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

Optimal. Leaf size=85 \[ \frac{2}{3} a^2 x^{3/2} (a B+3 A b)-\frac{2 a^3 A}{3 x^{3/2}}+\frac{2}{15} b^2 x^{15/2} (3 a B+A b)+\frac{2}{3} a b x^{9/2} (a B+A b)+\frac{2}{21} b^3 B x^{21/2} \]

[Out]

(-2*a^3*A)/(3*x^(3/2)) + (2*a^2*(3*A*b + a*B)*x^(3/2))/3 + (2*a*b*(A*b + a*B)*x^(9/2))/3 + (2*b^2*(A*b + 3*a*B
)*x^(15/2))/15 + (2*b^3*B*x^(21/2))/21

________________________________________________________________________________________

Rubi [A]  time = 0.0448236, antiderivative size = 85, 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} \[ \frac{2}{3} a^2 x^{3/2} (a B+3 A b)-\frac{2 a^3 A}{3 x^{3/2}}+\frac{2}{15} b^2 x^{15/2} (3 a B+A b)+\frac{2}{3} a b x^{9/2} (a B+A b)+\frac{2}{21} b^3 B x^{21/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*A)/(3*x^(3/2)) + (2*a^2*(3*A*b + a*B)*x^(3/2))/3 + (2*a*b*(A*b + a*B)*x^(9/2))/3 + (2*b^2*(A*b + 3*a*B
)*x^(15/2))/15 + (2*b^3*B*x^(21/2))/21

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 )^3 \left (A+B x^3\right )}{x^{5/2}} \, dx &=\int \left (\frac{a^3 A}{x^{5/2}}+a^2 (3 A b+a B) \sqrt{x}+3 a b (A b+a B) x^{7/2}+b^2 (A b+3 a B) x^{13/2}+b^3 B x^{19/2}\right ) \, dx\\ &=-\frac{2 a^3 A}{3 x^{3/2}}+\frac{2}{3} a^2 (3 A b+a B) x^{3/2}+\frac{2}{3} a b (A b+a B) x^{9/2}+\frac{2}{15} b^2 (A b+3 a B) x^{15/2}+\frac{2}{21} b^3 B x^{21/2}\\ \end{align*}

Mathematica [A]  time = 0.0227491, size = 77, normalized size = 0.91 \[ \frac{2 \left (35 a^2 b x^3 \left (3 A+B x^3\right )-35 a^3 \left (A-B x^3\right )+7 a b^2 x^6 \left (5 A+3 B x^3\right )+b^3 x^9 \left (7 A+5 B x^3\right )\right )}{105 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-35*a^3*(A - B*x^3) + 35*a^2*b*x^3*(3*A + B*x^3) + 7*a*b^2*x^6*(5*A + 3*B*x^3) + b^3*x^9*(7*A + 5*B*x^3)))
/(105*x^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 80, normalized size = 0.9 \begin{align*} -{\frac{-10\,{b}^{3}B{x}^{12}-14\,{x}^{9}{b}^{3}A-42\,{x}^{9}a{b}^{2}B-70\,{x}^{6}a{b}^{2}A-70\,{x}^{6}{a}^{2}bB-210\,{x}^{3}A{a}^{2}b-70\,{x}^{3}B{a}^{3}+70\,{a}^{3}A}{105}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(-5*B*b^3*x^12-7*A*b^3*x^9-21*B*a*b^2*x^9-35*A*a*b^2*x^6-35*B*a^2*b*x^6-105*A*a^2*b*x^3-35*B*a^3*x^3+35
*A*a^3)/x^(3/2)

________________________________________________________________________________________

Maxima [A]  time = 0.966754, size = 99, normalized size = 1.16 \begin{align*} \frac{2}{21} \, B b^{3} x^{\frac{21}{2}} + \frac{2}{15} \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{\frac{15}{2}} + \frac{2}{3} \,{\left (B a^{2} b + A a b^{2}\right )} x^{\frac{9}{2}} - \frac{2 \, A a^{3}}{3 \, x^{\frac{3}{2}}} + \frac{2}{3} \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*B*b^3*x^(21/2) + 2/15*(3*B*a*b^2 + A*b^3)*x^(15/2) + 2/3*(B*a^2*b + A*a*b^2)*x^(9/2) - 2/3*A*a^3/x^(3/2)
+ 2/3*(B*a^3 + 3*A*a^2*b)*x^(3/2)

________________________________________________________________________________________

Fricas [A]  time = 1.84283, size = 171, normalized size = 2.01 \begin{align*} \frac{2 \,{\left (5 \, B b^{3} x^{12} + 7 \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{9} + 35 \,{\left (B a^{2} b + A a b^{2}\right )} x^{6} - 35 \, A a^{3} + 35 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3}\right )}}{105 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(5*B*b^3*x^12 + 7*(3*B*a*b^2 + A*b^3)*x^9 + 35*(B*a^2*b + A*a*b^2)*x^6 - 35*A*a^3 + 35*(B*a^3 + 3*A*a^2*
b)*x^3)/x^(3/2)

________________________________________________________________________________________

Sympy [A]  time = 35.2066, size = 112, normalized size = 1.32 \begin{align*} - \frac{2 A a^{3}}{3 x^{\frac{3}{2}}} + 2 A a^{2} b x^{\frac{3}{2}} + \frac{2 A a b^{2} x^{\frac{9}{2}}}{3} + \frac{2 A b^{3} x^{\frac{15}{2}}}{15} + \frac{2 B a^{3} x^{\frac{3}{2}}}{3} + \frac{2 B a^{2} b x^{\frac{9}{2}}}{3} + \frac{2 B a b^{2} x^{\frac{15}{2}}}{5} + \frac{2 B b^{3} x^{\frac{21}{2}}}{21} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A*a**3/(3*x**(3/2)) + 2*A*a**2*b*x**(3/2) + 2*A*a*b**2*x**(9/2)/3 + 2*A*b**3*x**(15/2)/15 + 2*B*a**3*x**(3/
2)/3 + 2*B*a**2*b*x**(9/2)/3 + 2*B*a*b**2*x**(15/2)/5 + 2*B*b**3*x**(21/2)/21

________________________________________________________________________________________

Giac [A]  time = 1.11934, size = 104, normalized size = 1.22 \begin{align*} \frac{2}{21} \, B b^{3} x^{\frac{21}{2}} + \frac{2}{5} \, B a b^{2} x^{\frac{15}{2}} + \frac{2}{15} \, A b^{3} x^{\frac{15}{2}} + \frac{2}{3} \, B a^{2} b x^{\frac{9}{2}} + \frac{2}{3} \, A a b^{2} x^{\frac{9}{2}} + \frac{2}{3} \, B a^{3} x^{\frac{3}{2}} + 2 \, A a^{2} b x^{\frac{3}{2}} - \frac{2 \, A a^{3}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*B*b^3*x^(21/2) + 2/5*B*a*b^2*x^(15/2) + 2/15*A*b^3*x^(15/2) + 2/3*B*a^2*b*x^(9/2) + 2/3*A*a*b^2*x^(9/2) +
2/3*B*a^3*x^(3/2) + 2*A*a^2*b*x^(3/2) - 2/3*A*a^3/x^(3/2)