Integrand size = 26, antiderivative size = 139 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {(b B-A c) x^7}{b c \sqrt {b x^2+c x^4}}+\frac {8 b (6 b B-5 A c) \sqrt {b x^2+c x^4}}{15 c^4 x}-\frac {4 (6 b B-5 A c) x \sqrt {b x^2+c x^4}}{15 c^3}+\frac {(6 b B-5 A c) x^3 \sqrt {b x^2+c x^4}}{5 b c^2} \] Output:
-(-A*c+B*b)*x^7/b/c/(c*x^4+b*x^2)^(1/2)+8/15*b*(-5*A*c+6*B*b)*(c*x^4+b*x^2 )^(1/2)/c^4/x-4/15*(-5*A*c+6*B*b)*x*(c*x^4+b*x^2)^(1/2)/c^3+1/5*(-5*A*c+6* B*b)*x^3*(c*x^4+b*x^2)^(1/2)/b/c^2
Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.59 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x \left (48 b^3 B-8 b^2 c \left (5 A-3 B x^2\right )+c^3 x^4 \left (5 A+3 B x^2\right )-2 b c^2 x^2 \left (10 A+3 B x^2\right )\right )}{15 c^4 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
Output:
(x*(48*b^3*B - 8*b^2*c*(5*A - 3*B*x^2) + c^3*x^4*(5*A + 3*B*x^2) - 2*b*c^2 *x^2*(10*A + 3*B*x^2)))/(15*c^4*Sqrt[x^2*(b + c*x^2)])
Time = 0.53 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1943, 1421, 1421, 1420}
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.
\(\displaystyle \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {(6 b B-5 A c) \int \frac {x^6}{\sqrt {c x^4+b x^2}}dx}{b c}-\frac {x^7 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {(6 b B-5 A c) \left (\frac {x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {4 b \int \frac {x^4}{\sqrt {c x^4+b x^2}}dx}{5 c}\right )}{b c}-\frac {x^7 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {(6 b B-5 A c) \left (\frac {x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {4 b \left (\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \int \frac {x^2}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{5 c}\right )}{b c}-\frac {x^7 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1420 |
\(\displaystyle \frac {\left (\frac {x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {4 b \left (\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x}\right )}{5 c}\right ) (6 b B-5 A c)}{b c}-\frac {x^7 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
Input:
Int[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
Output:
-(((b*B - A*c)*x^7)/(b*c*Sqrt[b*x^2 + c*x^4])) + ((6*b*B - 5*A*c)*((x^3*Sq rt[b*x^2 + c*x^4])/(5*c) - (4*b*((-2*b*Sqrt[b*x^2 + c*x^4])/(3*c^2*x) + (x *Sqrt[b*x^2 + c*x^4])/(3*c)))/(5*c)))/(b*c)
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(2*c*(p + 1))), x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && EqQ[m + 2*p - 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && IGtQ[Simplify[(m + 2*p - 1)/2], 0] && NeQ[m + 4*p + 1, 0]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1))) Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 ] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
Time = 1.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-3 B \,c^{3} x^{6}-5 A \,c^{3} x^{4}+6 B b \,c^{2} x^{4}+20 A b \,c^{2} x^{2}-24 x^{2} B \,b^{2} c +40 A \,b^{2} c -48 B \,b^{3}\right ) x^{3}}{15 c^{4} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(91\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-3 B \,c^{3} x^{6}-5 A \,c^{3} x^{4}+6 B b \,c^{2} x^{4}+20 A b \,c^{2} x^{2}-24 x^{2} B \,b^{2} c +40 A \,b^{2} c -48 B \,b^{3}\right ) x^{3}}{15 c^{4} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(91\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-3 B \,c^{3} x^{6}-5 A \,c^{3} x^{4}+6 B b \,c^{2} x^{4}+20 A b \,c^{2} x^{2}-24 x^{2} B \,b^{2} c +40 A \,b^{2} c -48 B \,b^{3}\right ) x^{3}}{15 c^{4} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(91\) |
trager | \(-\frac {\left (-3 B \,c^{3} x^{6}-5 A \,c^{3} x^{4}+6 B b \,c^{2} x^{4}+20 A b \,c^{2} x^{2}-24 x^{2} B \,b^{2} c +40 A \,b^{2} c -48 B \,b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 \left (c \,x^{2}+b \right ) c^{4} x}\) | \(93\) |
