3.1760 \(\int (a+b x)^2 (a c+(b c+a d) x+b d x^2) \, dx\)

Optimal. Leaf size=38 \[ \frac{(a+b x)^4 (b c-a d)}{4 b^2}+\frac{d (a+b x)^5}{5 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0156864, 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} \[ \frac{(a+b x)^4 (b c-a d)}{4 b^2}+\frac{d (a+b x)^5}{5 b^2} \]

Antiderivative was successfully verified.

[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 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)^2 \left (a c+(b c+a d) x+b d x^2\right ) \, dx &=\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]  time = 0.010389, size = 67, normalized size = 1.76 \[ \frac{1}{2} a^2 x^2 (a d+3 b c)+a^3 c x+\frac{1}{4} b^2 x^4 (3 a d+b c)+a b x^3 (a d+b c)+\frac{1}{5} b^3 d x^5 \]

Antiderivative was successfully verified.

[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

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Maple [B]  time = 0.041, size = 94, normalized size = 2.5 \begin{align*}{\frac{{b}^{3}d{x}^{5}}{5}}+{\frac{ \left ( 2\,a{b}^{2}d+{b}^{2} \left ( ad+bc \right ) \right ){x}^{4}}{4}}+{\frac{ \left ({a}^{2}bd+2\,ab \left ( ad+bc \right ) +ac{b}^{2} \right ){x}^{3}}{3}}+{\frac{ \left ({a}^{2} \left ( ad+bc \right ) +2\,{a}^{2}bc \right ){x}^{2}}{2}}+{a}^{3}cx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.16033, size = 93, normalized size = 2.45 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [B]  time = 1.6618, size = 163, normalized size = 4.29 \begin{align*} \frac{1}{5} x^{5} d b^{3} + \frac{1}{4} x^{4} c b^{3} + \frac{3}{4} x^{4} d b^{2} a + x^{3} c b^{2} a + x^{3} d b a^{2} + \frac{3}{2} x^{2} c b a^{2} + \frac{1}{2} x^{2} d a^{3} + x c a^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.084755, size = 73, normalized size = 1.92 \begin{align*} a^{3} c x + \frac{b^{3} d x^{5}}{5} + x^{4} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B]  time = 1.17151, size = 97, normalized size = 2.55 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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