### 3.1436 $$\int (b d+2 c d x)^5 (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=121 $\frac{2 d^5 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )^{p+1}}{(p+2) (p+3)}+\frac{2 d^5 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{p+1}}{(p+1) (p+2) (p+3)}+\frac{d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{p+1}}{p+3}$

[Out]

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

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Rubi [A]  time = 0.0730149, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {692, 629} $\frac{2 d^5 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )^{p+1}}{(p+2) (p+3)}+\frac{2 d^5 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{p+1}}{(p+1) (p+2) (p+3)}+\frac{d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{p+1}}{p+3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^5*(a + b*x + c*x^2)^p,x]

[Out]

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

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rule 629

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

Rubi steps

\begin{align*} \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx &=\frac{d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{1+p}}{3+p}+\frac{\left (2 \left (b^2-4 a c\right ) d^2\right ) \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^p \, dx}{3+p}\\ &=\frac{2 \left (b^2-4 a c\right ) d^5 (b+2 c x)^2 \left (a+b x+c x^2\right )^{1+p}}{(2+p) (3+p)}+\frac{d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{1+p}}{3+p}+\frac{\left (2 \left (b^2-4 a c\right )^2 d^4\right ) \int (b d+2 c d x) \left (a+b x+c x^2\right )^p \, dx}{(2+p) (3+p)}\\ &=\frac{2 \left (b^2-4 a c\right )^2 d^5 \left (a+b x+c x^2\right )^{1+p}}{(1+p) (2+p) (3+p)}+\frac{2 \left (b^2-4 a c\right ) d^5 (b+2 c x)^2 \left (a+b x+c x^2\right )^{1+p}}{(2+p) (3+p)}+\frac{d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{1+p}}{3+p}\\ \end{align*}

Mathematica [A]  time = 0.100672, size = 145, normalized size = 1.2 $\frac{d^5 (a+x (b+c x))^{p+1} \left (16 c^2 \left (2 a^2-2 a c (p+1) x^2+c^2 \left (p^2+3 p+2\right ) x^4\right )-8 b^2 c \left (a (p+3)-c \left (3 p^2+10 p+7\right ) x^2\right )-32 b c^2 (p+1) x \left (a-c (p+2) x^2\right )+8 b^3 c \left (p^2+4 p+3\right ) x+b^4 \left (p^2+5 p+6\right )\right )}{(p+1) (p+2) (p+3)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^5*(a + b*x + c*x^2)^p,x]

[Out]

(d^5*(a + x*(b + c*x))^(1 + p)*(b^4*(6 + 5*p + p^2) + 8*b^3*c*(3 + 4*p + p^2)*x - 32*b*c^2*(1 + p)*x*(a - c*(2
+ p)*x^2) - 8*b^2*c*(a*(3 + p) - c*(7 + 10*p + 3*p^2)*x^2) + 16*c^2*(2*a^2 - 2*a*c*(1 + p)*x^2 + c^2*(2 + 3*p
+ p^2)*x^4)))/((1 + p)*(2 + p)*(3 + p))

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Maple [A]  time = 0.051, size = 233, normalized size = 1.9 \begin{align*}{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{1+p}{d}^{5} \left ( 16\,{c}^{4}{p}^{2}{x}^{4}+32\,b{c}^{3}{p}^{2}{x}^{3}+48\,{c}^{4}p{x}^{4}+24\,{b}^{2}{c}^{2}{p}^{2}{x}^{2}+96\,b{c}^{3}p{x}^{3}+32\,{c}^{4}{x}^{4}-32\,a{c}^{3}p{x}^{2}+8\,{b}^{3}c{p}^{2}x+80\,{b}^{2}{c}^{2}p{x}^{2}+64\,b{c}^{3}{x}^{3}-32\,ab{c}^{2}px-32\,{x}^{2}a{c}^{3}+{b}^{4}{p}^{2}+32\,{b}^{3}cpx+56\,{x}^{2}{b}^{2}{c}^{2}-8\,a{b}^{2}cp-32\,ba{c}^{2}x+5\,{b}^{4}p+24\,{b}^{3}cx+32\,{a}^{2}{c}^{2}-24\,ac{b}^{2}+6\,{b}^{4} \right ) }{{p}^{3}+6\,{p}^{2}+11\,p+6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x)

[Out]

(c*x^2+b*x+a)^(1+p)*d^5*(16*c^4*p^2*x^4+32*b*c^3*p^2*x^3+48*c^4*p*x^4+24*b^2*c^2*p^2*x^2+96*b*c^3*p*x^3+32*c^4
*x^4-32*a*c^3*p*x^2+8*b^3*c*p^2*x+80*b^2*c^2*p*x^2+64*b*c^3*x^3-32*a*b*c^2*p*x-32*a*c^3*x^2+b^4*p^2+32*b^3*c*p
*x+56*b^2*c^2*x^2-8*a*b^2*c*p-32*a*b*c^2*x+5*b^4*p+24*b^3*c*x+32*a^2*c^2-24*a*b^2*c+6*b^4)/(p^3+6*p^2+11*p+6)

