3.143 \(\int \sqrt{x} (A+B x) (b x+c x^2) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0152832, 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} \[ \frac{2}{7} x^{7/2} (A c+b B)+\frac{2}{5} A b x^{5/2}+\frac{2}{9} B c x^{9/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0112639, size = 33, normalized size = 0.85 \[ \frac{2}{315} x^{5/2} (9 A (7 b+5 c x)+5 B x (9 b+7 c x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 28, normalized size = 0.7 \begin{align*}{\frac{70\,Bc{x}^{2}+90\,Acx+90\,bBx+126\,Ab}{315}{x}^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)*x^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.06012, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{9} \, B c x^{\frac{9}{2}} + \frac{2}{5} \, A b x^{\frac{5}{2}} + \frac{2}{7} \,{\left (B b + A c\right )} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)*x^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.7753, size = 84, normalized size = 2.15 \begin{align*} \frac{2}{315} \,{\left (35 \, B c x^{4} + 63 \, A b x^{2} + 45 \,{\left (B b + A c\right )} x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)*x^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.90773, size = 37, normalized size = 0.95 \begin{align*} \frac{2 A b x^{\frac{5}{2}}}{5} + \frac{2 B c x^{\frac{9}{2}}}{9} + \frac{2 x^{\frac{7}{2}} \left (A c + B b\right )}{7} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)*x**(1/2),x)

[Out]

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

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Giac [A]  time = 1.12371, size = 39, normalized size = 1. \begin{align*} \frac{2}{9} \, B c x^{\frac{9}{2}} + \frac{2}{7} \, B b x^{\frac{7}{2}} + \frac{2}{7} \, A c x^{\frac{7}{2}} + \frac{2}{5} \, A b x^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)*x^(1/2),x, algorithm="giac")

[Out]

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