Integrand size = 26, antiderivative size = 243 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx=\frac {2 c \arctan \left (\frac {\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}-\frac {2 c \arctan \left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \]
2*c*arctan(x*(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(e*x^2+d)^(1/2)/(b-(-4 *a*c+b^2)^(1/2))^(1/2))/(-4*a*c+b^2)^(1/2)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)) )^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)-2*c*arctan(x*(2*c*d-e*(b+(-4*a*c+b^2) ^(1/2)))^(1/2)/(e*x^2+d)^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))/(-4*a*c+b^2)^ (1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.14 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx=-2 e^{3/2} \text {RootSum}\left [c d^4-4 c d^3 \text {$\#$1}+4 b d^2 e \text {$\#$1}+6 c d^2 \text {$\#$1}^2-8 b d e \text {$\#$1}^2+16 a e^2 \text {$\#$1}^2-4 c d \text {$\#$1}^3+4 b e \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {\log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}}{-c d^3+b d^2 e+3 c d^2 \text {$\#$1}-4 b d e \text {$\#$1}+8 a e^2 \text {$\#$1}-3 c d \text {$\#$1}^2+3 b e \text {$\#$1}^2+c \text {$\#$1}^3}\&\right ] \]
-2*e^(3/2)*RootSum[c*d^4 - 4*c*d^3*#1 + 4*b*d^2*e*#1 + 6*c*d^2*#1^2 - 8*b* d*e*#1^2 + 16*a*e^2*#1^2 - 4*c*d*#1^3 + 4*b*e*#1^3 + c*#1^4 & , (Log[d + 2 *e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1)/(-(c*d^3) + b*d^2*e + 3*c*d ^2*#1 - 4*b*d*e*#1 + 8*a*e^2*#1 - 3*c*d*#1^2 + 3*b*e*#1^2 + c*#1^3) & ]
Time = 0.33 (sec) , antiderivative size = 243, 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 = {1488, 291, 218}
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.
\(\displaystyle \int \frac {1}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 1488 |
\(\displaystyle \frac {2 c \int \frac {1}{\left (2 c x^2+b-\sqrt {b^2-4 a c}\right ) \sqrt {e x^2+d}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{\left (2 c x^2+b+\sqrt {b^2-4 a c}\right ) \sqrt {e x^2+d}}dx}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {2 c \int \frac {1}{-\frac {\left (\left (b-\sqrt {b^2-4 a c}\right ) e-2 c d\right ) x^2}{e x^2+d}+b-\sqrt {b^2-4 a c}}d\frac {x}{\sqrt {e x^2+d}}}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{-\frac {\left (\left (b+\sqrt {b^2-4 a c}\right ) e-2 c d\right ) x^2}{e x^2+d}+b+\sqrt {b^2-4 a c}}d\frac {x}{\sqrt {e x^2+d}}}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 c \arctan \left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}-\frac {2 c \arctan \left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{\sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\) |
(2*c*ArcTan[(Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c ]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]) - (2*c*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])] )/(Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^ 2 - 4*a*c])*e])
3.4.90.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb ol] :> With[{r = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/r) Int[(d + e*x^2)^q/(b - r + 2*c*x^2), x], x] - Simp[2*(c/r) Int[(d + e*x^2)^q/(b + r + 2*c*x^2), x], x]] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && !IntegerQ[q]
Time = 0.38 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.92
method | result | size |
default | \(-\frac {\sqrt {2}\, \left (\frac {\left (b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) \arctan \left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (-2 a e +b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) a}}\right )}{\sqrt {\left (-2 a e +b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) a}}-\frac {\left (-b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) \operatorname {arctanh}\left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (2 a e -b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) a}}\right )}{\sqrt {\left (2 a e -b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) a}}\right )}{2 \sqrt {-d^{2} \left (4 a c -b^{2}\right )}}\) | \(224\) |
pseudoelliptic | \(-\frac {\sqrt {2}\, \left (\frac {\left (b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) \arctan \left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (-2 a e +b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) a}}\right )}{\sqrt {\left (-2 a e +b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) a}}-\frac {\left (-b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) \operatorname {arctanh}\left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (2 a e -b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) a}}\right )}{\sqrt {\left (2 a e -b d +\sqrt {-d^{2} \left (4 a c -b^{2}\right )}\right ) a}}\right )}{2 \sqrt {-d^{2} \left (4 a c -b^{2}\right )}}\) | \(224\) |
-1/2*2^(1/2)/(-d^2*(4*a*c-b^2))^(1/2)*((b*d+(-d^2*(4*a*c-b^2))^(1/2))/((-2 *a*e+b*d+(-d^2*(4*a*c-b^2))^(1/2))*a)^(1/2)*arctan(a/x*(e*x^2+d)^(1/2)*2^( 1/2)/((-2*a*e+b*d+(-d^2*(4*a*c-b^2))^(1/2))*a)^(1/2))-(-b*d+(-d^2*(4*a*c-b ^2))^(1/2))/((2*a*e-b*d+(-d^2*(4*a*c-b^2))^(1/2))*a)^(1/2)*arctanh(a/x*(e* x^2+d)^(1/2)*2^(1/2)/((2*a*e-b*d+(-d^2*(4*a*c-b^2))^(1/2))*a)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 4557 vs. \(2 (203) = 406\).
