Integrand size = 26, antiderivative size = 131 \[ \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {8 b^2 (6 b B-7 A c) \sqrt {b x^2+c x^4}}{105 c^4 x}+\frac {4 b (6 b B-7 A c) x \sqrt {b x^2+c x^4}}{105 c^3}-\frac {(6 b B-7 A c) x^3 \sqrt {b x^2+c x^4}}{35 c^2}+\frac {B x^5 \sqrt {b x^2+c x^4}}{7 c} \] Output:
-8/105*b^2*(-7*A*c+6*B*b)*(c*x^4+b*x^2)^(1/2)/c^4/x+4/105*b*(-7*A*c+6*B*b) *x*(c*x^4+b*x^2)^(1/2)/c^3-1/35*(-7*A*c+6*B*b)*x^3*(c*x^4+b*x^2)^(1/2)/c^2 +1/7*B*x^5*(c*x^4+b*x^2)^(1/2)/c
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.65 \[ \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-48 b^3 B+8 b^2 c \left (7 A+3 B x^2\right )+3 c^3 x^4 \left (7 A+5 B x^2\right )-2 b c^2 x^2 \left (14 A+9 B x^2\right )\right )}{105 c^4 x} \] Input:
Integrate[(x^6*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(-48*b^3*B + 8*b^2*c*(7*A + 3*B*x^2) + 3*c^3*x^4*(7 *A + 5*B*x^2) - 2*b*c^2*x^2*(14*A + 9*B*x^2)))/(105*c^4*x)
Time = 0.49 (sec) , antiderivative size = 127, 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 = {1945, 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.
\(\displaystyle \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {B x^5 \sqrt {b x^2+c x^4}}{7 c}-\frac {(6 b B-7 A c) \int \frac {x^6}{\sqrt {c x^4+b x^2}}dx}{7 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {B x^5 \sqrt {b x^2+c x^4}}{7 c}-\frac {(6 b B-7 A c) \left (\frac {x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {4 b \int \frac {x^4}{\sqrt {c x^4+b x^2}}dx}{5 c}\right )}{7 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {B x^5 \sqrt {b x^2+c x^4}}{7 c}-\frac {(6 b B-7 A c) \left (\frac {x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {4 b \left (\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \int \frac {x^2}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{5 c}\right )}{7 c}\) |
\(\Big \downarrow \) 1420 |
\(\displaystyle \frac {B x^5 \sqrt {b x^2+c x^4}}{7 c}-\frac {\left (\frac {x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {4 b \left (\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x}\right )}{5 c}\right ) (6 b B-7 A c)}{7 c}\) |
Input:
Int[(x^6*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]
Output:
(B*x^5*Sqrt[b*x^2 + c*x^4])/(7*c) - ((6*b*B - 7*A*c)*((x^3*Sqrt[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)))/(7*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[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 1.00 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64
method | result | size |
trager | \(\frac {\left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 x^{2} B \,b^{2} c +56 A \,b^{2} c -48 B \,b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{4} x}\) | \(84\) |
gosper | \(\frac {\left (c \,x^{2}+b \right ) \left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 x^{2} B \,b^{2} c +56 A \,b^{2} c -48 B \,b^{3}\right ) x}{105 c^{4} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(89\) |
default | \(\frac {\left (c \,x^{2}+b \right ) \left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 x^{2} B \,b^{2} c +56 A \,b^{2} c -48 B \,b^{3}\right ) x}{105 c^{4} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(89\) |
risch | \(\frac {x \left (c \,x^{2}+b \right ) \left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 x^{2} B \,b^{2} c +56 A \,b^{2} c -48 B \,b^{3}\right )}{105 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c^{4}}\) | \(89\) |
orering | \(\frac {\left (c \,x^{2}+b \right ) \left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 x^{2} B \,b^{2} c +56 A \,b^{2} c -48 B \,b^{3}\right ) x}{105 c^{4} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(89\) |
