Integrand size = 26, antiderivative size = 85 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx=\frac {2}{11} A b^3 x^{11/2}+\frac {2}{15} b^2 (b B+3 A c) x^{15/2}+\frac {6}{19} b c (b B+A c) x^{19/2}+\frac {2}{23} c^2 (3 b B+A c) x^{23/2}+\frac {2}{27} B c^3 x^{27/2} \] Output:
2/11*A*b^3*x^(11/2)+2/15*b^2*(3*A*c+B*b)*x^(15/2)+6/19*b*c*(A*c+B*b)*x^(19 /2)+2/23*c^2*(A*c+3*B*b)*x^(23/2)+2/27*B*c^3*x^(27/2)
Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx=\frac {2 x^{11/2} \left (27 A \left (2185 b^3+4807 b^2 c x^2+3795 b c^2 x^4+1045 c^3 x^6\right )+11 B x^2 \left (3933 b^3+9315 b^2 c x^2+7695 b c^2 x^4+2185 c^3 x^6\right )\right )}{648945} \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(3/2),x]
Output:
(2*x^(11/2)*(27*A*(2185*b^3 + 4807*b^2*c*x^2 + 3795*b*c^2*x^4 + 1045*c^3*x ^6) + 11*B*x^2*(3933*b^3 + 9315*b^2*c*x^2 + 7695*b*c^2*x^4 + 2185*c^3*x^6) ))/648945
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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^{9/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (A b^3 x^{9/2}+b^2 x^{13/2} (3 A c+b B)+c^2 x^{21/2} (A c+3 b B)+3 b c x^{17/2} (A c+b B)+B c^3 x^{25/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{11} A b^3 x^{11/2}+\frac {2}{15} b^2 x^{15/2} (3 A c+b B)+\frac {2}{23} c^2 x^{23/2} (A c+3 b B)+\frac {6}{19} b c x^{19/2} (A c+b B)+\frac {2}{27} B c^3 x^{27/2}\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(3/2),x]
Output:
(2*A*b^3*x^(11/2))/11 + (2*b^2*(b*B + 3*A*c)*x^(15/2))/15 + (6*b*c*(b*B + A*c)*x^(19/2))/19 + (2*c^2*(3*b*B + A*c)*x^(23/2))/23 + (2*B*c^3*x^(27/2)) /27
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 | \(\frac {2 B \,c^{3} x^{\frac {27}{2}}}{27}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {23}{2}}}{23}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {19}{2}}}{19}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {15}{2}}}{15}+\frac {2 A \,b^{3} x^{\frac {11}{2}}}{11}\) | \(76\) |
default | \(\frac {2 B \,c^{3} x^{\frac {27}{2}}}{27}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {23}{2}}}{23}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {19}{2}}}{19}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {15}{2}}}{15}+\frac {2 A \,b^{3} x^{\frac {11}{2}}}{11}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {11}{2}} \left (24035 B \,c^{3} x^{8}+28215 A \,c^{3} x^{6}+84645 B b \,c^{2} x^{6}+102465 A b \,c^{2} x^{4}+102465 x^{4} B \,b^{2} c +129789 A \,b^{2} c \,x^{2}+43263 x^{2} B \,b^{3}+58995 A \,b^{3}\right )}{648945}\) | \(80\) |
trager | \(\frac {2 x^{\frac {11}{2}} \left (24035 B \,c^{3} x^{8}+28215 A \,c^{3} x^{6}+84645 B b \,c^{2} x^{6}+102465 A b \,c^{2} x^{4}+102465 x^{4} B \,b^{2} c +129789 A \,b^{2} c \,x^{2}+43263 x^{2} B \,b^{3}+58995 A \,b^{3}\right )}{648945}\) | \(80\) |
risch | \(\frac {2 x^{\frac {11}{2}} \left (24035 B \,c^{3} x^{8}+28215 A \,c^{3} x^{6}+84645 B b \,c^{2} x^{6}+102465 A b \,c^{2} x^{4}+102465 x^{4} B \,b^{2} c +129789 A \,b^{2} c \,x^{2}+43263 x^{2} B \,b^{3}+58995 A \,b^{3}\right )}{648945}\) | \(80\) |
orering | \(\frac {2 \left (24035 B \,c^{3} x^{8}+28215 A \,c^{3} x^{6}+84645 B b \,c^{2} x^{6}+102465 A b \,c^{2} x^{4}+102465 x^{4} B \,b^{2} c +129789 A \,b^{2} c \,x^{2}+43263 x^{2} B \,b^{3}+58995 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{3}}{648945 \sqrt {x}\, \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^(3/2),x,method=_RETURNVERBOSE)
Output:
