\(\int \frac {x^4 (a+b \csc ^{-1}(c x))}{(d+e x^2)^2} \, dx\) [107]

Optimal result

Integrand size = 21, antiderivative size = 803 \[ \int \frac {x^4 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=-\frac {d \left (a+b \csc ^{-1}(c x)\right )}{4 e^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}+\frac {d \left (a+b \csc ^{-1}(c x)\right )}{4 e^2 \left (\sqrt {-d} \sqrt {e}+\frac {d}{x}\right )}+\frac {x \left (a+b \csc ^{-1}(c x)\right )}{e^2}+\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c e^2}+\frac {b \sqrt {d} \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 e^2 \sqrt {c^2 d+e}}+\frac {b \sqrt {d} \text {arctanh}\left (\frac {c^2 d+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 e^2 \sqrt {c^2 d+e}}-\frac {3 \sqrt {-d} \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac {i c \sqrt {-d} e^{i \csc ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}+\frac {3 \sqrt {-d} \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac {i c \sqrt {-d} e^{i \csc ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}-\frac {3 \sqrt {-d} \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac {i c \sqrt {-d} e^{i \csc ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}+\frac {3 \sqrt {-d} \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac {i c \sqrt {-d} e^{i \csc ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}-\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,-\frac {i c \sqrt {-d} e^{i \csc ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}+\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {i c \sqrt {-d} e^{i \csc ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}-\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,-\frac {i c \sqrt {-d} e^{i \csc ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}+\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {i c \sqrt {-d} e^{i \csc ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 e^{5/2}} \] Output:

-1/4*d*(a+b*arccsc(c*x))/e^2/((-d)^(1/2)*e^(1/2)-d/x)+1/4*d*(a+b*arccsc(c* 
x))/e^2/((-d)^(1/2)*e^(1/2)+d/x)+x*(a+b*arccsc(c*x))/e^2+b*arctanh((1-1/c^ 
2/x^2)^(1/2))/c/e^2+1/4*b*d^(1/2)*arctanh((c^2*d-(-d)^(1/2)*e^(1/2)/x)/c/d 
^(1/2)/(c^2*d+e)^(1/2)/(1-1/c^2/x^2)^(1/2))/e^2/(c^2*d+e)^(1/2)+1/4*b*d^(1 
/2)*arctanh((c^2*d+(-d)^(1/2)*e^(1/2)/x)/c/d^(1/2)/(c^2*d+e)^(1/2)/(1-1/c^ 
2/x^2)^(1/2))/e^2/(c^2*d+e)^(1/2)-3/4*(-d)^(1/2)*(a+b*arccsc(c*x))*ln(1-I* 
c*(-d)^(1/2)*(I/c/x+(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2)))/e^(5/2 
)+3/4*(-d)^(1/2)*(a+b*arccsc(c*x))*ln(1+I*c*(-d)^(1/2)*(I/c/x+(1-1/c^2/x^2 
)^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2)))/e^(5/2)-3/4*(-d)^(1/2)*(a+b*arccsc(c*x 
))*ln(1-I*c*(-d)^(1/2)*(I/c/x+(1-1/c^2/x^2)^(1/2))/(e^(1/2)+(c^2*d+e)^(1/2 
)))/e^(5/2)+3/4*(-d)^(1/2)*(a+b*arccsc(c*x))*ln(1+I*c*(-d)^(1/2)*(I/c/x+(1 
-1/c^2/x^2)^(1/2))/(e^(1/2)+(c^2*d+e)^(1/2)))/e^(5/2)-3/4*I*b*(-d)^(1/2)*p 
olylog(2,-I*c*(-d)^(1/2)*(I/c/x+(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1 
/2)))/e^(5/2)+3/4*I*b*(-d)^(1/2)*polylog(2,I*c*(-d)^(1/2)*(I/c/x+(1-1/c^2/ 
x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2)))/e^(5/2)-3/4*I*b*(-d)^(1/2)*polylog( 
2,-I*c*(-d)^(1/2)*(I/c/x+(1-1/c^2/x^2)^(1/2))/(e^(1/2)+(c^2*d+e)^(1/2)))/e 
^(5/2)+3/4*I*b*(-d)^(1/2)*polylog(2,I*c*(-d)^(1/2)*(I/c/x+(1-1/c^2/x^2)^(1 
/2))/(e^(1/2)+(c^2*d+e)^(1/2)))/e^(5/2)
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1634\) vs. \(2(803)=1606\).