risch | \(-\frac {\left (-3 B \,c^{2} x^{4}-5 A \,c^{2} x^{2}+9 x^{2} B b c +25 A b c -33 B \,b^{2}\right ) \left (c \,x^{2}+b \right ) x}{15 c^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {\left (A c -B b \right ) b^{2} x}{c^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(96\) |
Input:
int(x^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/15*(c*x^2+b)*(-3*B*c^3*x^6-5*A*c^3*x^4+6*B*b*c^2*x^4+20*A*b*c^2*x^2-24* B*b^2*c*x^2+40*A*b^2*c-48*B*b^3)*x^3/c^4/(c*x^4+b*x^2)^(3/2)
Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.67 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (3 \, B c^{3} x^{6} - {\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + 48 \, B b^{3} - 40 \, A b^{2} c + 4 \, {\left (6 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, {\left (c^{5} x^{3} + b c^{4} x\right )}} \] Input:
integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
1/15*(3*B*c^3*x^6 - (6*B*b*c^2 - 5*A*c^3)*x^4 + 48*B*b^3 - 40*A*b^2*c + 4* (6*B*b^2*c - 5*A*b*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^5*x^3 + b*c^4*x)
\[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{8} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**8*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(x**8*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)
Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.59 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (c^{2} x^{4} - 4 \, b c x^{2} - 8 \, b^{2}\right )} A}{3 \, \sqrt {c x^{2} + b} c^{3}} + \frac {{\left (c^{3} x^{6} - 2 \, b c^{2} x^{4} + 8 \, b^{2} c x^{2} + 16 \, b^{3}\right )} B}{5 \, \sqrt {c x^{2} + b} c^{4}} \] Input:
integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
1/3*(c^2*x^4 - 4*b*c*x^2 - 8*b^2)*A/(sqrt(c*x^2 + b)*c^3) + 1/5*(c^3*x^6 - 2*b*c^2*x^4 + 8*b^2*c*x^2 + 16*b^3)*B/(sqrt(c*x^2 + b)*c^4)
Time = 0.24 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {8 \, {\left (6 \, B b^{3} - 5 \, A b^{2} c\right )} \mathrm {sgn}\left (x\right )}{15 \, \sqrt {b} c^{4}} + \frac {B b^{3} - A b^{2} c}{\sqrt {c x^{2} + b} c^{4} \mathrm {sgn}\left (x\right )} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B c^{16} - 15 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b c^{16} + 45 \, \sqrt {c x^{2} + b} B b^{2} c^{16} + 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c^{17} - 30 \, \sqrt {c x^{2} + b} A b c^{17}}{15 \, c^{20} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
-8/15*(6*B*b^3 - 5*A*b^2*c)*sgn(x)/(sqrt(b)*c^4) + (B*b^3 - A*b^2*c)/(sqrt (c*x^2 + b)*c^4*sgn(x)) + 1/15*(3*(c*x^2 + b)^(5/2)*B*c^16 - 15*(c*x^2 + b )^(3/2)*B*b*c^16 + 45*sqrt(c*x^2 + b)*B*b^2*c^16 + 5*(c*x^2 + b)^(3/2)*A*c ^17 - 30*sqrt(c*x^2 + b)*A*b*c^17)/(c^20*sgn(x))
Time = 9.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.66 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (48\,B\,b^3+24\,B\,b^2\,c\,x^2-40\,A\,b^2\,c-6\,B\,b\,c^2\,x^4-20\,A\,b\,c^2\,x^2+3\,B\,c^3\,x^6+5\,A\,c^3\,x^4\right )}{15\,c^4\,x\,\left (c\,x^2+b\right )} \] Input:
int((x^8*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
Output:
((b*x^2 + c*x^4)^(1/2)*(48*B*b^3 + 5*A*c^3*x^4 + 3*B*c^3*x^6 - 40*A*b^2*c - 20*A*b*c^2*x^2 + 24*B*b^2*c*x^2 - 6*B*b*c^2*x^4))/(15*c^4*x*(b + c*x^2))
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.60 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (3 b \,c^{3} x^{6}+5 a \,c^{3} x^{4}-6 b^{2} c^{2} x^{4}-20 a b \,c^{2} x^{2}+24 b^{3} c \,x^{2}-40 a \,b^{2} c +48 b^{4}\right )}{15 c^{4} \left (c \,x^{2}+b \right )} \] Input:
int(x^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
Output:
(sqrt(b + c*x**2)*( - 40*a*b**2*c - 20*a*b*c**2*x**2 + 5*a*c**3*x**4 + 48* b**4 + 24*b**3*c*x**2 - 6*b**2*c**2*x**4 + 3*b*c**3*x**6))/(15*c**4*(b + c *x**2))