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

Optimal. Leaf size=39 \[ \frac {2}{11} x^{11/2} (a B+A b)+\frac {2}{7} a A x^{7/2}+\frac {2}{15} b B x^{15/2} \]

[Out]

2/7*a*A*x^(7/2)+2/11*(A*b+B*a)*x^(11/2)+2/15*b*B*x^(15/2)

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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} \[ \frac {2}{11} x^{11/2} (a B+A b)+\frac {2}{7} a A x^{7/2}+\frac {2}{15} b B x^{15/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*x^(7/2))/7 + (2*(A*b + a*B)*x^(11/2))/11 + (2*b*B*x^(15/2))/15

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^2\right ) \left (A+B x^2\right ) \, dx &=\int \left (a A x^{5/2}+(A b+a B) x^{9/2}+b B x^{13/2}\right ) \, dx\\ &=\frac {2}{7} a A x^{7/2}+\frac {2}{11} (A b+a B) x^{11/2}+\frac {2}{15} b B x^{15/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.85 \[ \frac {2 x^{7/2} \left (105 x^2 (a B+A b)+165 a A+77 b B x^4\right )}{1155} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(7/2)*(165*a*A + 105*(A*b + a*B)*x^2 + 77*b*B*x^4))/1155

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fricas [A]  time = 0.46, size = 32, normalized size = 0.82 \[ \frac {2}{1155} \, {\left (77 \, B b x^{7} + 105 \, {\left (B a + A b\right )} x^{5} + 165 \, A a x^{3}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(77*B*b*x^7 + 105*(B*a + A*b)*x^5 + 165*A*a*x^3)*sqrt(x)

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giac [A]  time = 0.32, size = 29, normalized size = 0.74 \[ \frac {2}{15} \, B b x^{\frac {15}{2}} + \frac {2}{11} \, B a x^{\frac {11}{2}} + \frac {2}{11} \, A b x^{\frac {11}{2}} + \frac {2}{7} \, A a x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*B*b*x^(15/2) + 2/11*B*a*x^(11/2) + 2/11*A*b*x^(11/2) + 2/7*A*a*x^(7/2)

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maple [A]  time = 0.01, size = 32, normalized size = 0.82 \[ \frac {2 \left (77 B b \,x^{4}+105 A b \,x^{2}+105 B a \,x^{2}+165 A a \right ) x^{\frac {7}{2}}}{1155} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*x^(7/2)*(77*B*b*x^4+105*A*b*x^2+105*B*a*x^2+165*A*a)

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maxima [A]  time = 1.06, size = 27, normalized size = 0.69 \[ \frac {2}{15} \, B b x^{\frac {15}{2}} + \frac {2}{11} \, {\left (B a + A b\right )} x^{\frac {11}{2}} + \frac {2}{7} \, A a x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*B*b*x^(15/2) + 2/11*(B*a + A*b)*x^(11/2) + 2/7*A*a*x^(7/2)

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mupad [B]  time = 0.19, size = 31, normalized size = 0.79 \[ \frac {2\,x^{7/2}\,\left (165\,A\,a+105\,A\,b\,x^2+105\,B\,a\,x^2+77\,B\,b\,x^4\right )}{1155} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x^(7/2)*(165*A*a + 105*A*b*x^2 + 105*B*a*x^2 + 77*B*b*x^4))/1155

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sympy [A]  time = 5.30, size = 46, normalized size = 1.18 \[ \frac {2 A a x^{\frac {7}{2}}}{7} + \frac {2 A b x^{\frac {11}{2}}}{11} + \frac {2 B a x^{\frac {11}{2}}}{11} + \frac {2 B b x^{\frac {15}{2}}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a*x**(7/2)/7 + 2*A*b*x**(11/2)/11 + 2*B*a*x**(11/2)/11 + 2*B*b*x**(15/2)/15

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