Time = 6.05 (sec) , antiderivative size = 1634, normalized size of antiderivative = 2.03 \[ \int \frac {x^4 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(a*x)/e^2 + (a*d*x)/(2*e^2*(d + e*x^2)) - (3*a*Sqrt[d]*ArcTan[(Sqrt[e]*x)/ 
Sqrt[d]])/(2*e^(5/2)) + b*(-1/4*(d*(-(ArcCsc[c*x]/((-I)*Sqrt[d]*Sqrt[e] + 
e*x)) + (I*(ArcSin[1/(c*x)]/Sqrt[e] - Log[(2*Sqrt[d]*Sqrt[e]*(Sqrt[e] + c* 
((-I)*c*Sqrt[d] - Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x))/(Sqrt[-(c^ 
2*d) - e]*(Sqrt[d] + I*Sqrt[e]*x))]/Sqrt[-(c^2*d) - e]))/Sqrt[d]))/e^2 - ( 
d*(-(ArcCsc[c*x]/(I*Sqrt[d]*Sqrt[e] + e*x)) - (I*(ArcSin[1/(c*x)]/Sqrt[e] 
- Log[(2*Sqrt[d]*Sqrt[e]*(-Sqrt[e] + c*((-I)*c*Sqrt[d] + Sqrt[-(c^2*d) - e 
]*Sqrt[1 - 1/(c^2*x^2)])*x))/(Sqrt[-(c^2*d) - e]*(Sqrt[d] - I*Sqrt[e]*x))] 
/Sqrt[-(c^2*d) - e]))/Sqrt[d]))/(4*e^2) + (3*Sqrt[d]*(Pi^2 - 4*Pi*ArcCsc[c 
*x] + 8*ArcCsc[c*x]^2 - 32*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2 
]]*ArcTan[(((-I)*c*Sqrt[d] + Sqrt[e])*Cot[(Pi + 2*ArcCsc[c*x])/4])/Sqrt[c^ 
2*d + e]] + (4*I)*Pi*Log[1 + (Sqrt[e] - Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*A 
rcCsc[c*x]))] - (8*I)*ArcCsc[c*x]*Log[1 + (Sqrt[e] - Sqrt[c^2*d + e])/(c*S 
qrt[d]*E^(I*ArcCsc[c*x]))] + (16*I)*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d] 
)]/Sqrt[2]]*Log[1 + (Sqrt[e] - Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x 
]))] + (4*I)*Pi*Log[1 + (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc 
[c*x]))] - (8*I)*ArcCsc[c*x]*Log[1 + (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d 
]*E^(I*ArcCsc[c*x]))] - (16*I)*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sq 
rt[2]]*Log[1 + (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] 
+ (8*I)*ArcCsc[c*x]*Log[1 - E^((2*I)*ArcCsc[c*x])] - (4*I)*Pi*Log[Sqrt[...
 

Rubi [A] (verified)

Time = 2.86 (sec) , antiderivative size = 863, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5764, 5232, 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.

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

\(\Big \downarrow \) 5764

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

\(\Big \downarrow \) 5232

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x \left (a+b \arcsin \left (\frac {1}{c x}\right )\right )}{e^2}-\frac {3 \sqrt {-d} \log \left (1-\frac {i c \sqrt {-d} e^{i \arcsin \left (\frac {1}{c x}\right )}}{\sqrt {e}-\sqrt {d c^2+e}}\right ) \left (a+b \arcsin \left (\frac {1}{c x}\right )\right )}{4 e^{5/2}}+\frac {3 \sqrt {-d} \log \left (\frac {i \sqrt {-d} e^{i \arcsin \left (\frac {1}{c x}\right )} c}{\sqrt {e}-\sqrt {d c^2+e}}+1\right ) \left (a+b \arcsin \left (\frac {1}{c x}\right )\right )}{4 e^{5/2}}-\frac {3 \sqrt {-d} \log \left (1-\frac {i c \sqrt {-d} e^{i \arcsin \left (\frac {1}{c x}\right )}}{\sqrt {e}+\sqrt {d c^2+e}}\right ) \left (a+b \arcsin \left (\frac {1}{c x}\right )\right )}{4 e^{5/2}}+\frac {3 \sqrt {-d} \log \left (\frac {i \sqrt {-d} e^{i \arcsin \left (\frac {1}{c x}\right )} c}{\sqrt {e}+\sqrt {d c^2+e}}+1\right ) \left (a+b \arcsin \left (\frac {1}{c x}\right )\right )}{4 e^{5/2}}-\frac {d \left (a+b \arcsin \left (\frac {1}{c x}\right )\right )}{4 e^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}+\frac {d \left (a+b \arcsin \left (\frac {1}{c x}\right )\right )}{4 e^2 \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )}+\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c e^2}+\frac {b \sqrt {d} \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {d c^2+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 e^2 \sqrt {d c^2+e}}+\frac {b \sqrt {d} \text {arctanh}\left (\frac {d c^2+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {d c^2+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 e^2 \sqrt {d c^2+e}}-\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,-\frac {i c \sqrt {-d} e^{i \arcsin \left (\frac {1}{c x}\right )}}{\sqrt {e}-\sqrt {d c^2+e}}\right )}{4 e^{5/2}}+\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {i c \sqrt {-d} e^{i \arcsin \left (\frac {1}{c x}\right )}}{\sqrt {e}-\sqrt {d c^2+e}}\right )}{4 e^{5/2}}-\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,-\frac {i c \sqrt {-d} e^{i \arcsin \left (\frac {1}{c x}\right )}}{\sqrt {e}+\sqrt {d c^2+e}}\right )}{4 e^{5/2}}+\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {i c \sqrt {-d} e^{i \arcsin \left (\frac {1}{c x}\right )}}{\sqrt {e}+\sqrt {d c^2+e}}\right )}{4 e^{5/2}}\)

