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

Optimal. Leaf size=61 \[ \frac {(a+b x)^5 (A b-2 a B)}{5 b^3}-\frac {a (a+b x)^4 (A b-a B)}{4 b^3}+\frac {B (a+b x)^6}{6 b^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {76} \begin {gather*} \frac {(a+b x)^5 (A b-2 a B)}{5 b^3}-\frac {a (a+b x)^4 (A b-a B)}{4 b^3}+\frac {B (a+b x)^6}{6 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)^3*(A + B*x),x]

[Out]

-(a*(A*b - a*B)*(a + b*x)^4)/(4*b^3) + ((A*b - 2*a*B)*(a + b*x)^5)/(5*b^3) + (B*(a + b*x)^6)/(6*b^3)

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 69, normalized size = 1.13 \begin {gather*} \frac {1}{60} x^2 \left (10 a^3 (3 A+2 B x)+15 a^2 b x (4 A+3 B x)+9 a b^2 x^2 (5 A+4 B x)+2 b^3 x^3 (6 A+5 B x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x)^3*(A + B*x),x]

[Out]

(x^2*(10*a^3*(3*A + 2*B*x) + 15*a^2*b*x*(4*A + 3*B*x) + 9*a*b^2*x^2*(5*A + 4*B*x) + 2*b^3*x^3*(6*A + 5*B*x)))/
60

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x (a+b x)^3 (A+B x) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x*(a + b*x)^3*(A + B*x),x]

[Out]

IntegrateAlgebraic[x*(a + b*x)^3*(A + B*x), x]

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fricas [A]  time = 0.76, size = 76, normalized size = 1.25 \begin {gather*} \frac {1}{6} x^{6} b^{3} B + \frac {3}{5} x^{5} b^{2} a B + \frac {1}{5} x^{5} b^{3} A + \frac {3}{4} x^{4} b a^{2} B + \frac {3}{4} x^{4} b^{2} a A + \frac {1}{3} x^{3} a^{3} B + x^{3} b a^{2} A + \frac {1}{2} x^{2} a^{3} A \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*b^3*B + 3/5*x^5*b^2*a*B + 1/5*x^5*b^3*A + 3/4*x^4*b*a^2*B + 3/4*x^4*b^2*a*A + 1/3*x^3*a^3*B + x^3*b*a^
2*A + 1/2*x^2*a^3*A

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giac [A]  time = 1.25, size = 76, normalized size = 1.25 \begin {gather*} \frac {1}{6} \, B b^{3} x^{6} + \frac {3}{5} \, B a b^{2} x^{5} + \frac {1}{5} \, A b^{3} x^{5} + \frac {3}{4} \, B a^{2} b x^{4} + \frac {3}{4} \, A a b^{2} x^{4} + \frac {1}{3} \, B a^{3} x^{3} + A a^{2} b x^{3} + \frac {1}{2} \, A a^{3} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*b^3*x^6 + 3/5*B*a*b^2*x^5 + 1/5*A*b^3*x^5 + 3/4*B*a^2*b*x^4 + 3/4*A*a*b^2*x^4 + 1/3*B*a^3*x^3 + A*a^2*b*
x^3 + 1/2*A*a^3*x^2

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maple [A]  time = 0.00, size = 76, normalized size = 1.25 \begin {gather*} \frac {B \,b^{3} x^{6}}{6}+\frac {A \,a^{3} x^{2}}{2}+\frac {\left (b^{3} A +3 a \,b^{2} B \right ) x^{5}}{5}+\frac {\left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{4}}{4}+\frac {\left (3 a^{2} b A +a^{3} B \right ) x^{3}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^3*B*x^6+1/5*(A*b^3+3*B*a*b^2)*x^5+1/4*(3*A*a*b^2+3*B*a^2*b)*x^4+1/3*(3*A*a^2*b+B*a^3)*x^3+1/2*a^3*A*x^2

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maxima [A]  time = 1.05, size = 73, normalized size = 1.20 \begin {gather*} \frac {1}{6} \, B b^{3} x^{6} + \frac {1}{2} \, A a^{3} x^{2} + \frac {1}{5} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{5} + \frac {3}{4} \, {\left (B a^{2} b + A a b^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*b^3*x^6 + 1/2*A*a^3*x^2 + 1/5*(3*B*a*b^2 + A*b^3)*x^5 + 3/4*(B*a^2*b + A*a*b^2)*x^4 + 1/3*(B*a^3 + 3*A*a
^2*b)*x^3

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mupad [B]  time = 0.03, size = 68, normalized size = 1.11 \begin {gather*} x^3\,\left (\frac {B\,a^3}{3}+A\,b\,a^2\right )+x^5\,\left (\frac {A\,b^3}{5}+\frac {3\,B\,a\,b^2}{5}\right )+\frac {A\,a^3\,x^2}{2}+\frac {B\,b^3\,x^6}{6}+\frac {3\,a\,b\,x^4\,\left (A\,b+B\,a\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(A + B*x)*(a + b*x)^3,x)

[Out]

x^3*((B*a^3)/3 + A*a^2*b) + x^5*((A*b^3)/5 + (3*B*a*b^2)/5) + (A*a^3*x^2)/2 + (B*b^3*x^6)/6 + (3*a*b*x^4*(A*b
+ B*a))/4

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sympy [A]  time = 0.13, size = 80, normalized size = 1.31 \begin {gather*} \frac {A a^{3} x^{2}}{2} + \frac {B b^{3} x^{6}}{6} + x^{5} \left (\frac {A b^{3}}{5} + \frac {3 B a b^{2}}{5}\right ) + x^{4} \left (\frac {3 A a b^{2}}{4} + \frac {3 B a^{2} b}{4}\right ) + x^{3} \left (A a^{2} b + \frac {B a^{3}}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**3*x**2/2 + B*b**3*x**6/6 + x**5*(A*b**3/5 + 3*B*a*b**2/5) + x**4*(3*A*a*b**2/4 + 3*B*a**2*b/4) + x**3*(A*
a**2*b + B*a**3/3)

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