3.145 \(\int \frac{(a+b x^3)^2 (A+B x^3)}{x^{5/2}} \, dx\)

Optimal. Leaf size=63 \[ -\frac{2 a^2 A}{3 x^{3/2}}+\frac{2}{9} b x^{9/2} (2 a B+A b)+\frac{2}{3} a x^{3/2} (a B+2 A b)+\frac{2}{15} b^2 B x^{15/2} \]

[Out]

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

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Rubi [A]  time = 0.0301588, antiderivative size = 63, normalized size of antiderivative = 1., 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} \[ -\frac{2 a^2 A}{3 x^{3/2}}+\frac{2}{9} b x^{9/2} (2 a B+A b)+\frac{2}{3} a x^{3/2} (a B+2 A b)+\frac{2}{15} b^2 B x^{15/2} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*(A + B*x^3))/x^(5/2),x]

[Out]

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

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^3\right )^2 \left (A+B x^3\right )}{x^{5/2}} \, dx &=\int \left (\frac{a^2 A}{x^{5/2}}+a (2 A b+a B) \sqrt{x}+b (A b+2 a B) x^{7/2}+b^2 B x^{13/2}\right ) \, dx\\ &=-\frac{2 a^2 A}{3 x^{3/2}}+\frac{2}{3} a (2 A b+a B) x^{3/2}+\frac{2}{9} b (A b+2 a B) x^{9/2}+\frac{2}{15} b^2 B x^{15/2}\\ \end{align*}

Mathematica [A]  time = 0.0156948, size = 57, normalized size = 0.9 \[ \frac{-30 a^2 \left (A-B x^3\right )+20 a b x^3 \left (3 A+B x^3\right )+2 b^2 x^6 \left (5 A+3 B x^3\right )}{45 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*(A + B*x^3))/x^(5/2),x]

[Out]

(-30*a^2*(A - B*x^3) + 20*a*b*x^3*(3*A + B*x^3) + 2*b^2*x^6*(5*A + 3*B*x^3))/(45*x^(3/2))

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Maple [A]  time = 0.008, size = 56, normalized size = 0.9 \begin{align*} -{\frac{-6\,B{b}^{2}{x}^{9}-10\,A{b}^{2}{x}^{6}-20\,B{x}^{6}ab-60\,aAb{x}^{3}-30\,B{x}^{3}{a}^{2}+30\,{a}^{2}A}{45}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(B*x^3+A)/x^(5/2),x)

[Out]

-2/45*(-3*B*b^2*x^9-5*A*b^2*x^6-10*B*a*b*x^6-30*A*a*b*x^3-15*B*a^2*x^3+15*A*a^2)/x^(3/2)

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Maxima [A]  time = 0.948136, size = 69, normalized size = 1.1 \begin{align*} \frac{2}{15} \, B b^{2} x^{\frac{15}{2}} + \frac{2}{9} \,{\left (2 \, B a b + A b^{2}\right )} x^{\frac{9}{2}} + \frac{2}{3} \,{\left (B a^{2} + 2 \, A a b\right )} x^{\frac{3}{2}} - \frac{2 \, A a^{2}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.74359, size = 124, normalized size = 1.97 \begin{align*} \frac{2 \,{\left (3 \, B b^{2} x^{9} + 5 \,{\left (2 \, B a b + A b^{2}\right )} x^{6} + 15 \,{\left (B a^{2} + 2 \, A a b\right )} x^{3} - 15 \, A a^{2}\right )}}{45 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 9.44909, size = 80, normalized size = 1.27 \begin{align*} - \frac{2 A a^{2}}{3 x^{\frac{3}{2}}} + \frac{4 A a b x^{\frac{3}{2}}}{3} + \frac{2 A b^{2} x^{\frac{9}{2}}}{9} + \frac{2 B a^{2} x^{\frac{3}{2}}}{3} + \frac{4 B a b x^{\frac{9}{2}}}{9} + \frac{2 B b^{2} x^{\frac{15}{2}}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(B*x**3+A)/x**(5/2),x)

[Out]

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

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Giac [A]  time = 1.10721, size = 72, normalized size = 1.14 \begin{align*} \frac{2}{15} \, B b^{2} x^{\frac{15}{2}} + \frac{4}{9} \, B a b x^{\frac{9}{2}} + \frac{2}{9} \, A b^{2} x^{\frac{9}{2}} + \frac{2}{3} \, B a^{2} x^{\frac{3}{2}} + \frac{4}{3} \, A a b x^{\frac{3}{2}} - \frac{2 \, A a^{2}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^(5/2),x, algorithm="giac")

[Out]

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