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

Optimal. Leaf size=63 \[ \frac{2}{3} a^2 A x^{3/2}+\frac{2}{7} b x^{7/2} (2 a B+A b)+\frac{2}{5} a x^{5/2} (a B+2 A b)+\frac{2}{9} b^2 B x^{9/2} \]

[Out]

(2*a^2*A*x^(3/2))/3 + (2*a*(2*A*b + a*B)*x^(5/2))/5 + (2*b*(A*b + 2*a*B)*x^(7/2))/7 + (2*b^2*B*x^(9/2))/9

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Rubi [A]  time = 0.0290496, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {27, 76} \[ \frac{2}{3} a^2 A x^{3/2}+\frac{2}{7} b x^{7/2} (2 a B+A b)+\frac{2}{5} a x^{5/2} (a B+2 A b)+\frac{2}{9} b^2 B x^{9/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*A*x^(3/2))/3 + (2*a*(2*A*b + a*B)*x^(5/2))/5 + (2*b*(A*b + 2*a*B)*x^(7/2))/7 + (2*b^2*B*x^(9/2))/9

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

Mathematica [A]  time = 0.0181285, size = 52, normalized size = 0.83 \[ \frac{2}{315} x^{3/2} \left (21 a^2 (5 A+3 B x)+18 a b x (7 A+5 B x)+5 b^2 x^2 (9 A+7 B x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(21*a^2*(5*A + 3*B*x) + 18*a*b*x*(7*A + 5*B*x) + 5*b^2*x^2*(9*A + 7*B*x)))/315

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Maple [A]  time = 0.005, size = 52, normalized size = 0.8 \begin{align*}{\frac{70\,{b}^{2}B{x}^{3}+90\,A{b}^{2}{x}^{2}+180\,B{x}^{2}ab+252\,aAbx+126\,{a}^{2}Bx+210\,A{a}^{2}}{315}{x}^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*x^(3/2)*(35*B*b^2*x^3+45*A*b^2*x^2+90*B*a*b*x^2+126*A*a*b*x+63*B*a^2*x+105*A*a^2)

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Maxima [A]  time = 1.12236, size = 69, normalized size = 1.1 \begin{align*} \frac{2}{9} \, B b^{2} x^{\frac{9}{2}} + \frac{2}{3} \, A a^{2} x^{\frac{3}{2}} + \frac{2}{7} \,{\left (2 \, B a b + A b^{2}\right )} x^{\frac{7}{2}} + \frac{2}{5} \,{\left (B a^{2} + 2 \, A a b\right )} x^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*B*b^2*x^(9/2) + 2/3*A*a^2*x^(3/2) + 2/7*(2*B*a*b + A*b^2)*x^(7/2) + 2/5*(B*a^2 + 2*A*a*b)*x^(5/2)

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Fricas [A]  time = 1.51602, size = 132, normalized size = 2.1 \begin{align*} \frac{2}{315} \,{\left (35 \, B b^{2} x^{4} + 105 \, A a^{2} x + 45 \,{\left (2 \, B a b + A b^{2}\right )} x^{3} + 63 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*B*b^2*x^4 + 105*A*a^2*x + 45*(2*B*a*b + A*b^2)*x^3 + 63*(B*a^2 + 2*A*a*b)*x^2)*sqrt(x)

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Sympy [A]  time = 2.49036, size = 66, normalized size = 1.05 \begin{align*} \frac{2 A a^{2} x^{\frac{3}{2}}}{3} + \frac{2 B b^{2} x^{\frac{9}{2}}}{9} + \frac{2 x^{\frac{7}{2}} \left (A b^{2} + 2 B a b\right )}{7} + \frac{2 x^{\frac{5}{2}} \left (2 A a b + B a^{2}\right )}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*x**(3/2)/3 + 2*B*b**2*x**(9/2)/9 + 2*x**(7/2)*(A*b**2 + 2*B*a*b)/7 + 2*x**(5/2)*(2*A*a*b + B*a**2)/5

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Giac [A]  time = 1.13791, size = 72, normalized size = 1.14 \begin{align*} \frac{2}{9} \, B b^{2} x^{\frac{9}{2}} + \frac{4}{7} \, B a b x^{\frac{7}{2}} + \frac{2}{7} \, A b^{2} x^{\frac{7}{2}} + \frac{2}{5} \, B a^{2} x^{\frac{5}{2}} + \frac{4}{5} \, A a b x^{\frac{5}{2}} + \frac{2}{3} \, A a^{2} x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*B*b^2*x^(9/2) + 4/7*B*a*b*x^(7/2) + 2/7*A*b^2*x^(7/2) + 2/5*B*a^2*x^(5/2) + 4/5*A*a*b*x^(5/2) + 2/3*A*a^2*
x^(3/2)