Integrand size = 27, antiderivative size = 124 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2+c x^4}} \, dx=-\frac {A \sqrt {a+b x^2+c x^4}}{4 a x^4}+\frac {(3 A b-4 a B) \sqrt {a+b x^2+c x^4}}{8 a^2 x^2}-\frac {\left (3 A b^2-4 a b B-4 a A c\right ) \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{5/2}} \]
-1/16*(-4*A*a*c+3*A*b^2-4*B*a*b)*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b* x^2+a)^(1/2))/a^(5/2)-1/4*A*(c*x^4+b*x^2+a)^(1/2)/a/x^4+1/8*(3*A*b-4*B*a)* (c*x^4+b*x^2+a)^(1/2)/a^2/x^2
Time = 0.57 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2+c x^4}} \, dx=\frac {\sqrt {a} \sqrt {a+b x^2+c x^4} \left (3 A b x^2-2 a \left (A+2 B x^2\right )\right )+3 A b^2 x^4 \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )+4 a (b B+A c) x^4 \text {arctanh}\left (\frac {-\sqrt {c} x^2+\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{8 a^{5/2} x^4} \]
(Sqrt[a]*Sqrt[a + b*x^2 + c*x^4]*(3*A*b*x^2 - 2*a*(A + 2*B*x^2)) + 3*A*b^2 *x^4*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]] + 4*a*(b*B + A*c)*x^4*ArcTanh[(-(Sqrt[c]*x^2) + Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/(8* a^(5/2)*x^4)
Time = 0.31 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1578, 1237, 27, 1228, 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.
\(\displaystyle \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^6 \sqrt {c x^4+b x^2+a}}dx^2\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {2 A c x^2+3 A b-4 a B}{2 x^4 \sqrt {c x^4+b x^2+a}}dx^2}{2 a}-\frac {A \sqrt {a+b x^2+c x^4}}{2 a x^4}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {2 A c x^2+3 A b-4 a B}{x^4 \sqrt {c x^4+b x^2+a}}dx^2}{4 a}-\frac {A \sqrt {a+b x^2+c x^4}}{2 a x^4}\right )\) |
\(\Big \downarrow \) 1228 |
\(\displaystyle \frac {1}{2} \left (-\frac {-\frac {\left (-4 a A c-4 a b B+3 A b^2\right ) \int \frac {1}{x^2 \sqrt {c x^4+b x^2+a}}dx^2}{2 a}-\frac {(3 A b-4 a B) \sqrt {a+b x^2+c x^4}}{a x^2}}{4 a}-\frac {A \sqrt {a+b x^2+c x^4}}{2 a x^4}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\left (-4 a A c-4 a b B+3 A b^2\right ) \int \frac {1}{4 a-x^4}d\frac {b x^2+2 a}{\sqrt {c x^4+b x^2+a}}}{a}-\frac {(3 A b-4 a B) \sqrt {a+b x^2+c x^4}}{a x^2}}{4 a}-\frac {A \sqrt {a+b x^2+c x^4}}{2 a x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\left (-4 a A c-4 a b B+3 A b^2\right ) \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 a^{3/2}}-\frac {(3 A b-4 a B) \sqrt {a+b x^2+c x^4}}{a x^2}}{4 a}-\frac {A \sqrt {a+b x^2+c x^4}}{2 a x^4}\right )\) |
(-1/2*(A*Sqrt[a + b*x^2 + c*x^4])/(a*x^4) - (-(((3*A*b - 4*a*B)*Sqrt[a + b *x^2 + c*x^4])/(a*x^2)) + ((3*A*b^2 - 4*a*b*B - 4*a*A*c)*ArcTanh[(2*a + b* x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(2*a^(3/2)))/(4*a))/2
3.2.74.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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/(((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_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(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, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(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[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (-3 A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 a^{2} x^{4}}+\frac {\left (4 A a c -3 A \,b^{2}+4 a b B \right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 a^{\frac {5}{2}}}\) | \(98\) |
pseudoelliptic | \(\frac {\frac {\left (\left (a c -\frac {3 b^{2}}{4}\right ) A +a b B \right ) x^{4} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}+\frac {3 \left (\frac {2 \left (-2 B \,x^{2}-A \right ) a^{\frac {3}{2}}}{3}+A \sqrt {a}\, b \,x^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}}{a^{\frac {5}{2}} x^{4}}\) | \(104\) |
default | \(B \left (-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\right )+A \left (-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{4 a \,x^{4}}+\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 a^{2} x^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 a^{\frac {5}{2}}}+\frac {c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\right )\) | \(194\) |
elliptic | \(-\frac {B \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a \,x^{2}}+\frac {B b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}-\frac {A \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 a \,x^{4}}+\frac {3 A b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 a^{2} x^{2}}-\frac {3 A \,b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 a^{\frac {5}{2}}}+\frac {A c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\) | \(194\) |
-1/8*(c*x^4+b*x^2+a)^(1/2)*(-3*A*b*x^2+4*B*a*x^2+2*A*a)/a^2/x^4+1/16*(4*A* a*c-3*A*b^2+4*B*a*b)/a^(5/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2) )/x^2)
Time = 0.35 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.06 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2+c x^4}} \, dx=\left [\frac {{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} \sqrt {a} x^{4} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, A a^{2} + {\left (4 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )}}{32 \, a^{3} x^{4}}, -\frac {{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, A a^{2} + {\left (4 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )}}{16 \, a^{3} x^{4}}\right ] \]
[1/32*((4*B*a*b - 3*A*b^2 + 4*A*a*c)*sqrt(a)*x^4*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)*(2*A*a^2 + (4*B*a^2 - 3*A*a*b)*x^2))/(a^3*x^4 ), -1/16*((4*B*a*b - 3*A*b^2 + 4*A*a*c)*sqrt(-a)*x^4*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)*(2*A*a^2 + (4*B*a^2 - 3*A*a*b)*x^2))/(a^3*x^4)]
\[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {A + B x^{2}}{x^{5} \sqrt {a + b x^{2} + c x^{4}}}\, dx \]
Exception generated. \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2+c x^4}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (106) = 212\).
Time = 0.31 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.73 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2+c x^4}} \, dx=-\frac {{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{2}} + \frac {4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} B a b - 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} A b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} A a c + 8 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} B a^{2} \sqrt {c} - 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} B a^{2} b + 5 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} A a b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} A a^{2} c - 8 \, B a^{3} \sqrt {c} + 8 \, A a^{2} b \sqrt {c}}{8 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{2} a^{2}} \]
-1/8*(4*B*a*b - 3*A*b^2 + 4*A*a*c)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x ^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) + 1/8*(4*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x ^2 + a))^3*B*a*b - 3*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*A*b^2 + 4*( sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*A*a*c + 8*(sqrt(c)*x^2 - sqrt(c*x ^4 + b*x^2 + a))^2*B*a^2*sqrt(c) - 4*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a ))*B*a^2*b + 5*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*A*a*b^2 + 4*(sqrt(c )*x^2 - sqrt(c*x^4 + b*x^2 + a))*A*a^2*c - 8*B*a^3*sqrt(c) + 8*A*a^2*b*sqr t(c))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^2*a^2)
Timed out. \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {B\,x^2+A}{x^5\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \]