Integrand size = 26, antiderivative size = 85 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{15} A b^3 x^{15/2}+\frac {2}{19} b^2 (b B+3 A c) x^{19/2}+\frac {6}{23} b c (b B+A c) x^{23/2}+\frac {2}{27} c^2 (3 b B+A c) x^{27/2}+\frac {2}{31} B c^3 x^{31/2} \] Output:
2/15*A*b^3*x^(15/2)+2/19*b^2*(3*A*c+B*b)*x^(19/2)+6/23*b*c*(A*c+B*b)*x^(23 /2)+2/27*c^2*(A*c+3*B*b)*x^(27/2)+2/31*B*c^3*x^(31/2)
Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2 x^{15/2} \left (31 A \left (3933 b^3+9315 b^2 c x^2+7695 b c^2 x^4+2185 c^3 x^6\right )+15 B x^2 \left (6417 b^3+15903 b^2 c x^2+13547 b c^2 x^4+3933 c^3 x^6\right )\right )}{1828845} \] Input:
Integrate[Sqrt[x]*(A + B*x^2)*(b*x^2 + c*x^4)^3,x]
Output:
(2*x^(15/2)*(31*A*(3933*b^3 + 9315*b^2*c*x^2 + 7695*b*c^2*x^4 + 2185*c^3*x ^6) + 15*B*x^2*(6417*b^3 + 15903*b^2*c*x^2 + 13547*b*c^2*x^4 + 3933*c^3*x^ 6)))/1828845
Time = 0.35 (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.
\(\displaystyle \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^{13/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (A b^3 x^{13/2}+b^2 x^{17/2} (3 A c+b B)+c^2 x^{25/2} (A c+3 b B)+3 b c x^{21/2} (A c+b B)+B c^3 x^{29/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{15} A b^3 x^{15/2}+\frac {2}{19} b^2 x^{19/2} (3 A c+b B)+\frac {2}{27} c^2 x^{27/2} (A c+3 b B)+\frac {6}{23} b c x^{23/2} (A c+b B)+\frac {2}{31} B c^3 x^{31/2}\) |
Input:
Int[Sqrt[x]*(A + B*x^2)*(b*x^2 + c*x^4)^3,x]
Output:
(2*A*b^3*x^(15/2))/15 + (2*b^2*(b*B + 3*A*c)*x^(19/2))/19 + (6*b*c*(b*B + A*c)*x^(23/2))/23 + (2*c^2*(3*b*B + A*c)*x^(27/2))/27 + (2*B*c^3*x^(31/2)) /31
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 | \(\frac {2 B \,c^{3} x^{\frac {31}{2}}}{31}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {27}{2}}}{27}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {23}{2}}}{23}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {19}{2}}}{19}+\frac {2 A \,b^{3} x^{\frac {15}{2}}}{15}\) | \(76\) |
default | \(\frac {2 B \,c^{3} x^{\frac {31}{2}}}{31}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {27}{2}}}{27}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {23}{2}}}{23}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {19}{2}}}{19}+\frac {2 A \,b^{3} x^{\frac {15}{2}}}{15}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {15}{2}} \left (58995 B \,c^{3} x^{8}+67735 A \,c^{3} x^{6}+203205 B b \,c^{2} x^{6}+238545 A b \,c^{2} x^{4}+238545 x^{4} B \,b^{2} c +288765 A \,b^{2} c \,x^{2}+96255 x^{2} B \,b^{3}+121923 A \,b^{3}\right )}{1828845}\) | \(80\) |
trager | \(\frac {2 x^{\frac {15}{2}} \left (58995 B \,c^{3} x^{8}+67735 A \,c^{3} x^{6}+203205 B b \,c^{2} x^{6}+238545 A b \,c^{2} x^{4}+238545 x^{4} B \,b^{2} c +288765 A \,b^{2} c \,x^{2}+96255 x^{2} B \,b^{3}+121923 A \,b^{3}\right )}{1828845}\) | \(80\) |
risch | \(\frac {2 x^{\frac {15}{2}} \left (58995 B \,c^{3} x^{8}+67735 A \,c^{3} x^{6}+203205 B b \,c^{2} x^{6}+238545 A b \,c^{2} x^{4}+238545 x^{4} B \,b^{2} c +288765 A \,b^{2} c \,x^{2}+96255 x^{2} B \,b^{3}+121923 A \,b^{3}\right )}{1828845}\) | \(80\) |
orering | \(\frac {2 \left (58995 B \,c^{3} x^{8}+67735 A \,c^{3} x^{6}+203205 B b \,c^{2} x^{6}+238545 A b \,c^{2} x^{4}+238545 x^{4} B \,b^{2} c +288765 A \,b^{2} c \,x^{2}+96255 x^{2} B \,b^{3}+121923 A \,b^{3}\right ) x^{\frac {3}{2}} \left (c \,x^{4}+b \,x^{2}\right )^{3}}{1828845 \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int(x^(1/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
