3.15.38 \(\int (b d+2 c d x)^3 (a+b x+c x^2)^p \, dx\) [1438]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.02, 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 = {706, 643} \begin {gather*} \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} \end {gather*}

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 643

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 706

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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.47, size = 58, normalized size = 0.85 \begin {gather*} \frac {d^3 (a+x (b+c x))^{1+p} \left (b^2 (2+p)+4 b c (1+p) x+4 c \left (-a+c (1+p) x^2\right )\right )}{(1+p) (2+p)} \end {gather*}

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))

________________________________________________________________________________________

Maple [A]
time = 0.74, size = 74, normalized size = 1.09

method	result	size
gosper	\(-\frac {\left (c \,x^{2}+b x +a \right )^{1+p} \left (-4 c^{2} p \,x^{2}-4 b c p x -4 c^{2} x^{2}-b^{2} p -4 b c x +4 a c -2 b^{2}\right ) d^{3}}{p^{2}+3 p +2}\)	\(74\)
risch	\(-\frac {d^{3} \left (-4 p \,c^{3} x^{4}-8 p b \,c^{2} x^{3}-4 c^{3} x^{4}-4 a \,c^{2} p \,x^{2}-5 b^{2} c p \,x^{2}-8 b \,c^{2} x^{3}-4 a b c p x -b^{3} p x -6 b^{2} c \,x^{2}-a \,b^{2} p -2 b^{3} x +4 a^{2} c -2 a \,b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{p}}{\left (2+p \right ) \left (1+p \right )}\)	\(133\)
norman	\(\frac {b \,d^{3} \left (4 a c p +b^{2} p +2 b^{2}\right ) x \,{\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{p^{2}+3 p +2}+\frac {c \,d^{3} \left (4 a c p +5 b^{2} p +6 b^{2}\right ) x^{2} {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{p^{2}+3 p +2}-\frac {a \,d^{3} \left (-b^{2} p +4 a c -2 b^{2}\right ) {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{p^{2}+3 p +2}+\frac {4 c^{3} d^{3} x^{4} {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{2+p}+\frac {8 c^{2} b \,d^{3} x^{3} {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{2+p}\)	\(204\)

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,method=_RETURNVERBOSE)

[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)

