Integrand size = 22, antiderivative size = 77 \[ \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx=\frac {\sqrt {-a+b x^2+c x^4}}{2 a x^2}-\frac {b \arctan \left (\frac {2 a-b x^2}{2 \sqrt {a} \sqrt {-a+b x^2+c x^4}}\right )}{4 a^{3/2}} \] Output:
1/2*(c*x^4+b*x^2-a)^(1/2)/a/x^2-1/4*b*arctan(1/2*(-b*x^2+2*a)/a^(1/2)/(c*x ^4+b*x^2-a)^(1/2))/a^(3/2)
Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx=\frac {\sqrt {-a+b x^2+c x^4}}{2 a x^2}-\frac {b \arctan \left (\frac {\sqrt {c} x^2-\sqrt {-a+b x^2+c x^4}}{\sqrt {a}}\right )}{2 a^{3/2}} \] Input:
Integrate[1/(x^3*Sqrt[-a + b*x^2 + c*x^4]),x]
Output:
Sqrt[-a + b*x^2 + c*x^4]/(2*a*x^2) - (b*ArcTan[(Sqrt[c]*x^2 - Sqrt[-a + b* x^2 + c*x^4])/Sqrt[a]])/(2*a^(3/2))
Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1434, 1157, 1154, 217}
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.
\(\displaystyle \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \sqrt {c x^4+b x^2-a}}dx^2\) |
\(\Big \downarrow \) 1157 |
\(\displaystyle \frac {1}{2} \left (\frac {b \int \frac {1}{x^2 \sqrt {c x^4+b x^2-a}}dx^2}{2 a}+\frac {\sqrt {-a+b x^2+c x^4}}{a x^2}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (\frac {\sqrt {-a+b x^2+c x^4}}{a x^2}-\frac {b \int \frac {1}{-x^4-4 a}d\left (-\frac {2 a-b x^2}{\sqrt {c x^4+b x^2-a}}\right )}{a}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {\sqrt {-a+b x^2+c x^4}}{a x^2}-\frac {b \arctan \left (\frac {2 a-b x^2}{2 \sqrt {a} \sqrt {-a+b x^2+c x^4}}\right )}{2 a^{3/2}}\right )\) |
Input:
Int[1/(x^3*Sqrt[-a + b*x^2 + c*x^4]),x]
Output:
(Sqrt[-a + b*x^2 + c*x^4]/(a*x^2) - (b*ArcTan[(2*a - b*x^2)/(2*Sqrt[a]*Sqr t[-a + b*x^2 + c*x^4])])/(2*a^(3/2)))/2
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d ^2 - b*d*e + a*e^2))), x] + Simp[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.15 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}-a}}{2 a \,x^{2}}-\frac {b \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{4 a \sqrt {-a}}\) | \(74\) |
elliptic | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}-a}}{2 a \,x^{2}}-\frac {b \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{4 a \sqrt {-a}}\) | \(74\) |
pseudoelliptic | \(\frac {b \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right ) x^{2}-2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{4 x^{2} \left (-a \right )^{\frac {3}{2}}}\) | \(77\) |
risch | \(-\frac {-c \,x^{4}-b \,x^{2}+a}{2 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}-a}}-\frac {b \ln \left (\frac {-2 a +b \,x^{2}+2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{4 a \sqrt {-a}}\) | \(88\) |
Input:
int(1/x^3/(c*x^4+b*x^2-a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*(c*x^4+b*x^2-a)^(1/2)/a/x^2-1/4*b/a/(-a)^(1/2)*ln((-2*a+b*x^2+2*(-a)^( 1/2)*(c*x^4+b*x^2-a)^(1/2))/x^2)
Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.44 \[ \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx=\left [-\frac {\sqrt {-a} b x^{2} \log \left (\frac {{\left (b^{2} - 4 \, a c\right )} x^{4} - 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} - a} {\left (b x^{2} - 2 \, a\right )} \sqrt {-a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, \sqrt {c x^{4} + b x^{2} - a} a}{8 \, a^{2} x^{2}}, \frac {\sqrt {a} b x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} - a} {\left (b x^{2} - 2 \, a\right )} \sqrt {a}}{2 \, {\left (a c x^{4} + a b x^{2} - a^{2}\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2} - a} a}{4 \, a^{2} x^{2}}\right ] \] Input:
integrate(1/x^3/(c*x^4+b*x^2-a)^(1/2),x, algorithm="fricas")
Output:
[-1/8*(sqrt(-a)*b*x^2*log(((b^2 - 4*a*c)*x^4 - 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^2 - a)*(b*x^2 - 2*a)*sqrt(-a) + 8*a^2)/x^4) - 4*sqrt(c*x^4 + b*x^2 - a )*a)/(a^2*x^2), 1/4*(sqrt(a)*b*x^2*arctan(1/2*sqrt(c*x^4 + b*x^2 - a)*(b*x ^2 - 2*a)*sqrt(a)/(a*c*x^4 + a*b*x^2 - a^2)) + 2*sqrt(c*x^4 + b*x^2 - a)*a )/(a^2*x^2)]
\[ \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx=\int \frac {1}{x^{3} \sqrt {- a + b x^{2} + c x^{4}}}\, dx \] Input:
integrate(1/x**3/(c*x**4+b*x**2-a)**(1/2),x)
Output:
Integral(1/(x**3*sqrt(-a + b*x**2 + c*x**4)), x)
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx=-\frac {b \arcsin \left (-\frac {b}{\sqrt {b^{2} + 4 \, a c}} + \frac {2 \, a}{\sqrt {b^{2} + 4 \, a c} x^{2}}\right )}{4 \, a^{\frac {3}{2}}} + \frac {\sqrt {c x^{4} + b x^{2} - a}}{2 \, a x^{2}} \] Input:
integrate(1/x^3/(c*x^4+b*x^2-a)^(1/2),x, algorithm="maxima")
Output:
-1/4*b*arcsin(-b/sqrt(b^2 + 4*a*c) + 2*a/(sqrt(b^2 + 4*a*c)*x^2))/a^(3/2) + 1/2*sqrt(c*x^4 + b*x^2 - a)/(a*x^2)
Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.44 \[ \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx=\frac {b \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}}{\sqrt {a}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}\right )} b - 2 \, a \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}\right )}^{2} + a\right )} a} \] Input:
integrate(1/x^3/(c*x^4+b*x^2-a)^(1/2),x, algorithm="giac")
Output:
1/2*b*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 - a))/sqrt(a))/a^(3/2) - 1 /2*((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 - a))*b - 2*a*sqrt(c))/(((sqrt(c)*x^ 2 - sqrt(c*x^4 + b*x^2 - a))^2 + a)*a)
Time = 17.80 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx=\frac {\sqrt {c\,x^4+b\,x^2-a}}{2\,a\,x^2}-\frac {b\,\mathrm {atanh}\left (\frac {a-\frac {b\,x^2}{2}}{\sqrt {-a}\,\sqrt {c\,x^4+b\,x^2-a}}\right )}{4\,{\left (-a\right )}^{3/2}} \] Input:
int(1/(x^3*(b*x^2 - a + c*x^4)^(1/2)),x)
Output:
(b*x^2 - a + c*x^4)^(1/2)/(2*a*x^2) - (b*atanh((a - (b*x^2)/2)/((-a)^(1/2) *(b*x^2 - a + c*x^4)^(1/2))))/(4*(-a)^(3/2))
\[ \int \frac {1}{x^3 \sqrt {-a+b x^2+c x^4}} \, dx=\int \frac {1}{\sqrt {c \,x^{4}+b \,x^{2}-a}\, x^{3}}d x \] Input:
int(1/x^3/(c*x^4+b*x^2-a)^(1/2),x)
Output:
int(1/(sqrt( - a + b*x**2 + c*x**4)*x**3),x)