3.321 \(\int x^{3/2} (a+b x) (A+B x) \, dx\)

Optimal. Leaf size=39 \[ \frac{2}{7} x^{7/2} (a B+A b)+\frac{2}{5} a A x^{5/2}+\frac{2}{9} b B x^{9/2} \]

[Out]

(2*a*A*x^(5/2))/5 + (2*(A*b + a*B)*x^(7/2))/7 + (2*b*B*x^(9/2))/9

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Rubi [A]  time = 0.0144146, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {76} \[ \frac{2}{7} x^{7/2} (a B+A b)+\frac{2}{5} a A x^{5/2}+\frac{2}{9} b B x^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a + b*x)*(A + B*x),x]

[Out]

(2*a*A*x^(5/2))/5 + (2*(A*b + a*B)*x^(7/2))/7 + (2*b*B*x^(9/2))/9

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^{3/2} (a+b x) (A+B x) \, dx &=\int \left (a A x^{3/2}+(A b+a B) x^{5/2}+b B x^{7/2}\right ) \, dx\\ &=\frac{2}{5} a A x^{5/2}+\frac{2}{7} (A b+a B) x^{7/2}+\frac{2}{9} b B x^{9/2}\\ \end{align*}

Mathematica [A]  time = 0.010721, size = 33, normalized size = 0.85 \[ \frac{2}{315} x^{5/2} (9 a (7 A+5 B x)+5 b x (9 A+7 B x)) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a + b*x)*(A + B*x),x]

[Out]

(2*x^(5/2)*(9*a*(7*A + 5*B*x) + 5*b*x*(9*A + 7*B*x)))/315

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Maple [A]  time = 0.003, size = 28, normalized size = 0.7 \begin{align*}{\frac{70\,bB{x}^{2}+90\,Abx+90\,Bax+126\,Aa}{315}{x}^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x+a)*(B*x+A),x)

[Out]

2/315*x^(5/2)*(35*B*b*x^2+45*A*b*x+45*B*a*x+63*A*a)

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Maxima [A]  time = 1.65389, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{9} \, B b x^{\frac{9}{2}} + \frac{2}{5} \, A a x^{\frac{5}{2}} + \frac{2}{7} \,{\left (B a + A b\right )} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+a)*(B*x+A),x, algorithm="maxima")

[Out]

2/9*B*b*x^(9/2) + 2/5*A*a*x^(5/2) + 2/7*(B*a + A*b)*x^(7/2)

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Fricas [A]  time = 2.53521, size = 84, normalized size = 2.15 \begin{align*} \frac{2}{315} \,{\left (35 \, B b x^{4} + 63 \, A a x^{2} + 45 \,{\left (B a + A b\right )} x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+a)*(B*x+A),x, algorithm="fricas")

[Out]

2/315*(35*B*b*x^4 + 63*A*a*x^2 + 45*(B*a + A*b)*x^3)*sqrt(x)

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Sympy [A]  time = 1.39554, size = 46, normalized size = 1.18 \begin{align*} \frac{2 A a x^{\frac{5}{2}}}{5} + \frac{2 A b x^{\frac{7}{2}}}{7} + \frac{2 B a x^{\frac{7}{2}}}{7} + \frac{2 B b x^{\frac{9}{2}}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(b*x+a)*(B*x+A),x)

[Out]

2*A*a*x**(5/2)/5 + 2*A*b*x**(7/2)/7 + 2*B*a*x**(7/2)/7 + 2*B*b*x**(9/2)/9

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Giac [A]  time = 1.18707, size = 39, normalized size = 1. \begin{align*} \frac{2}{9} \, B b x^{\frac{9}{2}} + \frac{2}{7} \, B a x^{\frac{7}{2}} + \frac{2}{7} \, A b x^{\frac{7}{2}} + \frac{2}{5} \, A a x^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+a)*(B*x+A),x, algorithm="giac")

[Out]

2/9*B*b*x^(9/2) + 2/7*B*a*x^(7/2) + 2/7*A*b*x^(7/2) + 2/5*A*a*x^(5/2)