Integrand size = 19, antiderivative size = 108 \[ \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx=-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}+\frac {6 c \sqrt {b x^2+c x^4}}{35 b^2 x^6}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{35 b^3 x^4}+\frac {16 c^3 \sqrt {b x^2+c x^4}}{35 b^4 x^2} \] Output:
-1/7*(c*x^4+b*x^2)^(1/2)/b/x^8+6/35*c*(c*x^4+b*x^2)^(1/2)/b^2/x^6-8/35*c^2 *(c*x^4+b*x^2)^(1/2)/b^3/x^4+16/35*c^3*(c*x^4+b*x^2)^(1/2)/b^4/x^2
Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-5 b^3+6 b^2 c x^2-8 b c^2 x^4+16 c^3 x^6\right )}{35 b^4 x^8} \] Input:
Integrate[1/(x^7*Sqrt[b*x^2 + c*x^4]),x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(-5*b^3 + 6*b^2*c*x^2 - 8*b*c^2*x^4 + 16*c^3*x^6))/ (35*b^4*x^8)
Time = 0.43 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1423, 1423, 1423, 1422}
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.
\(\displaystyle \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {6 c \int \frac {1}{x^5 \sqrt {c x^4+b x^2}}dx}{7 b}-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {6 c \left (-\frac {4 c \int \frac {1}{x^3 \sqrt {c x^4+b x^2}}dx}{5 b}-\frac {\sqrt {b x^2+c x^4}}{5 b x^6}\right )}{7 b}-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {6 c \left (-\frac {4 c \left (-\frac {2 c \int \frac {1}{x \sqrt {c x^4+b x^2}}dx}{3 b}-\frac {\sqrt {b x^2+c x^4}}{3 b x^4}\right )}{5 b}-\frac {\sqrt {b x^2+c x^4}}{5 b x^6}\right )}{7 b}-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}\) |
\(\Big \downarrow \) 1422 |
\(\displaystyle -\frac {6 c \left (-\frac {4 c \left (\frac {2 c \sqrt {b x^2+c x^4}}{3 b^2 x^2}-\frac {\sqrt {b x^2+c x^4}}{3 b x^4}\right )}{5 b}-\frac {\sqrt {b x^2+c x^4}}{5 b x^6}\right )}{7 b}-\frac {\sqrt {b x^2+c x^4}}{7 b x^8}\) |
Input:
Int[1/(x^7*Sqrt[b*x^2 + c*x^4]),x]
Output:
-1/7*Sqrt[b*x^2 + c*x^4]/(b*x^8) - (6*c*(-1/5*Sqrt[b*x^2 + c*x^4]/(b*x^6) - (4*c*(-1/3*Sqrt[b*x^2 + c*x^4]/(b*x^4) + (2*c*Sqrt[b*x^2 + c*x^4])/(3*b^ 2*x^2)))/(5*b)))/(7*b)
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(-d)*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(2*b*(p + 1))), x] /; FreeQ[{b , c, d, m, p}, x] && !IntegerQ[p] && EqQ[m + 4*p + 3, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( (m + 4*p + 3)/(b*d^2*(m + 2*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] && ILtQ[Simplify[(m + 4*p + 3)/2], 0] && NeQ[m + 2*p + 1, 0]
Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.50
method | result | size |
trager | \(-\frac {\left (-16 c^{3} x^{6}+8 b \,c^{2} x^{4}-6 b^{2} c \,x^{2}+5 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{35 b^{4} x^{8}}\) | \(54\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+8 b \,c^{2} x^{4}-6 b^{2} c \,x^{2}+5 b^{3}\right )}{35 x^{6} b^{4} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(61\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+8 b \,c^{2} x^{4}-6 b^{2} c \,x^{2}+5 b^{3}\right )}{35 x^{6} b^{4} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(61\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+8 b \,c^{2} x^{4}-6 b^{2} c \,x^{2}+5 b^{3}\right )}{35 x^{6} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b^{4}}\) | \(61\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+8 b \,c^{2} x^{4}-6 b^{2} c \,x^{2}+5 b^{3}\right )}{35 x^{6} b^{4} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(61\) |
