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

Optimal. Leaf size=39 \[ \frac{2}{9} x^{9/2} (A c+b B)+\frac{2}{7} A b x^{7/2}+\frac{2}{11} B c x^{11/2} \]

[Out]

(2*A*b*x^(7/2))/7 + (2*(b*B + A*c)*x^(9/2))/9 + (2*B*c*x^(11/2))/11

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Rubi [A]  time = 0.0163165, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {765} \[ \frac{2}{9} x^{9/2} (A c+b B)+\frac{2}{7} A b x^{7/2}+\frac{2}{11} B c x^{11/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(A + B*x)*(b*x + c*x^2),x]

[Out]

(2*A*b*x^(7/2))/7 + (2*(b*B + A*c)*x^(9/2))/9 + (2*B*c*x^(11/2))/11

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int x^{3/2} (A+B x) \left (b x+c x^2\right ) \, dx &=\int \left (A b x^{5/2}+(b B+A c) x^{7/2}+B c x^{9/2}\right ) \, dx\\ &=\frac{2}{7} A b x^{7/2}+\frac{2}{9} (b B+A c) x^{9/2}+\frac{2}{11} B c x^{11/2}\\ \end{align*}

Mathematica [A]  time = 0.0090838, size = 33, normalized size = 0.85 \[ \frac{2}{693} x^{7/2} (11 A (9 b+7 c x)+7 B x (11 b+9 c x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(7/2)*(11*A*(9*b + 7*c*x) + 7*B*x*(11*b + 9*c*x)))/693

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Maple [A]  time = 0.004, size = 28, normalized size = 0.7 \begin{align*}{\frac{126\,Bc{x}^{2}+154\,Acx+154\,bBx+198\,Ab}{693}{x}^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/693*x^(7/2)*(63*B*c*x^2+77*A*c*x+77*B*b*x+99*A*b)

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Maxima [A]  time = 1.02077, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{11} \, B c x^{\frac{11}{2}} + \frac{2}{7} \, A b x^{\frac{7}{2}} + \frac{2}{9} \,{\left (B b + A c\right )} x^{\frac{9}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*B*c*x^(11/2) + 2/7*A*b*x^(7/2) + 2/9*(B*b + A*c)*x^(9/2)

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Fricas [A]  time = 1.7995, size = 84, normalized size = 2.15 \begin{align*} \frac{2}{693} \,{\left (63 \, B c x^{5} + 99 \, A b x^{3} + 77 \,{\left (B b + A c\right )} x^{4}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/693*(63*B*c*x^5 + 99*A*b*x^3 + 77*(B*b + A*c)*x^4)*sqrt(x)

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Sympy [A]  time = 1.81431, size = 46, normalized size = 1.18 \begin{align*} \frac{2 A b x^{\frac{7}{2}}}{7} + \frac{2 A c x^{\frac{9}{2}}}{9} + \frac{2 B b x^{\frac{9}{2}}}{9} + \frac{2 B c x^{\frac{11}{2}}}{11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*b*x**(7/2)/7 + 2*A*c*x**(9/2)/9 + 2*B*b*x**(9/2)/9 + 2*B*c*x**(11/2)/11

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Giac [A]  time = 1.17017, size = 39, normalized size = 1. \begin{align*} \frac{2}{11} \, B c x^{\frac{11}{2}} + \frac{2}{9} \, B b x^{\frac{9}{2}} + \frac{2}{9} \, A c x^{\frac{9}{2}} + \frac{2}{7} \, A b x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*B*c*x^(11/2) + 2/9*B*b*x^(9/2) + 2/9*A*c*x^(9/2) + 2/7*A*b*x^(7/2)