Integrand size = 26, antiderivative size = 85 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx=\frac {2}{9} A b^3 x^{9/2}+\frac {2}{13} b^2 (b B+3 A c) x^{13/2}+\frac {6}{17} b c (b B+A c) x^{17/2}+\frac {2}{21} c^2 (3 b B+A c) x^{21/2}+\frac {2}{25} B c^3 x^{25/2} \] Output:
2/9*A*b^3*x^(9/2)+2/13*b^2*(3*A*c+B*b)*x^(13/2)+6/17*b*c*(A*c+B*b)*x^(17/2 )+2/21*c^2*(A*c+3*B*b)*x^(21/2)+2/25*B*c^3*x^(25/2)
Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx=\frac {2 x^{9/2} \left (25 A \left (1547 b^3+3213 b^2 c x^2+2457 b c^2 x^4+663 c^3 x^6\right )+9 B x^2 \left (2975 b^3+6825 b^2 c x^2+5525 b c^2 x^4+1547 c^3 x^6\right )\right )}{348075} \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(5/2),x]
Output:
(2*x^(9/2)*(25*A*(1547*b^3 + 3213*b^2*c*x^2 + 2457*b*c^2*x^4 + 663*c^3*x^6 ) + 9*B*x^2*(2975*b^3 + 6825*b^2*c*x^2 + 5525*b*c^2*x^4 + 1547*c^3*x^6)))/ 348075
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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^{7/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (A b^3 x^{7/2}+b^2 x^{11/2} (3 A c+b B)+c^2 x^{19/2} (A c+3 b B)+3 b c x^{15/2} (A c+b B)+B c^3 x^{23/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{9} A b^3 x^{9/2}+\frac {2}{13} b^2 x^{13/2} (3 A c+b B)+\frac {2}{21} c^2 x^{21/2} (A c+3 b B)+\frac {6}{17} b c x^{17/2} (A c+b B)+\frac {2}{25} B c^3 x^{25/2}\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(5/2),x]
Output:
(2*A*b^3*x^(9/2))/9 + (2*b^2*(b*B + 3*A*c)*x^(13/2))/13 + (6*b*c*(b*B + A* c)*x^(17/2))/17 + (2*c^2*(3*b*B + A*c)*x^(21/2))/21 + (2*B*c^3*x^(25/2))/2 5
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 | \(\frac {2 B \,c^{3} x^{\frac {25}{2}}}{25}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {21}{2}}}{21}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {13}{2}}}{13}+\frac {2 A \,b^{3} x^{\frac {9}{2}}}{9}\) | \(76\) |
default | \(\frac {2 B \,c^{3} x^{\frac {25}{2}}}{25}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {21}{2}}}{21}+\frac {2 \left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (3 A \,b^{2} c +B \,b^{3}\right ) x^{\frac {13}{2}}}{13}+\frac {2 A \,b^{3} x^{\frac {9}{2}}}{9}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {9}{2}} \left (13923 B \,c^{3} x^{8}+16575 A \,c^{3} x^{6}+49725 B b \,c^{2} x^{6}+61425 A b \,c^{2} x^{4}+61425 x^{4} B \,b^{2} c +80325 A \,b^{2} c \,x^{2}+26775 x^{2} B \,b^{3}+38675 A \,b^{3}\right )}{348075}\) | \(80\) |
trager | \(\frac {2 x^{\frac {9}{2}} \left (13923 B \,c^{3} x^{8}+16575 A \,c^{3} x^{6}+49725 B b \,c^{2} x^{6}+61425 A b \,c^{2} x^{4}+61425 x^{4} B \,b^{2} c +80325 A \,b^{2} c \,x^{2}+26775 x^{2} B \,b^{3}+38675 A \,b^{3}\right )}{348075}\) | \(80\) |
risch | \(\frac {2 x^{\frac {9}{2}} \left (13923 B \,c^{3} x^{8}+16575 A \,c^{3} x^{6}+49725 B b \,c^{2} x^{6}+61425 A b \,c^{2} x^{4}+61425 x^{4} B \,b^{2} c +80325 A \,b^{2} c \,x^{2}+26775 x^{2} B \,b^{3}+38675 A \,b^{3}\right )}{348075}\) | \(80\) |
orering | \(\frac {2 \left (13923 B \,c^{3} x^{8}+16575 A \,c^{3} x^{6}+49725 B b \,c^{2} x^{6}+61425 A b \,c^{2} x^{4}+61425 x^{4} B \,b^{2} c +80325 A \,b^{2} c \,x^{2}+26775 x^{2} B \,b^{3}+38675 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{3}}{348075 x^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^(5/2),x,method=_RETURNVERBOSE)
Output:
