3.4.97 \(\int \frac {a+b \arcsin (c+d x^2)}{x^4} \, dx\) [397]

3.4.97.1 Optimal result

Integrand size = 16, antiderivative size = 284 \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^4} \, dx=-\frac {2 b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{3 \left (1-c^2\right ) x}-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}-\frac {2 b d^{3/2} \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} E\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{1+c}\right )}{3 \sqrt {1-c} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}+\frac {2 b d^{3/2} \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{1+c}\right )}{3 \sqrt {1-c} \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \]

output

1/3*(-a-b*arcsin(d*x^2+c))/x^3-2/3*b*d^(3/2)*EllipticE(x*d^(1/2)/(1-c)^(1/ 
2),((-1+c)/(1+c))^(1/2))*(1-d*x^2/(1-c))^(1/2)*(1+d*x^2/(1+c))^(1/2)/(1-c) 
^(1/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)+2/3*b*d^(3/2)*EllipticF(x*d^(1/2)/ 
(1-c)^(1/2),((-1+c)/(1+c))^(1/2))*(1-d*x^2/(1-c))^(1/2)*(1+d*x^2/(1+c))^(1 
/2)/(1-c)^(1/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)-2/3*b*d*(-d^2*x^4-2*c*d*x 
^2-c^2+1)^(1/2)/(-c^2+1)/x
 

3.4.97.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^4} \, dx=-\frac {a}{3 x^3}+\frac {2 b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{3 \left (-1+c^2\right ) x}-\frac {b \arcsin \left (c+d x^2\right )}{3 x^3}+\frac {2 i b (1-c) d^2 \sqrt {1-\frac {d x^2}{-1-c}} \sqrt {1-\frac {d x^2}{1-c}} \left (E\left (i \text {arcsinh}\left (\sqrt {-\frac {d}{-1-c}} x\right )|\frac {-1-c}{1-c}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {d}{-1-c}} x\right ),\frac {-1-c}{1-c}\right )\right )}{3 (-1+c) (1+c) \sqrt {-\frac {d}{-1-c}} \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \]

input

Integrate[(a + b*ArcSin[c + d*x^2])/x^4,x]
 

output

-1/3*a/x^3 + (2*b*d*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(3*(-1 + c^2)*x) 
- (b*ArcSin[c + d*x^2])/(3*x^3) + (((2*I)/3)*b*(1 - c)*d^2*Sqrt[1 - (d*x^2 
)/(-1 - c)]*Sqrt[1 - (d*x^2)/(1 - c)]*(EllipticE[I*ArcSinh[Sqrt[-(d/(-1 - 
c))]*x], (-1 - c)/(1 - c)] - EllipticF[I*ArcSinh[Sqrt[-(d/(-1 - c))]*x], ( 
-1 - c)/(1 - c)]))/((-1 + c)*(1 + c)*Sqrt[-(d/(-1 - c))]*Sqrt[1 - c^2 - 2* 
c*d*x^2 - d^2*x^4])
 

3.4.97.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.85, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5341, 27, 1443, 25, 27, 1460, 389, 321, 327}

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.

\(\displaystyle \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^4} \, dx\)

\(\Big \downarrow \) 5341

\(\displaystyle \frac {1}{3} b \int \frac {2 d}{x^2 \sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} b d \int \frac {1}{x^2 \sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 1443

\(\displaystyle \frac {2}{3} b d \left (\frac {\int -\frac {d^2 x^2}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{1-c^2}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2}{3} b d \left (-\frac {\int \frac {d^2 x^2}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{1-c^2}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} b d \left (-\frac {d^2 \int \frac {x^2}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{1-c^2}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 1460

\(\displaystyle \frac {2}{3} b d \left (-\frac {d^2 \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \int \frac {x^2}{\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1}}dx}{\left (1-c^2\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 389

\(\displaystyle \frac {2}{3} b d \left (-\frac {d^2 \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left (\frac {(c+1) \int \frac {\sqrt {\frac {d x^2}{c+1}+1}}{\sqrt {1-\frac {d x^2}{1-c}}}dx}{d}-\frac {(c+1) \int \frac {1}{\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1}}dx}{d}\right )}{\left (1-c^2\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {2}{3} b d \left (-\frac {d^2 \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left (\frac {(c+1) \int \frac {\sqrt {\frac {d x^2}{c+1}+1}}{\sqrt {1-\frac {d x^2}{1-c}}}dx}{d}-\frac {\sqrt {1-c} (c+1) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{c+1}\right )}{d^{3/2}}\right )}{\left (1-c^2\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {2}{3} b d \left (-\frac {d^2 \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left (\frac {\sqrt {1-c} (c+1) E\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{d^{3/2}}-\frac {\sqrt {1-c} (c+1) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{c+1}\right )}{d^{3/2}}\right )}{\left (1-c^2\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{3 x^3}\)

