3.1899 \(\int (a+b x) (d+e x) (a^2+2 a b x+b^2 x^2) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0166393, 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 = {27, 43} \[ \frac{(a+b x)^4 (b d-a e)}{4 b^2}+\frac{e (a+b x)^5}{5 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 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) (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx &=\int (a+b x)^3 (d+e x) \, dx\\ &=\int \left (\frac{(b d-a e) (a+b x)^3}{b}+\frac{e (a+b x)^4}{b}\right ) \, dx\\ &=\frac{(b d-a e) (a+b x)^4}{4 b^2}+\frac{e (a+b x)^5}{5 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0110288, size = 67, normalized size = 1.76 \[ \frac{1}{2} a^2 x^2 (a e+3 b d)+a^3 d x+\frac{1}{4} b^2 x^4 (3 a e+b d)+a b x^3 (a e+b d)+\frac{1}{5} b^3 e x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.002, size = 94, normalized size = 2.5 \begin{align*}{\frac{{b}^{3}e{x}^{5}}{5}}+{\frac{ \left ( \left ( ae+bd \right ){b}^{2}+2\,{b}^{2}ea \right ){x}^{4}}{4}}+{\frac{ \left ({b}^{2}da+2\, \left ( ae+bd \right ) ab+be{a}^{2} \right ){x}^{3}}{3}}+{\frac{ \left ( 2\,{a}^{2}db+ \left ( ae+bd \right ){a}^{2} \right ){x}^{2}}{2}}+{a}^{3}dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.963005, size = 93, normalized size = 2.45 \begin{align*} \frac{1}{5} \, b^{3} e x^{5} + a^{3} d x + \frac{1}{4} \,{\left (b^{3} d + 3 \, a b^{2} e\right )} x^{4} +{\left (a b^{2} d + a^{2} b e\right )} x^{3} + \frac{1}{2} \,{\left (3 \, a^{2} b d + a^{3} e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.29428, size = 163, normalized size = 4.29 \begin{align*} \frac{1}{5} x^{5} e b^{3} + \frac{1}{4} x^{4} d b^{3} + \frac{3}{4} x^{4} e b^{2} a + x^{3} d b^{2} a + x^{3} e b a^{2} + \frac{3}{2} x^{2} d b a^{2} + \frac{1}{2} x^{2} e a^{3} + x d a^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.070982, size = 73, normalized size = 1.92 \begin{align*} a^{3} d x + \frac{b^{3} e x^{5}}{5} + x^{4} \left (\frac{3 a b^{2} e}{4} + \frac{b^{3} d}{4}\right ) + x^{3} \left (a^{2} b e + a b^{2} d\right ) + x^{2} \left (\frac{a^{3} e}{2} + \frac{3 a^{2} b d}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.22442, size = 103, normalized size = 2.71 \begin{align*} \frac{1}{5} \, b^{3} x^{5} e + \frac{1}{4} \, b^{3} d x^{4} + \frac{3}{4} \, a b^{2} x^{4} e + a b^{2} d x^{3} + a^{2} b x^{3} e + \frac{3}{2} \, a^{2} b d x^{2} + \frac{1}{2} \, a^{3} x^{2} e + a^{3} d x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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