Integrand size = 26, antiderivative size = 68 \[ \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx=-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3}-\frac {(2 b B-A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}} \] Output:
-1/2*A*(c*x^4+b*x^2)^(1/2)/b/x^3-1/2*(-A*c+2*B*b)*arctanh(b^(1/2)*x/(c*x^4 +b*x^2)^(1/2))/b^(3/2)
Time = 0.15 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.26 \[ \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx=\frac {-A \sqrt {b} \left (b+c x^2\right )-(2 b B-A c) x^2 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{2 b^{3/2} x \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(A + B*x^2)/(x^2*Sqrt[b*x^2 + c*x^4]),x]
Output:
(-(A*Sqrt[b]*(b + c*x^2)) - (2*b*B - A*c)*x^2*Sqrt[b + c*x^2]*ArcTanh[Sqrt [b + c*x^2]/Sqrt[b]])/(2*b^(3/2)*x*Sqrt[x^2*(b + c*x^2)])
Time = 0.35 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1944, 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.
\(\displaystyle \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle \frac {(2 b B-A c) \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle -\frac {(2 b B-A c) \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {(2 b B-A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3}\) |
Input:
Int[(A + B*x^2)/(x^2*Sqrt[b*x^2 + c*x^4]),x]
Output:
-1/2*(A*Sqrt[b*x^2 + c*x^4])/(b*x^3) - ((2*b*B - A*c)*ArcTanh[(Sqrt[b]*x)/ Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2))
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[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 0.69 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.32
method | result | size |
risch | \(-\frac {A \left (c \,x^{2}+b \right )}{2 b x \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (A c -2 B b \right ) \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) x \sqrt {c \,x^{2}+b}}{2 b^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(90\) |
default | \(-\frac {\sqrt {c \,x^{2}+b}\, \left (2 B \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b^{2} x^{2}-A \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b c \,x^{2}+A \sqrt {c \,x^{2}+b}\, b^{\frac {3}{2}}\right )}{2 x \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {5}{2}}}\) | \(105\) |
Input:
int((B*x^2+A)/x^2/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/2/b*A*(c*x^2+b)/x/(x^2*(c*x^2+b))^(1/2)+1/2*(A*c-2*B*b)/b^(3/2)*ln((2*b +2*b^(1/2)*(c*x^2+b)^(1/2))/x)*x/(x^2*(c*x^2+b))^(1/2)*(c*x^2+b)^(1/2)
Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.16 \[ \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx=\left [-\frac {{\left (2 \, B b - A c\right )} \sqrt {b} x^{3} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} A b}{4 \, b^{2} x^{3}}, \frac {{\left (2 \, B b - A c\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) - \sqrt {c x^{4} + b x^{2}} A b}{2 \, b^{2} x^{3}}\right ] \] Input:
integrate((B*x^2+A)/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
[-1/4*((2*B*b - A*c)*sqrt(b)*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)*A*b)/(b^2*x^3), 1/2*((2*B*b - A*c )*sqrt(-b)*x^3*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) - sqrt(c*x^4 + b *x^2)*A*b)/(b^2*x^3)]
\[ \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx=\int \frac {A + B x^{2}}{x^{2} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate((B*x**2+A)/x**2/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral((A + B*x**2)/(x**2*sqrt(x**2*(b + c*x**2))), x)
\[ \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2}} x^{2}} \,d x } \] Input:
integrate((B*x^2+A)/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2)*x^2), x)
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx=\frac {c {\left (\frac {{\left (2 \, B b - A c\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b c} - \frac {\sqrt {c x^{2} + b} A}{b c x^{2}}\right )}}{2 \, \mathrm {sgn}\left (x\right )} \] Input:
integrate((B*x^2+A)/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
1/2*c*((2*B*b - A*c)*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b*c) - sqr t(c*x^2 + b)*A/(b*c*x^2))/sgn(x)
Timed out. \[ \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx=\int \frac {B\,x^2+A}{x^2\,\sqrt {c\,x^4+b\,x^2}} \,d x \] Input:
int((A + B*x^2)/(x^2*(b*x^2 + c*x^4)^(1/2)),x)
Output:
int((A + B*x^2)/(x^2*(b*x^2 + c*x^4)^(1/2)), x)
Time = 0.20 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.15 \[ \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx=\frac {-\sqrt {c \,x^{2}+b}\, a b -\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c \,x^{2}+2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x^{2}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c \,x^{2}-2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x^{2}}{2 b^{2} x^{2}} \] Input:
int((B*x^2+A)/x^2/(c*x^4+b*x^2)^(1/2),x)
Output:
( - sqrt(b + c*x**2)*a*b - sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt( c)*x)/sqrt(b))*a*c*x**2 + 2*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt (c)*x)/sqrt(b))*b**2*x**2 + sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) + sqrt (c)*x)/sqrt(b))*a*c*x**2 - 2*sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) + sqr t(c)*x)/sqrt(b))*b**2*x**2)/(2*b**2*x**2)