\(\int \frac {1}{x^4 (a+b \csc ^{-1}(c x))} \, dx\) [38]

Optimal result

Integrand size = 14, antiderivative size = 117 \[ \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx=-\frac {c^3 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b}-\frac {c^3 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}+\frac {c^3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b} \] Output:

-1/4*c^3*cos(a/b)*Ci(a/b+arccsc(c*x))/b+1/4*c^3*cos(3*a/b)*Ci(3*a/b+3*arcc 
sc(c*x))/b-1/4*c^3*sin(a/b)*Si(a/b+arccsc(c*x))/b+1/4*c^3*sin(3*a/b)*Si(3* 
a/b+3*arccsc(c*x))/b
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx=-\frac {c^3 \left (\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )-\cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\csc ^{-1}(c x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )-\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\csc ^{-1}(c x)\right )\right )\right )}{4 b} \] Input:

Integrate[1/(x^4*(a + b*ArcCsc[c*x])),x]
 

Output:

-1/4*(c^3*(Cos[a/b]*CosIntegral[a/b + ArcCsc[c*x]] - Cos[(3*a)/b]*CosInteg 
ral[3*(a/b + ArcCsc[c*x])] + Sin[a/b]*SinIntegral[a/b + ArcCsc[c*x]] - Sin 
[(3*a)/b]*SinIntegral[3*(a/b + ArcCsc[c*x])]))/b
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5746, 4906, 2009}

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.

\(\displaystyle \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx\)

\(\Big \downarrow \) 5746

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 2009

\(\displaystyle -c^3 \left (\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}-\frac {\cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}-\frac {\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b}\right )\)

Input:

Int[1/(x^4*(a + b*ArcCsc[c*x])),x]
 

Output:

-(c^3*((Cos[a/b]*CosIntegral[a/b + ArcCsc[c*x]])/(4*b) - (Cos[(3*a)/b]*Cos 
Integral[(3*a)/b + 3*ArcCsc[c*x]])/(4*b) + (Sin[a/b]*SinIntegral[a/b + Arc 
Csc[c*x]])/(4*b) - (Sin[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcCsc[c*x]])/(4* 
b)))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x 
]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG 
tQ[p, 0]
 

rule 5746

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[- 
(c^(m + 1))^(-1)   Subst[Int[(a + b*x)^n*Csc[x]^(m + 1)*Cot[x], x], x, ArcC 
sc[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n 
, 0] || LtQ[m, -1])
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(c^{3} \left (-\frac {\operatorname {Si}\left (\frac {a}{b}+\operatorname {arccsc}\left (c x \right )\right ) \sin \left (\frac {a}{b}\right )}{4 b}-\frac {\operatorname {Ci}\left (\frac {a}{b}+\operatorname {arccsc}\left (c x \right )\right ) \cos \left (\frac {a}{b}\right )}{4 b}+\frac {\operatorname {Si}\left (\frac {3 a}{b}+3 \,\operatorname {arccsc}\left (c x \right )\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}+\frac {\operatorname {Ci}\left (\frac {3 a}{b}+3 \,\operatorname {arccsc}\left (c x \right )\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}\right )\)	\(102\)
default	\(c^{3} \left (-\frac {\operatorname {Si}\left (\frac {a}{b}+\operatorname {arccsc}\left (c x \right )\right ) \sin \left (\frac {a}{b}\right )}{4 b}-\frac {\operatorname {Ci}\left (\frac {a}{b}+\operatorname {arccsc}\left (c x \right )\right ) \cos \left (\frac {a}{b}\right )}{4 b}+\frac {\operatorname {Si}\left (\frac {3 a}{b}+3 \,\operatorname {arccsc}\left (c x \right )\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}+\frac {\operatorname {Ci}\left (\frac {3 a}{b}+3 \,\operatorname {arccsc}\left (c x \right )\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}\right )\)	\(102\)

Input:

int(1/x^4/(a+b*arccsc(c*x)),x,method=_RETURNVERBOSE)
 

Output:

c^3*(-1/4*Si(a/b+arccsc(c*x))*sin(a/b)/b-1/4*Ci(a/b+arccsc(c*x))*cos(a/b)/ 
b+1/4*Si(3*a/b+3*arccsc(c*x))*sin(3*a/b)/b+1/4*Ci(3*a/b+3*arccsc(c*x))*cos 
(3*a/b)/b)
 

Fricas [F]

\[ \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx=\int { \frac {1}{{\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} x^{4}} \,d x } \] Input:

integrate(1/x^4/(a+b*arccsc(c*x)),x, algorithm="fricas")
 

Output:

integral(1/(b*x^4*arccsc(c*x) + a*x^4), x)
 

Sympy [F]

\[ \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx=\int \frac {1}{x^{4} \left (a + b \operatorname {acsc}{\left (c x \right )}\right )}\, dx \] Input:

integrate(1/x**4/(a+b*acsc(c*x)),x)
 

Output:

Integral(1/(x**4*(a + b*acsc(c*x))), x)
 

Maxima [F]

\[ \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx=\int { \frac {1}{{\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} x^{4}} \,d x } \] Input:

integrate(1/x^4/(a+b*arccsc(c*x)),x, algorithm="maxima")
 

Output:

integrate(1/((b*arccsc(c*x) + a)*x^4), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.71 \[ \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx=\frac {1}{4} \, {\left (\frac {4 \, c^{2} \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (\frac {1}{c x}\right )\right )}{b} + \frac {4 \, c^{2} \cos \left (\frac {a}{b}\right )^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (\frac {1}{c x}\right )\right )}{b} - \frac {3 \, c^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (\frac {1}{c x}\right )\right )}{b} - \frac {c^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (\frac {1}{c x}\right )\right )}{b} - \frac {c^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (\frac {1}{c x}\right )\right )}{b} - \frac {c^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (\frac {1}{c x}\right )\right )}{b}\right )} c \] Input:

integrate(1/x^4/(a+b*arccsc(c*x)),x, algorithm="giac")
 

Output:

1/4*(4*c^2*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(1/(c*x)))/b + 4*c^2*co 
s(a/b)^2*sin(a/b)*sin_integral(3*a/b + 3*arcsin(1/(c*x)))/b - 3*c^2*cos(a/ 
b)*cos_integral(3*a/b + 3*arcsin(1/(c*x)))/b - c^2*cos(a/b)*cos_integral(a 
/b + arcsin(1/(c*x)))/b - c^2*sin(a/b)*sin_integral(3*a/b + 3*arcsin(1/(c* 
x)))/b - c^2*sin(a/b)*sin_integral(a/b + arcsin(1/(c*x)))/b)*c
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx=\int \frac {1}{x^4\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right )} \,d x \] Input:

int(1/(x^4*(a + b*asin(1/(c*x)))),x)
 

Output:

int(1/(x^4*(a + b*asin(1/(c*x)))), x)
 

Reduce [F]

\[ \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx=\int \frac {1}{\mathit {acsc} \left (c x \right ) b \,x^{4}+a \,x^{4}}d x \] Input:

int(1/x^4/(a+b*acsc(c*x)),x)
                                                                                    
                                                                                    
 

Output:

int(1/(acsc(c*x)*b*x**4 + a*x**4),x)