Time = 6.17 (sec) , antiderivative size = 4557, normalized size of antiderivative = 18.75 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
1/4*sqrt(1/2)*sqrt(-(b*c*d - (b^2 - 2*a*c)*e - ((a*b^2*c - 4*a^2*c^2)*d^2 - (a*b^3 - 4*a^2*b*c)*d*e + (a^2*b^2 - 4*a^3*c)*e^2)*sqrt((c^2*d^2 - 2*b*c *d*e + b^2*e^2)/((a^2*b^2*c^2 - 4*a^3*c^3)*d^4 - 2*(a^2*b^3*c - 4*a^3*b*c^ 2)*d^3*e + (a^2*b^4 - 2*a^3*b^2*c - 8*a^4*c^2)*d^2*e^2 - 2*(a^3*b^3 - 4*a^ 4*b*c)*d*e^3 + (a^4*b^2 - 4*a^5*c)*e^4)))/((a*b^2*c - 4*a^2*c^2)*d^2 - (a* b^3 - 4*a^2*b*c)*d*e + (a^2*b^2 - 4*a^3*c)*e^2))*log(-(2*a*c^2*d^2 - 2*a*b *c*d*e + ((a*b^2*c^2 - 4*a^2*c^3)*d^3 - (a*b^3*c - 4*a^2*b*c^2)*d^2*e + (a ^2*b^2*c - 4*a^3*c^2)*d*e^2)*x^2*sqrt((c^2*d^2 - 2*b*c*d*e + b^2*e^2)/((a^ 2*b^2*c^2 - 4*a^3*c^3)*d^4 - 2*(a^2*b^3*c - 4*a^3*b*c^2)*d^3*e + (a^2*b^4 - 2*a^3*b^2*c - 8*a^4*c^2)*d^2*e^2 - 2*(a^3*b^3 - 4*a^4*b*c)*d*e^3 + (a^4* b^2 - 4*a^5*c)*e^4)) - (b*c^2*d^2 + 4*a*b*c*e^2 - (b^2*c + 4*a*c^2)*d*e)*x ^2 + 2*sqrt(1/2)*sqrt(e*x^2 + d)*((2*(a^2*b^2*c^2 - 4*a^3*c^3)*d^3 - 3*(a^ 2*b^3*c - 4*a^3*b*c^2)*d^2*e + (a^2*b^4 - 2*a^3*b^2*c - 8*a^4*c^2)*d*e^2 - (a^3*b^3 - 4*a^4*b*c)*e^3)*x*sqrt((c^2*d^2 - 2*b*c*d*e + b^2*e^2)/((a^2*b ^2*c^2 - 4*a^3*c^3)*d^4 - 2*(a^2*b^3*c - 4*a^3*b*c^2)*d^3*e + (a^2*b^4 - 2 *a^3*b^2*c - 8*a^4*c^2)*d^2*e^2 - 2*(a^3*b^3 - 4*a^4*b*c)*d*e^3 + (a^4*b^2 - 4*a^5*c)*e^4)) - ((a*b^2*c - 4*a^2*c^2)*d*e - (a*b^3 - 4*a^2*b*c)*e^2)* x)*sqrt(-(b*c*d - (b^2 - 2*a*c)*e - ((a*b^2*c - 4*a^2*c^2)*d^2 - (a*b^3 - 4*a^2*b*c)*d*e + (a^2*b^2 - 4*a^3*c)*e^2)*sqrt((c^2*d^2 - 2*b*c*d*e + b^2* e^2)/((a^2*b^2*c^2 - 4*a^3*c^3)*d^4 - 2*(a^2*b^3*c - 4*a^3*b*c^2)*d^3*e...
\[ \int \frac {1}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx=\int \frac {1}{\sqrt {d + e x^{2}} \left (a + b x^{2} + c x^{4}\right )}\, dx \]
\[ \int \frac {1}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} \sqrt {e x^{2} + d}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx=\int \frac {1}{\sqrt {e\,x^2+d}\,\left (c\,x^4+b\,x^2+a\right )} \,d x \]