p^{2} + 13 \, b^{3} p + 6 \, b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*b^2*d^2*p^2 + 6*a^2*b^2*d^2 - 4*a^3*b*d*e + a^4*e^2 + (2*b^4*e^2*p^2 + 5*b^4*e^2*p + 3*b^4*e^2)*x^4
+ 4*(2*b^4*d*e + a*b^3*e^2 + (b^4*d*e + a*b^3*e^2)*p^2 + (3*b^4*d*e + 2*a*b^3*e^2)*p)*x^3 + (6*b^4*d^2 + 12*a
*b^3*d*e + 2*(b^4*d^2 + 4*a*b^3*d*e + a^2*b^2*e^2)*p^2 + (7*b^4*d^2 + 22*a*b^3*d*e + a^2*b^2*e^2)*p)*x^2 + (7*
a^2*b^2*d^2 - 2*a^3*b*d*e)*p + 2*(6*a*b^3*d^2 + 2*(a*b^3*d^2 + a^2*b^2*d*e)*p^2 + (7*a*b^3*d^2 + 4*a^2*b^2*d*e
- a^3*b*e^2)*p)*x)*(b^2*x^2 + 2*a*b*x + a^2)^p/(2*b^3*p^3 + 9*b^3*p^2 + 13*b^3*p + 6*b^3)

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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)**2*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.1598, size = 1219, normalized size = 9.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^2*(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^4*p^2*x^4*e^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*b^4*d*p^2*x^3*e + 2*(b^2*x^
2 + 2*a*b*x + a^2)^p*b^4*d^2*p^2*x^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^3*p^2*x^3*e^2 + 5*(b^2*x^2 + 2*a*b*x
+ a^2)^p*b^4*p*x^4*e^2 + 8*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^3*d*p^2*x^2*e + 12*(b^2*x^2 + 2*a*b*x + a^2)^p*b^4*
d*p*x^3*e + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^3*d^2*p^2*x + 7*(b^2*x^2 + 2*a*b*x + a^2)^p*b^4*d^2*p*x^2 + 2*(b
^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^2*p^2*x^2*e^2 + 8*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^3*p*x^3*e^2 + 3*(b^2*x^2 + 2
*a*b*x + a^2)^p*b^4*x^4*e^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^2*d*p^2*x*e + 22*(b^2*x^2 + 2*a*b*x + a^2)^p
*a*b^3*d*p*x^2*e + 8*(b^2*x^2 + 2*a*b*x + a^2)^p*b^4*d*x^3*e + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^2*d^2*p^2 +
14*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^3*d^2*p*x + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*b^4*d^2*x^2 + (b^2*x^2 + 2*a*b*x
+ a^2)^p*a^2*b^2*p*x^2*e^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^3*x^3*e^2 + 8*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*
b^2*d*p*x*e + 12*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^3*d*x^2*e + 7*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^2*d^2*p + 12*
(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^3*d^2*x - 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^3*b*p*x*e^2 - 2*(b^2*x^2 + 2*a*b*x +
a^2)^p*a^3*b*d*p*e + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^2*d^2 - 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a^3*b*d*e + (b
^2*x^2 + 2*a*b*x + a^2)^p*a^4*e^2)/(2*b^3*p^3 + 9*b^3*p^2 + 13*b^3*p + 6*b^3)