pseudoelliptic | \(\frac {16 c^{4} x^{8}+8 b \,c^{3} x^{6}-2 b^{2} c^{2} x^{4}+x^{2} b^{3} c -5 b^{4}}{35 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, x^{6} b^{4}}\) | \(64\) |
Input:
int(1/x^7/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/35*(-16*c^3*x^6+8*b*c^2*x^4-6*b^2*c*x^2+5*b^3)/b^4/x^8*(c*x^4+b*x^2)^(1 /2)
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx=\frac {{\left (16 \, c^{3} x^{6} - 8 \, b c^{2} x^{4} + 6 \, b^{2} c x^{2} - 5 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, b^{4} x^{8}} \] Input:
integrate(1/x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
1/35*(16*c^3*x^6 - 8*b*c^2*x^4 + 6*b^2*c*x^2 - 5*b^3)*sqrt(c*x^4 + b*x^2)/ (b^4*x^8)
\[ \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx=\int \frac {1}{x^{7} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(1/x**7/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(1/(x**7*sqrt(x**2*(b + c*x**2))), x)
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx=\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{35 \, b^{4} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{35 \, b^{3} x^{4}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} c}{35 \, b^{2} x^{6}} - \frac {\sqrt {c x^{4} + b x^{2}}}{7 \, b x^{8}} \] Input:
integrate(1/x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
16/35*sqrt(c*x^4 + b*x^2)*c^3/(b^4*x^2) - 8/35*sqrt(c*x^4 + b*x^2)*c^2/(b^ 3*x^4) + 6/35*sqrt(c*x^4 + b*x^2)*c/(b^2*x^6) - 1/7*sqrt(c*x^4 + b*x^2)/(b *x^8)
Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx=\frac {32 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} - 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b + 7 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{2} - b^{3}\right )} c^{\frac {7}{2}}}{35 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{7} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
32/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b))^6 - 21*(sqrt(c)*x - sqrt(c*x^2 + b ))^4*b + 7*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^2 - b^3)*c^(7/2)/(((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^7*sgn(x))
Time = 18.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx=\frac {6\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,b^2\,x^6}-\frac {\sqrt {c\,x^4+b\,x^2}}{7\,b\,x^8}-\frac {8\,c^2\,\sqrt {c\,x^4+b\,x^2}}{35\,b^3\,x^4}+\frac {16\,c^3\,\sqrt {c\,x^4+b\,x^2}}{35\,b^4\,x^2} \] Input:
int(1/(x^7*(b*x^2 + c*x^4)^(1/2)),x)
Output:
(6*c*(b*x^2 + c*x^4)^(1/2))/(35*b^2*x^6) - (b*x^2 + c*x^4)^(1/2)/(7*b*x^8) - (8*c^2*(b*x^2 + c*x^4)^(1/2))/(35*b^3*x^4) + (16*c^3*(b*x^2 + c*x^4)^(1 /2))/(35*b^4*x^2)
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^7 \sqrt {b x^2+c x^4}} \, dx=\frac {-5 \sqrt {c \,x^{2}+b}\, b^{3}+6 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{2}-8 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}+16 \sqrt {c \,x^{2}+b}\, c^{3} x^{6}-16 \sqrt {c}\, c^{3} x^{7}}{35 b^{4} x^{7}} \] Input:
int(1/x^7/(c*x^4+b*x^2)^(1/2),x)
Output:
( - 5*sqrt(b + c*x**2)*b**3 + 6*sqrt(b + c*x**2)*b**2*c*x**2 - 8*sqrt(b + c*x**2)*b*c**2*x**4 + 16*sqrt(b + c*x**2)*c**3*x**6 - 16*sqrt(c)*c**3*x**7 )/(35*b**4*x**7)