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

Optimal. Leaf size=39 \[ \frac{2}{7} x^{7/2} (a B+A b)+\frac{2}{3} a A x^{3/2}+\frac{2}{11} b B x^{11/2} \]

[Out]

(2*a*A*x^(3/2))/3 + (2*(A*b + a*B)*x^(7/2))/7 + (2*b*B*x^(11/2))/11

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Rubi [A]  time = 0.0153497, 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 = {448} \[ \frac{2}{7} x^{7/2} (a B+A b)+\frac{2}{3} a A x^{3/2}+\frac{2}{11} b B x^{11/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*x^(3/2))/3 + (2*(A*b + a*B)*x^(7/2))/7 + (2*b*B*x^(11/2))/11

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.0146051, size = 33, normalized size = 0.85 \[ \frac{2}{231} x^{3/2} \left (33 x^2 (a B+A b)+77 a A+21 b B x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(77*a*A + 33*(A*b + a*B)*x^2 + 21*b*B*x^4))/231

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Maple [A]  time = 0.002, size = 32, normalized size = 0.8 \begin{align*}{\frac{42\,bB{x}^{4}+66\,A{x}^{2}b+66\,B{x}^{2}a+154\,Aa}{231}{x}^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/231*x^(3/2)*(21*B*b*x^4+33*A*b*x^2+33*B*a*x^2+77*A*a)

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Maxima [A]  time = 1.02638, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{11} \, B b x^{\frac{11}{2}} + \frac{2}{7} \,{\left (B a + A b\right )} x^{\frac{7}{2}} + \frac{2}{3} \, A a x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*B*b*x^(11/2) + 2/7*(B*a + A*b)*x^(7/2) + 2/3*A*a*x^(3/2)

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Fricas [A]  time = 0.889685, size = 81, normalized size = 2.08 \begin{align*} \frac{2}{231} \,{\left (21 \, B b x^{5} + 33 \,{\left (B a + A b\right )} x^{3} + 77 \, A a x\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/231*(21*B*b*x^5 + 33*(B*a + A*b)*x^3 + 77*A*a*x)*sqrt(x)

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Sympy [A]  time = 1.67964, size = 37, normalized size = 0.95 \begin{align*} \frac{2 A a x^{\frac{3}{2}}}{3} + \frac{2 B b x^{\frac{11}{2}}}{11} + \frac{2 x^{\frac{7}{2}} \left (A b + B a\right )}{7} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a*x**(3/2)/3 + 2*B*b*x**(11/2)/11 + 2*x**(7/2)*(A*b + B*a)/7

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Giac [A]  time = 1.1176, size = 39, normalized size = 1. \begin{align*} \frac{2}{11} \, B b x^{\frac{11}{2}} + \frac{2}{7} \, B a x^{\frac{7}{2}} + \frac{2}{7} \, A b x^{\frac{7}{2}} + \frac{2}{3} \, A a x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*B*b*x^(11/2) + 2/7*B*a*x^(7/2) + 2/7*A*b*x^(7/2) + 2/3*A*a*x^(3/2)