Integrand size = 29, antiderivative size = 232 \[ \int \frac {1}{x^2 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {a+c x^4}}{a x}+\frac {\sqrt {c} x \sqrt {a+c x^4}}{a \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+c x^4}}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a+c x^4}} \]
-(c*x^4+a)^(1/2)/a/x+x*c^(1/2)*(c*x^4+a)^(1/2)/a/(a^(1/2)+x^2*c^(1/2))-c^( 1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^( 1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^ 2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(3/4)/(c*x^4+a)^(1/ 2)+1/2*c^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^( 1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*( a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(3/4)/(c* x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.21 \[ \int \frac {1}{x^2 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {c x^4}{a}\right )}{x \sqrt {a+c x^4}} \]
-((Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[-1/4, 1/2, 3/4, -((c*x^4)/a)])/(x *Sqrt[a + c*x^4]))
Time = 0.30 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4, 847, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {a+(2 b-2 (b+1)+2) x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 4 |
\(\displaystyle \int \frac {1}{x^2 \sqrt {a+c x^4}}dx\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {c \int \frac {x^2}{\sqrt {c x^4+a}}dx}{a}-\frac {\sqrt {a+c x^4}}{a x}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {c \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {c}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+a}}dx}{\sqrt {c}}\right )}{a}-\frac {\sqrt {a+c x^4}}{a x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {c}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}\right )}{a}-\frac {\sqrt {a+c x^4}}{a x}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {c \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}\right )}{a}-\frac {\sqrt {a+c x^4}}{a x}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {c \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {a+c x^4}}-\frac {x \sqrt {a+c x^4}}{\sqrt {a}+\sqrt {c} x^2}}{\sqrt {c}}\right )}{a}-\frac {\sqrt {a+c x^4}}{a x}\) |
-(Sqrt[a + c*x^4]/(a*x)) + (c*(-((-((x*Sqrt[a + c*x^4])/(Sqrt[a] + Sqrt[c] *x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt [c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(1/4)*Sqrt[a + c*x^4]))/Sqrt[c]) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/( Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/( 2*c^(3/4)*Sqrt[a + c*x^4])))/a
3.11.9.3.1 Defintions of rubi rules used
Int[(u_.)*((a_.) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(a + c*x^(2*n))^p, x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[b, 0]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Result contains complex when optimal does not.
Time = 0.48 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.50
method | result | size |
default | \(-\frac {\sqrt {c \,x^{4}+a}}{a x}+\frac {i \sqrt {c}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) | \(115\) |
risch | \(-\frac {\sqrt {c \,x^{4}+a}}{a x}+\frac {i \sqrt {c}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) | \(115\) |
elliptic | \(-\frac {\sqrt {c \,x^{4}+a}}{a x}+\frac {i \sqrt {c}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) | \(115\) |
-(c*x^4+a)^(1/2)/a/x+I*c^(1/2)/a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1 /2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*(El lipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/ 2),I))
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^2 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {a} x \left (-\frac {c}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - \sqrt {a} x \left (-\frac {c}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + \sqrt {c x^{4} + a}}{a x} \]
-(sqrt(a)*x*(-c/a)^(3/4)*elliptic_e(arcsin(x*(-c/a)^(1/4)), -1) - sqrt(a)* x*(-c/a)^(3/4)*elliptic_f(arcsin(x*(-c/a)^(1/4)), -1) + sqrt(c*x^4 + a))/( a*x)
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.17 \[ \int \frac {1}{x^2 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} \]
\[ \int \frac {1}{x^2 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + a} x^{2}} \,d x } \]
\[ \int \frac {1}{x^2 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + a} x^{2}} \,d x } \]
Time = 13.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.17 \[ \int \frac {1}{x^2 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {\frac {a}{c\,x^4}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {a}{c\,x^4}\right )}{3\,x\,\sqrt {c\,x^4+a}} \]