3.344 \(\int x^{5/2} (a+b x^2) (A+B x^2) \, dx\)

Optimal. Leaf size=39 \[ \frac{2}{11} x^{11/2} (a B+A b)+\frac{2}{7} a A x^{7/2}+\frac{2}{15} b B x^{15/2} \]

[Out]

(2*a*A*x^(7/2))/7 + (2*(A*b + a*B)*x^(11/2))/11 + (2*b*B*x^(15/2))/15

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Rubi [A]  time = 0.0159631, 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} \[ \frac{2}{11} x^{11/2} (a B+A b)+\frac{2}{7} a A x^{7/2}+\frac{2}{15} b B x^{15/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(a + b*x^2)*(A + B*x^2),x]

[Out]

(2*a*A*x^(7/2))/7 + (2*(A*b + a*B)*x^(11/2))/11 + (2*b*B*x^(15/2))/15

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^{5/2} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx &=\int \left (a A x^{5/2}+(A b+a B) x^{9/2}+b B x^{13/2}\right ) \, dx\\ &=\frac{2}{7} a A x^{7/2}+\frac{2}{11} (A b+a B) x^{11/2}+\frac{2}{15} b B x^{15/2}\\ \end{align*}

Mathematica [A]  time = 0.0145277, size = 33, normalized size = 0.85 \[ \frac{2 x^{7/2} \left (105 x^2 (a B+A b)+165 a A+77 b B x^4\right )}{1155} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(a + b*x^2)*(A + B*x^2),x]

[Out]

(2*x^(7/2)*(165*a*A + 105*(A*b + a*B)*x^2 + 77*b*B*x^4))/1155

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Maple [A]  time = 0.003, size = 32, normalized size = 0.8 \begin{align*}{\frac{154\,bB{x}^{4}+210\,A{x}^{2}b+210\,B{x}^{2}a+330\,Aa}{1155}{x}^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x^2+a)*(B*x^2+A),x)

[Out]

2/1155*x^(7/2)*(77*B*b*x^4+105*A*b*x^2+105*B*a*x^2+165*A*a)

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Maxima [A]  time = 1.04101, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{15} \, B b x^{\frac{15}{2}} + \frac{2}{11} \,{\left (B a + A b\right )} x^{\frac{11}{2}} + \frac{2}{7} \, A a x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x^2+a)*(B*x^2+A),x, algorithm="maxima")

[Out]

2/15*B*b*x^(15/2) + 2/11*(B*a + A*b)*x^(11/2) + 2/7*A*a*x^(7/2)

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Fricas [A]  time = 0.852479, size = 88, normalized size = 2.26 \begin{align*} \frac{2}{1155} \,{\left (77 \, B b x^{7} + 105 \,{\left (B a + A b\right )} x^{5} + 165 \, A a x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x^2+a)*(B*x^2+A),x, algorithm="fricas")

[Out]

2/1155*(77*B*b*x^7 + 105*(B*a + A*b)*x^5 + 165*A*a*x^3)*sqrt(x)

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Sympy [A]  time = 5.66925, size = 46, normalized size = 1.18 \begin{align*} \frac{2 A a x^{\frac{7}{2}}}{7} + \frac{2 A b x^{\frac{11}{2}}}{11} + \frac{2 B a x^{\frac{11}{2}}}{11} + \frac{2 B b x^{\frac{15}{2}}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(b*x**2+a)*(B*x**2+A),x)

[Out]

2*A*a*x**(7/2)/7 + 2*A*b*x**(11/2)/11 + 2*B*a*x**(11/2)/11 + 2*B*b*x**(15/2)/15

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Giac [A]  time = 1.13482, size = 39, normalized size = 1. \begin{align*} \frac{2}{15} \, B b x^{\frac{15}{2}} + \frac{2}{11} \, B a x^{\frac{11}{2}} + \frac{2}{11} \, A b x^{\frac{11}{2}} + \frac{2}{7} \, A a x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x^2+a)*(B*x^2+A),x, algorithm="giac")

[Out]

2/15*B*b*x^(15/2) + 2/11*B*a*x^(11/2) + 2/11*A*b*x^(11/2) + 2/7*A*a*x^(7/2)