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

Optimal. Leaf size=39 \[ \frac {2}{11} x^{11/2} (a B+A b)+\frac {2}{5} a A x^{5/2}+\frac {2}{17} b B x^{17/2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {448} \[ \frac {2}{11} x^{11/2} (a B+A b)+\frac {2}{5} a A x^{5/2}+\frac {2}{17} b B x^{17/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.85 \[ \frac {2}{935} x^{5/2} \left (85 x^3 (a B+A b)+187 a A+55 b B x^6\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(187*a*A + 85*(A*b + a*B)*x^3 + 55*b*B*x^6))/935

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fricas [A]  time = 0.56, size = 32, normalized size = 0.82 \[ \frac {2}{935} \, {\left (55 \, B b x^{8} + 85 \, {\left (B a + A b\right )} x^{5} + 187 \, A a x^{2}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/935*(55*B*b*x^8 + 85*(B*a + A*b)*x^5 + 187*A*a*x^2)*sqrt(x)

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giac [A]  time = 0.18, size = 29, normalized size = 0.74 \[ \frac {2}{17} \, B b x^{\frac {17}{2}} + \frac {2}{11} \, B a x^{\frac {11}{2}} + \frac {2}{11} \, A b x^{\frac {11}{2}} + \frac {2}{5} \, A a x^{\frac {5}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 32, normalized size = 0.82 \[ \frac {2 \left (55 B b \,x^{6}+85 A b \,x^{3}+85 B a \,x^{3}+187 A a \right ) x^{\frac {5}{2}}}{935} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/935*x^(5/2)*(55*B*b*x^6+85*A*b*x^3+85*B*a*x^3+187*A*a)

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maxima [A]  time = 0.45, size = 27, normalized size = 0.69 \[ \frac {2}{17} \, B b x^{\frac {17}{2}} + \frac {2}{11} \, {\left (B a + A b\right )} x^{\frac {11}{2}} + \frac {2}{5} \, A a x^{\frac {5}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 31, normalized size = 0.79 \[ \frac {2\,x^{5/2}\,\left (187\,A\,a+85\,A\,b\,x^3+85\,B\,a\,x^3+55\,B\,b\,x^6\right )}{935} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x^(5/2)*(187*A*a + 85*A*b*x^3 + 85*B*a*x^3 + 55*B*b*x^6))/935

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sympy [A]  time = 6.74, size = 46, normalized size = 1.18 \[ \frac {2 A a x^{\frac {5}{2}}}{5} + \frac {2 A b x^{\frac {11}{2}}}{11} + \frac {2 B a x^{\frac {11}{2}}}{11} + \frac {2 B b x^{\frac {17}{2}}}{17} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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