3.3.90 \(\int \frac {\text {arcsinh}(a x^2)}{x^4} \, dx\) [290]

3.3.90.1 Optimal result

Integrand size = 10, antiderivative size = 197 \[ \int \frac {\text {arcsinh}\left (a x^2\right )}{x^4} \, dx=-\frac {2 a \sqrt {1+a^2 x^4}}{3 x}+\frac {2 a^2 x \sqrt {1+a^2 x^4}}{3 \left (1+a x^2\right )}-\frac {\text {arcsinh}\left (a x^2\right )}{3 x^3}-\frac {2 a^{3/2} \left (1+a x^2\right ) \sqrt {\frac {1+a^2 x^4}{\left (1+a x^2\right )^2}} E\left (2 \arctan \left (\sqrt {a} x\right )|\frac {1}{2}\right )}{3 \sqrt {1+a^2 x^4}}+\frac {a^{3/2} \left (1+a x^2\right ) \sqrt {\frac {1+a^2 x^4}{\left (1+a x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {a} x\right ),\frac {1}{2}\right )}{3 \sqrt {1+a^2 x^4}} \]

output

-1/3*arcsinh(a*x^2)/x^3-2/3*a*(a^2*x^4+1)^(1/2)/x+2/3*a^2*x*(a^2*x^4+1)^(1 
/2)/(a*x^2+1)-2/3*a^(3/2)*(a*x^2+1)*(cos(2*arctan(x*a^(1/2)))^2)^(1/2)/cos 
(2*arctan(x*a^(1/2)))*EllipticE(sin(2*arctan(x*a^(1/2))),1/2*2^(1/2))*((a^ 
2*x^4+1)/(a*x^2+1)^2)^(1/2)/(a^2*x^4+1)^(1/2)+1/3*a^(3/2)*(a*x^2+1)*(cos(2 
*arctan(x*a^(1/2)))^2)^(1/2)/cos(2*arctan(x*a^(1/2)))*EllipticF(sin(2*arct 
an(x*a^(1/2))),1/2*2^(1/2))*((a^2*x^4+1)/(a*x^2+1)^2)^(1/2)/(a^2*x^4+1)^(1 
/2)
 

3.3.90.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.45 \[ \int \frac {\text {arcsinh}\left (a x^2\right )}{x^4} \, dx=\frac {1}{3} \left (-\frac {2 a \sqrt {1+a^2 x^4}}{x}-\frac {\text {arcsinh}\left (a x^2\right )}{x^3}+\frac {2 a^2 \left (E\left (\left .i \text {arcsinh}\left (\sqrt {i a} x\right )\right |-1\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {i a} x\right ),-1\right )\right )}{\sqrt {i a}}\right ) \]

input

Integrate[ArcSinh[a*x^2]/x^4,x]
 

output

((-2*a*Sqrt[1 + a^2*x^4])/x - ArcSinh[a*x^2]/x^3 + (2*a^2*(EllipticE[I*Arc 
Sinh[Sqrt[I*a]*x], -1] - EllipticF[I*ArcSinh[Sqrt[I*a]*x], -1]))/Sqrt[I*a] 
)/3
 

3.3.90.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6290, 27, 847, 834, 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.

\(\displaystyle \int \frac {\text {arcsinh}\left (a x^2\right )}{x^4} \, dx\)

\(\Big \downarrow \) 6290

\(\displaystyle \frac {1}{3} \int \frac {2 a}{x^2 \sqrt {a^2 x^4+1}}dx-\frac {\text {arcsinh}\left (a x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} a \int \frac {1}{x^2 \sqrt {a^2 x^4+1}}dx-\frac {\text {arcsinh}\left (a x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {2}{3} a \left (a^2 \int \frac {x^2}{\sqrt {a^2 x^4+1}}dx-\frac {\sqrt {a^2 x^4+1}}{x}\right )-\frac {\text {arcsinh}\left (a x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {2}{3} a \left (a^2 \left (\frac {\int \frac {1}{\sqrt {a^2 x^4+1}}dx}{a}-\frac {\int \frac {1-a x^2}{\sqrt {a^2 x^4+1}}dx}{a}\right )-\frac {\sqrt {a^2 x^4+1}}{x}\right )-\frac {\text {arcsinh}\left (a x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{3} a \left (a^2 \left (\frac {\left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {a} x\right ),\frac {1}{2}\right )}{2 a^{3/2} \sqrt {a^2 x^4+1}}-\frac {\int \frac {1-a x^2}{\sqrt {a^2 x^4+1}}dx}{a}\right )-\frac {\sqrt {a^2 x^4+1}}{x}\right )-\frac {\text {arcsinh}\left (a x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2}{3} a \left (a^2 \left (\frac {\left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {a} x\right ),\frac {1}{2}\right )}{2 a^{3/2} \sqrt {a^2 x^4+1}}-\frac {\frac {\left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {a} x\right )|\frac {1}{2}\right )}{\sqrt {a} \sqrt {a^2 x^4+1}}-\frac {x \sqrt {a^2 x^4+1}}{a x^2+1}}{a}\right )-\frac {\sqrt {a^2 x^4+1}}{x}\right )-\frac {\text {arcsinh}\left (a x^2\right )}{3 x^3}\)

input

Int[ArcSinh[a*x^2]/x^4,x]
 

output

-1/3*ArcSinh[a*x^2]/x^3 + (2*a*(-(Sqrt[1 + a^2*x^4]/x) + a^2*(-((-((x*Sqrt 
[1 + a^2*x^4])/(1 + a*x^2)) + ((1 + a*x^2)*Sqrt[(1 + a^2*x^4)/(1 + a*x^2)^ 
2]*EllipticE[2*ArcTan[Sqrt[a]*x], 1/2])/(Sqrt[a]*Sqrt[1 + a^2*x^4]))/a) + 
((1 + a*x^2)*Sqrt[(1 + a^2*x^4)/(1 + a*x^2)^2]*EllipticF[2*ArcTan[Sqrt[a]* 
x], 1/2])/(2*a^(3/2)*Sqrt[1 + a^2*x^4]))))/3
 

