3.108 \(\int (a+b x)^3 (A+B x) \, dx\)

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

[Out]

((A*b - a*B)*(a + b*x)^4)/(4*b^2) + (B*(a + b*x)^5)/(5*b^2)

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Rubi [A]  time = 0.014005, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {43} \[ \frac{(a+b x)^4 (A b-a B)}{4 b^2}+\frac{B (a+b x)^5}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3*(A + B*x),x]

[Out]

((A*b - a*B)*(a + b*x)^4)/(4*b^2) + (B*(a + b*x)^5)/(5*b^2)

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

Mathematica [A]  time = 0.0092553, size = 67, normalized size = 1.76 \[ \frac{1}{2} a^2 x^2 (a B+3 A b)+a^3 A x+\frac{1}{4} b^2 x^4 (3 a B+A b)+a b x^3 (a B+A b)+\frac{1}{5} b^3 B x^5 \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3*(A + B*x),x]

[Out]

a^3*A*x + (a^2*(3*A*b + a*B)*x^2)/2 + a*b*(A*b + a*B)*x^3 + (b^2*(A*b + 3*a*B)*x^4)/4 + (b^3*B*x^5)/5

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Maple [B]  time = 0.001, size = 73, normalized size = 1.9 \begin{align*}{\frac{{b}^{3}B{x}^{5}}{5}}+{\frac{ \left ({b}^{3}A+3\,a{b}^{2}B \right ){x}^{4}}{4}}+{\frac{ \left ( 3\,a{b}^{2}A+3\,{a}^{2}bB \right ){x}^{3}}{3}}+{\frac{ \left ( 3\,{a}^{2}bA+{a}^{3}B \right ){x}^{2}}{2}}+{a}^{3}Ax \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(B*x+A),x)

[Out]

1/5*b^3*B*x^5+1/4*(A*b^3+3*B*a*b^2)*x^4+1/3*(3*A*a*b^2+3*B*a^2*b)*x^3+1/2*(3*A*a^2*b+B*a^3)*x^2+a^3*A*x

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Maxima [B]  time = 1.01911, size = 93, normalized size = 2.45 \begin{align*} \frac{1}{5} \, B b^{3} x^{5} + A a^{3} x + \frac{1}{4} \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{4} +{\left (B a^{2} b + A a b^{2}\right )} x^{3} + \frac{1}{2} \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A),x, algorithm="maxima")

[Out]

1/5*B*b^3*x^5 + A*a^3*x + 1/4*(3*B*a*b^2 + A*b^3)*x^4 + (B*a^2*b + A*a*b^2)*x^3 + 1/2*(B*a^3 + 3*A*a^2*b)*x^2

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Fricas [B]  time = 1.47611, size = 163, normalized size = 4.29 \begin{align*} \frac{1}{5} x^{5} b^{3} B + \frac{3}{4} x^{4} b^{2} a B + \frac{1}{4} x^{4} b^{3} A + x^{3} b a^{2} B + x^{3} b^{2} a A + \frac{1}{2} x^{2} a^{3} B + \frac{3}{2} x^{2} b a^{2} A + x a^{3} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A),x, algorithm="fricas")

[Out]

1/5*x^5*b^3*B + 3/4*x^4*b^2*a*B + 1/4*x^4*b^3*A + x^3*b*a^2*B + x^3*b^2*a*A + 1/2*x^2*a^3*B + 3/2*x^2*b*a^2*A
+ x*a^3*A

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Sympy [B]  time = 0.080009, size = 73, normalized size = 1.92 \begin{align*} A a^{3} x + \frac{B b^{3} x^{5}}{5} + x^{4} \left (\frac{A b^{3}}{4} + \frac{3 B a b^{2}}{4}\right ) + x^{3} \left (A a b^{2} + B a^{2} b\right ) + x^{2} \left (\frac{3 A a^{2} b}{2} + \frac{B a^{3}}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(B*x+A),x)

[Out]

A*a**3*x + B*b**3*x**5/5 + x**4*(A*b**3/4 + 3*B*a*b**2/4) + x**3*(A*a*b**2 + B*a**2*b) + x**2*(3*A*a**2*b/2 +
B*a**3/2)

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Giac [B]  time = 1.19566, size = 97, normalized size = 2.55 \begin{align*} \frac{1}{5} \, B b^{3} x^{5} + \frac{3}{4} \, B a b^{2} x^{4} + \frac{1}{4} \, A b^{3} x^{4} + B a^{2} b x^{3} + A a b^{2} x^{3} + \frac{1}{2} \, B a^{3} x^{2} + \frac{3}{2} \, A a^{2} b x^{2} + A a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A),x, algorithm="giac")

[Out]

1/5*B*b^3*x^5 + 3/4*B*a*b^2*x^4 + 1/4*A*b^3*x^4 + B*a^2*b*x^3 + A*a*b^2*x^3 + 1/2*B*a^3*x^2 + 3/2*A*a^2*b*x^2
+ A*a^3*x