\(\int \frac {a+b \arcsin (c+d x^2)}{x^6} \, dx\) [273]

Optimal result

Integrand size = 16, antiderivative size = 355 \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx=-\frac {2 b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right ) x^3}-\frac {8 b c d^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right )^2 x}-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}-\frac {8 b c d^{5/2} \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} E\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{1+c}\right )}{15 \sqrt {1-c} \left (1-c^2\right ) \sqrt {1-c^2-2 c d x^2-d^2 x^4}}+\frac {2 b (1+3 c) d^{5/2} \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{1+c}\right )}{15 \sqrt {1-c} \left (1-c^2\right ) \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \] Output:

-2/15*b*d*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/(-c^2+1)/x^3-8/15*b*c*d^2*(-d^2 
*x^4-2*c*d*x^2-c^2+1)^(1/2)/(-c^2+1)^2/x-1/5*(a+b*arcsin(d*x^2+c))/x^5-8/1 
5*b*c*d^(5/2)*(1-d*x^2/(1-c))^(1/2)*(1+d*x^2/(1+c))^(1/2)*EllipticE(d^(1/2 
)*x/(1-c)^(1/2),(-(1-c)/(1+c))^(1/2))/(1-c)^(1/2)/(-c^2+1)/(-d^2*x^4-2*c*d 
*x^2-c^2+1)^(1/2)+2/15*b*(1+3*c)*d^(5/2)*(1-d*x^2/(1-c))^(1/2)*(1+d*x^2/(1 
+c))^(1/2)*EllipticF(d^(1/2)*x/(1-c)^(1/2),(-(1-c)/(1+c))^(1/2))/(1-c)^(1/ 
2)/(-c^2+1)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx=\frac {\sqrt {\frac {d}{1+c}} \left (-3 a \left (-1+c^2\right )^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4}+2 b d x^2 \left (-1-c^4+2 c^3 d x^2+d^2 x^4+c^2 \left (2+7 d^2 x^4\right )+c \left (-2 d x^2+4 d^3 x^6\right )\right )-3 b \left (-1+c^2\right )^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4} \arcsin \left (c+d x^2\right )\right )+8 i b (-1+c) c d^3 x^5 \sqrt {\frac {-1+c+d x^2}{-1+c}} \sqrt {\frac {1+c+d x^2}{1+c}} E\left (i \text {arcsinh}\left (\sqrt {\frac {d}{1+c}} x\right )|\frac {1+c}{-1+c}\right )-2 i b \left (1-4 c+3 c^2\right ) d^3 x^5 \sqrt {\frac {-1+c+d x^2}{-1+c}} \sqrt {\frac {1+c+d x^2}{1+c}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {d}{1+c}} x\right ),\frac {1+c}{-1+c}\right )}{15 \left (-1+c^2\right )^2 \sqrt {\frac {d}{1+c}} x^5 \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \] Input:

Integrate[(a + b*ArcSin[c + d*x^2])/x^6,x]
 

Output:

(Sqrt[d/(1 + c)]*(-3*a*(-1 + c^2)^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4] + 
2*b*d*x^2*(-1 - c^4 + 2*c^3*d*x^2 + d^2*x^4 + c^2*(2 + 7*d^2*x^4) + c*(-2* 
d*x^2 + 4*d^3*x^6)) - 3*b*(-1 + c^2)^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4] 
*ArcSin[c + d*x^2]) + (8*I)*b*(-1 + c)*c*d^3*x^5*Sqrt[(-1 + c + d*x^2)/(-1 
 + c)]*Sqrt[(1 + c + d*x^2)/(1 + c)]*EllipticE[I*ArcSinh[Sqrt[d/(1 + c)]*x 
], (1 + c)/(-1 + c)] - (2*I)*b*(1 - 4*c + 3*c^2)*d^3*x^5*Sqrt[(-1 + c + d* 
x^2)/(-1 + c)]*Sqrt[(1 + c + d*x^2)/(1 + c)]*EllipticF[I*ArcSinh[Sqrt[d/(1 
 + c)]*x], (1 + c)/(-1 + c)])/(15*(-1 + c^2)^2*Sqrt[d/(1 + c)]*x^5*Sqrt[1 
- c^2 - 2*c*d*x^2 - d^2*x^4])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.85, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {5341, 27, 1443, 27, 1604, 25, 27, 1514, 399, 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.

\(\displaystyle \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx\)

\(\Big \downarrow \) 5341

\(\displaystyle \frac {1}{5} b \int \frac {2 d}{x^4 \sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{5} b d \int \frac {1}{x^4 \sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 1443

\(\displaystyle \frac {2}{5} b d \left (\frac {\int \frac {d \left (d x^2+4 c\right )}{x^2 \sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{5} b d \left (\frac {d \int \frac {d x^2+4 c}{x^2 \sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 1604

\(\displaystyle \frac {2}{5} b d \left (\frac {d \left (-\frac {\int -\frac {d \left (-c^2-4 d x^2 c+1\right )}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{1-c^2}-\frac {4 c \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2}{5} b d \left (\frac {d \left (\frac {\int \frac {d \left (-c^2-4 d x^2 c+1\right )}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{1-c^2}-\frac {4 c \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{5} b d \left (\frac {d \left (\frac {d \int \frac {-c^2-4 d x^2 c+1}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{1-c^2}-\frac {4 c \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 1514

