\(\int (a+b x) (a c+(b c+a d) x+b d x^2) \, dx\) [1761]

Optimal result

Integrand size = 25, antiderivative size = 38 \[ \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {(b c-a d) (a+b x)^3}{3 b^2}+\frac {d (a+b x)^4}{4 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {640, 45} \[ \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {(a+b x)^3 (b c-a d)}{3 b^2}+\frac {d (a+b x)^4}{4 b^2} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {1}{12} x \left (6 a^2 (2 c+d x)+4 a b x (3 c+2 d x)+b^2 x^2 (4 c+3 d x)\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {b^{2} d \,x^{4}}{4}+\left (\frac {2}{3} a b d +\frac {1}{3} b^{2} c \right ) x^{3}+\left (\frac {1}{2} a^{2} d +a b c \right ) x^{2}+a^{2} c x\)	\(48\)
gosper	\(\frac {x \left (3 d \,x^{3} b^{2}+8 x^{2} a b d +4 b^{2} c \,x^{2}+6 a^{2} d x +12 a b c x +12 a^{2} c \right )}{12}\)	\(50\)
risch	\(\frac {1}{4} b^{2} d \,x^{4}+\frac {2}{3} x^{3} a b d +\frac {1}{3} b^{2} c \,x^{3}+\frac {1}{2} a^{2} d \,x^{2}+a b c \,x^{2}+a^{2} c x\)	\(50\)
parallelrisch	\(\frac {1}{4} b^{2} d \,x^{4}+\frac {2}{3} x^{3} a b d +\frac {1}{3} b^{2} c \,x^{3}+\frac {1}{2} a^{2} d \,x^{2}+a b c \,x^{2}+a^{2} c x\)	\(50\)
default	\(\frac {b^{2} d \,x^{4}}{4}+\frac {\left (a b d +b \left (a d +b c \right )\right ) x^{3}}{3}+\frac {\left (a \left (a d +b c \right )+a b c \right ) x^{2}}{2}+a^{2} c x\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {1}{4} \, b^{2} d x^{4} + a^{2} c x + \frac {1}{3} \, {\left (b^{2} c + 2 \, a b d\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a b c + a^{2} d\right )} x^{2} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right ) \, dx=a^{2} c x + \frac {b^{2} d x^{4}}{4} + x^{3} \cdot \left (\frac {2 a b d}{3} + \frac {b^{2} c}{3}\right ) + x^{2} \left (\frac {a^{2} d}{2} + a b c\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {1}{4} \, b^{2} d x^{4} + a^{2} c x + \frac {1}{3} \, {\left (b^{2} c + 2 \, a b d\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a b c + a^{2} d\right )} x^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {1}{4} \, b^{2} d x^{4} + \frac {1}{3} \, b^{2} c x^{3} + \frac {2}{3} \, a b d x^{3} + a b c x^{2} + \frac {1}{2} \, a^{2} d x^{2} + a^{2} c x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right ) \, dx=x^2\,\left (\frac {d\,a^2}{2}+b\,c\,a\right )+x^3\,\left (\frac {c\,b^2}{3}+\frac {2\,a\,d\,b}{3}\right )+\frac {b^2\,d\,x^4}{4}+a^2\,c\,x \]

[In]

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

[Out]

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