3.134 \(\int \sqrt{x} (a+b x^3) (A+B x^3) \, dx\)

Optimal. Leaf size=39 \[ \frac{2}{9} x^{9/2} (a B+A b)+\frac{2}{3} a A x^{3/2}+\frac{2}{15} b B x^{15/2} \]

[Out]

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

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Rubi [A]  time = 0.0160388, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {448} \[ \frac{2}{9} x^{9/2} (a B+A b)+\frac{2}{3} a A x^{3/2}+\frac{2}{15} b B x^{15/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*x^3)*(A + B*x^3),x]

[Out]

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

Mathematica [A]  time = 0.0131893, size = 33, normalized size = 0.85 \[ \frac{2}{45} x^{3/2} \left (5 x^3 (a B+A b)+15 a A+3 b B x^6\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a + b*x^3)*(A + B*x^3),x]

[Out]

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

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Maple [A]  time = 0.003, size = 32, normalized size = 0.8 \begin{align*}{\frac{6\,bB{x}^{6}+10\,A{x}^{3}b+10\,B{x}^{3}a+30\,Aa}{45}{x}^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)*(B*x^3+A)*x^(1/2),x)

[Out]

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

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Maxima [A]  time = 0.960754, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{15} \, B b x^{\frac{15}{2}} + \frac{2}{9} \,{\left (B a + A b\right )} x^{\frac{9}{2}} + \frac{2}{3} \, A a x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)*x^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.69488, size = 77, normalized size = 1.97 \begin{align*} \frac{2}{45} \,{\left (3 \, B b x^{7} + 5 \,{\left (B a + A b\right )} x^{4} + 15 \, A a x\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)*x^(1/2),x, algorithm="fricas")

[Out]

2/45*(3*B*b*x^7 + 5*(B*a + A*b)*x^4 + 15*A*a*x)*sqrt(x)

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Sympy [A]  time = 2.52409, size = 46, normalized size = 1.18 \begin{align*} \frac{2 A a x^{\frac{3}{2}}}{3} + \frac{2 A b x^{\frac{9}{2}}}{9} + \frac{2 B a x^{\frac{9}{2}}}{9} + \frac{2 B b x^{\frac{15}{2}}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)*(B*x**3+A)*x**(1/2),x)

[Out]

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

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Giac [A]  time = 1.09066, size = 39, normalized size = 1. \begin{align*} \frac{2}{15} \, B b x^{\frac{15}{2}} + \frac{2}{9} \, B a x^{\frac{9}{2}} + \frac{2}{9} \, A b x^{\frac{9}{2}} + \frac{2}{3} \, A a x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)*x^(1/2),x, algorithm="giac")

[Out]

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