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

Optimal. Leaf size=39 \[ \frac{2}{9} x^{9/2} (a B+A b)+\frac{2}{5} a A x^{5/2}+\frac{2}{13} b B x^{13/2} \]

[Out]

(2*a*A*x^(5/2))/5 + (2*(A*b + a*B)*x^(9/2))/9 + (2*b*B*x^(13/2))/13

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Rubi [A]  time = 0.015995, 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} \[ \frac{2}{9} x^{9/2} (a B+A b)+\frac{2}{5} a A x^{5/2}+\frac{2}{13} b B x^{13/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*x^(5/2))/5 + (2*(A*b + a*B)*x^(9/2))/9 + (2*b*B*x^(13/2))/13

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 x^{3/2} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx &=\int \left (a A x^{3/2}+(A b+a B) x^{7/2}+b B x^{11/2}\right ) \, dx\\ &=\frac{2}{5} a A x^{5/2}+\frac{2}{9} (A b+a B) x^{9/2}+\frac{2}{13} b B x^{13/2}\\ \end{align*}

Mathematica [A]  time = 0.0133352, size = 33, normalized size = 0.85 \[ \frac{2}{585} x^{5/2} \left (65 x^2 (a B+A b)+117 a A+45 b B x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(117*a*A + 65*(A*b + a*B)*x^2 + 45*b*B*x^4))/585

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Maple [A]  time = 0.003, size = 32, normalized size = 0.8 \begin{align*}{\frac{90\,bB{x}^{4}+130\,A{x}^{2}b+130\,B{x}^{2}a+234\,Aa}{585}{x}^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/585*x^(5/2)*(45*B*b*x^4+65*A*b*x^2+65*B*a*x^2+117*A*a)

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Maxima [A]  time = 1.05933, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{13} \, B b x^{\frac{13}{2}} + \frac{2}{9} \,{\left (B a + A b\right )} x^{\frac{9}{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^2+a)*(B*x^2+A),x, algorithm="maxima")

[Out]

2/13*B*b*x^(13/2) + 2/9*(B*a + A*b)*x^(9/2) + 2/5*A*a*x^(5/2)

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Fricas [A]  time = 0.881758, size = 85, normalized size = 2.18 \begin{align*} \frac{2}{585} \,{\left (45 \, B b x^{6} + 65 \,{\left (B a + A b\right )} x^{4} + 117 \, A a x^{2}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/585*(45*B*b*x^6 + 65*(B*a + A*b)*x^4 + 117*A*a*x^2)*sqrt(x)

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Sympy [A]  time = 2.56887, size = 46, normalized size = 1.18 \begin{align*} \frac{2 A a x^{\frac{5}{2}}}{5} + \frac{2 A b x^{\frac{9}{2}}}{9} + \frac{2 B a x^{\frac{9}{2}}}{9} + \frac{2 B b x^{\frac{13}{2}}}{13} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a*x**(5/2)/5 + 2*A*b*x**(9/2)/9 + 2*B*a*x**(9/2)/9 + 2*B*b*x**(13/2)/13

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Giac [A]  time = 1.21152, size = 39, normalized size = 1. \begin{align*} \frac{2}{13} \, B b x^{\frac{13}{2}} + \frac{2}{9} \, B a x^{\frac{9}{2}} + \frac{2}{9} \, A b x^{\frac{9}{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^2+a)*(B*x^2+A),x, algorithm="giac")

[Out]

2/13*B*b*x^(13/2) + 2/9*B*a*x^(9/2) + 2/9*A*b*x^(9/2) + 2/5*A*a*x^(5/2)