\(\int (a+b x)^2 (a c+(b c+a d) x+b d x^2) \, dx\) [1760]

Optimal result

Integrand size = 27, antiderivative size = 38 \[ \int (a+b x)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {(b c-a d) (a+b x)^4}{4 b^2}+\frac {d (a+b x)^5}{5 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {640, 45} \[ \int (a+b x)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {(a+b x)^4 (b c-a d)}{4 b^2}+\frac {d (a+b x)^5}{5 b^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.76 \[ \int (a+b x)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx=a^3 c x+\frac {1}{2} a^2 (3 b c+a d) x^2+a b (b c+a d) x^3+\frac {1}{4} b^2 (b c+3 a d) x^4+\frac {1}{5} b^3 d x^5 \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(69\) vs. \(2(34)=68\).

Time = 2.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.84

method	result	size
norman	\(\frac {b^{3} d \,x^{5}}{5}+\left (\frac {3}{4} a \,b^{2} d +\frac {1}{4} b^{3} c \right ) x^{4}+\left (d \,a^{2} b +a \,b^{2} c \right ) x^{3}+\left (\frac {1}{2} a^{3} d +\frac {3}{2} a^{2} b c \right ) x^{2}+a^{3} c x\)	\(70\)
risch	\(\frac {1}{5} b^{3} d \,x^{5}+\frac {3}{4} a \,b^{2} d \,x^{4}+\frac {1}{4} b^{3} c \,x^{4}+a^{2} b d \,x^{3}+a \,b^{2} c \,x^{3}+\frac {1}{2} a^{3} d \,x^{2}+\frac {3}{2} a^{2} b c \,x^{2}+a^{3} c x\)	\(73\)
parallelrisch	\(\frac {1}{5} b^{3} d \,x^{5}+\frac {3}{4} a \,b^{2} d \,x^{4}+\frac {1}{4} b^{3} c \,x^{4}+a^{2} b d \,x^{3}+a \,b^{2} c \,x^{3}+\frac {1}{2} a^{3} d \,x^{2}+\frac {3}{2} a^{2} b c \,x^{2}+a^{3} c x\)	\(73\)
gosper	\(\frac {x \left (4 d \,x^{4} b^{3}+15 a \,b^{2} d \,x^{3}+5 b^{3} c \,x^{3}+20 a^{2} b d \,x^{2}+20 a \,b^{2} c \,x^{2}+10 a^{3} d x +30 a^{2} b c x +20 c \,a^{3}\right )}{20}\)	\(74\)
default	\(\frac {b^{3} d \,x^{5}}{5}+\frac {\left (2 a \,b^{2} d +b^{2} \left (a d +b c \right )\right ) x^{4}}{4}+\frac {\left (d \,a^{2} b +2 a b \left (a d +b c \right )+a \,b^{2} c \right ) x^{3}}{3}+\frac {\left (a^{2} \left (a d +b c \right )+2 a^{2} b c \right ) x^{2}}{2}+a^{3} c x\)	\(94\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.82 \[ \int (a+b x)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {1}{5} \, b^{3} d x^{5} + a^{3} c x + \frac {1}{4} \, {\left (b^{3} c + 3 \, a b^{2} d\right )} x^{4} + {\left (a b^{2} c + a^{2} b d\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} b c + a^{3} d\right )} x^{2} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (32) = 64\).

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.92 \[ \int (a+b x)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx=a^{3} c x + \frac {b^{3} d x^{5}}{5} + x^{4} \cdot \left (\frac {3 a b^{2} d}{4} + \frac {b^{3} c}{4}\right ) + x^{3} \left (a^{2} b d + a b^{2} c\right ) + x^{2} \left (\frac {a^{3} d}{2} + \frac {3 a^{2} b c}{2}\right ) \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.82 \[ \int (a+b x)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {1}{5} \, b^{3} d x^{5} + a^{3} c x + \frac {1}{4} \, {\left (b^{3} c + 3 \, a b^{2} d\right )} x^{4} + {\left (a b^{2} c + a^{2} b d\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} b c + a^{3} d\right )} x^{2} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (34) = 68\).

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.89 \[ \int (a+b x)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx=\frac {1}{5} \, b^{3} d x^{5} + \frac {1}{4} \, b^{3} c x^{4} + \frac {3}{4} \, a b^{2} d x^{4} + a b^{2} c x^{3} + a^{2} b d x^{3} + \frac {3}{2} \, a^{2} b c x^{2} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.66 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.71 \[ \int (a+b x)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx=x^4\,\left (\frac {c\,b^3}{4}+\frac {3\,a\,d\,b^2}{4}\right )+x^2\,\left (\frac {d\,a^3}{2}+\frac {3\,b\,c\,a^2}{2}\right )+\frac {b^3\,d\,x^5}{5}+a^3\,c\,x+a\,b\,x^3\,\left (a\,d+b\,c\right ) \]

[In]

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

[Out]

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