________________________________________________________________________________________

Maxima [A]
time = 0.31, size = 123, normalized size = 1.81 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (68) = 136\).
time = 3.06, size = 151, normalized size = 2.22 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 877 vs. \(2 (61) = 122\).
time = 115.06, size = 877, normalized size = 12.90 \begin {gather*} \begin {cases} \frac {4 a c d^{3} \log {\left (\frac {b}{2 c} + x - \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{a + b x + c x^{2}} + \frac {4 a c d^{3} \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{a + b x + c x^{2}} + \frac {4 a c d^{3}}{a + b x + c x^{2}} - \frac {b^{2} d^{3}}{a + b x + c x^{2}} + \frac {4 b c d^{3} x \log {\left (\frac {b}{2 c} + x - \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{a + b x + c x^{2}} + \frac {4 b c d^{3} x \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{a + b x + c x^{2}} + \frac {4 c^{2} d^{3} x^{2} \log {\left (\frac {b}{2 c} + x - \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{a + b x + c x^{2}} + \frac {4 c^{2} d^{3} x^{2} \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{a + b x + c x^{2}} & \text {for}\: p = -2 \\- 4 a c d^{3} \log {\left (\frac {b}{2 c} + x - \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )} - 4 a c d^{3} \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )} + b^{2} d^{3} \log {\left (\frac {b}{2 c} + x - \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )} + b^{2} d^{3} \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )} + 4 b c d^{3} x + 4 c^{2} d^{3} x^{2} & \text {for}\: p = -1 \\- \frac {4 a^{2} c d^{3} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {a b^{2} d^{3} p \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {2 a b^{2} d^{3} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {4 a b c d^{3} p x \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {4 a c^{2} d^{3} p x^{2} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {b^{3} d^{3} p x \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {2 b^{3} d^{3} x \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {5 b^{2} c d^{3} p x^{2} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {6 b^{2} c d^{3} x^{2} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {8 b c^{2} d^{3} p x^{3} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {8 b c^{2} d^{3} x^{3} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {4 c^{3} d^{3} p x^{4} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} + \frac {4 c^{3} d^{3} x^{4} \left (a + b x + c x^{2}\right )^{p}}{p^{2} + 3 p + 2} & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((4*a*c*d**3*log(b/(2*c) + x - sqrt(-4*a*c + b**2)/(2*c))/(a + b*x + c*x**2) + 4*a*c*d**3*log(b/(2*c)
+ x + sqrt(-4*a*c + b**2)/(2*c))/(a + b*x + c*x**2) + 4*a*c*d**3/(a + b*x + c*x**2) - b**2*d**3/(a + b*x + c*
x**2) + 4*b*c*d**3*x*log(b/(2*c) + x - sqrt(-4*a*c + b**2)/(2*c))/(a + b*x + c*x**2) + 4*b*c*d**3*x*log(b/(2*c
) + x + sqrt(-4*a*c + b**2)/(2*c))/(a + b*x + c*x**2) + 4*c**2*d**3*x**2*log(b/(2*c) + x - sqrt(-4*a*c + b**2)
/(2*c))/(a + b*x + c*x**2) + 4*c**2*d**3*x**2*log(b/(2*c) + x + sqrt(-4*a*c + b**2)/(2*c))/(a + b*x + c*x**2),
Eq(p, -2)), (-4*a*c*d**3*log(b/(2*c) + x - sqrt(-4*a*c + b**2)/(2*c)) - 4*a*c*d**3*log(b/(2*c) + x + sqrt(-4*
a*c + b**2)/(2*c)) + b**2*d**3*log(b/(2*c) + x - sqrt(-4*a*c + b**2)/(2*c)) + b**2*d**3*log(b/(2*c) + x + sqrt
(-4*a*c + b**2)/(2*c)) + 4*b*c*d**3*x + 4*c**2*d**3*x**2, Eq(p, -1)), (-4*a**2*c*d**3*(a + b*x + c*x**2)**p/(p
**2 + 3*p + 2) + a*b**2*d**3*p*(a + b*x + c*x**2)**p/(p**2 + 3*p + 2) + 2*a*b**2*d**3*(a + b*x + c*x**2)**p/(p
**2 + 3*p + 2) + 4*a*b*c*d**3*p*x*(a + b*x + c*x**2)**p/(p**2 + 3*p + 2) + 4*a*c**2*d**3*p*x**2*(a + b*x + c*x
**2)**p/(p**2 + 3*p + 2) + b**3*d**3*p*x*(a + b*x + c*x**2)**p/(p**2 + 3*p + 2) + 2*b**3*d**3*x*(a + b*x + c*x
**2)**p/(p**2 + 3*p + 2) + 5*b**2*c*d**3*p*x**2*(a + b*x + c*x**2)**p/(p**2 + 3*p + 2) + 6*b**2*c*d**3*x**2*(a
+ b*x + c*x**2)**p/(p**2 + 3*p + 2) + 8*b*c**2*d**3*p*x**3*(a + b*x + c*x**2)**p/(p**2 + 3*p + 2) + 8*b*c**2*
d**3*x**3*(a + b*x + c*x**2)**p/(p**2 + 3*p + 2) + 4*c**3*d**3*p*x**4*(a + b*x + c*x**2)**p/(p**2 + 3*p + 2) +
4*c**3*d**3*x**4*(a + b*x + c*x**2)**p/(p**2 + 3*p + 2), True))

________________________________________________________________________________________

Giac [A]
time = 1.70, size = 88, normalized size = 1.29 \begin {gather*} \frac {4 \, {\left (c x^{2} + b x + a\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} c d^{3}}{p + 2} + \frac {{\left (c x^{2} + b x + a\right )}^{p + 1} b^{2} d^{3}}{p + 1} - \frac {4 \, {\left (c x^{2} + b x + a\right )}^{p + 1} a c d^{3}}{p + 1} \end {gather*}

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

________________________________________________________________________________________

Mupad [B]
time = 0.65, size = 160, normalized size = 2.35 \begin {gather*} {\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 ) \end {gather*}

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))

________________________________________________________________________________________