Integrand size = 8, antiderivative size = 165 \[ \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx=-\frac {1}{a x}-\frac {2 \sqrt {1+\frac {1}{a^2 x^4}}}{\left (a+\frac {1}{x^2}\right ) x}+\sqrt {1+\frac {1}{a^2 x^4}} x+\frac {2 \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{a^{3/2} \sqrt {1+\frac {1}{a^2 x^4}}}-\frac {\sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\sqrt {a} x\right ),\frac {1}{2}\right )}{a^{3/2} \sqrt {1+\frac {1}{a^2 x^4}}} \] Output:
-1/a/x-2*(1+1/a^2/x^4)^(1/2)/(a+1/x^2)/x+(1+1/a^2/x^4)^(1/2)*x+2*((a^2+1/x ^4)/(a+1/x^2)^2)^(1/2)*(a+1/x^2)*EllipticE(sin(2*arccot(a^(1/2)*x)),1/2*2^ (1/2))/a^(3/2)/(1+1/a^2/x^4)^(1/2)-((a^2+1/x^4)/(a+1/x^2)^2)^(1/2)*(a+1/x^ 2)*InverseJacobiAM(2*arccot(a^(1/2)*x),1/2*2^(1/2))/a^(3/2)/(1+1/a^2/x^4)^ (1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.58 \[ \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx=\frac {\sqrt {2} e^{\text {csch}^{-1}\left (a x^2\right )} \sqrt {\frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{-1+e^{2 \text {csch}^{-1}\left (a x^2\right )}}} x \left (-3+4 \sqrt {1-e^{2 \text {csch}^{-1}\left (a x^2\right )}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )\right )}{3 \sqrt {a x^2}} \] Input:
Integrate[E^ArcCsch[a*x^2],x]
Output:
(Sqrt[2]*E^ArcCsch[a*x^2]*Sqrt[E^ArcCsch[a*x^2]/(-1 + E^(2*ArcCsch[a*x^2]) )]*x*(-3 + 4*Sqrt[1 - E^(2*ArcCsch[a*x^2])]*Hypergeometric2F1[3/4, 3/2, 7/ 4, E^(2*ArcCsch[a*x^2])]))/(3*Sqrt[a*x^2])
Time = 0.58 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6885, 15, 773, 809, 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.
\(\displaystyle \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx\) |
\(\Big \downarrow \) 6885 |
\(\displaystyle \int \sqrt {1+\frac {1}{x^4 a^2}}dx+\frac {\int \frac {1}{x^2}dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \int \sqrt {1+\frac {1}{x^4 a^2}}dx-\frac {1}{a x}\) |
\(\Big \downarrow \) 773 |
\(\displaystyle -\int \sqrt {1+\frac {1}{x^4 a^2}} x^2d\frac {1}{x}-\frac {1}{a x}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle -\frac {2 \int \frac {1}{\sqrt {1+\frac {1}{x^4 a^2}} x^2}d\frac {1}{x}}{a^2}+x \sqrt {\frac {1}{a^2 x^4}+1}-\frac {1}{a x}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle -\frac {2 \left (a \int \frac {1}{\sqrt {1+\frac {1}{x^4 a^2}}}d\frac {1}{x}-a \int \frac {a-\frac {1}{x^2}}{a \sqrt {1+\frac {1}{x^4 a^2}}}d\frac {1}{x}\right )}{a^2}+x \sqrt {\frac {1}{a^2 x^4}+1}-\frac {1}{a x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \left (a \int \frac {1}{\sqrt {1+\frac {1}{x^4 a^2}}}d\frac {1}{x}-\int \frac {a-\frac {1}{x^2}}{\sqrt {1+\frac {1}{x^4 a^2}}}d\frac {1}{x}\right )}{a^2}+x \sqrt {\frac {1}{a^2 x^4}+1}-\frac {1}{a x}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {2 \left (\frac {\sqrt {a} \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{\sqrt {a} x}\right ),\frac {1}{2}\right )}{2 \sqrt {\frac {1}{a^2 x^4}+1}}-\int \frac {a-\frac {1}{x^2}}{\sqrt {1+\frac {1}{x^4 a^2}}}d\frac {1}{x}\right )}{a^2}+x \sqrt {\frac {1}{a^2 x^4}+1}-\frac {1}{a x}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle -\frac {2 \left (\frac {\sqrt {a} \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{\sqrt {a} x}\right ),\frac {1}{2}\right )}{2 \sqrt {\frac {1}{a^2 x^4}+1}}-\frac {\sqrt {a} \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) E\left (2 \arctan \left (\frac {1}{\sqrt {a} x}\right )|\frac {1}{2}\right )}{\sqrt {\frac {1}{a^2 x^4}+1}}+\frac {a^2 \sqrt {\frac {1}{a^2 x^4}+1}}{x \left (a+\frac {1}{x^2}\right )}\right )}{a^2}+x \sqrt {\frac {1}{a^2 x^4}+1}-\frac {1}{a x}\) |