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Maxima [B]  time = 1.28906, size = 398, normalized size = 3.29 \begin{align*} \frac{{\left (16 \,{\left (p^{2} + 3 \, p + 2\right )} c^{5} d^{5} x^{6} + 48 \,{\left (p^{2} + 3 \, p + 2\right )} b c^{4} d^{5} x^{5} +{\left (p^{2} + 5 \, p + 6\right )} a b^{4} d^{5} - 8 \, a^{2} b^{2} c d^{5}{\left (p + 3\right )} + 32 \, a^{3} c^{2} d^{5} + 8 \,{\left ({\left (7 \, p^{2} + 22 \, p + 15\right )} b^{2} c^{3} d^{5} + 2 \,{\left (p^{2} + p\right )} a c^{4} d^{5}\right )} x^{4} + 16 \,{\left ({\left (2 \, p^{2} + 7 \, p + 5\right )} b^{3} c^{2} d^{5} + 2 \,{\left (p^{2} + p\right )} a b c^{3} d^{5}\right )} x^{3} +{\left ({\left (9 \, p^{2} + 37 \, p + 30\right )} b^{4} c d^{5} + 8 \,{\left (3 \, p^{2} + 5 \, p\right )} a b^{2} c^{2} d^{5} - 32 \, a^{2} c^{3} d^{5} p\right )} x^{2} +{\left ({\left (p^{2} + 5 \, p + 6\right )} b^{5} d^{5} + 8 \,{\left (p^{2} + 3 \, p\right )} a b^{3} c d^{5} - 32 \, a^{2} b c^{2} d^{5} p\right )} x\right )}{\left (c x^{2} + b x + a\right )}^{p}}{p^{3} + 6 \, p^{2} + 11 \, p + 6} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

(16*(p^2 + 3*p + 2)*c^5*d^5*x^6 + 48*(p^2 + 3*p + 2)*b*c^4*d^5*x^5 + (p^2 + 5*p + 6)*a*b^4*d^5 - 8*a^2*b^2*c*d
^5*(p + 3) + 32*a^3*c^2*d^5 + 8*((7*p^2 + 22*p + 15)*b^2*c^3*d^5 + 2*(p^2 + p)*a*c^4*d^5)*x^4 + 16*((2*p^2 + 7
*p + 5)*b^3*c^2*d^5 + 2*(p^2 + p)*a*b*c^3*d^5)*x^3 + ((9*p^2 + 37*p + 30)*b^4*c*d^5 + 8*(3*p^2 + 5*p)*a*b^2*c^
2*d^5 - 32*a^2*c^3*d^5*p)*x^2 + ((p^2 + 5*p + 6)*b^5*d^5 + 8*(p^2 + 3*p)*a*b^3*c*d^5 - 32*a^2*b*c^2*d^5*p)*x)*
(c*x^2 + b*x + a)^p/(p^3 + 6*p^2 + 11*p + 6)

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Fricas [B]  time = 2.22443, size = 828, normalized size = 6.84 \begin{align*} \frac{{\left (a b^{4} d^{5} p^{2} +{\left (5 \, a b^{4} - 8 \, a^{2} b^{2} c\right )} d^{5} p + 16 \,{\left (c^{5} d^{5} p^{2} + 3 \, c^{5} d^{5} p + 2 \, c^{5} d^{5}\right )} x^{6} + 2 \,{\left (3 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} d^{5} + 48 \,{\left (b c^{4} d^{5} p^{2} + 3 \, b c^{4} d^{5} p + 2 \, b c^{4} d^{5}\right )} x^{5} + 8 \,{\left (15 \, b^{2} c^{3} d^{5} +{\left (7 \, b^{2} c^{3} + 2 \, a c^{4}\right )} d^{5} p^{2} + 2 \,{\left (11 \, b^{2} c^{3} + a c^{4}\right )} d^{5} p\right )} x^{4} + 16 \,{\left (5 \, b^{3} c^{2} d^{5} + 2 \,{\left (b^{3} c^{2} + a b c^{3}\right )} d^{5} p^{2} +{\left (7 \, b^{3} c^{2} + 2 \, a b c^{3}\right )} d^{5} p\right )} x^{3} +{\left (30 \, b^{4} c d^{5} + 3 \,{\left (3 \, b^{4} c + 8 \, a b^{2} c^{2}\right )} d^{5} p^{2} +{\left (37 \, b^{4} c + 40 \, a b^{2} c^{2} - 32 \, a^{2} c^{3}\right )} d^{5} p\right )} x^{2} +{\left (6 \, b^{5} d^{5} +{\left (b^{5} + 8 \, a b^{3} c\right )} d^{5} p^{2} +{\left (5 \, b^{5} + 24 \, a b^{3} c - 32 \, a^{2} b c^{2}\right )} d^{5} p\right )} x\right )}{\left (c x^{2} + b x + a\right )}^{p}}{p^{3} + 6 \, p^{2} + 11 \, p + 6} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

