\(\int (b d+2 c d x)^3 (a+b x+c x^2) \, dx\) [1110]

Optimal result

Integrand size = 22, antiderivative size = 45 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right ) \, dx=-\frac {\left (b^2-4 a c\right ) d^3 (b+2 c x)^4}{32 c^2}+\frac {d^3 (b+2 c x)^6}{48 c^2} \]

[Out]

-1/32*(-4*a*c+b^2)*d^3*(2*c*x+b)^4/c^2+1/48*d^3*(2*c*x+b)^6/c^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {697} \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right ) \, dx=\frac {d^3 (b+2 c x)^6}{48 c^2}-\frac {d^3 \left (b^2-4 a c\right ) (b+2 c x)^4}{32 c^2} \]

[In]

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

[Out]

-1/32*((b^2 - 4*a*c)*d^3*(b + 2*c*x)^4)/c^2 + (d^3*(b + 2*c*x)^6)/(48*c^2)

Rule 697

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (-b^2+4 a c\right ) (b d+2 c d x)^3}{4 c}+\frac {(b d+2 c d x)^5}{4 c d^2}\right ) \, dx \\ & = -\frac {\left (b^2-4 a c\right ) d^3 (b+2 c x)^4}{32 c^2}+\frac {d^3 (b+2 c x)^6}{48 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right ) \, dx=\frac {1}{6} d^3 x (b+c x) \left (6 a \left (b^2+2 b c x+2 c^2 x^2\right )+x \left (3 b^3+11 b^2 c x+16 b c^2 x^2+8 c^3 x^3\right )\right ) \]

[In]

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

[Out]

(d^3*x*(b + c*x)*(6*a*(b^2 + 2*b*c*x + 2*c^2*x^2) + x*(3*b^3 + 11*b^2*c*x + 16*b*c^2*x^2 + 8*c^3*x^3)))/6

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02

method	result	size
default	\(-d^{3} \left (-\frac {4 c \left (c \,x^{2}+b x +a \right )^{3}}{3}+\frac {\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2}}{2}\right )\)	\(46\)
gosper	\(\frac {x \left (8 c^{4} x^{5}+24 b \,c^{3} x^{4}+12 x^{3} c^{3} a +27 b^{2} c^{2} x^{3}+24 a b \,c^{2} x^{2}+14 x^{2} b^{3} c +18 a \,b^{2} c x +3 b^{4} x +6 a \,b^{3}\right ) d^{3}}{6}\)	\(84\)
norman	\(\left (2 a \,c^{3} d^{3}+\frac {9}{2} b^{2} c^{2} d^{3}\right ) x^{4}+\left (4 a b \,c^{2} d^{3}+\frac {7}{3} b^{3} c \,d^{3}\right ) x^{3}+\left (3 a \,b^{2} c \,d^{3}+\frac {1}{2} b^{4} d^{3}\right ) x^{2}+a \,d^{3} x \,b^{3}+\frac {4 d^{3} c^{4} x^{6}}{3}+4 c^{3} b \,d^{3} x^{5}\)	\(106\)
parallelrisch	\(\frac {4}{3} d^{3} c^{4} x^{6}+4 c^{3} b \,d^{3} x^{5}+2 d^{3} a \,c^{3} x^{4}+\frac {9}{2} d^{3} b^{2} c^{2} x^{4}+4 d^{3} a b \,c^{2} x^{3}+\frac {7}{3} d^{3} b^{3} c \,x^{3}+3 d^{3} a \,b^{2} c \,x^{2}+\frac {1}{2} d^{3} b^{4} x^{2}+a \,d^{3} x \,b^{3}\)	\(109\)
risch	\(\frac {4}{3} d^{3} c^{4} x^{6}+4 c^{3} b \,d^{3} x^{5}+2 d^{3} a \,c^{3} x^{4}+\frac {9}{2} d^{3} b^{2} c^{2} x^{4}+4 d^{3} a b \,c^{2} x^{3}+\frac {7}{3} d^{3} b^{3} c \,x^{3}+3 d^{3} a \,b^{2} c \,x^{2}-\frac {2}{3} a^{3} c \,d^{3}+\frac {1}{2} d^{3} b^{4} x^{2}+a \,d^{3} x \,b^{3}+\frac {1}{2} d^{3} a^{2} b^{2}\)	\(129\)

[In]

int((2*c*d*x+b*d)^3*(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (41) = 82\).

Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.16 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right ) \, dx=\frac {4}{3} \, c^{4} d^{3} x^{6} + 4 \, b c^{3} d^{3} x^{5} + a b^{3} d^{3} x + \frac {1}{2} \, {\left (9 \, b^{2} c^{2} + 4 \, a c^{3}\right )} d^{3} x^{4} + \frac {1}{3} \, {\left (7 \, b^{3} c + 12 \, a b c^{2}\right )} d^{3} x^{3} + \frac {1}{2} \, {\left (b^{4} + 6 \, a b^{2} c\right )} d^{3} x^{2} \]

[In]

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

[Out]

4/3*c^4*d^3*x^6 + 4*b*c^3*d^3*x^5 + a*b^3*d^3*x + 1/2*(9*b^2*c^2 + 4*a*c^3)*d^3*x^4 + 1/3*(7*b^3*c + 12*a*b*c^
2)*d^3*x^3 + 1/2*(b^4 + 6*a*b^2*c)*d^3*x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (41) = 82\).

Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.53 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right ) \, dx=a b^{3} d^{3} x + 4 b c^{3} d^{3} x^{5} + \frac {4 c^{4} d^{3} x^{6}}{3} + x^{4} \cdot \left (2 a c^{3} d^{3} + \frac {9 b^{2} c^{2} d^{3}}{2}\right ) + x^{3} \cdot \left (4 a b c^{2} d^{3} + \frac {7 b^{3} c d^{3}}{3}\right ) + x^{2} \cdot \left (3 a b^{2} c d^{3} + \frac {b^{4} d^{3}}{2}\right ) \]

[In]

integrate((2*c*d*x+b*d)**3*(c*x**2+b*x+a),x)

[Out]

a*b**3*d**3*x + 4*b*c**3*d**3*x**5 + 4*c**4*d**3*x**6/3 + x**4*(2*a*c**3*d**3 + 9*b**2*c**2*d**3/2) + x**3*(4*
a*b*c**2*d**3 + 7*b**3*c*d**3/3) + x**2*(3*a*b**2*c*d**3 + b**4*d**3/2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (41) = 82\).

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.16 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right ) \, dx=\frac {4}{3} \, c^{4} d^{3} x^{6} + 4 \, b c^{3} d^{3} x^{5} + a b^{3} d^{3} x + \frac {1}{2} \, {\left (9 \, b^{2} c^{2} + 4 \, a c^{3}\right )} d^{3} x^{4} + \frac {1}{3} \, {\left (7 \, b^{3} c + 12 \, a b c^{2}\right )} d^{3} x^{3} + \frac {1}{2} \, {\left (b^{4} + 6 \, a b^{2} c\right )} d^{3} x^{2} \]

[In]

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

[Out]

4/3*c^4*d^3*x^6 + 4*b*c^3*d^3*x^5 + a*b^3*d^3*x + 1/2*(9*b^2*c^2 + 4*a*c^3)*d^3*x^4 + 1/3*(7*b^3*c + 12*a*b*c^
2)*d^3*x^3 + 1/2*(b^4 + 6*a*b^2*c)*d^3*x^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.62 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right ) \, dx={\left (c d x^{2} + b d x\right )} a b^{2} d^{2} + \frac {1}{2} \, {\left (c d x^{2} + b d x\right )}^{2} b^{2} d + 2 \, {\left (c d x^{2} + b d x\right )}^{2} a c d + \frac {4}{3} \, {\left (c d x^{2} + b d x\right )}^{3} c \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.07 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right ) \, dx=\frac {4\,c^4\,d^3\,x^6}{3}+\frac {b^2\,d^3\,x^2\,\left (b^2+6\,a\,c\right )}{2}+4\,b\,c^3\,d^3\,x^5+\frac {c^2\,d^3\,x^4\,\left (9\,b^2+4\,a\,c\right )}{2}+a\,b^3\,d^3\,x+\frac {b\,c\,d^3\,x^3\,\left (7\,b^2+12\,a\,c\right )}{3} \]

[In]

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

[Out]

(4*c^4*d^3*x^6)/3 + (b^2*d^3*x^2*(6*a*c + b^2))/2 + 4*b*c^3*d^3*x^5 + (c^2*d^3*x^4*(4*a*c + 9*b^2))/2 + a*b^3*
d^3*x + (b*c*d^3*x^3*(12*a*c + 7*b^2))/3