Integrand size = 26, antiderivative size = 85 \[ \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{17} A b^3 x^{17/2}+\frac {2}{21} b^2 (b B+3 A c) x^{21/2}+\frac {6}{25} b c (b B+A c) x^{25/2}+\frac {2}{29} c^2 (3 b B+A c) x^{29/2}+\frac {2}{33} B c^3 x^{33/2} \] Output:
2/17*A*b^3*x^(17/2)+2/21*b^2*(3*A*c+B*b)*x^(21/2)+6/25*b*c*(A*c+B*b)*x^(25 /2)+2/29*c^2*(A*c+3*B*b)*x^(29/2)+2/33*B*c^3*x^(33/2)
Time = 0.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14 \[ \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2 \left (167475 A b^3 x^{17/2}+135575 b^3 B x^{21/2}+406725 A b^2 c x^{21/2}+341649 b^2 B c x^{25/2}+341649 A b c^2 x^{25/2}+294525 b B c^2 x^{29/2}+98175 A c^3 x^{29/2}+86275 B c^3 x^{33/2}\right )}{2847075} \] Input:
Integrate[x^(3/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x]
Output:
(2*(167475*A*b^3*x^(17/2) + 135575*b^3*B*x^(21/2) + 406725*A*b^2*c*x^(21/2 ) + 341649*b^2*B*c*x^(25/2) + 341649*A*b*c^2*x^(25/2) + 294525*b*B*c^2*x^( 29/2) + 98175*A*c^3*x^(29/2) + 86275*B*c^3*x^(33/2)))/2847075
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.
\(\displaystyle \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^{15/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (A b^3 x^{15/2}+b^2 x^{19/2} (3 A c+b B)+c^2 x^{27/2} (A c+3 b B)+3 b c x^{23/2} (A c+b B)+B c^3 x^{31/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{17} A b^3 x^{17/2}+\frac {2}{21} b^2 x^{21/2} (3 A c+b B)+\frac {2}{29} c^2 x^{29/2} (A c+3 b B)+\frac {6}{25} b c x^{25/2} (A c+b B)+\frac {2}{33} B c^3 x^{33/2}\) |
Input:
Int[x^(3/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x]
Output:
(2*A*b^3*x^(17/2))/17 + (2*b^2*(b*B + 3*A*c)*x^(21/2))/21 + (6*b*c*(b*B + A*c)*x^(25/2))/25 + (2*c^2*(3*b*B + A*c)*x^(29/2))/29 + (2*B*c^3*x^(33/2)) /33
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 | \(\frac {2 B \,c^{3} x^{\frac {33}{2}}}{33}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {29}{2}}}{29}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {25}{2}}}{25}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {21}{2}}}{21}+\frac {2 A \,b^{3} x^{\frac {17}{2}}}{17}\) | \(76\) |
default | \(\frac {2 B \,c^{3} x^{\frac {33}{2}}}{33}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {29}{2}}}{29}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {25}{2}}}{25}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {21}{2}}}{21}+\frac {2 A \,b^{3} x^{\frac {17}{2}}}{17}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {17}{2}} \left (86275 B \,c^{3} x^{8}+98175 A \,c^{3} x^{6}+294525 B b \,c^{2} x^{6}+341649 A b \,c^{2} x^{4}+341649 x^{4} B \,b^{2} c +406725 A \,b^{2} c \,x^{2}+135575 x^{2} B \,b^{3}+167475 A \,b^{3}\right )}{2847075}\) | \(80\) |
trager | \(\frac {2 x^{\frac {17}{2}} \left (86275 B \,c^{3} x^{8}+98175 A \,c^{3} x^{6}+294525 B b \,c^{2} x^{6}+341649 A b \,c^{2} x^{4}+341649 x^{4} B \,b^{2} c +406725 A \,b^{2} c \,x^{2}+135575 x^{2} B \,b^{3}+167475 A \,b^{3}\right )}{2847075}\) | \(80\) |
risch | \(\frac {2 x^{\frac {17}{2}} \left (86275 B \,c^{3} x^{8}+98175 A \,c^{3} x^{6}+294525 B b \,c^{2} x^{6}+341649 A b \,c^{2} x^{4}+341649 x^{4} B \,b^{2} c +406725 A \,b^{2} c \,x^{2}+135575 x^{2} B \,b^{3}+167475 A \,b^{3}\right )}{2847075}\) | \(80\) |
orering | \(\frac {2 \left (86275 B \,c^{3} x^{8}+98175 A \,c^{3} x^{6}+294525 B b \,c^{2} x^{6}+341649 A b \,c^{2} x^{4}+341649 x^{4} B \,b^{2} c +406725 A \,b^{2} c \,x^{2}+135575 x^{2} B \,b^{3}+167475 A \,b^{3}\right ) x^{\frac {5}{2}} \left (c \,x^{4}+b \,x^{2}\right )^{3}}{2847075 \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int(x^(3/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
