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]

-2/3*a*A/x^(3/2)+2/5*b*B*x^(5/2)+2*(A*b+B*a)*x^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, 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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.44, size = 29, normalized size = 0.78 \[ \frac {2 \, {\left (3 \, B b x^{4} + 15 \, {\left (B a + A b\right )} x^{2} - 5 \, A a\right )}}{15 \, x^{\frac {3}{2}}} \]

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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giac [A]  time = 0.33, size = 29, normalized size = 0.78 \[ \frac {2}{5} \, B b x^{\frac {5}{2}} + 2 \, B a \sqrt {x} + 2 \, A b \sqrt {x} - \frac {2 \, A a}{3 \, x^{\frac {3}{2}}} \]

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)

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maple [A]  time = 0.00, size = 32, normalized size = 0.86 \[ -\frac {2 \left (-3 B b \,x^{4}-15 A b \,x^{2}-15 B a \,x^{2}+5 A a \right )}{15 x^{\frac {3}{2}}} \]

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 = 1.03, size = 27, normalized size = 0.73 \[ \frac {2}{5} \, B b x^{\frac {5}{2}} + 2 \, {\left (B a + A b\right )} \sqrt {x} - \frac {2 \, A a}{3 \, x^{\frac {3}{2}}} \]

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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mupad [B]  time = 0.04, size = 31, normalized size = 0.84 \[ \frac {30\,A\,b\,x^2-10\,A\,a+30\,B\,a\,x^2+6\,B\,b\,x^4}{15\,x^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(30*A*b*x^2 - 10*A*a + 30*B*a*x^2 + 6*B*b*x^4)/(15*x^(3/2))

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sympy [A]  time = 1.19, size = 42, normalized size = 1.14 \[ - \frac {2 A a}{3 x^{\frac {3}{2}}} + 2 A b \sqrt {x} + 2 B a \sqrt {x} + \frac {2 B b x^{\frac {5}{2}}}{5} \]

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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