3.1438 \(\int (b d+2 c d x)^3 (a+b x+c x^2)^p \, dx\)

Optimal. Leaf size=68 \[ \frac{d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{p+1}}{(p+1) (p+2)}+\frac{d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{p+1}}{p+2} \]

[Out]

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

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Rubi [A]  time = 0.0266789, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {692, 629} \[ \frac{d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{p+1}}{(p+1) (p+2)}+\frac{d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{p+1}}{p+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0498266, size = 58, normalized size = 0.85 \[ \frac{d^3 (a+x (b+c x))^{p+1} \left (4 c \left (c (p+1) x^2-a\right )+b^2 (p+2)+4 b c (p+1) x\right )}{(p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 74, normalized size = 1.1 \begin{align*} -{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{1+p} \left ( -4\,{c}^{2}p{x}^{2}-4\,bcpx-4\,{c}^{2}{x}^{2}-{b}^{2}p-4\,bcx+4\,ac-2\,{b}^{2} \right ){d}^{3}}{{p}^{2}+3\,p+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.21346, size = 166, normalized size = 2.44 \begin{align*} \frac{{\left (4 \, c^{3} d^{3}{\left (p + 1\right )} x^{4} + 8 \, b c^{2} d^{3}{\left (p + 1\right )} x^{3} + a b^{2} d^{3}{\left (p + 2\right )} - 4 \, a^{2} c d^{3} +{\left (b^{2} c d^{3}{\left (5 \, p + 6\right )} + 4 \, a c^{2} d^{3} p\right )} x^{2} +{\left (b^{3} d^{3}{\left (p + 2\right )} + 4 \, a b c d^{3} p\right )} x\right )}{\left (c x^{2} + b x + a\right )}^{p}}{p^{2} + 3 \, p + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.18462, size = 309, normalized size = 4.54 \begin{align*} \frac{{\left (a b^{2} d^{3} p + 4 \,{\left (c^{3} d^{3} p + c^{3} d^{3}\right )} x^{4} + 2 \,{\left (a b^{2} - 2 \, a^{2} c\right )} d^{3} + 8 \,{\left (b c^{2} d^{3} p + b c^{2} d^{3}\right )} x^{3} +{\left (6 \, b^{2} c d^{3} +{\left (5 \, b^{2} c + 4 \, a c^{2}\right )} d^{3} p\right )} x^{2} +{\left (2 \, b^{3} d^{3} +{\left (b^{3} + 4 \, a b c\right )} d^{3} p\right )} x\right )}{\left (c x^{2} + b x + a\right )}^{p}}{p^{2} + 3 \, p + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*b^2*d^3*p + 4*(c^3*d^3*p + c^3*d^3)*x^4 + 2*(a*b^2 - 2*a^2*c)*d^3 + 8*(b*c^2*d^3*p + b*c^2*d^3)*x^3 + (6*b^
2*c*d^3 + (5*b^2*c + 4*a*c^2)*d^3*p)*x^2 + (2*b^3*d^3 + (b^3 + 4*a*b*c)*d^3*p)*x)*(c*x^2 + b*x + a)^p/(p^2 + 3
*p + 2)

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

[Out]

Timed out

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Giac [B]  time = 1.22823, size = 417, normalized size = 6.13 \begin{align*} \frac{4 \,{\left (c x^{2} + b x + a\right )}^{p} c^{3} d^{3} p x^{4} + 8 \,{\left (c x^{2} + b x + a\right )}^{p} b c^{2} d^{3} p x^{3} + 4 \,{\left (c x^{2} + b x + a\right )}^{p} c^{3} d^{3} x^{4} + 5 \,{\left (c x^{2} + b x + a\right )}^{p} b^{2} c d^{3} p x^{2} + 4 \,{\left (c x^{2} + b x + a\right )}^{p} a c^{2} d^{3} p x^{2} + 8 \,{\left (c x^{2} + b x + a\right )}^{p} b c^{2} d^{3} x^{3} +{\left (c x^{2} + b x + a\right )}^{p} b^{3} d^{3} p x + 4 \,{\left (c x^{2} + b x + a\right )}^{p} a b c d^{3} p x + 6 \,{\left (c x^{2} + b x + a\right )}^{p} b^{2} c d^{3} x^{2} +{\left (c x^{2} + b x + a\right )}^{p} a b^{2} d^{3} p + 2 \,{\left (c x^{2} + b x + a\right )}^{p} b^{3} d^{3} x + 2 \,{\left (c x^{2} + b x + a\right )}^{p} a b^{2} d^{3} - 4 \,{\left (c x^{2} + b x + a\right )}^{p} a^{2} c d^{3}}{p^{2} + 3 \, p + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

(4*(c*x^2 + b*x + a)^p*c^3*d^3*p*x^4 + 8*(c*x^2 + b*x + a)^p*b*c^2*d^3*p*x^3 + 4*(c*x^2 + b*x + a)^p*c^3*d^3*x
^4 + 5*(c*x^2 + b*x + a)^p*b^2*c*d^3*p*x^2 + 4*(c*x^2 + b*x + a)^p*a*c^2*d^3*p*x^2 + 8*(c*x^2 + b*x + a)^p*b*c
^2*d^3*x^3 + (c*x^2 + b*x + a)^p*b^3*d^3*p*x + 4*(c*x^2 + b*x + a)^p*a*b*c*d^3*p*x + 6*(c*x^2 + b*x + a)^p*b^2
*c*d^3*x^2 + (c*x^2 + b*x + a)^p*a*b^2*d^3*p + 2*(c*x^2 + b*x + a)^p*b^3*d^3*x + 2*(c*x^2 + b*x + a)^p*a*b^2*d
^3 - 4*(c*x^2 + b*x + a)^p*a^2*c*d^3)/(p^2 + 3*p + 2)