Integrand size = 28, antiderivative size = 59 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}+\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}} \] Output:
-1/2*(c*x^4+b*x^2)^(1/2)/b/x^3+1/2*c*arctanh(b^(1/2)*x/(c*x^4+b*x^2)^(1/2) )/b^(3/2)
Time = 0.01 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {-\sqrt {b} \left (b+c x^2\right )+c x^2 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{2 b^{3/2} x \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[1/(x^2*Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4]),x]
Output:
(-(Sqrt[b]*(b + c*x^2)) + c*x^2*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b + c*x^2]/Sq rt[b]])/(2*b^(3/2)*x*Sqrt[x^2*(b + c*x^2)])
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3, 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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {2 a-2 (a+1)+b x^2+c x^4+2}} \, dx\) |
\(\Big \downarrow \) 3 |
\(\displaystyle \int \frac {1}{x^2 \sqrt {b x^2+c x^4}}dx\) |
\(\Big \downarrow \) 1430 |
\(\displaystyle -\frac {c \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle \frac {c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\) |
Input:
Int[1/(x^2*Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4]),x]
Output:
-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))
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 = 0.58 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.24
method | result | size |
default | \(-\frac {\sqrt {c \,x^{2}+b}\, \left (-c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b \,x^{2}+\sqrt {c \,x^{2}+b}\, b^{\frac {3}{2}}\right )}{2 x \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {5}{2}}}\) | \(73\) |
risch | \(-\frac {c \,x^{2}+b}{2 b x \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) x \sqrt {c \,x^{2}+b}}{2 b^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(82\) |
Input:
int(1/x^2/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/2/x*(c*x^2+b)^(1/2)*(-c*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*b*x^2+(c*x^ 2+b)^(1/2)*b^(3/2))/(c*x^4+b*x^2)^(1/2)/b^(5/2)
Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.17 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\left [\frac {\sqrt {b} c x^{3} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} b}{4 \, b^{2} x^{3}}, -\frac {\sqrt {-b} c x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} b}{2 \, b^{2} x^{3}}\right ] \] Input:
integrate(1/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
[1/4*(sqrt(b)*c*x^3*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)*b)/(b^2*x^3), -1/2*(sqrt(-b)*c*x^3*arctan(sqrt (c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + sqrt(c*x^4 + b*x^2)*b)/(b^2*x^3)]
\[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\int \frac {1}{x^{2} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(1/x**2/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(1/(x**2*sqrt(x**2*(b + c*x**2))), x)
\[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(c*x^4 + b*x^2)*x^2), x)
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=-\frac {c {\left (\frac {\arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {\sqrt {c x^{2} + b}}{b c x^{2}}\right )}}{2 \, \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
-1/2*c*(arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b) + sqrt(c*x^2 + b)/(b *c*x^2))/sgn(x)
Time = 17.36 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=-\frac {\left (\frac {\sqrt {c}\,x^2\,\sqrt {c+\frac {b}{x^2}}}{2\,b}+\frac {c^{3/2}\,x^3\,\mathrm {asin}\left (\frac {\sqrt {b}\,1{}\mathrm {i}}{\sqrt {c}\,x}\right )\,1{}\mathrm {i}}{2\,b^{3/2}}\right )\,\sqrt {\frac {b}{c\,x^2}+1}}{x\,\sqrt {c\,x^4+b\,x^2}} \] Input:
int(1/(x^2*(b*x^2 + c*x^4)^(1/2)),x)
Output:
-(((c^(1/2)*x^2*(c + b/x^2)^(1/2))/(2*b) + (c^(3/2)*x^3*asin((b^(1/2)*1i)/ (c^(1/2)*x))*1i)/(2*b^(3/2)))*(b/(c*x^2) + 1)^(1/2))/(x*(b*x^2 + c*x^4)^(1 /2))
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {-\sqrt {c \,x^{2}+b}\, b -\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c \,x^{2}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c \,x^{2}}{2 b^{2} x^{2}} \] Input:
int(1/x^2/(c*x^4+b*x^2)^(1/2),x)
Output:
( - sqrt(b + c*x**2)*b - sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt(c) *x)/sqrt(b))*c*x**2 + sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x) /sqrt(b))*c*x**2)/(2*b**2*x**2)