\(\displaystyle \frac {2}{5} b d \left (\frac {d \left (\frac {d \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \int \frac {-c^2-4 d x^2 c+1}{\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1}}dx}{\left (1-c^2\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {4 c \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {2}{5} b d \left (\frac {d \left (\frac {d \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left ((c+1) (3 c+1) \int \frac {1}{\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1}}dx-4 c (c+1) \int \frac {\sqrt {\frac {d x^2}{c+1}+1}}{\sqrt {1-\frac {d x^2}{1-c}}}dx\right )}{\left (1-c^2\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {4 c \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {2}{5} b d \left (\frac {d \left (\frac {d \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left (\frac {\sqrt {1-c} (c+1) (3 c+1) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{c+1}\right )}{\sqrt {d}}-4 c (c+1) \int \frac {\sqrt {\frac {d x^2}{c+1}+1}}{\sqrt {1-\frac {d x^2}{1-c}}}dx\right )}{\left (1-c^2\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {4 c \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {2}{5} b d \left (\frac {d \left (\frac {d \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left (\frac {\sqrt {1-c} (c+1) (3 c+1) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{c+1}\right )}{\sqrt {d}}-\frac {4 \sqrt {1-c} c (c+1) E\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{\sqrt {d}}\right )}{\left (1-c^2\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {4 c \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{\left (1-c^2\right ) x}\right )}{3 \left (1-c^2\right )}-\frac {\sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x^3}\right )-\frac {a+b \arcsin \left (c+d x^2\right )}{5 x^5}\)

Input:

Int[(a + b*ArcSin[c + d*x^2])/x^6,x]
 

Output:

-1/5*(a + b*ArcSin[c + d*x^2])/x^5 + (2*b*d*(-1/3*Sqrt[1 - c^2 - 2*c*d*x^2 
 - d^2*x^4]/((1 - c^2)*x^3) + (d*((-4*c*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4 
])/((1 - c^2)*x) + (d*Sqrt[1 - (d*x^2)/(1 - c)]*Sqrt[1 + (d*x^2)/(1 + c)]* 
((-4*Sqrt[1 - c]*c*(1 + c)*EllipticE[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 
 - c)/(1 + c))])/Sqrt[d] + (Sqrt[1 - c]*(1 + c)*(1 + 3*c)*EllipticF[ArcSin 
[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 + c))])/Sqrt[d]))/((1 - c^2)*Sqrt[ 
1 - c^2 - 2*c*d*x^2 - d^2*x^4])))/(3*(1 - c^2))))/5
 

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 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 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 1514

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> 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[(d + e*x^2)/(Sqrt[1 
+ 2*c*(x^2/(b - q))]*Sqrt[1 + 2*c*(x^2/(b + q))]), x], x]] /; FreeQ[{a, b, 
c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[c/a]
 

rule 1604

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

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]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 346, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {a}{5 x^{5}}+b \left (-\frac {\arcsin \left (d \,x^{2}+c \right )}{5 x^{5}}+\frac {2 d \left (\frac {\sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{3 \left (c^{2}-1\right ) x^{3}}-\frac {4 d c \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{3 \left (c^{2}-1\right )^{2} x}-\frac {d^{2} \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )}{3 \left (c^{2}-1\right ) \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}+\frac {8 d^{3} c \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{3 \left (c^{2}-1\right )^{2} \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 c d +2 d \right )}\right )}{5}\right )\)	\(346\)
parts	\(-\frac {a}{5 x^{5}}+b \left (-\frac {\arcsin \left (d \,x^{2}+c \right )}{5 x^{5}}+\frac {2 d \left (\frac {\sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{3 \left (c^{2}-1\right ) x^{3}}-\frac {4 d c \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{3 \left (c^{2}-1\right )^{2} x}-\frac {d^{2} \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )}{3 \left (c^{2}-1\right ) \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}+\frac {8 d^{3} c \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{3 \left (c^{2}-1\right )^{2} \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 c d +2 d \right )}\right )}{5}\right )\)	\(346\)

Input:

int((a+b*arcsin(d*x^2+c))/x^6,x,method=_RETURNVERBOSE)
 

Output:

-1/5*a/x^5+b*(-1/5/x^5*arcsin(d*x^2+c)+2/5*d*(1/3/(c^2-1)*(-d^2*x^4-2*c*d* 
x^2-c^2+1)^(1/2)/x^3-4/3*d*c/(c^2-1)^2*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/x- 
1/3*d^2/(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)*EllipticF(x*(-d/(-1+c))^(1/2),(-1+2* 
c/(1+c))^(1/2))+8/3*d^3*c/(c^2-1)^2*(-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)))))
 

Fricas [F]

\[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx=\int { \frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{6}} \,d x } \] Input:

integrate((a+b*arcsin(d*x^2+c))/x^6,x, algorithm="fricas")
 

Output:

integral((b*arcsin(d*x^2 + c) + a)/x^6, x)
 

Sympy [F]

\[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx=\int \frac {a + b \operatorname {asin}{\left (c + d x^{2} \right )}}{x^{6}}\, dx \] Input:

integrate((a+b*asin(d*x**2+c))/x**6,x)
 

Output:

Integral((a + b*asin(c + d*x**2))/x**6, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*arcsin(d*x^2+c))/x^6,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
 

Giac [F]

\[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx=\int { \frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{6}} \,d x } \] Input:

integrate((a+b*arcsin(d*x^2+c))/x^6,x, algorithm="giac")
 

Output:

integrate((b*arcsin(d*x^2 + c) + a)/x^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx=\int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x^6} \,d x \] Input:

int((a + b*asin(c + d*x^2))/x^6,x)
 

Output:

int((a + b*asin(c + d*x^2))/x^6, x)
 

Reduce [F]

\[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^6} \, dx=\frac {5 \left (\int \frac {\mathit {asin} \left (d \,x^{2}+c \right )}{x^{6}}d x \right ) b \,x^{5}-a}{5 x^{5}} \] Input:

int((a+b*asin(d*x^2+c))/x^6,x)
 

Output:

(5*int(asin(c + d*x**2)/x**6,x)*b*x**5 - a)/(5*x**5)