3.2.38 \(\int \sqrt {x} (a+b x^3)^2 (A+B x^3) \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{3} a^2 A x^{3/2}+\frac {2}{15} b x^{15/2} (2 a B+A b)+\frac {2}{9} a x^{9/2} (a B+2 A b)+\frac {2}{21} b^2 B x^{21/2} \]

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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}{3} a^2 A x^{3/2}+\frac {2}{15} b x^{15/2} (2 a B+A b)+\frac {2}{9} a x^{9/2} (a B+2 A b)+\frac {2}{21} b^2 B x^{21/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.11, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{315} x^{3/2} \left (105 a^2 A+21 b x^6 (2 a B+A b)+35 a x^3 (a B+2 A b)+15 b^2 B x^9\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(105*a^2*A + 35*a*(2*A*b + a*B)*x^3 + 21*b*(A*b + 2*a*B)*x^6 + 15*b^2*B*x^9))/315

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IntegrateAlgebraic [A]  time = 0.04, size = 69, normalized size = 1.10 \begin {gather*} \frac {2}{315} \left (105 a^2 A x^{3/2}+35 a^2 B x^{9/2}+70 a A b x^{9/2}+42 a b B x^{15/2}+21 A b^2 x^{15/2}+15 b^2 B x^{21/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(105*a^2*A*x^(3/2) + 70*a*A*b*x^(9/2) + 35*a^2*B*x^(9/2) + 21*A*b^2*x^(15/2) + 42*a*b*B*x^(15/2) + 15*b^2*B
*x^(21/2)))/315

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fricas [A]  time = 0.63, size = 54, normalized size = 0.86 \begin {gather*} \frac {2}{315} \, {\left (15 \, B b^{2} x^{10} + 21 \, {\left (2 \, B a b + A b^{2}\right )} x^{7} + 35 \, {\left (B a^{2} + 2 \, A a b\right )} x^{4} + 105 \, A a^{2} x\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(15*B*b^2*x^10 + 21*(2*B*a*b + A*b^2)*x^7 + 35*(B*a^2 + 2*A*a*b)*x^4 + 105*A*a^2*x)*sqrt(x)

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giac [A]  time = 0.16, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{21} \, B b^{2} x^{\frac {21}{2}} + \frac {4}{15} \, B a b x^{\frac {15}{2}} + \frac {2}{15} \, A b^{2} x^{\frac {15}{2}} + \frac {2}{9} \, B a^{2} x^{\frac {9}{2}} + \frac {4}{9} \, A a b x^{\frac {9}{2}} + \frac {2}{3} \, A a^{2} 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^(1/2),x, algorithm="giac")

[Out]

2/21*B*b^2*x^(21/2) + 4/15*B*a*b*x^(15/2) + 2/15*A*b^2*x^(15/2) + 2/9*B*a^2*x^(9/2) + 4/9*A*a*b*x^(9/2) + 2/3*
A*a^2*x^(3/2)

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maple [A]  time = 0.04, size = 56, normalized size = 0.89 \begin {gather*} \frac {2 \left (15 b^{2} B \,x^{9}+21 A \,b^{2} x^{6}+42 B a b \,x^{6}+70 A a b \,x^{3}+35 B \,a^{2} x^{3}+105 a^{2} A \right ) x^{\frac {3}{2}}}{315} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*x^(3/2)*(15*B*b^2*x^9+21*A*b^2*x^6+42*B*a*b*x^6+70*A*a*b*x^3+35*B*a^2*x^3+105*A*a^2)

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maxima [A]  time = 0.54, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{21} \, B b^{2} x^{\frac {21}{2}} + \frac {2}{15} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {15}{2}} + \frac {2}{9} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {9}{2}} + \frac {2}{3} \, A a^{2} 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^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \begin {gather*} x^{9/2}\,\left (\frac {2\,B\,a^2}{9}+\frac {4\,A\,b\,a}{9}\right )+x^{15/2}\,\left (\frac {2\,A\,b^2}{15}+\frac {4\,B\,a\,b}{15}\right )+\frac {2\,A\,a^2\,x^{3/2}}{3}+\frac {2\,B\,b^2\,x^{21/2}}{21} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(9/2)*((2*B*a^2)/9 + (4*A*a*b)/9) + x^(15/2)*((2*A*b^2)/15 + (4*B*a*b)/15) + (2*A*a^2*x^(3/2))/3 + (2*B*b^2*
x^(21/2))/21

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sympy [A]  time = 5.31, size = 80, normalized size = 1.27 \begin {gather*} \frac {2 A a^{2} x^{\frac {3}{2}}}{3} + \frac {4 A a b x^{\frac {9}{2}}}{9} + \frac {2 A b^{2} x^{\frac {15}{2}}}{15} + \frac {2 B a^{2} x^{\frac {9}{2}}}{9} + \frac {4 B a b x^{\frac {15}{2}}}{15} + \frac {2 B b^{2} x^{\frac {21}{2}}}{21} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*x**(3/2)/3 + 4*A*a*b*x**(9/2)/9 + 2*A*b**2*x**(15/2)/15 + 2*B*a**2*x**(9/2)/9 + 4*B*a*b*x**(15/2)/15
+ 2*B*b**2*x**(21/2)/21

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