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

Optimal. Leaf size=61 \[ 2 a^2 A \sqrt {x}+\frac {2}{9} b x^{9/2} (2 a B+A b)+\frac {2}{5} a x^{5/2} (a B+2 A b)+\frac {2}{13} b^2 B x^{13/2} \]

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {448} \begin {gather*} 2 a^2 A \sqrt {x}+\frac {2}{9} b x^{9/2} (2 a B+A b)+\frac {2}{5} a x^{5/2} (a B+2 A b)+\frac {2}{13} b^2 B x^{13/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*a^2*A*Sqrt[x] + (2*a*(2*A*b + a*B)*x^(5/2))/5 + (2*b*(A*b + 2*a*B)*x^(9/2))/9 + (2*b^2*B*x^(13/2))/13

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 )^2 \left (A+B x^2\right )}{\sqrt {x}} \, dx &=\int \left (\frac {a^2 A}{\sqrt {x}}+a (2 A b+a B) x^{3/2}+b (A b+2 a B) x^{7/2}+b^2 B x^{11/2}\right ) \, dx\\ &=2 a^2 A \sqrt {x}+\frac {2}{5} a (2 A b+a B) x^{5/2}+\frac {2}{9} b (A b+2 a B) x^{9/2}+\frac {2}{13} b^2 B x^{13/2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.87 \begin {gather*} \frac {2}{585} \sqrt {x} \left (585 a^2 A+65 b x^4 (2 a B+A b)+117 a x^2 (a B+2 A b)+45 b^2 B x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(585*a^2*A + 117*a*(2*A*b + a*B)*x^2 + 65*b*(A*b + 2*a*B)*x^4 + 45*b^2*B*x^6))/585

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IntegrateAlgebraic [A]  time = 0.03, size = 69, normalized size = 1.13 \begin {gather*} \frac {2}{585} \left (585 a^2 A \sqrt {x}+117 a^2 B x^{5/2}+234 a A b x^{5/2}+130 a b B x^{9/2}+65 A b^2 x^{9/2}+45 b^2 B x^{13/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(585*a^2*A*Sqrt[x] + 234*a*A*b*x^(5/2) + 117*a^2*B*x^(5/2) + 65*A*b^2*x^(9/2) + 130*a*b*B*x^(9/2) + 45*b^2*
B*x^(13/2)))/585

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fricas [A]  time = 1.34, size = 53, normalized size = 0.87 \begin {gather*} \frac {2}{585} \, {\left (45 \, B b^{2} x^{6} + 65 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + 585 \, A a^{2} + 117 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/585*(45*B*b^2*x^6 + 65*(2*B*a*b + A*b^2)*x^4 + 585*A*a^2 + 117*(B*a^2 + 2*A*a*b)*x^2)*sqrt(x)

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giac [A]  time = 0.26, size = 53, normalized size = 0.87 \begin {gather*} \frac {2}{13} \, B b^{2} x^{\frac {13}{2}} + \frac {4}{9} \, B a b x^{\frac {9}{2}} + \frac {2}{9} \, A b^{2} x^{\frac {9}{2}} + \frac {2}{5} \, B a^{2} x^{\frac {5}{2}} + \frac {4}{5} \, A a b x^{\frac {5}{2}} + 2 \, A a^{2} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/13*B*b^2*x^(13/2) + 4/9*B*a*b*x^(9/2) + 2/9*A*b^2*x^(9/2) + 2/5*B*a^2*x^(5/2) + 4/5*A*a*b*x^(5/2) + 2*A*a^2*
sqrt(x)

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maple [A]  time = 0.01, size = 56, normalized size = 0.92 \begin {gather*} \frac {2 \left (45 B \,b^{2} x^{6}+65 A \,b^{2} x^{4}+130 B a b \,x^{4}+234 A a b \,x^{2}+117 B \,a^{2} x^{2}+585 a^{2} A \right ) \sqrt {x}}{585} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/585*x^(1/2)*(45*B*b^2*x^6+65*A*b^2*x^4+130*B*a*b*x^4+234*A*a*b*x^2+117*B*a^2*x^2+585*A*a^2)

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maxima [A]  time = 0.96, size = 51, normalized size = 0.84 \begin {gather*} \frac {2}{13} \, B b^{2} x^{\frac {13}{2}} + \frac {2}{9} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {9}{2}} + 2 \, A a^{2} \sqrt {x} + \frac {2}{5} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/13*B*b^2*x^(13/2) + 2/9*(2*B*a*b + A*b^2)*x^(9/2) + 2*A*a^2*sqrt(x) + 2/5*(B*a^2 + 2*A*a*b)*x^(5/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.84 \begin {gather*} x^{5/2}\,\left (\frac {2\,B\,a^2}{5}+\frac {4\,A\,b\,a}{5}\right )+x^{9/2}\,\left (\frac {2\,A\,b^2}{9}+\frac {4\,B\,a\,b}{9}\right )+2\,A\,a^2\,\sqrt {x}+\frac {2\,B\,b^2\,x^{13/2}}{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(5/2)*((2*B*a^2)/5 + (4*A*a*b)/5) + x^(9/2)*((2*A*b^2)/9 + (4*B*a*b)/9) + 2*A*a^2*x^(1/2) + (2*B*b^2*x^(13/2
))/13

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sympy [A]  time = 2.11, size = 78, normalized size = 1.28 \begin {gather*} 2 A a^{2} \sqrt {x} + \frac {4 A a b x^{\frac {5}{2}}}{5} + \frac {2 A b^{2} x^{\frac {9}{2}}}{9} + \frac {2 B a^{2} x^{\frac {5}{2}}}{5} + \frac {4 B a b x^{\frac {9}{2}}}{9} + \frac {2 B b^{2} x^{\frac {13}{2}}}{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*sqrt(x) + 4*A*a*b*x**(5/2)/5 + 2*A*b**2*x**(9/2)/9 + 2*B*a**2*x**(5/2)/5 + 4*B*a*b*x**(9/2)/9 + 2*B*b
**2*x**(13/2)/13

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