### 3.1759 $$\int (a+b x)^3 (a c+(b c+a d) x+b d x^2) \, dx$$

Optimal. Leaf size=38 $\frac{(a+b x)^5 (b c-a d)}{5 b^2}+\frac{d (a+b x)^6}{6 b^2}$

[Out]

((b*c - a*d)*(a + b*x)^5)/(5*b^2) + (d*(a + b*x)^6)/(6*b^2)

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Rubi [A]  time = 0.0178423, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.074, Rules used = {626, 43} $\frac{(a+b x)^5 (b c-a d)}{5 b^2}+\frac{d (a+b x)^6}{6 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((b*c - a*d)*(a + b*x)^5)/(5*b^2) + (d*(a + b*x)^6)/(6*b^2)

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^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]

Rule 43

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])

Rubi steps

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

Mathematica [B]  time = 0.0180026, size = 84, normalized size = 2.21 $\frac{1}{30} x \left (15 a^2 b^2 x^2 (4 c+3 d x)+20 a^3 b x (3 c+2 d x)+15 a^4 (2 c+d x)+6 a b^3 x^3 (5 c+4 d x)+b^4 x^4 (6 c+5 d x)\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x*(15*a^4*(2*c + d*x) + 20*a^3*b*x*(3*c + 2*d*x) + 15*a^2*b^2*x^2*(4*c + 3*d*x) + 6*a*b^3*x^3*(5*c + 4*d*x) +
b^4*x^4*(6*c + 5*d*x)))/30

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Maple [B]  time = 0.039, size = 133, normalized size = 3.5 \begin{align*}{\frac{{b}^{4}d{x}^{6}}{6}}+{\frac{ \left ( 3\,ad{b}^{3}+{b}^{3} \left ( ad+bc \right ) \right ){x}^{5}}{5}}+{\frac{ \left ( 3\,{b}^{2}{a}^{2}d+3\,{b}^{2}a \left ( ad+bc \right ) +a{b}^{3}c \right ){x}^{4}}{4}}+{\frac{ \left ({a}^{3}bd+3\,b{a}^{2} \left ( ad+bc \right ) +3\,{b}^{2}{a}^{2}c \right ){x}^{3}}{3}}+{\frac{ \left ({a}^{3} \left ( ad+bc \right ) +3\,b{a}^{3}c \right ){x}^{2}}{2}}+{a}^{4}cx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.09863, size = 130, normalized size = 3.42 \begin{align*} \frac{1}{6} \, b^{4} d x^{6} + a^{4} c x + \frac{1}{5} \,{\left (b^{4} c + 4 \, a b^{3} d\right )} x^{5} + \frac{1}{2} \,{\left (2 \, a b^{3} c + 3 \, a^{2} b^{2} d\right )} x^{4} + \frac{2}{3} \,{\left (3 \, a^{2} b^{2} c + 2 \, a^{3} b d\right )} x^{3} + \frac{1}{2} \,{\left (4 \, a^{3} b c + a^{4} d\right )} x^{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.53655, size = 217, normalized size = 5.71 \begin{align*} \frac{1}{6} x^{6} d b^{4} + \frac{1}{5} x^{5} c b^{4} + \frac{4}{5} x^{5} d b^{3} a + x^{4} c b^{3} a + \frac{3}{2} x^{4} d b^{2} a^{2} + 2 x^{3} c b^{2} a^{2} + \frac{4}{3} x^{3} d b a^{3} + 2 x^{2} c b a^{3} + \frac{1}{2} x^{2} d a^{4} + x c a^{4} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*d*b^4 + 1/5*x^5*c*b^4 + 4/5*x^5*d*b^3*a + x^4*c*b^3*a + 3/2*x^4*d*b^2*a^2 + 2*x^3*c*b^2*a^2 + 4/3*x^3*
d*b*a^3 + 2*x^2*c*b*a^3 + 1/2*x^2*d*a^4 + x*c*a^4

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Sympy [B]  time = 0.140041, size = 100, normalized size = 2.63 \begin{align*} a^{4} c x + \frac{b^{4} d x^{6}}{6} + x^{5} \left (\frac{4 a b^{3} d}{5} + \frac{b^{4} c}{5}\right ) + x^{4} \left (\frac{3 a^{2} b^{2} d}{2} + a b^{3} c\right ) + x^{3} \left (\frac{4 a^{3} b d}{3} + 2 a^{2} b^{2} c\right ) + x^{2} \left (\frac{a^{4} d}{2} + 2 a^{3} b c\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.16292, size = 131, normalized size = 3.45 \begin{align*} \frac{1}{6} \, b^{4} d x^{6} + \frac{1}{5} \, b^{4} c x^{5} + \frac{4}{5} \, a b^{3} d x^{5} + a b^{3} c x^{4} + \frac{3}{2} \, a^{2} b^{2} d x^{4} + 2 \, a^{2} b^{2} c x^{3} + \frac{4}{3} \, a^{3} b d x^{3} + 2 \, a^{3} b c x^{2} + \frac{1}{2} \, a^{4} d x^{2} + a^{4} c x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^4*d*x^6 + 1/5*b^4*c*x^5 + 4/5*a*b^3*d*x^5 + a*b^3*c*x^4 + 3/2*a^2*b^2*d*x^4 + 2*a^2*b^2*c*x^3 + 4/3*a^3*
b*d*x^3 + 2*a^3*b*c*x^2 + 1/2*a^4*d*x^2 + a^4*c*x