Input:
Int[E^ArcCsch[a*x^2],x]
Output:
-(1/(a*x)) + Sqrt[1 + 1/(a^2*x^4)]*x - (2*((a^2*Sqrt[1 + 1/(a^2*x^4)])/((a + x^(-2))*x) - (Sqrt[a]*Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))* EllipticE[2*ArcTan[1/(Sqrt[a]*x)], 1/2])/Sqrt[1 + 1/(a^2*x^4)] + (Sqrt[a]* Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))*EllipticF[2*ArcTan[1/(Sqr t[a]*x)], 1/2])/(2*Sqrt[1 + 1/(a^2*x^4)])))/a^2
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^ 2, x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && !IntegerQ[p]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ [n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntB inomialQ[a, b, c, n, m, p, x]
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[((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]
Int[E^ArcCsch[(a_.)*(x_)^(p_.)], x_Symbol] :> Simp[1/a Int[1/x^p, x], x] + Int[Sqrt[1 + 1/(a^2*x^(2*p))], x] /; FreeQ[{a, p}, x]
Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {\sqrt {\frac {a^{2} x^{4}+1}{a^{2} x^{4}}}\, x \left (-\sqrt {i a}\, a^{2} x^{4}+2 i \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, x \operatorname {EllipticF}\left (x \sqrt {i a}, i\right ) a -2 i \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, x \operatorname {EllipticE}\left (x \sqrt {i a}, i\right ) a -\sqrt {i a}\right )}{\left (a^{2} x^{4}+1\right ) \sqrt {i a}}-\frac {1}{a x}\) | \(146\) |
Input:
int(1/a/x^2+(1+1/a^2/x^4)^(1/2),x,method=_RETURNVERBOSE)
Output:
((a^2*x^4+1)/a^2/x^4)^(1/2)*x*(-(I*a)^(1/2)*a^2*x^4+2*I*(1-I*a*x^2)^(1/2)* (1+I*a*x^2)^(1/2)*x*EllipticF(x*(I*a)^(1/2),I)*a-2*I*(1-I*a*x^2)^(1/2)*(1+ I*a*x^2)^(1/2)*x*EllipticE(x*(I*a)^(1/2),I)*a-(I*a)^(1/2))/(a^2*x^4+1)/(I* a)^(1/2)-1/a/x
\[ \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx=\int { \sqrt {\frac {1}{a^{2} x^{4}} + 1} + \frac {1}{a x^{2}} \,d x } \] Input:
integrate(1/a/x^2+(1+1/a^2/x^4)^(1/2),x, algorithm="fricas")
Output:
integral((a*x^2*sqrt((a^2*x^4 + 1)/(a^2*x^4)) + 1)/(a*x^2), x)
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.25 \[ \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx=- \frac {x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} - \frac {1}{a x} \] Input:
integrate(1/a/x**2+(1+1/a**2/x**4)**(1/2),x)
Output:
-x*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), exp_polar(I*pi)/(a**2*x**4))/(4 *gamma(3/4)) - 1/(a*x)
\[ \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx=\int { \sqrt {\frac {1}{a^{2} x^{4}} + 1} + \frac {1}{a x^{2}} \,d x } \] Input:
integrate(1/a/x^2+(1+1/a^2/x^4)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(a^2*x^4 + 1)/x^2, x)/a - 1/(a*x)
\[ \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx=\int { \sqrt {\frac {1}{a^{2} x^{4}} + 1} + \frac {1}{a x^{2}} \,d x } \] Input:
integrate(1/a/x^2+(1+1/a^2/x^4)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(1/(a^2*x^4) + 1) + 1/(a*x^2), x)
Time = 24.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.15 \[ \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx=x\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},-\frac {1}{4};\ \frac {3}{4};\ -\frac {1}{a^2\,x^4}\right )-\frac {1}{a\,x} \] Input:
int((1/(a^2*x^4) + 1)^(1/2) + 1/(a*x^2),x)
Output:
x*hypergeom([-1/2, -1/4], 3/4, -1/(a^2*x^4)) - 1/(a*x)
\[ \int e^{\text {csch}^{-1}\left (a x^2\right )} \, dx=\frac {\sqrt {a^{2} x^{4}+1}+2 \left (\int \frac {\sqrt {a^{2} x^{4}+1}}{a^{2} x^{6}+x^{2}}d x \right ) x -1}{a x} \] Input:
int(1/a/x^2+(1+1/a^2/x^4)^(1/2),x)
Output:
(sqrt(a**2*x**4 + 1) + 2*int(sqrt(a**2*x**4 + 1)/(a**2*x**6 + x**2),x)*x - 1)/(a*x)