\(\int \frac {a+b \arcsin (c+d x^2)}{x^5} \, dx\) [266]

Optimal result

Integrand size = 16, antiderivative size = 137 \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^5} \, dx=-\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (1-c^2\right ) x^2}-\frac {a+b \arcsin \left (c+d x^2\right )}{4 x^4}-\frac {b c d^2 \text {arctanh}\left (\frac {1-c^2-c d x^2}{\sqrt {1-c^2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\right )}{4 \left (1-c^2\right )^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^5} \, dx=-\frac {a}{4 x^4}+\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (-1+c^2\right ) x^2}-\frac {b \arcsin \left (c+d x^2\right )}{4 x^4}+\frac {b c d^2 \arctan \left (\frac {\sqrt {-d^2} x^2-\sqrt {1-c^2-2 c d x^2-d^2 x^4}}{\sqrt {-1+c^2}}\right )}{2 (-1+c) (1+c) \sqrt {-1+c^2}} \] Input:

Integrate[(a + b*ArcSin[c + d*x^2])/x^5,x]
 

Output:

-1/4*a/x^4 + (b*d*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(4*(-1 + c^2)*x^2) 
- (b*ArcSin[c + d*x^2])/(4*x^4) + (b*c*d^2*ArcTan[(Sqrt[-d^2]*x^2 - Sqrt[1 
 - c^2 - 2*c*d*x^2 - d^2*x^4])/Sqrt[-1 + c^2]])/(2*(-1 + c)*(1 + c)*Sqrt[- 
1 + c^2])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5341, 27, 1434, 1157, 1154, 219}

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.

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

\(\Big \downarrow \) 5341

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1157

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

Int[(a + b*ArcSin[c + d*x^2])/x^5,x]
 

Output:

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

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1157

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d 
^2 - b*d*e + a*e^2))), x] + Simp[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e^2)) 
  Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e 
, m, p}, x] && EqQ[m + 2*p + 3, 0]
 

rule 1434

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp 
[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
 

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 = 0.07 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {a}{4 x^{4}}-\frac {b \arcsin \left (d \,x^{2}+c \right )}{4 x^{4}}-\frac {b d \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{4 \left (-c^{2}+1\right ) x^{2}}-\frac {b c \,d^{2} \ln \left (\frac {-2 c^{2}+2-2 c d \,x^{2}+2 \sqrt {-c^{2}+1}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{x^{2}}\right )}{4 \left (-c^{2}+1\right )^{\frac {3}{2}}}\)	\(132\)
parts	\(-\frac {a}{4 x^{4}}-\frac {b \arcsin \left (d \,x^{2}+c \right )}{4 x^{4}}-\frac {b d \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{4 \left (-c^{2}+1\right ) x^{2}}-\frac {b c \,d^{2} \ln \left (\frac {-2 c^{2}+2-2 c d \,x^{2}+2 \sqrt {-c^{2}+1}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{x^{2}}\right )}{4 \left (-c^{2}+1\right )^{\frac {3}{2}}}\)	\(132\)

Input:

int((a+b*arcsin(d*x^2+c))/x^5,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.86 \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^5} \, dx=\left [-\frac {\sqrt {-c^{2} + 1} b c d^{2} x^{4} \log \left (\frac {{\left (2 \, c^{2} - 1\right )} d^{2} x^{4} + 2 \, c^{4} + 4 \, {\left (c^{3} - c\right )} d x^{2} - 2 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (c d x^{2} + c^{2} - 1\right )} \sqrt {-c^{2} + 1} - 4 \, c^{2} + 2}{x^{4}}\right ) + 2 \, a c^{4} - 2 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (b c^{2} - b\right )} d x^{2} - 4 \, a c^{2} + 2 \, {\left (b c^{4} - 2 \, b c^{2} + b\right )} \arcsin \left (d x^{2} + c\right ) + 2 \, a}{8 \, {\left (c^{4} - 2 \, c^{2} + 1\right )} x^{4}}, -\frac {\sqrt {c^{2} - 1} b c d^{2} x^{4} \arctan \left (\frac {\sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (c d x^{2} + c^{2} - 1\right )} \sqrt {c^{2} - 1}}{{\left (c^{2} - 1\right )} d^{2} x^{4} + c^{4} + 2 \, {\left (c^{3} - c\right )} d x^{2} - 2 \, c^{2} + 1}\right ) + a c^{4} - \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (b c^{2} - b\right )} d x^{2} - 2 \, a c^{2} + {\left (b c^{4} - 2 \, b c^{2} + b\right )} \arcsin \left (d x^{2} + c\right ) + a}{4 \, {\left (c^{4} - 2 \, c^{2} + 1\right )} x^{4}}\right ] \] Input:

integrate((a+b*arcsin(d*x^2+c))/x^5,x, algorithm="fricas")
 

Output:

[-1/8*(sqrt(-c^2 + 1)*b*c*d^2*x^4*log(((2*c^2 - 1)*d^2*x^4 + 2*c^4 + 4*(c^ 
3 - c)*d*x^2 - 2*sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1)*(c*d*x^2 + c^2 - 1)* 
sqrt(-c^2 + 1) - 4*c^2 + 2)/x^4) + 2*a*c^4 - 2*sqrt(-d^2*x^4 - 2*c*d*x^2 - 
 c^2 + 1)*(b*c^2 - b)*d*x^2 - 4*a*c^2 + 2*(b*c^4 - 2*b*c^2 + b)*arcsin(d*x 
^2 + c) + 2*a)/((c^4 - 2*c^2 + 1)*x^4), -1/4*(sqrt(c^2 - 1)*b*c*d^2*x^4*ar 
ctan(sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1)*(c*d*x^2 + c^2 - 1)*sqrt(c^2 - 1 
)/((c^2 - 1)*d^2*x^4 + c^4 + 2*(c^3 - c)*d*x^2 - 2*c^2 + 1)) + a*c^4 - sqr 
t(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1)*(b*c^2 - b)*d*x^2 - 2*a*c^2 + (b*c^4 - 2 
*b*c^2 + b)*arcsin(d*x^2 + c) + a)/((c^4 - 2*c^2 + 1)*x^4)]
 

Sympy [F]

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

integrate((a+b*asin(d*x**2+c))/x**5,x)
 

Output:

Integral((a + b*asin(c + d*x**2))/x**5, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^5} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*arcsin(d*x^2+c))/x^5,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]

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

integrate((a+b*arcsin(d*x^2+c))/x^5,x, algorithm="giac")
 

Output:

integrate((b*arcsin(d*x^2 + c) + a)/x^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^5} \, dx=\int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x^5} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \arcsin \left (c+d x^2\right )}{x^5} \, dx=\frac {4 \left (\int \frac {\mathit {asin} \left (d \,x^{2}+c \right )}{x^{5}}d x \right ) b \,x^{4}-a}{4 x^{4}} \] Input:

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

Output:

(4*int(asin(c + d*x**2)/x**5,x)*b*x**4 - a)/(4*x**4)