Integrand size = 26, antiderivative size = 104 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {(b B-A c) x^5}{b c \sqrt {b x^2+c x^4}}-\frac {2 (4 b B-3 A c) \sqrt {b x^2+c x^4}}{3 c^3 x}+\frac {(4 b B-3 A c) x \sqrt {b x^2+c x^4}}{3 b c^2} \] Output:
-(-A*c+B*b)*x^5/b/c/(c*x^4+b*x^2)^(1/2)-2/3*(-3*A*c+4*B*b)*(c*x^4+b*x^2)^( 1/2)/c^3/x+1/3*(-3*A*c+4*B*b)*x*(c*x^4+b*x^2)^(1/2)/b/c^2
Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.58 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x \left (-8 b^2 B+c^2 x^2 \left (3 A+B x^2\right )+b \left (6 A c-4 B c x^2\right )\right )}{3 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(x^6*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
Output:
(x*(-8*b^2*B + c^2*x^2*(3*A + B*x^2) + b*(6*A*c - 4*B*c*x^2)))/(3*c^3*Sqrt [x^2*(b + c*x^2)])
Time = 0.44 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1943, 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.
\(\displaystyle \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {(4 b B-3 A c) \int \frac {x^4}{\sqrt {c x^4+b x^2}}dx}{b c}-\frac {x^5 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {(4 b B-3 A c) \left (\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \int \frac {x^2}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{b c}-\frac {x^5 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1420 |
\(\displaystyle \frac {\left (\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x}\right ) (4 b B-3 A c)}{b c}-\frac {x^5 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
Input:
Int[(x^6*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
Output:
-(((b*B - A*c)*x^5)/(b*c*Sqrt[b*x^2 + c*x^4])) + ((4*b*B - 3*A*c)*((-2*b*S qrt[b*x^2 + c*x^4])/(3*c^2*x) + (x*Sqrt[b*x^2 + c*x^4])/(3*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 = 1.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(\frac {\left (c \,x^{2}+b \right ) \left (B \,c^{2} x^{4}+3 A \,c^{2} x^{2}-4 x^{2} B b c +6 A b c -8 B \,b^{2}\right ) x^{3}}{3 c^{3} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(66\) |
default | \(\frac {\left (c \,x^{2}+b \right ) \left (B \,c^{2} x^{4}+3 A \,c^{2} x^{2}-4 x^{2} B b c +6 A b c -8 B \,b^{2}\right ) x^{3}}{3 c^{3} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(66\) |
orering | \(\frac {\left (c \,x^{2}+b \right ) \left (B \,c^{2} x^{4}+3 A \,c^{2} x^{2}-4 x^{2} B b c +6 A b c -8 B \,b^{2}\right ) x^{3}}{3 c^{3} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(66\) |
trager | \(\frac {\left (B \,c^{2} x^{4}+3 A \,c^{2} x^{2}-4 x^{2} B b c +6 A b c -8 B \,b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3 \left (c \,x^{2}+b \right ) c^{3} x}\) | \(68\) |
risch | \(\frac {\left (B c \,x^{2}+3 A c -5 B b \right ) \left (c \,x^{2}+b \right ) x}{3 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {b \left (A c -B b \right ) x}{c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(70\) |
Input:
int(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/3*(c*x^2+b)*(B*c^2*x^4+3*A*c^2*x^2-4*B*b*c*x^2+6*A*b*c-8*B*b^2)*x^3/c^3/ (c*x^4+b*x^2)^(3/2)
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (B c^{2} x^{4} - 8 \, B b^{2} + 6 \, A b c - {\left (4 \, B b c - 3 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3 \, {\left (c^{4} x^{3} + b c^{3} x\right )}} \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
1/3*(B*c^2*x^4 - 8*B*b^2 + 6*A*b*c - (4*B*b*c - 3*A*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^4*x^3 + b*c^3*x)
\[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{6} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**6*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(x**6*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)
Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.57 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (c x^{2} + 2 \, b\right )} A}{\sqrt {c x^{2} + b} c^{2}} + \frac {{\left (c^{2} x^{4} - 4 \, b c x^{2} - 8 \, b^{2}\right )} B}{3 \, \sqrt {c x^{2} + b} c^{3}} \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
(c*x^2 + 2*b)*A/(sqrt(c*x^2 + b)*c^2) + 1/3*(c^2*x^4 - 4*b*c*x^2 - 8*b^2)* B/(sqrt(c*x^2 + b)*c^3)
Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {2 \, {\left (4 \, B b^{2} - 3 \, A b c\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {b} c^{3}} - \frac {B b^{2} - A b c}{\sqrt {c x^{2} + b} c^{3} \mathrm {sgn}\left (x\right )} + \frac {{\left (c x^{2} + b\right )}^{\frac {3}{2}} B c^{6} - 6 \, \sqrt {c x^{2} + b} B b c^{6} + 3 \, \sqrt {c x^{2} + b} A c^{7}}{3 \, c^{9} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
2/3*(4*B*b^2 - 3*A*b*c)*sgn(x)/(sqrt(b)*c^3) - (B*b^2 - A*b*c)/(sqrt(c*x^2 + b)*c^3*sgn(x)) + 1/3*((c*x^2 + b)^(3/2)*B*c^6 - 6*sqrt(c*x^2 + b)*B*b*c ^6 + 3*sqrt(c*x^2 + b)*A*c^7)/(c^9*sgn(x))
Time = 9.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.64 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-8\,B\,b^2-4\,B\,b\,c\,x^2+6\,A\,b\,c+B\,c^2\,x^4+3\,A\,c^2\,x^2\right )}{3\,c^3\,x\,\left (c\,x^2+b\right )} \] Input:
int((x^6*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
Output:
((b*x^2 + c*x^4)^(1/2)*(3*A*c^2*x^2 - 8*B*b^2 + B*c^2*x^4 + 6*A*b*c - 4*B* b*c*x^2))/(3*c^3*x*(b + c*x^2))
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.57 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (b \,c^{2} x^{4}+3 a \,c^{2} x^{2}-4 b^{2} c \,x^{2}+6 a b c -8 b^{3}\right )}{3 c^{3} \left (c \,x^{2}+b \right )} \] Input:
int(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
Output:
(sqrt(b + c*x**2)*(6*a*b*c + 3*a*c**2*x**2 - 8*b**3 - 4*b**2*c*x**2 + b*c* *2*x**4))/(3*c**3*(b + c*x**2))