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

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 63, 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}{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} \end {gather*}

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 = 0.07, size = 57, normalized size = 0.90 \begin {gather*} \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}} \end {gather*}

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))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.04, size = 59, normalized size = 0.94 \begin {gather*} \frac {2 \left (-15 a^2 A+15 a^2 B x^3+30 a A b x^3+10 a b B x^6+5 A b^2 x^6+3 b^2 B x^9\right )}{45 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 53, normalized size = 0.84 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 53, normalized size = 0.84 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 56, normalized size = 0.89 \begin {gather*} -\frac {2 \left (-3 b^{2} B \,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^{2} A \right )}{45 x^{\frac {3}{2}}} \end {gather*}

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)

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 51, normalized size = 0.81 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \begin {gather*} x^{3/2}\,\left (\frac {2\,B\,a^2}{3}+\frac {4\,A\,b\,a}{3}\right )+x^{9/2}\,\left (\frac {2\,A\,b^2}{9}+\frac {4\,B\,a\,b}{9}\right )-\frac {2\,A\,a^2}{3\,x^{3/2}}+\frac {2\,B\,b^2\,x^{15/2}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 11.74, size = 80, normalized size = 1.27 \begin {gather*} - \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 {gather*}

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

________________________________________________________________________________________