(a*b^4*d^5*p^2 + (5*a*b^4 - 8*a^2*b^2*c)*d^5*p + 16*(c^5*d^5*p^2 + 3*c^5*d^5*p + 2*c^5*d^5)*x^6 + 2*(3*a*b^4 -
12*a^2*b^2*c + 16*a^3*c^2)*d^5 + 48*(b*c^4*d^5*p^2 + 3*b*c^4*d^5*p + 2*b*c^4*d^5)*x^5 + 8*(15*b^2*c^3*d^5 + (
7*b^2*c^3 + 2*a*c^4)*d^5*p^2 + 2*(11*b^2*c^3 + a*c^4)*d^5*p)*x^4 + 16*(5*b^3*c^2*d^5 + 2*(b^3*c^2 + a*b*c^3)*d
^5*p^2 + (7*b^3*c^2 + 2*a*b*c^3)*d^5*p)*x^3 + (30*b^4*c*d^5 + 3*(3*b^4*c + 8*a*b^2*c^2)*d^5*p^2 + (37*b^4*c +
40*a*b^2*c^2 - 32*a^2*c^3)*d^5*p)*x^2 + (6*b^5*d^5 + (b^5 + 8*a*b^3*c)*d^5*p^2 + (5*b^5 + 24*a*b^3*c - 32*a^2*
b*c^2)*d^5*p)*x)*(c*x^2 + b*x + a)^p/(p^3 + 6*p^2 + 11*p + 6)

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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((2*c*d*x+b*d)**5*(c*x**2+b*x+a)**p,x)

[Out]

Timed out

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Giac [B]  time = 1.21724, size = 1184, normalized size = 9.79 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

(16*(c*x^2 + b*x + a)^p*c^5*d^5*p^2*x^6 + 48*(c*x^2 + b*x + a)^p*b*c^4*d^5*p^2*x^5 + 48*(c*x^2 + b*x + a)^p*c^
5*d^5*p*x^6 + 56*(c*x^2 + b*x + a)^p*b^2*c^3*d^5*p^2*x^4 + 16*(c*x^2 + b*x + a)^p*a*c^4*d^5*p^2*x^4 + 144*(c*x
^2 + b*x + a)^p*b*c^4*d^5*p*x^5 + 32*(c*x^2 + b*x + a)^p*c^5*d^5*x^6 + 32*(c*x^2 + b*x + a)^p*b^3*c^2*d^5*p^2*
x^3 + 32*(c*x^2 + b*x + a)^p*a*b*c^3*d^5*p^2*x^3 + 176*(c*x^2 + b*x + a)^p*b^2*c^3*d^5*p*x^4 + 16*(c*x^2 + b*x
+ a)^p*a*c^4*d^5*p*x^4 + 96*(c*x^2 + b*x + a)^p*b*c^4*d^5*x^5 + 9*(c*x^2 + b*x + a)^p*b^4*c*d^5*p^2*x^2 + 24*
(c*x^2 + b*x + a)^p*a*b^2*c^2*d^5*p^2*x^2 + 112*(c*x^2 + b*x + a)^p*b^3*c^2*d^5*p*x^3 + 32*(c*x^2 + b*x + a)^p
*a*b*c^3*d^5*p*x^3 + 120*(c*x^2 + b*x + a)^p*b^2*c^3*d^5*x^4 + (c*x^2 + b*x + a)^p*b^5*d^5*p^2*x + 8*(c*x^2 +
b*x + a)^p*a*b^3*c*d^5*p^2*x + 37*(c*x^2 + b*x + a)^p*b^4*c*d^5*p*x^2 + 40*(c*x^2 + b*x + a)^p*a*b^2*c^2*d^5*p
*x^2 - 32*(c*x^2 + b*x + a)^p*a^2*c^3*d^5*p*x^2 + 80*(c*x^2 + b*x + a)^p*b^3*c^2*d^5*x^3 + (c*x^2 + b*x + a)^p
*a*b^4*d^5*p^2 + 5*(c*x^2 + b*x + a)^p*b^5*d^5*p*x + 24*(c*x^2 + b*x + a)^p*a*b^3*c*d^5*p*x - 32*(c*x^2 + b*x
+ a)^p*a^2*b*c^2*d^5*p*x + 30*(c*x^2 + b*x + a)^p*b^4*c*d^5*x^2 + 5*(c*x^2 + b*x + a)^p*a*b^4*d^5*p - 8*(c*x^2
+ b*x + a)^p*a^2*b^2*c*d^5*p + 6*(c*x^2 + b*x + a)^p*b^5*d^5*x + 6*(c*x^2 + b*x + a)^p*a*b^4*d^5 - 24*(c*x^2
+ b*x + a)^p*a^2*b^2*c*d^5 + 32*(c*x^2 + b*x + a)^p*a^3*c^2*d^5)/(p^3 + 6*p^2 + 11*p + 6)