\(\int (b d+2 c d x)^2 (a+b x+c x^2) \, dx\) [1111]

Optimal result

Integrand size = 22, antiderivative size = 45 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right ) \, dx=-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3}{24 c^2}+\frac {d^2 (b+2 c x)^5}{40 c^2} \]

[Out]

-1/24*(-4*a*c+b^2)*d^2*(2*c*x+b)^3/c^2+1/40*d^2*(2*c*x+b)^5/c^2

Rubi [A] (verified)

Time = 0.03 (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} \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {d^2 (b+2 c x)^5}{40 c^2}-\frac {d^2 \left (b^2-4 a c\right ) (b+2 c x)^3}{24 c^2} \]

[In]

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

[Out]

-1/24*((b^2 - 4*a*c)*d^2*(b + 2*c*x)^3)/c^2 + (d^2*(b + 2*c*x)^5)/(40*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)^2}{4 c}+\frac {(b d+2 c d x)^4}{4 c d^2}\right ) \, dx \\ & = -\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3}{24 c^2}+\frac {d^2 (b+2 c x)^5}{40 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.42 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right ) \, dx=d^2 \left (a b^2 x+\frac {1}{2} b \left (b^2+4 a c\right ) x^2+\frac {1}{3} c \left (5 b^2+4 a c\right ) x^3+2 b c^2 x^4+\frac {4 c^3 x^5}{5}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.36

method	result	size
gosper	\(\frac {x \left (24 c^{3} x^{4}+60 b \,c^{2} x^{3}+40 a \,c^{2} x^{2}+50 b^{2} c \,x^{2}+60 a b c x +15 b^{3} x +30 a \,b^{2}\right ) d^{2}}{30}\)	\(61\)
norman	\(\left (\frac {4}{3} a \,c^{2} d^{2}+\frac {5}{3} b^{2} c \,d^{2}\right ) x^{3}+\left (2 a b c \,d^{2}+\frac {1}{2} b^{3} d^{2}\right ) x^{2}+a \,b^{2} d^{2} x +\frac {4 c^{3} d^{2} x^{5}}{5}+2 b \,c^{2} d^{2} x^{4}\)	\(78\)
default	\(\frac {4 c^{3} d^{2} x^{5}}{5}+2 b \,c^{2} d^{2} x^{4}+\frac {\left (4 a \,c^{2} d^{2}+5 b^{2} c \,d^{2}\right ) x^{3}}{3}+\frac {\left (4 a b c \,d^{2}+b^{3} d^{2}\right ) x^{2}}{2}+a \,b^{2} d^{2} x\)	\(79\)
risch	\(\frac {4}{5} c^{3} d^{2} x^{5}+2 b \,c^{2} d^{2} x^{4}+\frac {4}{3} a \,c^{2} d^{2} x^{3}+\frac {5}{3} b^{2} c \,d^{2} x^{3}+2 a b c \,d^{2} x^{2}+\frac {1}{2} d^{2} b^{3} x^{2}+a \,b^{2} d^{2} x\)	\(80\)
parallelrisch	\(\frac {4}{5} c^{3} d^{2} x^{5}+2 b \,c^{2} d^{2} x^{4}+\frac {4}{3} a \,c^{2} d^{2} x^{3}+\frac {5}{3} b^{2} c \,d^{2} x^{3}+2 a b c \,d^{2} x^{2}+\frac {1}{2} d^{2} b^{3} x^{2}+a \,b^{2} d^{2} x\)	\(80\)

[In]

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

[Out]

1/30*x*(24*c^3*x^4+60*b*c^2*x^3+40*a*c^2*x^2+50*b^2*c*x^2+60*a*b*c*x+15*b^3*x+30*a*b^2)*d^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {4}{5} \, c^{3} d^{2} x^{5} + 2 \, b c^{2} d^{2} x^{4} + a b^{2} d^{2} x + \frac {1}{3} \, {\left (5 \, b^{2} c + 4 \, a c^{2}\right )} d^{2} x^{3} + \frac {1}{2} \, {\left (b^{3} + 4 \, a b c\right )} d^{2} x^{2} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (41) = 82\).

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.89 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right ) \, dx=a b^{2} d^{2} x + 2 b c^{2} d^{2} x^{4} + \frac {4 c^{3} d^{2} x^{5}}{5} + x^{3} \cdot \left (\frac {4 a c^{2} d^{2}}{3} + \frac {5 b^{2} c d^{2}}{3}\right ) + x^{2} \cdot \left (2 a b c d^{2} + \frac {b^{3} d^{2}}{2}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {4}{5} \, c^{3} d^{2} x^{5} + 2 \, b c^{2} d^{2} x^{4} + a b^{2} d^{2} x + \frac {1}{3} \, {\left (5 \, b^{2} c + 4 \, a c^{2}\right )} d^{2} x^{3} + \frac {1}{2} \, {\left (b^{3} + 4 \, a b c\right )} d^{2} x^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.76 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {4}{5} \, c^{3} d^{2} x^{5} + 2 \, b c^{2} d^{2} x^{4} + \frac {5}{3} \, b^{2} c d^{2} x^{3} + \frac {4}{3} \, a c^{2} d^{2} x^{3} + \frac {1}{2} \, b^{3} d^{2} x^{2} + 2 \, a b c d^{2} x^{2} + a b^{2} d^{2} x \]

[In]

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

[Out]

4/5*c^3*d^2*x^5 + 2*b*c^2*d^2*x^4 + 5/3*b^2*c*d^2*x^3 + 4/3*a*c^2*d^2*x^3 + 1/2*b^3*d^2*x^2 + 2*a*b*c*d^2*x^2
+ a*b^2*d^2*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.53 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {4\,c^3\,d^2\,x^5}{5}+\frac {c\,d^2\,x^3\,\left (5\,b^2+4\,a\,c\right )}{3}+2\,b\,c^2\,d^2\,x^4+\frac {b\,d^2\,x^2\,\left (b^2+4\,a\,c\right )}{2}+a\,b^2\,d^2\,x \]

[In]

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

[Out]

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