Integrand size = 20, antiderivative size = 109 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x} \, dx=\frac {1}{2} \sqrt {a+b x^2+c x^4}-\frac {1}{2} \sqrt {a} \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )+\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c}} \]
-1/2*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))*a^(1/2)+1/4*b* arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(1/2)+1/2*(c*x^4+ b*x^2+a)^(1/2)
Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x} \, dx=\frac {1}{2} \sqrt {a+b x^2+c x^4}+\sqrt {a} \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )-\frac {b \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{4 \sqrt {c}} \]
Sqrt[a + b*x^2 + c*x^4]/2 + Sqrt[a]*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]] - (b*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4 ]])/(4*Sqrt[c])
Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1434, 1162, 25, 1269, 1092, 219, 1154, 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.
\(\displaystyle \int \frac {\sqrt {a+b x^2+c x^4}}{x} \, dx\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {c x^4+b x^2+a}}{x^2}dx^2\) |
\(\Big \downarrow \) 1162 |
\(\displaystyle \frac {1}{2} \left (\sqrt {a+b x^2+c x^4}-\frac {1}{2} \int -\frac {b x^2+2 a}{x^2 \sqrt {c x^4+b x^2+a}}dx^2\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {b x^2+2 a}{x^2 \sqrt {c x^4+b x^2+a}}dx^2+\sqrt {a+b x^2+c x^4}\right )\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (b \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2+2 a \int \frac {1}{x^2 \sqrt {c x^4+b x^2+a}}dx^2\right )+\sqrt {a+b x^2+c x^4}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (2 b \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}+2 a \int \frac {1}{x^2 \sqrt {c x^4+b x^2+a}}dx^2\right )+\sqrt {a+b x^2+c x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (2 a \int \frac {1}{x^2 \sqrt {c x^4+b x^2+a}}dx^2+\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{\sqrt {c}}\right )+\sqrt {a+b x^2+c x^4}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{\sqrt {c}}-4 a \int \frac {1}{4 a-x^4}d\frac {b x^2+2 a}{\sqrt {c x^4+b x^2+a}}\right )+\sqrt {a+b x^2+c x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{\sqrt {c}}-2 \sqrt {a} \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )\right )+\sqrt {a+b x^2+c x^4}\right )\) |
(Sqrt[a + b*x^2 + c*x^4] + (-2*Sqrt[a]*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sq rt[a + b*x^2 + c*x^4])] + (b*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x ^2 + c*x^4])])/Sqrt[c])/2)/2
3.10.24.3.1 Defintions of rubi rules used
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[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
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[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p/(e*(m + 2*p + 1)) Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x ] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || LtQ[m, 1]) && !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 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 = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2}+\frac {b \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 \sqrt {c}}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2}\) | \(91\) |
elliptic | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2}+\frac {b \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 \sqrt {c}}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2}\) | \(91\) |
pseudoelliptic | \(-\frac {\sqrt {a}\, \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) \sqrt {c}-\frac {b \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )}{2}+\frac {b \ln \left (2\right )}{2}-\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}}{2 \sqrt {c}}\) | \(106\) |
1/2*(c*x^4+b*x^2+a)^(1/2)+1/4*b*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^( 1/2))/c^(1/2)-1/2*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 = 0.30 (sec) , antiderivative size = 566, normalized size of antiderivative = 5.19 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x} \, dx=\left [\frac {b \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 2 \, \sqrt {a} c \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, \sqrt {c x^{4} + b x^{2} + a} c}{8 \, c}, -\frac {b \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - \sqrt {a} c \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2} + a} c}{4 \, c}, \frac {4 \, \sqrt {-a} c \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + b \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, \sqrt {c x^{4} + b x^{2} + a} c}{8 \, c}, \frac {2 \, \sqrt {-a} c \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) - b \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2} + a} c}{4 \, c}\right ] \]
[1/8*(b*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + 2*sqrt(a)*c*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)*c)/c, -1/4*(b*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) - sqrt(a)*c* 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) - 2*sqrt(c*x^4 + b*x^2 + a)*c)/c, 1/8*(4*sqrt(-a )*c*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)) + b*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + 4*sqrt(c*x^4 + b*x^2 + a)*c )/c, 1/4*(2*sqrt(-a)*c*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sq rt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) - b*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x ^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*sqrt(c*x^4 + b*x^2 + a)*c)/c]
\[ \int \frac {\sqrt {a+b x^2+c x^4}}{x} \, dx=\int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x}\, dx \]
Exception generated. \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Exception generated. \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Time = 12.98 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x} \, dx=\frac {\sqrt {c\,x^4+b\,x^2+a}}{2}-\frac {\sqrt {a}\,\ln \left (\frac {b}{2}+\frac {a}{x^2}+\frac {\sqrt {a}\,\sqrt {c\,x^4+b\,x^2+a}}{x^2}\right )}{2}+\frac {b\,\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )}{4\,\sqrt {c}} \]