2/27*B*c^3*x^(27/2)+2/23*(A*c^3+3*B*b*c^2)*x^(23/2)+2/19*(3*A*b*c^2+3*B*b^ 2*c)*x^(19/2)+2/15*(3*A*b^2*c+B*b^3)*x^(15/2)+2/11*A*b^3*x^(11/2)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx=\frac {2}{648945} \, {\left (24035 \, B c^{3} x^{13} + 28215 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{11} + 102465 \, {\left (B b^{2} c + A b c^{2}\right )} x^{9} + 58995 \, A b^{3} x^{5} + 43263 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{7}\right )} \sqrt {x} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(3/2),x, algorithm="fricas")
Output:
2/648945*(24035*B*c^3*x^13 + 28215*(3*B*b*c^2 + A*c^3)*x^11 + 102465*(B*b^ 2*c + A*b*c^2)*x^9 + 58995*A*b^3*x^5 + 43263*(B*b^3 + 3*A*b^2*c)*x^7)*sqrt (x)
Time = 1.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx=\frac {2 A b^{3} x^{\frac {11}{2}}}{11} + \frac {2 A b^{2} c x^{\frac {15}{2}}}{5} + \frac {6 A b c^{2} x^{\frac {19}{2}}}{19} + \frac {2 A c^{3} x^{\frac {23}{2}}}{23} + \frac {2 B b^{3} x^{\frac {15}{2}}}{15} + \frac {6 B b^{2} c x^{\frac {19}{2}}}{19} + \frac {6 B b c^{2} x^{\frac {23}{2}}}{23} + \frac {2 B c^{3} x^{\frac {27}{2}}}{27} \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**(3/2),x)
Output:
2*A*b**3*x**(11/2)/11 + 2*A*b**2*c*x**(15/2)/5 + 6*A*b*c**2*x**(19/2)/19 + 2*A*c**3*x**(23/2)/23 + 2*B*b**3*x**(15/2)/15 + 6*B*b**2*c*x**(19/2)/19 + 6*B*b*c**2*x**(23/2)/23 + 2*B*c**3*x**(27/2)/27
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx=\frac {2}{27} \, B c^{3} x^{\frac {27}{2}} + \frac {2}{23} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {23}{2}} + \frac {6}{19} \, {\left (B b^{2} c + A b c^{2}\right )} x^{\frac {19}{2}} + \frac {2}{11} \, A b^{3} x^{\frac {11}{2}} + \frac {2}{15} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac {15}{2}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(3/2),x, algorithm="maxima")
Output:
2/27*B*c^3*x^(27/2) + 2/23*(3*B*b*c^2 + A*c^3)*x^(23/2) + 6/19*(B*b^2*c + A*b*c^2)*x^(19/2) + 2/11*A*b^3*x^(11/2) + 2/15*(B*b^3 + 3*A*b^2*c)*x^(15/2 )
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx=\frac {2}{27} \, B c^{3} x^{\frac {27}{2}} + \frac {6}{23} \, B b c^{2} x^{\frac {23}{2}} + \frac {2}{23} \, A c^{3} x^{\frac {23}{2}} + \frac {6}{19} \, B b^{2} c x^{\frac {19}{2}} + \frac {6}{19} \, A b c^{2} x^{\frac {19}{2}} + \frac {2}{15} \, B b^{3} x^{\frac {15}{2}} + \frac {2}{5} \, A b^{2} c x^{\frac {15}{2}} + \frac {2}{11} \, A b^{3} x^{\frac {11}{2}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(3/2),x, algorithm="giac")
Output:
2/27*B*c^3*x^(27/2) + 6/23*B*b*c^2*x^(23/2) + 2/23*A*c^3*x^(23/2) + 6/19*B *b^2*c*x^(19/2) + 6/19*A*b*c^2*x^(19/2) + 2/15*B*b^3*x^(15/2) + 2/5*A*b^2* c*x^(15/2) + 2/11*A*b^3*x^(11/2)
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx=x^{15/2}\,\left (\frac {2\,B\,b^3}{15}+\frac {2\,A\,c\,b^2}{5}\right )+x^{23/2}\,\left (\frac {2\,A\,c^3}{23}+\frac {6\,B\,b\,c^2}{23}\right )+\frac {2\,A\,b^3\,x^{11/2}}{11}+\frac {2\,B\,c^3\,x^{27/2}}{27}+\frac {6\,b\,c\,x^{19/2}\,\left (A\,c+B\,b\right )}{19} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(3/2),x)
Output:
x^(15/2)*((2*B*b^3)/15 + (2*A*b^2*c)/5) + x^(23/2)*((2*A*c^3)/23 + (6*B*b* c^2)/23) + (2*A*b^3*x^(11/2))/11 + (2*B*c^3*x^(27/2))/27 + (6*b*c*x^(19/2) *(A*c + B*b))/19
Time = 0.21 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx=\frac {2 \sqrt {x}\, x^{5} \left (24035 b \,c^{3} x^{8}+28215 a \,c^{3} x^{6}+84645 b^{2} c^{2} x^{6}+102465 a b \,c^{2} x^{4}+102465 b^{3} c \,x^{4}+129789 a \,b^{2} c \,x^{2}+43263 b^{4} x^{2}+58995 a \,b^{3}\right )}{648945} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^(3/2),x)
Output:
(2*sqrt(x)*x**5*(58995*a*b**3 + 129789*a*b**2*c*x**2 + 102465*a*b*c**2*x** 4 + 28215*a*c**3*x**6 + 43263*b**4*x**2 + 102465*b**3*c*x**4 + 84645*b**2* c**2*x**6 + 24035*b*c**3*x**8))/648945