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

Optimal. Leaf size=63 \[ \frac {2}{7} a^2 A x^{7/2}+\frac {2}{15} b x^{15/2} (2 a B+A b)+\frac {2}{11} a x^{11/2} (a B+2 A b)+\frac {2}{19} b^2 B x^{19/2} \]

[Out]

2/7*a^2*A*x^(7/2)+2/11*a*(2*A*b+B*a)*x^(11/2)+2/15*b*(A*b+2*B*a)*x^(15/2)+2/19*b^2*B*x^(19/2)

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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} \[ \frac {2}{7} a^2 A x^{7/2}+\frac {2}{15} b x^{15/2} (2 a B+A b)+\frac {2}{11} a x^{11/2} (a B+2 A b)+\frac {2}{19} b^2 B x^{19/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*A*x^(7/2))/7 + (2*a*(2*A*b + a*B)*x^(11/2))/11 + (2*b*(A*b + 2*a*B)*x^(15/2))/15 + (2*b^2*B*x^(19/2))/1
9

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 )^2 \left (A+B x^2\right ) \, dx &=\int \left (a^2 A x^{5/2}+a (2 A b+a B) x^{9/2}+b (A b+2 a B) x^{13/2}+b^2 B x^{17/2}\right ) \, dx\\ &=\frac {2}{7} a^2 A x^{7/2}+\frac {2}{11} a (2 A b+a B) x^{11/2}+\frac {2}{15} b (A b+2 a B) x^{15/2}+\frac {2}{19} b^2 B x^{19/2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 63, normalized size = 1.00 \[ \frac {2}{7} a^2 A x^{7/2}+\frac {2}{15} b x^{15/2} (2 a B+A b)+\frac {2}{11} a x^{11/2} (a B+2 A b)+\frac {2}{19} b^2 B x^{19/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*A*x^(7/2))/7 + (2*a*(2*A*b + a*B)*x^(11/2))/11 + (2*b*(A*b + 2*a*B)*x^(15/2))/15 + (2*b^2*B*x^(19/2))/1
9

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fricas [A]  time = 0.43, size = 56, normalized size = 0.89 \[ \frac {2}{21945} \, {\left (1155 \, B b^{2} x^{9} + 1463 \, {\left (2 \, B a b + A b^{2}\right )} x^{7} + 3135 \, A a^{2} x^{3} + 1995 \, {\left (B a^{2} + 2 \, A a b\right )} x^{5}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21945*(1155*B*b^2*x^9 + 1463*(2*B*a*b + A*b^2)*x^7 + 3135*A*a^2*x^3 + 1995*(B*a^2 + 2*A*a*b)*x^5)*sqrt(x)

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giac [A]  time = 0.29, size = 53, normalized size = 0.84 \[ \frac {2}{19} \, B b^{2} x^{\frac {19}{2}} + \frac {4}{15} \, B a b x^{\frac {15}{2}} + \frac {2}{15} \, A b^{2} x^{\frac {15}{2}} + \frac {2}{11} \, B a^{2} x^{\frac {11}{2}} + \frac {4}{11} \, A a b x^{\frac {11}{2}} + \frac {2}{7} \, A a^{2} x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/19*B*b^2*x^(19/2) + 4/15*B*a*b*x^(15/2) + 2/15*A*b^2*x^(15/2) + 2/11*B*a^2*x^(11/2) + 4/11*A*a*b*x^(11/2) +
2/7*A*a^2*x^(7/2)

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maple [A]  time = 0.01, size = 56, normalized size = 0.89 \[ \frac {2 \left (1155 B \,b^{2} x^{6}+1463 A \,b^{2} x^{4}+2926 B a b \,x^{4}+3990 A a b \,x^{2}+1995 B \,a^{2} x^{2}+3135 a^{2} A \right ) x^{\frac {7}{2}}}{21945} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21945*x^(7/2)*(1155*B*b^2*x^6+1463*A*b^2*x^4+2926*B*a*b*x^4+3990*A*a*b*x^2+1995*B*a^2*x^2+3135*A*a^2)

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maxima [A]  time = 1.10, size = 51, normalized size = 0.81 \[ \frac {2}{19} \, B b^{2} x^{\frac {19}{2}} + \frac {2}{15} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {15}{2}} + \frac {2}{7} \, A a^{2} x^{\frac {7}{2}} + \frac {2}{11} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {11}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/19*B*b^2*x^(19/2) + 2/15*(2*B*a*b + A*b^2)*x^(15/2) + 2/7*A*a^2*x^(7/2) + 2/11*(B*a^2 + 2*A*a*b)*x^(11/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \[ x^{11/2}\,\left (\frac {2\,B\,a^2}{11}+\frac {4\,A\,b\,a}{11}\right )+x^{15/2}\,\left (\frac {2\,A\,b^2}{15}+\frac {4\,B\,a\,b}{15}\right )+\frac {2\,A\,a^2\,x^{7/2}}{7}+\frac {2\,B\,b^2\,x^{19/2}}{19} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(11/2)*((2*B*a^2)/11 + (4*A*a*b)/11) + x^(15/2)*((2*A*b^2)/15 + (4*B*a*b)/15) + (2*A*a^2*x^(7/2))/7 + (2*B*b
^2*x^(19/2))/19

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sympy [A]  time = 10.85, size = 80, normalized size = 1.27 \[ \frac {2 A a^{2} x^{\frac {7}{2}}}{7} + \frac {4 A a b x^{\frac {11}{2}}}{11} + \frac {2 A b^{2} x^{\frac {15}{2}}}{15} + \frac {2 B a^{2} x^{\frac {11}{2}}}{11} + \frac {4 B a b x^{\frac {15}{2}}}{15} + \frac {2 B b^{2} x^{\frac {19}{2}}}{19} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*x**(7/2)/7 + 4*A*a*b*x**(11/2)/11 + 2*A*b**2*x**(15/2)/15 + 2*B*a**2*x**(11/2)/11 + 4*B*a*b*x**(15/2)
/15 + 2*B*b**2*x**(19/2)/19

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