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

Optimal. Leaf size=37 \[ \frac {2}{3} x^{3/2} (a B+A b)-\frac {2 a A}{\sqrt {x}}+\frac {2}{7} b B x^{7/2} \]

[Out]

2/3*(A*b+B*a)*x^(3/2)+2/7*b*B*x^(7/2)-2*a*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} \[ \frac {2}{3} x^{3/2} (a B+A b)-\frac {2 a A}{\sqrt {x}}+\frac {2}{7} b B x^{7/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*A)/Sqrt[x] + (2*(A*b + a*B)*x^(3/2))/3 + (2*b*B*x^(7/2))/7

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

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.95 \[ \frac {2 \left (-21 a A+7 a B x^2+7 A b x^2+3 b B x^4\right )}{21 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-21*a*A + 7*A*b*x^2 + 7*a*B*x^2 + 3*b*B*x^4))/(21*Sqrt[x])

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fricas [A]  time = 0.42, size = 29, normalized size = 0.78 \[ \frac {2 \, {\left (3 \, B b x^{4} + 7 \, {\left (B a + A b\right )} x^{2} - 21 \, A a\right )}}{21 \, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*(3*B*b*x^4 + 7*(B*a + A*b)*x^2 - 21*A*a)/sqrt(x)

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giac [A]  time = 0.35, size = 29, normalized size = 0.78 \[ \frac {2}{7} \, B b x^{\frac {7}{2}} + \frac {2}{3} \, B a x^{\frac {3}{2}} + \frac {2}{3} \, A b x^{\frac {3}{2}} - \frac {2 \, A a}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 32, normalized size = 0.86 \[ -\frac {2 \left (-3 B b \,x^{4}-7 A b \,x^{2}-7 B a \,x^{2}+21 A a \right )}{21 \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*(-3*B*b*x^4-7*A*b*x^2-7*B*a*x^2+21*A*a)/x^(1/2)

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maxima [A]  time = 1.10, size = 27, normalized size = 0.73 \[ \frac {2}{7} \, B b x^{\frac {7}{2}} + \frac {2}{3} \, {\left (B a + A b\right )} x^{\frac {3}{2}} - \frac {2 \, A a}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.19, size = 31, normalized size = 0.84 \[ \frac {14\,A\,b\,x^2-42\,A\,a+14\,B\,a\,x^2+6\,B\,b\,x^4}{21\,\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(14*A*b*x^2 - 42*A*a + 14*B*a*x^2 + 6*B*b*x^4)/(21*x^(1/2))

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sympy [A]  time = 0.99, size = 44, normalized size = 1.19 \[ - \frac {2 A a}{\sqrt {x}} + \frac {2 A b x^{\frac {3}{2}}}{3} + \frac {2 B a x^{\frac {3}{2}}}{3} + \frac {2 B b x^{\frac {7}{2}}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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