3.132 \(\int x^{5/2} (a+b x^3) (A+B x^3) \, dx\)

Optimal. Leaf size=39 \[ \frac {2}{13} x^{13/2} (a B+A b)+\frac {2}{7} a A x^{7/2}+\frac {2}{19} b B x^{19/2} \]

[Out]

2/7*a*A*x^(7/2)+2/13*(A*b+B*a)*x^(13/2)+2/19*b*B*x^(19/2)

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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} \[ \frac {2}{13} x^{13/2} (a B+A b)+\frac {2}{7} a A x^{7/2}+\frac {2}{19} b B x^{19/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(a + b*x^3)*(A + B*x^3),x]

[Out]

(2*a*A*x^(7/2))/7 + (2*(A*b + a*B)*x^(13/2))/13 + (2*b*B*x^(19/2))/19

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^{5/2} \left (a+b x^3\right ) \left (A+B x^3\right ) \, dx &=\int \left (a A x^{5/2}+(A b+a B) x^{11/2}+b B x^{17/2}\right ) \, dx\\ &=\frac {2}{7} a A x^{7/2}+\frac {2}{13} (A b+a B) x^{13/2}+\frac {2}{19} b B x^{19/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.85 \[ \frac {2 x^{7/2} \left (133 x^3 (a B+A b)+247 a A+91 b B x^6\right )}{1729} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(a + b*x^3)*(A + B*x^3),x]

[Out]

(2*x^(7/2)*(247*a*A + 133*(A*b + a*B)*x^3 + 91*b*B*x^6))/1729

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fricas [A]  time = 0.61, size = 32, normalized size = 0.82 \[ \frac {2}{1729} \, {\left (91 \, B b x^{9} + 133 \, {\left (B a + A b\right )} x^{6} + 247 \, A a x^{3}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x^3+a)*(B*x^3+A),x, algorithm="fricas")

[Out]

2/1729*(91*B*b*x^9 + 133*(B*a + A*b)*x^6 + 247*A*a*x^3)*sqrt(x)

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giac [A]  time = 0.15, size = 29, normalized size = 0.74 \[ \frac {2}{19} \, B b x^{\frac {19}{2}} + \frac {2}{13} \, B a x^{\frac {13}{2}} + \frac {2}{13} \, A b x^{\frac {13}{2}} + \frac {2}{7} \, A a x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x^3+a)*(B*x^3+A),x, algorithm="giac")

[Out]

2/19*B*b*x^(19/2) + 2/13*B*a*x^(13/2) + 2/13*A*b*x^(13/2) + 2/7*A*a*x^(7/2)

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maple [A]  time = 0.04, size = 32, normalized size = 0.82 \[ \frac {2 \left (91 B b \,x^{6}+133 A b \,x^{3}+133 B a \,x^{3}+247 A a \right ) x^{\frac {7}{2}}}{1729} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x^3+a)*(B*x^3+A),x)

[Out]

2/1729*x^(7/2)*(91*B*b*x^6+133*A*b*x^3+133*B*a*x^3+247*A*a)

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maxima [A]  time = 0.47, size = 27, normalized size = 0.69 \[ \frac {2}{19} \, B b x^{\frac {19}{2}} + \frac {2}{13} \, {\left (B a + A b\right )} x^{\frac {13}{2}} + \frac {2}{7} \, A a x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x^3+a)*(B*x^3+A),x, algorithm="maxima")

[Out]

2/19*B*b*x^(19/2) + 2/13*(B*a + A*b)*x^(13/2) + 2/7*A*a*x^(7/2)

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mupad [B]  time = 2.56, size = 31, normalized size = 0.79 \[ \frac {2\,x^{7/2}\,\left (247\,A\,a+133\,A\,b\,x^3+133\,B\,a\,x^3+91\,B\,b\,x^6\right )}{1729} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(A + B*x^3)*(a + b*x^3),x)

[Out]

(2*x^(7/2)*(247*A*a + 133*A*b*x^3 + 133*B*a*x^3 + 91*B*b*x^6))/1729

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sympy [A]  time = 12.71, size = 46, normalized size = 1.18 \[ \frac {2 A a x^{\frac {7}{2}}}{7} + \frac {2 A b x^{\frac {13}{2}}}{13} + \frac {2 B a x^{\frac {13}{2}}}{13} + \frac {2 B b x^{\frac {19}{2}}}{19} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(b*x**3+a)*(B*x**3+A),x)

[Out]

2*A*a*x**(7/2)/7 + 2*A*b*x**(13/2)/13 + 2*B*a*x**(13/2)/13 + 2*B*b*x**(19/2)/19

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