Integrand size = 26, antiderivative size = 85 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx=\frac {2}{7} A b^3 x^{7/2}+\frac {2}{11} b^2 (b B+3 A c) x^{11/2}+\frac {2}{5} b c (b B+A c) x^{15/2}+\frac {2}{19} c^2 (3 b B+A c) x^{19/2}+\frac {2}{23} B c^3 x^{23/2} \] Output:
2/7*A*b^3*x^(7/2)+2/11*b^2*(3*A*c+B*b)*x^(11/2)+2/5*b*c*(A*c+B*b)*x^(15/2) +2/19*c^2*(A*c+3*B*b)*x^(19/2)+2/23*B*c^3*x^(23/2)
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx=\frac {2 x^{7/2} \left (23 A \left (1045 b^3+1995 b^2 c x^2+1463 b c^2 x^4+385 c^3 x^6\right )+7 B x^2 \left (2185 b^3+4807 b^2 c x^2+3795 b c^2 x^4+1045 c^3 x^6\right )\right )}{168245} \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(7/2),x]
Output:
(2*x^(7/2)*(23*A*(1045*b^3 + 1995*b^2*c*x^2 + 1463*b*c^2*x^4 + 385*c^3*x^6 ) + 7*B*x^2*(2185*b^3 + 4807*b^2*c*x^2 + 3795*b*c^2*x^4 + 1045*c^3*x^6)))/ 168245
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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^{5/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (A b^3 x^{5/2}+b^2 x^{9/2} (3 A c+b B)+c^2 x^{17/2} (A c+3 b B)+3 b c x^{13/2} (A c+b B)+B c^3 x^{21/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{7} A b^3 x^{7/2}+\frac {2}{11} b^2 x^{11/2} (3 A c+b B)+\frac {2}{19} c^2 x^{19/2} (A c+3 b B)+\frac {2}{5} b c x^{15/2} (A c+b B)+\frac {2}{23} B c^3 x^{23/2}\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(7/2),x]
Output:
(2*A*b^3*x^(7/2))/7 + (2*b^2*(b*B + 3*A*c)*x^(11/2))/11 + (2*b*c*(b*B + A* c)*x^(15/2))/5 + (2*c^2*(3*b*B + A*c)*x^(19/2))/19 + (2*B*c^3*x^(23/2))/23
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 | \(\frac {2 B \,c^{3} x^{\frac {23}{2}}}{23}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {19}{2}}}{19}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {11}{2}}}{11}+\frac {2 A \,b^{3} x^{\frac {7}{2}}}{7}\) | \(76\) |
default | \(\frac {2 B \,c^{3} x^{\frac {23}{2}}}{23}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {19}{2}}}{19}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {11}{2}}}{11}+\frac {2 A \,b^{3} x^{\frac {7}{2}}}{7}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {7}{2}} \left (7315 B \,c^{3} x^{8}+8855 A \,c^{3} x^{6}+26565 B b \,c^{2} x^{6}+33649 A b \,c^{2} x^{4}+33649 x^{4} B \,b^{2} c +45885 A \,b^{2} c \,x^{2}+15295 x^{2} B \,b^{3}+24035 A \,b^{3}\right )}{168245}\) | \(80\) |
trager | \(\frac {2 x^{\frac {7}{2}} \left (7315 B \,c^{3} x^{8}+8855 A \,c^{3} x^{6}+26565 B b \,c^{2} x^{6}+33649 A b \,c^{2} x^{4}+33649 x^{4} B \,b^{2} c +45885 A \,b^{2} c \,x^{2}+15295 x^{2} B \,b^{3}+24035 A \,b^{3}\right )}{168245}\) | \(80\) |
risch | \(\frac {2 x^{\frac {7}{2}} \left (7315 B \,c^{3} x^{8}+8855 A \,c^{3} x^{6}+26565 B b \,c^{2} x^{6}+33649 A b \,c^{2} x^{4}+33649 x^{4} B \,b^{2} c +45885 A \,b^{2} c \,x^{2}+15295 x^{2} B \,b^{3}+24035 A \,b^{3}\right )}{168245}\) | \(80\) |
orering | \(\frac {2 \left (7315 B \,c^{3} x^{8}+8855 A \,c^{3} x^{6}+26565 B b \,c^{2} x^{6}+33649 A b \,c^{2} x^{4}+33649 x^{4} B \,b^{2} c +45885 A \,b^{2} c \,x^{2}+15295 x^{2} B \,b^{3}+24035 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{3}}{168245 x^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^(7/2),x,method=_RETURNVERBOSE)
Output:
