3.4.28 \(\int \sqrt {x} (a+b x^2) (A+B x^2) \, dx\)

Optimal. Leaf size=39 \[ \frac {2}{7} x^{7/2} (a B+A b)+\frac {2}{3} a A x^{3/2}+\frac {2}{11} b B x^{11/2} \]

________________________________________________________________________________________

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} \begin {gather*} \frac {2}{7} x^{7/2} (a B+A b)+\frac {2}{3} a A x^{3/2}+\frac {2}{11} b B x^{11/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*x^(3/2))/3 + (2*(A*b + a*B)*x^(7/2))/7 + (2*b*B*x^(11/2))/11

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 33, normalized size = 0.85 \begin {gather*} \frac {2}{231} x^{3/2} \left (33 x^2 (a B+A b)+77 a A+21 b B x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(77*a*A + 33*(A*b + a*B)*x^2 + 21*b*B*x^4))/231

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.02, size = 41, normalized size = 1.05 \begin {gather*} \frac {2}{231} \left (77 a A x^{3/2}+33 a B x^{7/2}+33 A b x^{7/2}+21 b B x^{11/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(77*a*A*x^(3/2) + 33*A*b*x^(7/2) + 33*a*B*x^(7/2) + 21*b*B*x^(11/2)))/231

________________________________________________________________________________________

fricas [A]  time = 1.02, size = 30, normalized size = 0.77 \begin {gather*} \frac {2}{231} \, {\left (21 \, B b x^{5} + 33 \, {\left (B a + A b\right )} x^{3} + 77 \, A a x\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)*x^(1/2),x, algorithm="fricas")

[Out]

2/231*(21*B*b*x^5 + 33*(B*a + A*b)*x^3 + 77*A*a*x)*sqrt(x)

________________________________________________________________________________________

giac [A]  time = 0.30, size = 29, normalized size = 0.74 \begin {gather*} \frac {2}{11} \, B b x^{\frac {11}{2}} + \frac {2}{7} \, B a x^{\frac {7}{2}} + \frac {2}{7} \, A b x^{\frac {7}{2}} + \frac {2}{3} \, A a x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)*x^(1/2),x, algorithm="giac")

[Out]

2/11*B*b*x^(11/2) + 2/7*B*a*x^(7/2) + 2/7*A*b*x^(7/2) + 2/3*A*a*x^(3/2)

________________________________________________________________________________________

maple [A]  time = 0.00, size = 32, normalized size = 0.82 \begin {gather*} \frac {2 \left (21 B b \,x^{4}+33 A b \,x^{2}+33 B a \,x^{2}+77 A a \right ) x^{\frac {3}{2}}}{231} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(B*x^2+A)*x^(1/2),x)

[Out]

2/231*x^(3/2)*(21*B*b*x^4+33*A*b*x^2+33*B*a*x^2+77*A*a)

________________________________________________________________________________________

maxima [A]  time = 0.95, size = 27, normalized size = 0.69 \begin {gather*} \frac {2}{11} \, B b x^{\frac {11}{2}} + \frac {2}{7} \, {\left (B a + A b\right )} x^{\frac {7}{2}} + \frac {2}{3} \, A a x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)*x^(1/2),x, algorithm="maxima")

[Out]

2/11*B*b*x^(11/2) + 2/7*(B*a + A*b)*x^(7/2) + 2/3*A*a*x^(3/2)

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 31, normalized size = 0.79 \begin {gather*} \frac {2\,x^{3/2}\,\left (77\,A\,a+33\,A\,b\,x^2+33\,B\,a\,x^2+21\,B\,b\,x^4\right )}{231} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(A + B*x^2)*(a + b*x^2),x)

[Out]

(2*x^(3/2)*(77*A*a + 33*A*b*x^2 + 33*B*a*x^2 + 21*B*b*x^4))/231

________________________________________________________________________________________

sympy [A]  time = 1.99, size = 37, normalized size = 0.95 \begin {gather*} \frac {2 A a x^{\frac {3}{2}}}{3} + \frac {2 B b x^{\frac {11}{2}}}{11} + \frac {2 x^{\frac {7}{2}} \left (A b + B a\right )}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(B*x**2+A)*x**(1/2),x)

[Out]

2*A*a*x**(3/2)/3 + 2*B*b*x**(11/2)/11 + 2*x**(7/2)*(A*b + B*a)/7

________________________________________________________________________________________