2/25*B*c^3*x^(25/2)+2/21*(A*c^3+3*B*b*c^2)*x^(21/2)+2/17*(3*A*b*c^2+3*B*b^ 2*c)*x^(17/2)+2/13*(3*A*b^2*c+B*b^3)*x^(13/2)+2/9*A*b^3*x^(9/2)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx=\frac {2}{348075} \, {\left (13923 \, B c^{3} x^{12} + 16575 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{10} + 61425 \, {\left (B b^{2} c + A b c^{2}\right )} x^{8} + 38675 \, A b^{3} x^{4} + 26775 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{6}\right )} \sqrt {x} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(5/2),x, algorithm="fricas")
Output:
2/348075*(13923*B*c^3*x^12 + 16575*(3*B*b*c^2 + A*c^3)*x^10 + 61425*(B*b^2 *c + A*b*c^2)*x^8 + 38675*A*b^3*x^4 + 26775*(B*b^3 + 3*A*b^2*c)*x^6)*sqrt( x)
Time = 1.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx=\frac {2 A b^{3} x^{\frac {9}{2}}}{9} + \frac {6 A b^{2} c x^{\frac {13}{2}}}{13} + \frac {6 A b c^{2} x^{\frac {17}{2}}}{17} + \frac {2 A c^{3} x^{\frac {21}{2}}}{21} + \frac {2 B b^{3} x^{\frac {13}{2}}}{13} + \frac {6 B b^{2} c x^{\frac {17}{2}}}{17} + \frac {2 B b c^{2} x^{\frac {21}{2}}}{7} + \frac {2 B c^{3} x^{\frac {25}{2}}}{25} \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**(5/2),x)
Output:
2*A*b**3*x**(9/2)/9 + 6*A*b**2*c*x**(13/2)/13 + 6*A*b*c**2*x**(17/2)/17 + 2*A*c**3*x**(21/2)/21 + 2*B*b**3*x**(13/2)/13 + 6*B*b**2*c*x**(17/2)/17 + 2*B*b*c**2*x**(21/2)/7 + 2*B*c**3*x**(25/2)/25
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx=\frac {2}{25} \, B c^{3} x^{\frac {25}{2}} + \frac {2}{21} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {21}{2}} + \frac {6}{17} \, {\left (B b^{2} c + A b c^{2}\right )} x^{\frac {17}{2}} + \frac {2}{9} \, A b^{3} x^{\frac {9}{2}} + \frac {2}{13} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac {13}{2}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(5/2),x, algorithm="maxima")
Output:
2/25*B*c^3*x^(25/2) + 2/21*(3*B*b*c^2 + A*c^3)*x^(21/2) + 6/17*(B*b^2*c + A*b*c^2)*x^(17/2) + 2/9*A*b^3*x^(9/2) + 2/13*(B*b^3 + 3*A*b^2*c)*x^(13/2)
Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx=\frac {2}{25} \, B c^{3} x^{\frac {25}{2}} + \frac {2}{7} \, B b c^{2} x^{\frac {21}{2}} + \frac {2}{21} \, A c^{3} x^{\frac {21}{2}} + \frac {6}{17} \, B b^{2} c x^{\frac {17}{2}} + \frac {6}{17} \, A b c^{2} x^{\frac {17}{2}} + \frac {2}{13} \, B b^{3} x^{\frac {13}{2}} + \frac {6}{13} \, A b^{2} c x^{\frac {13}{2}} + \frac {2}{9} \, A b^{3} x^{\frac {9}{2}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^(5/2),x, algorithm="giac")
Output:
2/25*B*c^3*x^(25/2) + 2/7*B*b*c^2*x^(21/2) + 2/21*A*c^3*x^(21/2) + 6/17*B* b^2*c*x^(17/2) + 6/17*A*b*c^2*x^(17/2) + 2/13*B*b^3*x^(13/2) + 6/13*A*b^2* c*x^(13/2) + 2/9*A*b^3*x^(9/2)
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx=x^{13/2}\,\left (\frac {2\,B\,b^3}{13}+\frac {6\,A\,c\,b^2}{13}\right )+x^{21/2}\,\left (\frac {2\,A\,c^3}{21}+\frac {2\,B\,b\,c^2}{7}\right )+\frac {2\,A\,b^3\,x^{9/2}}{9}+\frac {2\,B\,c^3\,x^{25/2}}{25}+\frac {6\,b\,c\,x^{17/2}\,\left (A\,c+B\,b\right )}{17} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(5/2),x)
Output:
x^(13/2)*((2*B*b^3)/13 + (6*A*b^2*c)/13) + x^(21/2)*((2*A*c^3)/21 + (2*B*b *c^2)/7) + (2*A*b^3*x^(9/2))/9 + (2*B*c^3*x^(25/2))/25 + (6*b*c*x^(17/2)*( A*c + B*b))/17
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{5/2}} \, dx=\frac {2 \sqrt {x}\, x^{4} \left (13923 b \,c^{3} x^{8}+16575 a \,c^{3} x^{6}+49725 b^{2} c^{2} x^{6}+61425 a b \,c^{2} x^{4}+61425 b^{3} c \,x^{4}+80325 a \,b^{2} c \,x^{2}+26775 b^{4} x^{2}+38675 a \,b^{3}\right )}{348075} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^(5/2),x)
Output:
(2*sqrt(x)*x**4*(38675*a*b**3 + 80325*a*b**2*c*x**2 + 61425*a*b*c**2*x**4 + 16575*a*c**3*x**6 + 26775*b**4*x**2 + 61425*b**3*c*x**4 + 49725*b**2*c** 2*x**6 + 13923*b*c**3*x**8))/348075