2/23*B*c^3*x^(23/2)+2/19*(A*c^3+3*B*b*c^2)*x^(19/2)+2/15*(3*A*b*c^2+3*B*b^ 2*c)*x^(15/2)+2/11*(3*A*b^2*c+B*b^3)*x^(11/2)+2/7*A*b^3*x^(7/2)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx=\frac {2}{168245} \, {\left (7315 \, B c^{3} x^{11} + 8855 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + 33649 \, {\left (B b^{2} c + A b c^{2}\right )} x^{7} + 24035 \, A b^{3} x^{3} + 15295 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{5}\right )} \sqrt {x} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(7/2),x, algorithm="fricas")
Output:
2/168245*(7315*B*c^3*x^11 + 8855*(3*B*b*c^2 + A*c^3)*x^9 + 33649*(B*b^2*c + A*b*c^2)*x^7 + 24035*A*b^3*x^3 + 15295*(B*b^3 + 3*A*b^2*c)*x^5)*sqrt(x)
Time = 1.73 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx=\frac {2 A b^{3} x^{\frac {7}{2}}}{7} + \frac {6 A b^{2} c x^{\frac {11}{2}}}{11} + \frac {2 A b c^{2} x^{\frac {15}{2}}}{5} + \frac {2 A c^{3} x^{\frac {19}{2}}}{19} + \frac {2 B b^{3} x^{\frac {11}{2}}}{11} + \frac {2 B b^{2} c x^{\frac {15}{2}}}{5} + \frac {6 B b c^{2} x^{\frac {19}{2}}}{19} + \frac {2 B c^{3} x^{\frac {23}{2}}}{23} \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**(7/2),x)
Output:
2*A*b**3*x**(7/2)/7 + 6*A*b**2*c*x**(11/2)/11 + 2*A*b*c**2*x**(15/2)/5 + 2 *A*c**3*x**(19/2)/19 + 2*B*b**3*x**(11/2)/11 + 2*B*b**2*c*x**(15/2)/5 + 6* B*b*c**2*x**(19/2)/19 + 2*B*c**3*x**(23/2)/23
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx=\frac {2}{23} \, B c^{3} x^{\frac {23}{2}} + \frac {2}{19} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {19}{2}} + \frac {2}{5} \, {\left (B b^{2} c + A b c^{2}\right )} x^{\frac {15}{2}} + \frac {2}{7} \, A b^{3} x^{\frac {7}{2}} + \frac {2}{11} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac {11}{2}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(7/2),x, algorithm="maxima")
Output:
2/23*B*c^3*x^(23/2) + 2/19*(3*B*b*c^2 + A*c^3)*x^(19/2) + 2/5*(B*b^2*c + A *b*c^2)*x^(15/2) + 2/7*A*b^3*x^(7/2) + 2/11*(B*b^3 + 3*A*b^2*c)*x^(11/2)
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx=\frac {2}{23} \, B c^{3} x^{\frac {23}{2}} + \frac {6}{19} \, B b c^{2} x^{\frac {19}{2}} + \frac {2}{19} \, A c^{3} x^{\frac {19}{2}} + \frac {2}{5} \, B b^{2} c x^{\frac {15}{2}} + \frac {2}{5} \, A b c^{2} x^{\frac {15}{2}} + \frac {2}{11} \, B b^{3} x^{\frac {11}{2}} + \frac {6}{11} \, A b^{2} c x^{\frac {11}{2}} + \frac {2}{7} \, A b^{3} x^{\frac {7}{2}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(7/2),x, algorithm="giac")
Output:
2/23*B*c^3*x^(23/2) + 6/19*B*b*c^2*x^(19/2) + 2/19*A*c^3*x^(19/2) + 2/5*B* b^2*c*x^(15/2) + 2/5*A*b*c^2*x^(15/2) + 2/11*B*b^3*x^(11/2) + 6/11*A*b^2*c *x^(11/2) + 2/7*A*b^3*x^(7/2)
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx=x^{11/2}\,\left (\frac {2\,B\,b^3}{11}+\frac {6\,A\,c\,b^2}{11}\right )+x^{19/2}\,\left (\frac {2\,A\,c^3}{19}+\frac {6\,B\,b\,c^2}{19}\right )+\frac {2\,A\,b^3\,x^{7/2}}{7}+\frac {2\,B\,c^3\,x^{23/2}}{23}+\frac {2\,b\,c\,x^{15/2}\,\left (A\,c+B\,b\right )}{5} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(7/2),x)
Output:
x^(11/2)*((2*B*b^3)/11 + (6*A*b^2*c)/11) + x^(19/2)*((2*A*c^3)/19 + (6*B*b *c^2)/19) + (2*A*b^3*x^(7/2))/7 + (2*B*c^3*x^(23/2))/23 + (2*b*c*x^(15/2)* (A*c + B*b))/5
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx=\frac {2 \sqrt {x}\, x^{3} \left (7315 b \,c^{3} x^{8}+8855 a \,c^{3} x^{6}+26565 b^{2} c^{2} x^{6}+33649 a b \,c^{2} x^{4}+33649 b^{3} c \,x^{4}+45885 a \,b^{2} c \,x^{2}+15295 b^{4} x^{2}+24035 a \,b^{3}\right )}{168245} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^(7/2),x)
Output:
(2*sqrt(x)*x**3*(24035*a*b**3 + 45885*a*b**2*c*x**2 + 33649*a*b*c**2*x**4 + 8855*a*c**3*x**6 + 15295*b**4*x**2 + 33649*b**3*c*x**4 + 26565*b**2*c**2 *x**6 + 7315*b*c**3*x**8))/168245