Integrand size = 26, antiderivative size = 85 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{19} A b^3 x^{19/2}+\frac {2}{23} b^2 (b B+3 A c) x^{23/2}+\frac {2}{9} b c (b B+A c) x^{27/2}+\frac {2}{31} c^2 (3 b B+A c) x^{31/2}+\frac {2}{35} B c^3 x^{35/2} \] Output:
2/19*A*b^3*x^(19/2)+2/23*b^2*(3*A*c+B*b)*x^(23/2)+2/9*b*c*(A*c+B*b)*x^(27/ 2)+2/31*c^2*(A*c+3*B*b)*x^(31/2)+2/35*B*c^3*x^(35/2)
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{19} A b^3 x^{19/2}+\frac {2}{23} b^2 (b B+3 A c) x^{23/2}+\frac {2}{9} b c (b B+A c) x^{27/2}+\frac {2}{31} c^2 (3 b B+A c) x^{31/2}+\frac {2}{35} B c^3 x^{35/2} \] Input:
Integrate[x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x]
Output:
(2*A*b^3*x^(19/2))/19 + (2*b^2*(b*B + 3*A*c)*x^(23/2))/23 + (2*b*c*(b*B + A*c)*x^(27/2))/9 + (2*c^2*(3*b*B + A*c)*x^(31/2))/31 + (2*B*c^3*x^(35/2))/ 35
Time = 0.37 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {9, 355, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^{17/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (A b^3 x^{17/2}+b^2 x^{21/2} (3 A c+b B)+c^2 x^{29/2} (A c+3 b B)+3 b c x^{25/2} (A c+b B)+B c^3 x^{33/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{19} A b^3 x^{19/2}+\frac {2}{23} b^2 x^{23/2} (3 A c+b B)+\frac {2}{31} c^2 x^{31/2} (A c+3 b B)+\frac {2}{9} b c x^{27/2} (A c+b B)+\frac {2}{35} B c^3 x^{35/2}\) |
Input:
Int[x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x]
Output:
(2*A*b^3*x^(19/2))/19 + (2*b^2*(b*B + 3*A*c)*x^(23/2))/23 + (2*b*c*(b*B + A*c)*x^(27/2))/9 + (2*c^2*(3*b*B + A*c)*x^(31/2))/31 + (2*B*c^3*x^(35/2))/ 35
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Time = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {2 B \,c^{3} x^{\frac {35}{2}}}{35}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {31}{2}}}{31}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {27}{2}}}{27}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {23}{2}}}{23}+\frac {2 A \,b^{3} x^{\frac {19}{2}}}{19}\) | \(76\) |
default | \(\frac {2 B \,c^{3} x^{\frac {35}{2}}}{35}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {31}{2}}}{31}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {27}{2}}}{27}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {23}{2}}}{23}+\frac {2 A \,b^{3} x^{\frac {19}{2}}}{19}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {19}{2}} \left (121923 B \,c^{3} x^{8}+137655 A \,c^{3} x^{6}+412965 B b \,c^{2} x^{6}+474145 A b \,c^{2} x^{4}+474145 x^{4} B \,b^{2} c +556605 A \,b^{2} c \,x^{2}+185535 x^{2} B \,b^{3}+224595 A \,b^{3}\right )}{4267305}\) | \(80\) |
trager | \(\frac {2 x^{\frac {19}{2}} \left (121923 B \,c^{3} x^{8}+137655 A \,c^{3} x^{6}+412965 B b \,c^{2} x^{6}+474145 A b \,c^{2} x^{4}+474145 x^{4} B \,b^{2} c +556605 A \,b^{2} c \,x^{2}+185535 x^{2} B \,b^{3}+224595 A \,b^{3}\right )}{4267305}\) | \(80\) |
risch | \(\frac {2 x^{\frac {19}{2}} \left (121923 B \,c^{3} x^{8}+137655 A \,c^{3} x^{6}+412965 B b \,c^{2} x^{6}+474145 A b \,c^{2} x^{4}+474145 x^{4} B \,b^{2} c +556605 A \,b^{2} c \,x^{2}+185535 x^{2} B \,b^{3}+224595 A \,b^{3}\right )}{4267305}\) | \(80\) |
orering | \(\frac {2 \left (121923 B \,c^{3} x^{8}+137655 A \,c^{3} x^{6}+412965 B b \,c^{2} x^{6}+474145 A b \,c^{2} x^{4}+474145 x^{4} B \,b^{2} c +556605 A \,b^{2} c \,x^{2}+185535 x^{2} B \,b^{3}+224595 A \,b^{3}\right ) x^{\frac {7}{2}} \left (c \,x^{4}+b \,x^{2}\right )^{3}}{4267305 \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