Input:
int(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/105*(15*B*c^3*x^6+21*A*c^3*x^4-18*B*b*c^2*x^4-28*A*b*c^2*x^2+24*B*b^2*c* x^2+56*A*b^2*c-48*B*b^3)/c^4/x*(c*x^4+b*x^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.63 \[ \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {{\left (15 \, B c^{3} x^{6} - 3 \, {\left (6 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} - 48 \, B b^{3} + 56 \, A b^{2} c + 4 \, {\left (6 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{105 \, c^{4} x} \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
1/105*(15*B*c^3*x^6 - 3*(6*B*b*c^2 - 7*A*c^3)*x^4 - 48*B*b^3 + 56*A*b^2*c + 4*(6*B*b^2*c - 7*A*b*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^4*x)
\[ \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{6} \left (A + B x^{2}\right )}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(x**6*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x**6*(A + B*x**2)/sqrt(x**2*(b + c*x**2)), x)
Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.81 \[ \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {{\left (3 \, c^{3} x^{6} - b c^{2} x^{4} + 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} A}{15 \, \sqrt {c x^{2} + b} c^{3}} + \frac {{\left (5 \, c^{4} x^{8} - b c^{3} x^{6} + 2 \, b^{2} c^{2} x^{4} - 8 \, b^{3} c x^{2} - 16 \, b^{4}\right )} B}{35 \, \sqrt {c x^{2} + b} c^{4}} \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
1/15*(3*c^3*x^6 - b*c^2*x^4 + 4*b^2*c*x^2 + 8*b^3)*A/(sqrt(c*x^2 + b)*c^3) + 1/35*(5*c^4*x^8 - b*c^3*x^6 + 2*b^2*c^2*x^4 - 8*b^3*c*x^2 - 16*b^4)*B/( sqrt(c*x^2 + b)*c^4)
Time = 0.22 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.99 \[ \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {8 \, {\left (6 \, B b^{\frac {7}{2}} - 7 \, A b^{\frac {5}{2}} c\right )} \mathrm {sgn}\left (x\right )}{105 \, c^{4}} - \frac {{\left (B b^{3} - A b^{2} c\right )} \sqrt {c x^{2} + b}}{c^{4} \mathrm {sgn}\left (x\right )} + \frac {15 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B - 63 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b + 105 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{2} + 21 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A c - 70 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b c}{105 \, c^{4} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
8/105*(6*B*b^(7/2) - 7*A*b^(5/2)*c)*sgn(x)/c^4 - (B*b^3 - A*b^2*c)*sqrt(c* x^2 + b)/(c^4*sgn(x)) + 1/105*(15*(c*x^2 + b)^(7/2)*B - 63*(c*x^2 + b)^(5/ 2)*B*b + 105*(c*x^2 + b)^(3/2)*B*b^2 + 21*(c*x^2 + b)^(5/2)*A*c - 70*(c*x^ 2 + b)^(3/2)*A*b*c)/(c^4*sgn(x))
Time = 9.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.66 \[ \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {48\,B\,b^3-56\,A\,b^2\,c}{105\,c^4}-\frac {B\,x^6}{7\,c}-\frac {x^4\,\left (21\,A\,c^3-18\,B\,b\,c^2\right )}{105\,c^4}+\frac {4\,b\,x^2\,\left (7\,A\,c-6\,B\,b\right )}{105\,c^3}\right )}{x} \] Input:
int((x^6*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2),x)
Output:
-((b*x^2 + c*x^4)^(1/2)*((48*B*b^3 - 56*A*b^2*c)/(105*c^4) - (B*x^6)/(7*c) - (x^4*(21*A*c^3 - 18*B*b*c^2))/(105*c^4) + (4*b*x^2*(7*A*c - 6*B*b))/(10 5*c^3)))/x
Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.56 \[ \int \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (15 b \,c^{3} x^{6}+21 a \,c^{3} x^{4}-18 b^{2} c^{2} x^{4}-28 a b \,c^{2} x^{2}+24 b^{3} c \,x^{2}+56 a \,b^{2} c -48 b^{4}\right )}{105 c^{4}} \] Input:
int(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)
Output:
(sqrt(b + c*x**2)*(56*a*b**2*c - 28*a*b*c**2*x**2 + 21*a*c**3*x**4 - 48*b* *4 + 24*b**3*c*x**2 - 18*b**2*c**2*x**4 + 15*b*c**3*x**6))/(105*c**4)