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

Optimal. Leaf size=63 \[ \frac {2}{5} a^2 A x^{5/2}+\frac {2}{17} b x^{17/2} (2 a B+A b)+\frac {2}{11} a x^{11/2} (a B+2 A b)+\frac {2}{23} b^2 B x^{23/2} \]

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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}{5} a^2 A x^{5/2}+\frac {2}{17} b x^{17/2} (2 a B+A b)+\frac {2}{11} a x^{11/2} (a B+2 A b)+\frac {2}{23} b^2 B x^{23/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*A*x^(5/2))/5 + (2*a*(2*A*b + a*B)*x^(11/2))/11 + (2*b*(A*b + 2*a*B)*x^(17/2))/17 + (2*b^2*B*x^(23/2))/2
3

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 x^{3/2} \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx &=\int \left (a^2 A x^{3/2}+a (2 A b+a B) x^{9/2}+b (A b+2 a B) x^{15/2}+b^2 B x^{21/2}\right ) \, dx\\ &=\frac {2}{5} a^2 A x^{5/2}+\frac {2}{11} a (2 A b+a B) x^{11/2}+\frac {2}{17} b (A b+2 a B) x^{17/2}+\frac {2}{23} b^2 B x^{23/2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 53, normalized size = 0.84 \begin {gather*} \frac {2 x^{5/2} \left (4301 a^2 A+1265 b x^6 (2 a B+A b)+1955 a x^3 (a B+2 A b)+935 b^2 B x^9\right )}{21505} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(4301*a^2*A + 1955*a*(2*A*b + a*B)*x^3 + 1265*b*(A*b + 2*a*B)*x^6 + 935*b^2*B*x^9))/21505

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IntegrateAlgebraic [A]  time = 0.03, size = 69, normalized size = 1.10 \begin {gather*} \frac {2 \left (4301 a^2 A x^{5/2}+1955 a^2 B x^{11/2}+3910 a A b x^{11/2}+2530 a b B x^{17/2}+1265 A b^2 x^{17/2}+935 b^2 B x^{23/2}\right )}{21505} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(4301*a^2*A*x^(5/2) + 3910*a*A*b*x^(11/2) + 1955*a^2*B*x^(11/2) + 1265*A*b^2*x^(17/2) + 2530*a*b*B*x^(17/2)
+ 935*b^2*B*x^(23/2)))/21505

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fricas [A]  time = 0.78, size = 56, normalized size = 0.89 \begin {gather*} \frac {2}{21505} \, {\left (935 \, B b^{2} x^{11} + 1265 \, {\left (2 \, B a b + A b^{2}\right )} x^{8} + 1955 \, {\left (B a^{2} + 2 \, A a b\right )} x^{5} + 4301 \, A a^{2} x^{2}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21505*(935*B*b^2*x^11 + 1265*(2*B*a*b + A*b^2)*x^8 + 1955*(B*a^2 + 2*A*a*b)*x^5 + 4301*A*a^2*x^2)*sqrt(x)

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giac [A]  time = 0.16, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{23} \, B b^{2} x^{\frac {23}{2}} + \frac {4}{17} \, B a b x^{\frac {17}{2}} + \frac {2}{17} \, A b^{2} x^{\frac {17}{2}} + \frac {2}{11} \, B a^{2} x^{\frac {11}{2}} + \frac {4}{11} \, A a b x^{\frac {11}{2}} + \frac {2}{5} \, A a^{2} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/23*B*b^2*x^(23/2) + 4/17*B*a*b*x^(17/2) + 2/17*A*b^2*x^(17/2) + 2/11*B*a^2*x^(11/2) + 4/11*A*a*b*x^(11/2) +
2/5*A*a^2*x^(5/2)

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maple [A]  time = 0.04, size = 56, normalized size = 0.89 \begin {gather*} \frac {2 \left (935 b^{2} B \,x^{9}+1265 A \,b^{2} x^{6}+2530 B a b \,x^{6}+3910 A a b \,x^{3}+1955 B \,a^{2} x^{3}+4301 a^{2} A \right ) x^{\frac {5}{2}}}{21505} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21505*x^(5/2)*(935*B*b^2*x^9+1265*A*b^2*x^6+2530*B*a*b*x^6+3910*A*a*b*x^3+1955*B*a^2*x^3+4301*A*a^2)

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maxima [A]  time = 0.50, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{23} \, B b^{2} x^{\frac {23}{2}} + \frac {2}{17} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {17}{2}} + \frac {2}{11} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {11}{2}} + \frac {2}{5} \, A a^{2} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/23*B*b^2*x^(23/2) + 2/17*(2*B*a*b + A*b^2)*x^(17/2) + 2/11*(B*a^2 + 2*A*a*b)*x^(11/2) + 2/5*A*a^2*x^(5/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \begin {gather*} x^{11/2}\,\left (\frac {2\,B\,a^2}{11}+\frac {4\,A\,b\,a}{11}\right )+x^{17/2}\,\left (\frac {2\,A\,b^2}{17}+\frac {4\,B\,a\,b}{17}\right )+\frac {2\,A\,a^2\,x^{5/2}}{5}+\frac {2\,B\,b^2\,x^{23/2}}{23} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(11/2)*((2*B*a^2)/11 + (4*A*a*b)/11) + x^(17/2)*((2*A*b^2)/17 + (4*B*a*b)/17) + (2*A*a^2*x^(5/2))/5 + (2*B*b
^2*x^(23/2))/23

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sympy [A]  time = 21.96, size = 80, normalized size = 1.27 \begin {gather*} \frac {2 A a^{2} x^{\frac {5}{2}}}{5} + \frac {4 A a b x^{\frac {11}{2}}}{11} + \frac {2 A b^{2} x^{\frac {17}{2}}}{17} + \frac {2 B a^{2} x^{\frac {11}{2}}}{11} + \frac {4 B a b x^{\frac {17}{2}}}{17} + \frac {2 B b^{2} x^{\frac {23}{2}}}{23} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*x**(5/2)/5 + 4*A*a*b*x**(11/2)/11 + 2*A*b**2*x**(17/2)/17 + 2*B*a**2*x**(11/2)/11 + 4*B*a*b*x**(17/2)
/17 + 2*B*b**2*x**(23/2)/23

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