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]

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

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [A]  time = 0.05, 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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fricas [B]  time = 1.35, size = 151, normalized size = 2.22 \[ \frac {{\left (a b^{2} d^{3} p + 4 \, {\left (c^{3} d^{3} p + c^{3} d^{3}\right )} x^{4} + 2 \, {\left (a b^{2} - 2 \, a^{2} c\right )} d^{3} + 8 \, {\left (b c^{2} d^{3} p + b c^{2} d^{3}\right )} x^{3} + {\left (6 \, b^{2} c d^{3} + {\left (5 \, b^{2} c + 4 \, a c^{2}\right )} d^{3} p\right )} x^{2} + {\left (2 \, b^{3} d^{3} + {\left (b^{3} + 4 \, a b c\right )} d^{3} p\right )} x\right )} {\left (c x^{2} + b x + a\right )}^{p}}{p^{2} + 3 \, p + 2} \]

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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giac [B]  time = 0.19, size = 291, normalized size = 4.28 \[ \frac {a b^{2} d^{3} p \left (\frac {c d x^{2} + b d x + a d}{d}\right )^{p} + 2 \, a b^{2} d^{3} \left (\frac {c d x^{2} + b d x + a d}{d}\right )^{p} - 4 \, a^{2} c d^{3} \left (\frac {c d x^{2} + b d x + a d}{d}\right )^{p} + {\left (c d x^{2} + b d x\right )} b^{2} d^{2} p \left (\frac {c d x^{2} + b d x + a d}{d}\right )^{p} + 4 \, {\left (c d x^{2} + b d x\right )} a c d^{2} p \left (\frac {c d x^{2} + b d x + a d}{d}\right )^{p} + 2 \, {\left (c d x^{2} + b d x\right )} b^{2} d^{2} \left (\frac {c d x^{2} + b d x + a d}{d}\right )^{p} + 4 \, {\left (c d x^{2} + b d x\right )}^{2} c d p \left (\frac {c d x^{2} + b d x + a d}{d}\right )^{p} + 4 \, {\left (c d x^{2} + b d x\right )}^{2} c d \left (\frac {c d x^{2} + b d x + a d}{d}\right )^{p}}{p^{2} + 3 \, p + 2} \]

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]

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

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maple [A]  time = 0.05, size = 74, normalized size = 1.09 \[ -\frac {\left (-4 c^{2} p \,x^{2}-4 b c p x -4 c^{2} x^{2}-b^{2} p -4 b c x +4 a c -2 b^{2}\right ) d^{3} \left (c \,x^{2}+b x +a \right )^{p +1}}{p^{2}+3 p +2} \]

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)^(p+1)*(-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 = 1.83, size = 123, normalized size = 1.81 \[ \frac {{\left (4 \, c^{3} d^{3} {\left (p + 1\right )} x^{4} + 8 \, b c^{2} d^{3} {\left (p + 1\right )} x^{3} + a b^{2} d^{3} {\left (p + 2\right )} - 4 \, a^{2} c d^{3} + {\left (b^{2} c d^{3} {\left (5 \, p + 6\right )} + 4 \, a c^{2} d^{3} p\right )} x^{2} + {\left (b^{3} d^{3} {\left (p + 2\right )} + 4 \, a b c d^{3} p\right )} x\right )} {\left (c x^{2} + b x + a\right )}^{p}}{p^{2} + 3 \, p + 2} \]

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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mupad [B]  time = 0.65, size = 160, normalized size = 2.35 \[ {\left (c\,x^2+b\,x+a\right )}^p\,\left (\frac {a\,d^3\,\left (b^2\,p-4\,a\,c+2\,b^2\right )}{p^2+3\,p+2}+\frac {c\,d^3\,x^2\,\left (5\,b^2\,p+6\,b^2+4\,a\,c\,p\right )}{p^2+3\,p+2}+\frac {4\,c^3\,d^3\,x^4\,\left (p+1\right )}{p^2+3\,p+2}+\frac {b\,d^3\,x\,\left (b^2\,p+2\,b^2+4\,a\,c\,p\right )}{p^2+3\,p+2}+\frac {8\,b\,c^2\,d^3\,x^3\,\left (p+1\right )}{p^2+3\,p+2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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