Integrand size = 28, antiderivative size = 87 \[ \int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}+\frac {3 c \sqrt {b x^2+c x^4}}{8 b^2 x^3}-\frac {3 c^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{5/2}} \] Output:
-1/4*(c*x^4+b*x^2)^(1/2)/b/x^5+3/8*c*(c*x^4+b*x^2)^(1/2)/b^2/x^3-3/8*c^2*a rctanh(b^(1/2)*x/(c*x^4+b*x^2)^(1/2))/b^(5/2)
Time = 0.02 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\sqrt {b} \left (-2 b^2+b c x^2+3 c^2 x^4\right )-3 c^2 x^4 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{8 b^{5/2} x^3 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[1/(x^4*Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4]),x]
Output:
(Sqrt[b]*(-2*b^2 + b*c*x^2 + 3*c^2*x^4) - 3*c^2*x^4*Sqrt[b + c*x^2]*ArcTan h[Sqrt[b + c*x^2]/Sqrt[b]])/(8*b^(5/2)*x^3*Sqrt[x^2*(b + c*x^2)])
Time = 0.37 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3, 1430, 1430, 1400, 219}
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.
\(\displaystyle \int \frac {1}{x^4 \sqrt {2 a-2 (a+1)+b x^2+c x^4+2}} \, dx\) |
\(\Big \downarrow \) 3 |
\(\displaystyle \int \frac {1}{x^4 \sqrt {b x^2+c x^4}}dx\) |
\(\Big \downarrow \) 1430 |
\(\displaystyle -\frac {3 c \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\) |
\(\Big \downarrow \) 1430 |
\(\displaystyle -\frac {3 c \left (-\frac {c \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle -\frac {3 c \left (\frac {c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {3 c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\) |
Input:
Int[1/(x^4*Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4]),x]
Output:
-1/4*Sqrt[b*x^2 + c*x^4]/(b*x^5) - (3*c*(-1/2*Sqrt[b*x^2 + c*x^4]/(b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2))))/(4*b)
Int[(u_.)*((a_) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n + c*x^(2*n))^p, x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x ^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
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] && LtQ[m + 2*p + 1, 0 ]
Time = 0.68 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.08
method | result | size |
default | \(-\frac {\sqrt {c \,x^{2}+b}\, \left (3 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b \,c^{2} x^{4}-3 b^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, c \,x^{2}+2 \sqrt {c \,x^{2}+b}\, b^{\frac {5}{2}}\right )}{8 x^{3} \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {7}{2}}}\) | \(94\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (-3 c \,x^{2}+2 b \right )}{8 b^{2} x^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {3 c^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) x \sqrt {c \,x^{2}+b}}{8 b^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(94\) |
Input:
int(1/x^4/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/8/x^3*(c*x^2+b)^(1/2)*(3*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*b*c^2*x^4- 3*b^(3/2)*(c*x^2+b)^(1/2)*c*x^2+2*(c*x^2+b)^(1/2)*b^(5/2))/(c*x^4+b*x^2)^( 1/2)/b^(7/2)
Time = 0.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.82 \[ \int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\left [\frac {3 \, \sqrt {b} c^{2} x^{5} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (3 \, b c x^{2} - 2 \, b^{2}\right )}}{16 \, b^{3} x^{5}}, \frac {3 \, \sqrt {-b} c^{2} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (3 \, b c x^{2} - 2 \, b^{2}\right )}}{8 \, b^{3} x^{5}}\right ] \] Input:
integrate(1/x^4/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
[1/16*(3*sqrt(b)*c^2*x^5*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt( b))/x^3) + 2*sqrt(c*x^4 + b*x^2)*(3*b*c*x^2 - 2*b^2))/(b^3*x^5), 1/8*(3*sq rt(-b)*c^2*x^5*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + sqrt(c*x^4 + b *x^2)*(3*b*c*x^2 - 2*b^2))/(b^3*x^5)]
\[ \int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\int \frac {1}{x^{4} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(1/x**4/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(1/(x**4*sqrt(x**2*(b + c*x**2))), x)
\[ \int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(c*x^4 + b*x^2)*x^4), x)
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\frac {3 \, c^{3} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} c^{3} - 5 \, \sqrt {c x^{2} + b} b c^{3}}{b^{2} c^{2} x^{4}}}{8 \, c \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^4/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
1/8*(3*c^3*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(c*x^2 + b )^(3/2)*c^3 - 5*sqrt(c*x^2 + b)*b*c^3)/(b^2*c^2*x^4))/(c*sgn(x))
Timed out. \[ \int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\int \frac {1}{x^4\,\sqrt {c\,x^4+b\,x^2}} \,d x \] Input:
int(1/(x^4*(b*x^2 + c*x^4)^(1/2)),x)
Output:
int(1/(x^4*(b*x^2 + c*x^4)^(1/2)), x)
Time = 0.18 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, b^{2}+3 \sqrt {c \,x^{2}+b}\, b c \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c^{2} x^{4}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c^{2} x^{4}}{8 b^{3} x^{4}} \] Input:
int(1/x^4/(c*x^4+b*x^2)^(1/2),x)
Output:
( - 2*sqrt(b + c*x**2)*b**2 + 3*sqrt(b + c*x**2)*b*c*x**2 + 3*sqrt(b)*log( (sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*c**2*x**4 - 3*sqrt(b)*lo g((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*c**2*x**4)/(8*b**3*x** 4)