2/31*B*c^3*x^(31/2)+2/27*(A*c^3+3*B*b*c^2)*x^(27/2)+2/23*(3*A*b*c^2+3*B*b^ 2*c)*x^(23/2)+2/19*(3*A*b^2*c+B*b^3)*x^(19/2)+2/15*A*b^3*x^(15/2)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{1828845} \, {\left (58995 \, B c^{3} x^{15} + 67735 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{13} + 238545 \, {\left (B b^{2} c + A b c^{2}\right )} x^{11} + 121923 \, A b^{3} x^{7} + 96255 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{9}\right )} \sqrt {x} \] Input:
integrate(x^(1/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
2/1828845*(58995*B*c^3*x^15 + 67735*(3*B*b*c^2 + A*c^3)*x^13 + 238545*(B*b ^2*c + A*b*c^2)*x^11 + 121923*A*b^3*x^7 + 96255*(B*b^3 + 3*A*b^2*c)*x^9)*s qrt(x)
Time = 1.53 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2 A b^{3} x^{\frac {15}{2}}}{15} + \frac {2 B c^{3} x^{\frac {31}{2}}}{31} + \frac {2 x^{\frac {27}{2}} \left (A c^{3} + 3 B b c^{2}\right )}{27} + \frac {2 x^{\frac {23}{2}} \cdot \left (3 A b c^{2} + 3 B b^{2} c\right )}{23} + \frac {2 x^{\frac {19}{2}} \cdot \left (3 A b^{2} c + B b^{3}\right )}{19} \] Input:
integrate(x**(1/2)*(B*x**2+A)*(c*x**4+b*x**2)**3,x)
Output:
2*A*b**3*x**(15/2)/15 + 2*B*c**3*x**(31/2)/31 + 2*x**(27/2)*(A*c**3 + 3*B* b*c**2)/27 + 2*x**(23/2)*(3*A*b*c**2 + 3*B*b**2*c)/23 + 2*x**(19/2)*(3*A*b **2*c + B*b**3)/19
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{31} \, B c^{3} x^{\frac {31}{2}} + \frac {2}{27} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {27}{2}} + \frac {6}{23} \, {\left (B b^{2} c + A b c^{2}\right )} x^{\frac {23}{2}} + \frac {2}{15} \, A b^{3} x^{\frac {15}{2}} + \frac {2}{19} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac {19}{2}} \] Input:
integrate(x^(1/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
2/31*B*c^3*x^(31/2) + 2/27*(3*B*b*c^2 + A*c^3)*x^(27/2) + 6/23*(B*b^2*c + A*b*c^2)*x^(23/2) + 2/15*A*b^3*x^(15/2) + 2/19*(B*b^3 + 3*A*b^2*c)*x^(19/2 )
Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{31} \, B c^{3} x^{\frac {31}{2}} + \frac {2}{9} \, B b c^{2} x^{\frac {27}{2}} + \frac {2}{27} \, A c^{3} x^{\frac {27}{2}} + \frac {6}{23} \, B b^{2} c x^{\frac {23}{2}} + \frac {6}{23} \, A b c^{2} x^{\frac {23}{2}} + \frac {2}{19} \, B b^{3} x^{\frac {19}{2}} + \frac {6}{19} \, A b^{2} c x^{\frac {19}{2}} + \frac {2}{15} \, A b^{3} x^{\frac {15}{2}} \] Input:
integrate(x^(1/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
2/31*B*c^3*x^(31/2) + 2/9*B*b*c^2*x^(27/2) + 2/27*A*c^3*x^(27/2) + 6/23*B* b^2*c*x^(23/2) + 6/23*A*b*c^2*x^(23/2) + 2/19*B*b^3*x^(19/2) + 6/19*A*b^2* c*x^(19/2) + 2/15*A*b^3*x^(15/2)
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=x^{19/2}\,\left (\frac {2\,B\,b^3}{19}+\frac {6\,A\,c\,b^2}{19}\right )+x^{27/2}\,\left (\frac {2\,A\,c^3}{27}+\frac {2\,B\,b\,c^2}{9}\right )+\frac {2\,A\,b^3\,x^{15/2}}{15}+\frac {2\,B\,c^3\,x^{31/2}}{31}+\frac {6\,b\,c\,x^{23/2}\,\left (A\,c+B\,b\right )}{23} \] Input:
int(x^(1/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x)
Output:
x^(19/2)*((2*B*b^3)/19 + (6*A*b^2*c)/19) + x^(27/2)*((2*A*c^3)/27 + (2*B*b *c^2)/9) + (2*A*b^3*x^(15/2))/15 + (2*B*c^3*x^(31/2))/31 + (6*b*c*x^(23/2) *(A*c + B*b))/23
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2 \sqrt {x}\, x^{7} \left (58995 b \,c^{3} x^{8}+67735 a \,c^{3} x^{6}+203205 b^{2} c^{2} x^{6}+238545 a b \,c^{2} x^{4}+238545 b^{3} c \,x^{4}+288765 a \,b^{2} c \,x^{2}+96255 b^{4} x^{2}+121923 a \,b^{3}\right )}{1828845} \] Input:
int(x^(1/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x)
Output:
(2*sqrt(x)*x**7*(121923*a*b**3 + 288765*a*b**2*c*x**2 + 238545*a*b*c**2*x* *4 + 67735*a*c**3*x**6 + 96255*b**4*x**2 + 238545*b**3*c*x**4 + 203205*b** 2*c**2*x**6 + 58995*b*c**3*x**8))/1828845