3.3.90.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 761

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]
 

rule 834

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]
 

rule 847

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]
 

rule 1510

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]
 

rule 6290

Int[((a_.) + ArcSinh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*((a + b*ArcSinh[u])/(d*(m + 1))), x] - Simp[b/(d*(m + 
1))   Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/Sqrt[1 + u^2]), x], 
x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u 
, x] &&  !FunctionOfQ[(c + d*x)^(m + 1), u, x] &&  !FunctionOfExponentialQ[ 
u, x]
 

3.3.90.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.51

method	result	size
default	\(-\frac {\operatorname {arcsinh}\left (a \,x^{2}\right )}{3 x^{3}}+\frac {2 a \left (-\frac {\sqrt {a^{2} x^{4}+1}}{x}+\frac {i a \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i a}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i a}, i\right )\right )}{\sqrt {i a}\, \sqrt {a^{2} x^{4}+1}}\right )}{3}\)	\(101\)
parts	\(-\frac {\operatorname {arcsinh}\left (a \,x^{2}\right )}{3 x^{3}}+\frac {2 a \left (-\frac {\sqrt {a^{2} x^{4}+1}}{x}+\frac {i a \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i a}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i a}, i\right )\right )}{\sqrt {i a}\, \sqrt {a^{2} x^{4}+1}}\right )}{3}\)	\(101\)

input

int(arcsinh(a*x^2)/x^4,x,method=_RETURNVERBOSE)
 

output

-1/3*arcsinh(a*x^2)/x^3+2/3*a*(-(a^2*x^4+1)^(1/2)/x+I*a/(I*a)^(1/2)*(1-I*a 
*x^2)^(1/2)*(1+I*a*x^2)^(1/2)/(a^2*x^4+1)^(1/2)*(EllipticF(x*(I*a)^(1/2),I 
)-EllipticE(x*(I*a)^(1/2),I)))
 

3.3.90.5 Fricas [F]

\[ \int \frac {\text {arcsinh}\left (a x^2\right )}{x^4} \, dx=\int { \frac {\operatorname {arsinh}\left (a x^{2}\right )}{x^{4}} \,d x } \]

input

integrate(arcsinh(a*x^2)/x^4,x, algorithm="fricas")
 

output

integral(arcsinh(a*x^2)/x^4, x)
 

3.3.90.6 Sympy [F]

\[ \int \frac {\text {arcsinh}\left (a x^2\right )}{x^4} \, dx=\int \frac {\operatorname {asinh}{\left (a x^{2} \right )}}{x^{4}}\, dx \]

input

integrate(asinh(a*x**2)/x**4,x)
 

output

Integral(asinh(a*x**2)/x**4, x)
 

3.3.90.7 Maxima [F]

\[ \int \frac {\text {arcsinh}\left (a x^2\right )}{x^4} \, dx=\int { \frac {\operatorname {arsinh}\left (a x^{2}\right )}{x^{4}} \,d x } \]

input

integrate(arcsinh(a*x^2)/x^4,x, algorithm="maxima")
 

output

-1/12*I*sqrt(2)*a^(3/2)*(log(1/2*I*sqrt(2)*(2*a*x + sqrt(2)*sqrt(a))/sqrt( 
a) + 1) - log(-1/2*I*sqrt(2)*(2*a*x + sqrt(2)*sqrt(a))/sqrt(a) + 1)) - 1/1 
2*I*sqrt(2)*a^(3/2)*(log(1/2*I*sqrt(2)*(2*a*x - sqrt(2)*sqrt(a))/sqrt(a) + 
 1) - log(-1/2*I*sqrt(2)*(2*a*x - sqrt(2)*sqrt(a))/sqrt(a) + 1)) + 1/12*sq 
rt(2)*a^(3/2)*log(a*x^2 + sqrt(2)*sqrt(a)*x + 1) - 1/12*sqrt(2)*a^(3/2)*lo 
g(a*x^2 - sqrt(2)*sqrt(a)*x + 1) + 2*a*integrate(1/3/(a^3*x^8 + a*x^4 + (a 
^2*x^6 + x^2)*sqrt(a^2*x^4 + 1)), x) - 1/3*log(a*x^2 + sqrt(a^2*x^4 + 1))/ 
x^3
 

3.3.90.8 Giac [F]

\[ \int \frac {\text {arcsinh}\left (a x^2\right )}{x^4} \, dx=\int { \frac {\operatorname {arsinh}\left (a x^{2}\right )}{x^{4}} \,d x } \]

input

integrate(arcsinh(a*x^2)/x^4,x, algorithm="giac")
 

output

integrate(arcsinh(a*x^2)/x^4, x)
 

3.3.90.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arcsinh}\left (a x^2\right )}{x^4} \, dx=\int \frac {\mathrm {asinh}\left (a\,x^2\right )}{x^4} \,d x \]

input

int(asinh(a*x^2)/x^4,x)
 

output

int(asinh(a*x^2)/x^4, x)