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

Optimal. Leaf size=37 \[ 2 \sqrt{x} (a B+A b)-\frac{2 a A}{3 x^{3/2}}+\frac{2}{5} b B x^{5/2} \]

[Out]

(-2*a*A)/(3*x^(3/2)) + 2*(A*b + a*B)*Sqrt[x] + (2*b*B*x^(5/2))/5

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Rubi [A]  time = 0.016267, antiderivative size = 37, 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} \[ 2 \sqrt{x} (a B+A b)-\frac{2 a A}{3 x^{3/2}}+\frac{2}{5} b B x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*A)/(3*x^(3/2)) + 2*(A*b + a*B)*Sqrt[x] + (2*b*B*x^(5/2))/5

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

Mathematica [A]  time = 0.012476, size = 36, normalized size = 0.97 \[ \frac{2 \left (3 b x^2 \left (5 A+B x^2\right )-5 a \left (A-3 B x^2\right )\right )}{15 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-5*a*(A - 3*B*x^2) + 3*b*x^2*(5*A + B*x^2)))/(15*x^(3/2))

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Maple [A]  time = 0.003, size = 32, normalized size = 0.9 \begin{align*} -{\frac{-6\,bB{x}^{4}-30\,A{x}^{2}b-30\,B{x}^{2}a+10\,Aa}{15}{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^(5/2),x)

[Out]

-2/15*(-3*B*b*x^4-15*A*b*x^2-15*B*a*x^2+5*A*a)/x^(3/2)

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Maxima [A]  time = 1.03156, size = 36, normalized size = 0.97 \begin{align*} \frac{2}{5} \, B b x^{\frac{5}{2}} + 2 \,{\left (B a + A b\right )} \sqrt{x} - \frac{2 \, A a}{3 \, 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^(5/2),x, algorithm="maxima")

[Out]

2/5*B*b*x^(5/2) + 2*(B*a + A*b)*sqrt(x) - 2/3*A*a/x^(3/2)

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Fricas [A]  time = 0.663719, size = 74, normalized size = 2. \begin{align*} \frac{2 \,{\left (3 \, B b x^{4} + 15 \,{\left (B a + A b\right )} x^{2} - 5 \, A a\right )}}{15 \, 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^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*B*b*x^4 + 15*(B*a + A*b)*x^2 - 5*A*a)/x^(3/2)

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Sympy [A]  time = 1.25066, size = 42, normalized size = 1.14 \begin{align*} - \frac{2 A a}{3 x^{\frac{3}{2}}} + 2 A b \sqrt{x} + 2 B a \sqrt{x} + \frac{2 B b x^{\frac{5}{2}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A*a/(3*x**(3/2)) + 2*A*b*sqrt(x) + 2*B*a*sqrt(x) + 2*B*b*x**(5/2)/5

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Giac [A]  time = 1.16129, size = 39, normalized size = 1.05 \begin{align*} \frac{2}{5} \, B b x^{\frac{5}{2}} + 2 \, B a \sqrt{x} + 2 \, A b \sqrt{x} - \frac{2 \, A a}{3 \, 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^(5/2),x, algorithm="giac")

[Out]

2/5*B*b*x^(5/2) + 2*B*a*sqrt(x) + 2*A*b*sqrt(x) - 2/3*A*a/x^(3/2)