input

Int[(a + b*ArcSin[c + d*x^2])/x^4,x]
 

output

-1/3*(a + b*ArcSin[c + d*x^2])/x^3 + (2*b*d*(-(Sqrt[1 - c^2 - 2*c*d*x^2 - 
d^2*x^4]/((1 - c^2)*x)) - (d^2*Sqrt[1 - (d*x^2)/(1 - c)]*Sqrt[1 + (d*x^2)/ 
(1 + c)]*((Sqrt[1 - c]*(1 + c)*EllipticE[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], 
-((1 - c)/(1 + c))])/d^(3/2) - (Sqrt[1 - c]*(1 + c)*EllipticF[ArcSin[(Sqrt 
[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 + c))])/d^(3/2)))/((1 - c^2)*Sqrt[1 - c^ 
2 - 2*c*d*x^2 - d^2*x^4])))/3
 

3.4.97.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 389

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] 
 :> Simp[1/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] - Simp[a/b   Int 
[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && N 
eQ[b*c - a*d, 0] &&  !SimplerSqrtQ[-b/a, -d/c]
 

rule 1443

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1)/(a*d*(m + 1))), x] - Sim 
p[1/(a*d^2*(m + 1))   Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p + 5)* 
x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 
- 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1460

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[b^2 - 4*a*c, 2]}, Simp[Sqrt[1 + 2*c*(x^2/(b - q))]*(Sqrt[1 + 2*c*(x^2/( 
b + q))]/Sqrt[a + b*x^2 + c*x^4])   Int[x^2/(Sqrt[1 + 2*c*(x^2/(b - q))]*Sq 
rt[1 + 2*c*(x^2/(b + q))]), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a 
*c, 0] && NegQ[c/a]
 

rule 5341

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Sim 
p[(c + d*x)^(m + 1)*((a + b*ArcSin[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]
 

3.4.97.4 Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.73

method	result	size
default	\(-\frac {a}{3 x^{3}}+b \left (-\frac {\arcsin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {2 d \left (\frac {\sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{\left (c^{2}-1\right ) x}-\frac {2 d^{2} \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{\left (c^{2}-1\right ) \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 c d +2 d \right )}\right )}{3}\right )\)	\(207\)
parts	\(-\frac {a}{3 x^{3}}+b \left (-\frac {\arcsin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {2 d \left (\frac {\sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{\left (c^{2}-1\right ) x}-\frac {2 d^{2} \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{\left (c^{2}-1\right ) \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 c d +2 d \right )}\right )}{3}\right )\)	\(207\)

input

int((a+b*arcsin(d*x^2+c))/x^4,x,method=_RETURNVERBOSE)
 

output

-1/3*a/x^3+b*(-1/3/x^3*arcsin(d*x^2+c)+2/3*d*(1/(c^2-1)*(-d^2*x^4-2*c*d*x^ 
2-c^2+1)^(1/2)/x-2*d^2/(c^2-1)*(-c^2+1)/(-d/(-1+c))^(1/2)*(1+d/(-1+c)*x^2) 
^(1/2)*(1+d*x^2/(1+c))^(1/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/(-2*c*d+2*d) 
*(EllipticF(x*(-d/(-1+c))^(1/2),(-1+2*c/(1+c))^(1/2))-EllipticE(x*(-d/(-1+ 
c))^(1/2),(-1+2*c/(1+c))^(1/2)))))
 

3.4.97.5 Fricas [F]

\[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^4} \, dx=\int { \frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{4}} \,d x } \]

input

integrate((a+b*arcsin(d*x^2+c))/x^4,x, algorithm="fricas")
 

output

integral((b*arcsin(d*x^2 + c) + a)/x^4, x)
 

3.4.97.6 Sympy [F]

\[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^4} \, dx=\int \frac {a + b \operatorname {asin}{\left (c + d x^{2} \right )}}{x^{4}}\, dx \]

input

integrate((a+b*asin(d*x**2+c))/x**4,x)
 

output

Integral((a + b*asin(c + d*x**2))/x**4, x)
 

3.4.97.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^4} \, dx=\text {Exception raised: ValueError} \]

input

integrate((a+b*arcsin(d*x^2+c))/x^4,x, algorithm="maxima")
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(c-1>0)', see `assume?` for more 
details)Is
 

3.4.97.8 Giac [F]

\[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^4} \, dx=\int { \frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{4}} \,d x } \]

input

integrate((a+b*arcsin(d*x^2+c))/x^4,x, algorithm="giac")
 

output

integrate((b*arcsin(d*x^2 + c) + a)/x^4, x)
 

3.4.97.9 Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^4} \, dx=\int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x^4} \,d x \]

input

int((a + b*asin(c + d*x^2))/x^4,x)
 

output

int((a + b*asin(c + d*x^2))/x^4, x)