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

Optimal. Leaf size=61 \[ -\frac {2 a^2 A}{3 x^{3/2}}+\frac {2}{5} b x^{5/2} (2 a B+A b)+2 a \sqrt {x} (a B+2 A b)+\frac {2}{9} b^2 B x^{9/2} \]

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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*} -\frac {2 a^2 A}{3 x^{3/2}}+\frac {2}{5} b x^{5/2} (2 a B+A b)+2 a \sqrt {x} (a B+2 A b)+\frac {2}{9} b^2 B x^{9/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.93 \begin {gather*} \frac {-30 a^2 \left (A-3 B x^2\right )+36 a b x^2 \left (5 A+B x^2\right )+2 b^2 x^4 \left (9 A+5 B x^2\right )}{45 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*a^2*(A - 3*B*x^2) + 36*a*b*x^2*(5*A + B*x^2) + 2*b^2*x^4*(9*A + 5*B*x^2))/(45*x^(3/2))

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IntegrateAlgebraic [A]  time = 0.04, size = 59, normalized size = 0.97 \begin {gather*} \frac {2 \left (-15 a^2 A+45 a^2 B x^2+90 a A b x^2+18 a b B x^4+9 A b^2 x^4+5 b^2 B x^6\right )}{45 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-15*a^2*A + 90*a*A*b*x^2 + 45*a^2*B*x^2 + 9*A*b^2*x^4 + 18*a*b*B*x^4 + 5*b^2*B*x^6))/(45*x^(3/2))

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fricas [A]  time = 1.08, size = 53, normalized size = 0.87 \begin {gather*} \frac {2 \, {\left (5 \, B b^{2} x^{6} + 9 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} - 15 \, A a^{2} + 45 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}\right )}}{45 \, x^{\frac {3}{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^(5/2),x, algorithm="fricas")

[Out]

2/45*(5*B*b^2*x^6 + 9*(2*B*a*b + A*b^2)*x^4 - 15*A*a^2 + 45*(B*a^2 + 2*A*a*b)*x^2)/x^(3/2)

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giac [A]  time = 0.41, size = 53, normalized size = 0.87 \begin {gather*} \frac {2}{9} \, B b^{2} x^{\frac {9}{2}} + \frac {4}{5} \, B a b x^{\frac {5}{2}} + \frac {2}{5} \, A b^{2} x^{\frac {5}{2}} + 2 \, B a^{2} \sqrt {x} + 4 \, A a b \sqrt {x} - \frac {2 \, A a^{2}}{3 \, x^{\frac {3}{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^(5/2),x, algorithm="giac")

[Out]

2/9*B*b^2*x^(9/2) + 4/5*B*a*b*x^(5/2) + 2/5*A*b^2*x^(5/2) + 2*B*a^2*sqrt(x) + 4*A*a*b*sqrt(x) - 2/3*A*a^2/x^(3
/2)

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maple [A]  time = 0.01, size = 56, normalized size = 0.92 \begin {gather*} -\frac {2 \left (-5 B \,b^{2} x^{6}-9 A \,b^{2} x^{4}-18 B a b \,x^{4}-90 A a b \,x^{2}-45 B \,a^{2} x^{2}+15 a^{2} A \right )}{45 x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45*(-5*B*b^2*x^6-9*A*b^2*x^4-18*B*a*b*x^4-90*A*a*b*x^2-45*B*a^2*x^2+15*A*a^2)/x^(3/2)

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maxima [A]  time = 0.99, size = 51, normalized size = 0.84 \begin {gather*} \frac {2}{9} \, B b^{2} x^{\frac {9}{2}} + \frac {2}{5} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {5}{2}} - \frac {2 \, A a^{2}}{3 \, x^{\frac {3}{2}}} + 2 \, {\left (B a^{2} + 2 \, A a b\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^(5/2),x, algorithm="maxima")

[Out]

2/9*B*b^2*x^(9/2) + 2/5*(2*B*a*b + A*b^2)*x^(5/2) - 2/3*A*a^2/x^(3/2) + 2*(B*a^2 + 2*A*a*b)*sqrt(x)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.84 \begin {gather*} \sqrt {x}\,\left (2\,B\,a^2+4\,A\,b\,a\right )+x^{5/2}\,\left (\frac {2\,A\,b^2}{5}+\frac {4\,B\,a\,b}{5}\right )-\frac {2\,A\,a^2}{3\,x^{3/2}}+\frac {2\,B\,b^2\,x^{9/2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1/2)*(2*B*a^2 + 4*A*a*b) + x^(5/2)*((2*A*b^2)/5 + (4*B*a*b)/5) - (2*A*a^2)/(3*x^(3/2)) + (2*B*b^2*x^(9/2))/
9

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sympy [A]  time = 2.96, size = 76, normalized size = 1.25 \begin {gather*} - \frac {2 A a^{2}}{3 x^{\frac {3}{2}}} + 4 A a b \sqrt {x} + \frac {2 A b^{2} x^{\frac {5}{2}}}{5} + 2 B a^{2} \sqrt {x} + \frac {4 B a b x^{\frac {5}{2}}}{5} + \frac {2 B b^{2} x^{\frac {9}{2}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A*a**2/(3*x**(3/2)) + 4*A*a*b*sqrt(x) + 2*A*b**2*x**(5/2)/5 + 2*B*a**2*sqrt(x) + 4*B*a*b*x**(5/2)/5 + 2*B*b
**2*x**(9/2)/9

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