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

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

[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

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Rubi [A]  time = 0.0299877, antiderivative size = 63, 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} \[ \frac{2}{5} a^2 A x^{5/2}+\frac{2}{17} b x^{17/2} (2 a B+A b)+\frac{2}{11} a x^{11/2} (a B+2 A b)+\frac{2}{23} b^2 B x^{23/2} \]

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*}

Mathematica [A]  time = 0.027907, size = 53, normalized size = 0.84 \[ \frac{2 x^{5/2} \left (4301 a^2 A+1265 b x^6 (2 a B+A b)+1955 a x^3 (a B+2 A b)+935 b^2 B x^9\right )}{21505} \]

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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Maple [A]  time = 0.009, size = 56, normalized size = 0.9 \begin{align*}{\frac{1870\,B{b}^{2}{x}^{9}+2530\,A{b}^{2}{x}^{6}+5060\,B{x}^{6}ab+7820\,aAb{x}^{3}+3910\,B{x}^{3}{a}^{2}+8602\,{a}^{2}A}{21505}{x}^{{\frac{5}{2}}}} \end{align*}

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 = 0.937444, size = 69, normalized size = 1.1 \begin{align*} \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{align*}

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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Fricas [A]  time = 1.63661, size = 147, normalized size = 2.33 \begin{align*} \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{align*}

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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Sympy [A]  time = 18.7176, size = 80, normalized size = 1.27 \begin{align*} \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{align*}

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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Giac [A]  time = 1.12972, size = 72, normalized size = 1.14 \begin{align*} \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{align*}

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)