2/35*B*c^3*x^(35/2)+2/31*(A*c^3+3*B*b*c^2)*x^(31/2)+2/27*(3*A*b*c^2+3*B*b^ 2*c)*x^(27/2)+2/23*(3*A*b^2*c+B*b^3)*x^(23/2)+2/19*A*b^3*x^(19/2)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{4267305} \, {\left (121923 \, B c^{3} x^{17} + 137655 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{15} + 474145 \, {\left (B b^{2} c + A b c^{2}\right )} x^{13} + 224595 \, A b^{3} x^{9} + 185535 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{11}\right )} \sqrt {x} \] Input:
integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
2/4267305*(121923*B*c^3*x^17 + 137655*(3*B*b*c^2 + A*c^3)*x^15 + 474145*(B *b^2*c + A*b*c^2)*x^13 + 224595*A*b^3*x^9 + 185535*(B*b^3 + 3*A*b^2*c)*x^1 1)*sqrt(x)
Time = 2.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2 A b^{3} x^{\frac {19}{2}}}{19} + \frac {6 A b^{2} c x^{\frac {23}{2}}}{23} + \frac {2 A b c^{2} x^{\frac {27}{2}}}{9} + \frac {2 A c^{3} x^{\frac {31}{2}}}{31} + \frac {2 B b^{3} x^{\frac {23}{2}}}{23} + \frac {2 B b^{2} c x^{\frac {27}{2}}}{9} + \frac {6 B b c^{2} x^{\frac {31}{2}}}{31} + \frac {2 B c^{3} x^{\frac {35}{2}}}{35} \] Input:
integrate(x**(5/2)*(B*x**2+A)*(c*x**4+b*x**2)**3,x)
Output:
2*A*b**3*x**(19/2)/19 + 6*A*b**2*c*x**(23/2)/23 + 2*A*b*c**2*x**(27/2)/9 + 2*A*c**3*x**(31/2)/31 + 2*B*b**3*x**(23/2)/23 + 2*B*b**2*c*x**(27/2)/9 + 6*B*b*c**2*x**(31/2)/31 + 2*B*c**3*x**(35/2)/35
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{35} \, B c^{3} x^{\frac {35}{2}} + \frac {2}{31} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {31}{2}} + \frac {2}{9} \, {\left (B b^{2} c + A b c^{2}\right )} x^{\frac {27}{2}} + \frac {2}{19} \, A b^{3} x^{\frac {19}{2}} + \frac {2}{23} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac {23}{2}} \] Input:
integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
2/35*B*c^3*x^(35/2) + 2/31*(3*B*b*c^2 + A*c^3)*x^(31/2) + 2/9*(B*b^2*c + A *b*c^2)*x^(27/2) + 2/19*A*b^3*x^(19/2) + 2/23*(B*b^3 + 3*A*b^2*c)*x^(23/2)
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{35} \, B c^{3} x^{\frac {35}{2}} + \frac {6}{31} \, B b c^{2} x^{\frac {31}{2}} + \frac {2}{31} \, A c^{3} x^{\frac {31}{2}} + \frac {2}{9} \, B b^{2} c x^{\frac {27}{2}} + \frac {2}{9} \, A b c^{2} x^{\frac {27}{2}} + \frac {2}{23} \, B b^{3} x^{\frac {23}{2}} + \frac {6}{23} \, A b^{2} c x^{\frac {23}{2}} + \frac {2}{19} \, A b^{3} x^{\frac {19}{2}} \] Input:
integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
2/35*B*c^3*x^(35/2) + 6/31*B*b*c^2*x^(31/2) + 2/31*A*c^3*x^(31/2) + 2/9*B* b^2*c*x^(27/2) + 2/9*A*b*c^2*x^(27/2) + 2/23*B*b^3*x^(23/2) + 6/23*A*b^2*c *x^(23/2) + 2/19*A*b^3*x^(19/2)
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=x^{23/2}\,\left (\frac {2\,B\,b^3}{23}+\frac {6\,A\,c\,b^2}{23}\right )+x^{31/2}\,\left (\frac {2\,A\,c^3}{31}+\frac {6\,B\,b\,c^2}{31}\right )+\frac {2\,A\,b^3\,x^{19/2}}{19}+\frac {2\,B\,c^3\,x^{35/2}}{35}+\frac {2\,b\,c\,x^{27/2}\,\left (A\,c+B\,b\right )}{9} \] Input:
int(x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x)
Output:
x^(23/2)*((2*B*b^3)/23 + (6*A*b^2*c)/23) + x^(31/2)*((2*A*c^3)/31 + (6*B*b *c^2)/31) + (2*A*b^3*x^(19/2))/19 + (2*B*c^3*x^(35/2))/35 + (2*b*c*x^(27/2 )*(A*c + B*b))/9
Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2 \sqrt {x}\, x^{9} \left (121923 b \,c^{3} x^{8}+137655 a \,c^{3} x^{6}+412965 b^{2} c^{2} x^{6}+474145 a b \,c^{2} x^{4}+474145 b^{3} c \,x^{4}+556605 a \,b^{2} c \,x^{2}+185535 b^{4} x^{2}+224595 a \,b^{3}\right )}{4267305} \] Input:
int(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x)
Output:
(2*sqrt(x)*x**9*(224595*a*b**3 + 556605*a*b**2*c*x**2 + 474145*a*b*c**2*x* *4 + 137655*a*c**3*x**6 + 185535*b**4*x**2 + 474145*b**3*c*x**4 + 412965*b **2*c**2*x**6 + 121923*b*c**3*x**8))/4267305