2/33*B*c^3*x^(33/2)+2/29*(A*c^3+3*B*b*c^2)*x^(29/2)+2/25*(3*A*b*c^2+3*B*b^ 2*c)*x^(25/2)+2/21*(3*A*b^2*c+B*b^3)*x^(21/2)+2/17*A*b^3*x^(17/2)
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{2847075} \, {\left (86275 \, B c^{3} x^{16} + 98175 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{14} + 341649 \, {\left (B b^{2} c + A b c^{2}\right )} x^{12} + 167475 \, A b^{3} x^{8} + 135575 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{10}\right )} \sqrt {x} \] Input:
integrate(x^(3/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
2/2847075*(86275*B*c^3*x^16 + 98175*(3*B*b*c^2 + A*c^3)*x^14 + 341649*(B*b ^2*c + A*b*c^2)*x^12 + 167475*A*b^3*x^8 + 135575*(B*b^3 + 3*A*b^2*c)*x^10) *sqrt(x)
Time = 1.61 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2 A b^{3} x^{\frac {17}{2}}}{17} + \frac {2 A b^{2} c x^{\frac {21}{2}}}{7} + \frac {6 A b c^{2} x^{\frac {25}{2}}}{25} + \frac {2 A c^{3} x^{\frac {29}{2}}}{29} + \frac {2 B b^{3} x^{\frac {21}{2}}}{21} + \frac {6 B b^{2} c x^{\frac {25}{2}}}{25} + \frac {6 B b c^{2} x^{\frac {29}{2}}}{29} + \frac {2 B c^{3} x^{\frac {33}{2}}}{33} \] Input:
integrate(x**(3/2)*(B*x**2+A)*(c*x**4+b*x**2)**3,x)
Output:
2*A*b**3*x**(17/2)/17 + 2*A*b**2*c*x**(21/2)/7 + 6*A*b*c**2*x**(25/2)/25 + 2*A*c**3*x**(29/2)/29 + 2*B*b**3*x**(21/2)/21 + 6*B*b**2*c*x**(25/2)/25 + 6*B*b*c**2*x**(29/2)/29 + 2*B*c**3*x**(33/2)/33
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{33} \, B c^{3} x^{\frac {33}{2}} + \frac {2}{29} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {29}{2}} + \frac {6}{25} \, {\left (B b^{2} c + A b c^{2}\right )} x^{\frac {25}{2}} + \frac {2}{17} \, A b^{3} x^{\frac {17}{2}} + \frac {2}{21} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac {21}{2}} \] Input:
integrate(x^(3/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
2/33*B*c^3*x^(33/2) + 2/29*(3*B*b*c^2 + A*c^3)*x^(29/2) + 6/25*(B*b^2*c + A*b*c^2)*x^(25/2) + 2/17*A*b^3*x^(17/2) + 2/21*(B*b^3 + 3*A*b^2*c)*x^(21/2 )
Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2}{33} \, B c^{3} x^{\frac {33}{2}} + \frac {6}{29} \, B b c^{2} x^{\frac {29}{2}} + \frac {2}{29} \, A c^{3} x^{\frac {29}{2}} + \frac {6}{25} \, B b^{2} c x^{\frac {25}{2}} + \frac {6}{25} \, A b c^{2} x^{\frac {25}{2}} + \frac {2}{21} \, B b^{3} x^{\frac {21}{2}} + \frac {2}{7} \, A b^{2} c x^{\frac {21}{2}} + \frac {2}{17} \, A b^{3} x^{\frac {17}{2}} \] Input:
integrate(x^(3/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
2/33*B*c^3*x^(33/2) + 6/29*B*b*c^2*x^(29/2) + 2/29*A*c^3*x^(29/2) + 6/25*B *b^2*c*x^(25/2) + 6/25*A*b*c^2*x^(25/2) + 2/21*B*b^3*x^(21/2) + 2/7*A*b^2* c*x^(21/2) + 2/17*A*b^3*x^(17/2)
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=x^{21/2}\,\left (\frac {2\,B\,b^3}{21}+\frac {2\,A\,c\,b^2}{7}\right )+x^{29/2}\,\left (\frac {2\,A\,c^3}{29}+\frac {6\,B\,b\,c^2}{29}\right )+\frac {2\,A\,b^3\,x^{17/2}}{17}+\frac {2\,B\,c^3\,x^{33/2}}{33}+\frac {6\,b\,c\,x^{25/2}\,\left (A\,c+B\,b\right )}{25} \] Input:
int(x^(3/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x)
Output:
x^(21/2)*((2*B*b^3)/21 + (2*A*b^2*c)/7) + x^(29/2)*((2*A*c^3)/29 + (6*B*b* c^2)/29) + (2*A*b^3*x^(17/2))/17 + (2*B*c^3*x^(33/2))/33 + (6*b*c*x^(25/2) *(A*c + B*b))/25
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx=\frac {2 \sqrt {x}\, x^{8} \left (86275 b \,c^{3} x^{8}+98175 a \,c^{3} x^{6}+294525 b^{2} c^{2} x^{6}+341649 a b \,c^{2} x^{4}+341649 b^{3} c \,x^{4}+406725 a \,b^{2} c \,x^{2}+135575 b^{4} x^{2}+167475 a \,b^{3}\right )}{2847075} \] Input:
int(x^(3/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x)
Output:
(2*sqrt(x)*x**8*(167475*a*b**3 + 406725*a*b**2*c*x**2 + 341649*a*b*c**2*x* *4 + 98175*a*c**3*x**6 + 135575*b**4*x**2 + 341649*b**3*c*x**4 + 294525*b* *2*c**2*x**6 + 86275*b*c**3*x**8))/2847075