Integrand size = 26, antiderivative size = 57 \[ \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx=-\frac {A \sqrt {b x^2+c x^4}}{b x^2}+\frac {B \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}} \] Output:
-A*(c*x^4+b*x^2)^(1/2)/b/x^2+B*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^ (1/2)
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.35 \[ \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx=\frac {-A \sqrt {c} \left (b+c x^2\right )-b B x \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )}{b \sqrt {c} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(A + B*x^2)/(x*Sqrt[b*x^2 + c*x^4]),x]
Output:
(-(A*Sqrt[c]*(b + c*x^2)) - b*B*x*Sqrt[b + c*x^2]*Log[-(Sqrt[c]*x) + Sqrt[ b + c*x^2]])/(b*Sqrt[c]*Sqrt[x^2*(b + c*x^2)])
Time = 0.39 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1940, 1220, 1091, 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.
\(\displaystyle \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^2 \sqrt {c x^4+b x^2}}dx^2\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {1}{2} \left (B \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 A \sqrt {b x^2+c x^4}}{b x^2}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (2 B \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}-\frac {2 A \sqrt {b x^2+c x^4}}{b x^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {2 B \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}-\frac {2 A \sqrt {b x^2+c x^4}}{b x^2}\right )\) |
Input:
Int[(A + B*x^2)/(x*Sqrt[b*x^2 + c*x^4]),x]
Output:
((-2*A*Sqrt[b*x^2 + c*x^4])/(b*x^2) + (2*B*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^ 2 + c*x^4]])/Sqrt[c])/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_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
Time = 0.45 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18
method | result | size |
default | \(-\frac {\sqrt {c \,x^{2}+b}\, \left (-B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b x +A \sqrt {c \,x^{2}+b}\, \sqrt {c}\right )}{\sqrt {c \,x^{4}+b \,x^{2}}\, \sqrt {c}\, b}\) | \(67\) |
risch | \(-\frac {A \left (c \,x^{2}+b \right )}{b \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{\sqrt {c}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(72\) |
pseudoelliptic | \(-\frac {-\frac {B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b \,x^{2}}{2}+\frac {B \ln \left (2\right ) b \,x^{2}}{2}+A \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}}{\sqrt {c}\, b \,x^{2}}\) | \(78\) |
Input:
int((B*x^2+A)/x/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-(c*x^2+b)^(1/2)*(-B*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b*x+A*(c*x^2+b)^(1/2)*c ^(1/2))/(c*x^4+b*x^2)^(1/2)/c^(1/2)/b
Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.39 \[ \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx=\left [\frac {B b \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} A c}{2 \, b c x^{2}}, -\frac {B b \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} A c}{b c x^{2}}\right ] \] Input:
integrate((B*x^2+A)/x/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
[1/2*(B*b*sqrt(c)*x^2*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*sqrt(c*x^4 + b*x^2)*A*c)/(b*c*x^2), -(B*b*sqrt(-c)*x^2*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + sqrt(c*x^4 + b*x^2)*A*c)/(b*c*x^2)]
\[ \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx=\int \frac {A + B x^{2}}{x \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate((B*x**2+A)/x/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral((A + B*x**2)/(x*sqrt(x**2*(b + c*x**2))), x)
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx=\frac {B \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, \sqrt {c}} - \frac {\sqrt {c x^{4} + b x^{2}} A}{b x^{2}} \] Input:
integrate((B*x^2+A)/x/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
1/2*B*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/sqrt(c) - sqrt(c*x^ 4 + b*x^2)*A/(b*x^2)
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx=-\frac {B \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right )}{2 \, \sqrt {c} \mathrm {sgn}\left (x\right )} + \frac {2 \, A \sqrt {c}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )} \mathrm {sgn}\left (x\right )} \] Input:
integrate((B*x^2+A)/x/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
-1/2*B*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)/(sqrt(c)*sgn(x)) + 2*A*sqrt(c) /(((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)*sgn(x))
Time = 9.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx=\frac {B\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,\sqrt {c}}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{b\,x^2} \] Input:
int((A + B*x^2)/(x*(b*x^2 + c*x^4)^(1/2)),x)
Output:
(B*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(2*c^(1/2)) - (A*(b *x^2 + c*x^4)^(1/2))/(b*x^2)
Time = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2}{x \sqrt {b x^2+c x^4}} \, dx=\frac {-\sqrt {c \,x^{2}+b}\, a c +\sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x -\sqrt {c}\, a c x}{b c x} \] Input:
int((B*x^2+A)/x/(c*x^4+b*x^2)^(1/2),x)
Output:
( - sqrt(b + c*x**2)*a*c + sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt (b))*b**2*x - sqrt(c)*a*c*x)/(b*c*x)