\(\int \sqrt {x} (a+b x^2) (A+B x^2) \, dx\) [346]

Optimal result

Integrand size = 20, antiderivative size = 39 \[ \int \sqrt {x} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {2}{3} a A x^{3/2}+\frac {2}{7} (A b+a B) x^{7/2}+\frac {2}{11} b B x^{11/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \[ \int \sqrt {x} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {2}{7} x^{7/2} (a B+A b)+\frac {2}{3} a A x^{3/2}+\frac {2}{11} b B x^{11/2} \]

[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 459

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*} \text {integral}& = \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] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \sqrt {x} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {2}{231} x^{3/2} \left (77 a A+33 A b x^2+33 a B x^2+21 b B x^4\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {2 a A \,x^{\frac {3}{2}}}{3}+\frac {2 \left (A b +B a \right ) x^{\frac {7}{2}}}{7}+\frac {2 b B \,x^{\frac {11}{2}}}{11}\)	\(28\)
default	\(\frac {2 a A \,x^{\frac {3}{2}}}{3}+\frac {2 \left (A b +B a \right ) x^{\frac {7}{2}}}{7}+\frac {2 b B \,x^{\frac {11}{2}}}{11}\)	\(28\)
gosper	\(\frac {2 x^{\frac {3}{2}} \left (21 b B \,x^{4}+33 A b \,x^{2}+33 B a \,x^{2}+77 A a \right )}{231}\)	\(32\)
trager	\(\frac {2 x^{\frac {3}{2}} \left (21 b B \,x^{4}+33 A b \,x^{2}+33 B a \,x^{2}+77 A a \right )}{231}\)	\(32\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (21 b B \,x^{4}+33 A b \,x^{2}+33 B a \,x^{2}+77 A a \right )}{231}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.77 \[ \int \sqrt {x} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {2}{231} \, {\left (21 \, B b x^{5} + 33 \, {\left (B a + A b\right )} x^{3} + 77 \, A a x\right )} \sqrt {x} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \sqrt {x} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {2 A a x^{\frac {3}{2}}}{3} + \frac {2 B b x^{\frac {11}{2}}}{11} + \frac {2 x^{\frac {7}{2}} \left (A b + B a\right )}{7} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \sqrt {x} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {2}{11} \, B b x^{\frac {11}{2}} + \frac {2}{7} \, {\left (B a + A b\right )} x^{\frac {7}{2}} + \frac {2}{3} \, A a x^{\frac {3}{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \sqrt {x} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {2}{11} \, B b x^{\frac {11}{2}} + \frac {2}{7} \, B a x^{\frac {7}{2}} + \frac {2}{7} \, A b x^{\frac {7}{2}} + \frac {2}{3} \, A a x^{\frac {3}{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 4.89 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \sqrt {x} \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {2\,x^{3/2}\,\left (77\,A\,a+33\,A\,b\,x^2+33\,B\,a\,x^2+21\,B\,b\,x^4\right )}{231} \]

[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