3.347 \(\int \frac {(a+b x^2) (A+B x^2)}{\sqrt {x}} \, dx\)

Optimal. Leaf size=37 \[ \frac {2}{5} x^{5/2} (a B+A b)+2 a A \sqrt {x}+\frac {2}{9} b B x^{9/2} \]

[Out]

2/5*(A*b+B*a)*x^(5/2)+2/9*b*B*x^(9/2)+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} \[ \frac {2}{5} x^{5/2} (a B+A b)+2 a A \sqrt {x}+\frac {2}{9} b B x^{9/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*A*Sqrt[x] + (2*(A*b + a*B)*x^(5/2))/5 + (2*b*B*x^(9/2))/9

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.89 \[ \frac {2}{45} \sqrt {x} \left (9 x^2 (a B+A b)+45 a A+5 b B x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(45*a*A + 9*(A*b + a*B)*x^2 + 5*b*B*x^4))/45

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fricas [A]  time = 0.45, size = 29, normalized size = 0.78 \[ \frac {2}{45} \, {\left (5 \, B b x^{4} + 9 \, {\left (B a + A b\right )} x^{2} + 45 \, A a\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^(1/2),x, algorithm="fricas")

[Out]

2/45*(5*B*b*x^4 + 9*(B*a + A*b)*x^2 + 45*A*a)*sqrt(x)

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giac [A]  time = 0.31, size = 29, normalized size = 0.78 \[ \frac {2}{9} \, B b x^{\frac {9}{2}} + \frac {2}{5} \, B a x^{\frac {5}{2}} + \frac {2}{5} \, A b x^{\frac {5}{2}} + 2 \, A a \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 32, normalized size = 0.86 \[ \frac {2 \left (5 B b \,x^{4}+9 A b \,x^{2}+9 B a \,x^{2}+45 A a \right ) \sqrt {x}}{45} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(B*x^2+A)/x^(1/2),x)

[Out]

2/45*x^(1/2)*(5*B*b*x^4+9*A*b*x^2+9*B*a*x^2+45*A*a)

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maxima [A]  time = 1.06, size = 27, normalized size = 0.73 \[ \frac {2}{9} \, B b x^{\frac {9}{2}} + \frac {2}{5} \, {\left (B a + A b\right )} x^{\frac {5}{2}} + 2 \, A a \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.19, size = 31, normalized size = 0.84 \[ \frac {2\,\sqrt {x}\,\left (45\,A\,a+9\,A\,b\,x^2+9\,B\,a\,x^2+5\,B\,b\,x^4\right )}{45} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2))/x^(1/2),x)

[Out]

(2*x^(1/2)*(45*A*a + 9*A*b*x^2 + 9*B*a*x^2 + 5*B*b*x^4))/45

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sympy [A]  time = 0.78, size = 44, normalized size = 1.19 \[ 2 A a \sqrt {x} + \frac {2 A b x^{\frac {5}{2}}}{5} + \frac {2 B a x^{\frac {5}{2}}}{5} + \frac {2 B b x^{\frac {9}{2}}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(B*x**2+A)/x**(1/2),x)

[Out]

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

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