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

Optimal. Leaf size=63 \[ \frac {2}{5} a^2 A x^{5/2}+\frac {2}{13} b x^{13/2} (2 a B+A b)+\frac {2}{9} a x^{9/2} (a B+2 A b)+\frac {2}{17} b^2 B x^{17/2} \]

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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}{13} b x^{13/2} (2 a B+A b)+\frac {2}{9} a x^{9/2} (a B+2 A b)+\frac {2}{17} b^2 B x^{17/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*A*x^(5/2))/5 + (2*a*(2*A*b + a*B)*x^(9/2))/9 + (2*b*(A*b + 2*a*B)*x^(13/2))/13 + (2*b^2*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^2\right )^2 \left (A+B x^2\right ) \, dx &=\int \left (a^2 A x^{3/2}+a (2 A b+a B) x^{7/2}+b (A b+2 a B) x^{11/2}+b^2 B x^{15/2}\right ) \, dx\\ &=\frac {2}{5} a^2 A x^{5/2}+\frac {2}{9} a (2 A b+a B) x^{9/2}+\frac {2}{13} b (A b+2 a B) x^{13/2}+\frac {2}{17} b^2 B x^{17/2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.84 \begin {gather*} \frac {2 x^{5/2} \left (1989 a^2 A+765 b x^4 (2 a B+A b)+1105 a x^2 (a B+2 A b)+585 b^2 B x^6\right )}{9945} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(1989*a^2*A + 1105*a*(2*A*b + a*B)*x^2 + 765*b*(A*b + 2*a*B)*x^4 + 585*b^2*B*x^6))/9945

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IntegrateAlgebraic [A]  time = 0.03, size = 69, normalized size = 1.10 \begin {gather*} \frac {2 \left (1989 a^2 A x^{5/2}+1105 a^2 B x^{9/2}+2210 a A b x^{9/2}+1530 a b B x^{13/2}+765 A b^2 x^{13/2}+585 b^2 B x^{17/2}\right )}{9945} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1989*a^2*A*x^(5/2) + 2210*a*A*b*x^(9/2) + 1105*a^2*B*x^(9/2) + 765*A*b^2*x^(13/2) + 1530*a*b*B*x^(13/2) +
585*b^2*B*x^(17/2)))/9945

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fricas [A]  time = 1.33, size = 56, normalized size = 0.89 \begin {gather*} \frac {2}{9945} \, {\left (585 \, B b^{2} x^{8} + 765 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + 1989 \, A a^{2} x^{2} + 1105 \, {\left (B a^{2} + 2 \, A a b\right )} x^{4}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9945*(585*B*b^2*x^8 + 765*(2*B*a*b + A*b^2)*x^6 + 1989*A*a^2*x^2 + 1105*(B*a^2 + 2*A*a*b)*x^4)*sqrt(x)

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giac [A]  time = 0.31, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{17} \, B b^{2} x^{\frac {17}{2}} + \frac {4}{13} \, B a b x^{\frac {13}{2}} + \frac {2}{13} \, A b^{2} x^{\frac {13}{2}} + \frac {2}{9} \, B a^{2} x^{\frac {9}{2}} + \frac {4}{9} \, A a b x^{\frac {9}{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^2+a)^2*(B*x^2+A),x, algorithm="giac")

[Out]

2/17*B*b^2*x^(17/2) + 4/13*B*a*b*x^(13/2) + 2/13*A*b^2*x^(13/2) + 2/9*B*a^2*x^(9/2) + 4/9*A*a*b*x^(9/2) + 2/5*
A*a^2*x^(5/2)

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maple [A]  time = 0.01, size = 56, normalized size = 0.89 \begin {gather*} \frac {2 \left (585 B \,b^{2} x^{6}+765 A \,b^{2} x^{4}+1530 B a b \,x^{4}+2210 A a b \,x^{2}+1105 B \,a^{2} x^{2}+1989 a^{2} A \right ) x^{\frac {5}{2}}}{9945} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9945*x^(5/2)*(585*B*b^2*x^6+765*A*b^2*x^4+1530*B*a*b*x^4+2210*A*a*b*x^2+1105*B*a^2*x^2+1989*A*a^2)

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maxima [A]  time = 1.08, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{17} \, B b^{2} x^{\frac {17}{2}} + \frac {2}{13} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {13}{2}} + \frac {2}{5} \, A a^{2} x^{\frac {5}{2}} + \frac {2}{9} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {9}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/17*B*b^2*x^(17/2) + 2/13*(2*B*a*b + A*b^2)*x^(13/2) + 2/5*A*a^2*x^(5/2) + 2/9*(B*a^2 + 2*A*a*b)*x^(9/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \begin {gather*} x^{9/2}\,\left (\frac {2\,B\,a^2}{9}+\frac {4\,A\,b\,a}{9}\right )+x^{13/2}\,\left (\frac {2\,A\,b^2}{13}+\frac {4\,B\,a\,b}{13}\right )+\frac {2\,A\,a^2\,x^{5/2}}{5}+\frac {2\,B\,b^2\,x^{17/2}}{17} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(9/2)*((2*B*a^2)/9 + (4*A*a*b)/9) + x^(13/2)*((2*A*b^2)/13 + (4*B*a*b)/13) + (2*A*a^2*x^(5/2))/5 + (2*B*b^2*
x^(17/2))/17

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sympy [A]  time = 5.71, 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 {9}{2}}}{9} + \frac {2 A b^{2} x^{\frac {13}{2}}}{13} + \frac {2 B a^{2} x^{\frac {9}{2}}}{9} + \frac {4 B a b x^{\frac {13}{2}}}{13} + \frac {2 B b^{2} x^{\frac {17}{2}}}{17} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*x**(5/2)/5 + 4*A*a*b*x**(9/2)/9 + 2*A*b**2*x**(13/2)/13 + 2*B*a**2*x**(9/2)/9 + 4*B*a*b*x**(13/2)/13
+ 2*B*b**2*x**(17/2)/17

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