Input:

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

Output:

-1/4*(d*(a + b*ArcSin[1/(c*x)]))/(e^2*(Sqrt[-d]*Sqrt[e] - d/x)) + (d*(a + 
b*ArcSin[1/(c*x)]))/(4*e^2*(Sqrt[-d]*Sqrt[e] + d/x)) + (x*(a + b*ArcSin[1/ 
(c*x)]))/e^2 + (b*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]])/(c*e^2) + (b*Sqrt[d]*Arc 
Tanh[(c^2*d - (Sqrt[-d]*Sqrt[e])/x)/(c*Sqrt[d]*Sqrt[c^2*d + e]*Sqrt[1 - 1/ 
(c^2*x^2)])])/(4*e^2*Sqrt[c^2*d + e]) + (b*Sqrt[d]*ArcTanh[(c^2*d + (Sqrt[ 
-d]*Sqrt[e])/x)/(c*Sqrt[d]*Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])])/(4*e^2 
*Sqrt[c^2*d + e]) - (3*Sqrt[-d]*(a + b*ArcSin[1/(c*x)])*Log[1 - (I*c*Sqrt[ 
-d]*E^(I*ArcSin[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(4*e^(5/2)) + (3* 
Sqrt[-d]*(a + b*ArcSin[1/(c*x)])*Log[1 + (I*c*Sqrt[-d]*E^(I*ArcSin[1/(c*x) 
]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(4*e^(5/2)) - (3*Sqrt[-d]*(a + b*ArcSin[ 
1/(c*x)])*Log[1 - (I*c*Sqrt[-d]*E^(I*ArcSin[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2 
*d + e])])/(4*e^(5/2)) + (3*Sqrt[-d]*(a + b*ArcSin[1/(c*x)])*Log[1 + (I*c* 
Sqrt[-d]*E^(I*ArcSin[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(4*e^(5/2)) 
- (((3*I)/4)*b*Sqrt[-d]*PolyLog[2, ((-I)*c*Sqrt[-d]*E^(I*ArcSin[1/(c*x)])) 
/(Sqrt[e] - Sqrt[c^2*d + e])])/e^(5/2) + (((3*I)/4)*b*Sqrt[-d]*PolyLog[2, 
(I*c*Sqrt[-d]*E^(I*ArcSin[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^(5/2) 
 - (((3*I)/4)*b*Sqrt[-d]*PolyLog[2, ((-I)*c*Sqrt[-d]*E^(I*ArcSin[1/(c*x)]) 
)/(Sqrt[e] + Sqrt[c^2*d + e])])/e^(5/2) + (((3*I)/4)*b*Sqrt[-d]*PolyLog[2, 
 (I*c*Sqrt[-d]*E^(I*ArcSin[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^(5/2 
)
 

Defintions of rubi rules used

rule 2009

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

rule 5232

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcSin[c*x])^n, ( 
f*x)^m*(d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[c^2*d + 
 e, 0] && IGtQ[n, 0] && IntegerQ[p] && IntegerQ[m]
 

rule 5764

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_) 
^2)^(p_.), x_Symbol] :> -Subst[Int[(e + d*x^2)^p*((a + b*ArcSin[x/c])^n/x^( 
m + 2*(p + 1))), x], x, 1/x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[n, 0] 
&& IntegerQ[m] && IntegerQ[p]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 123.08 (sec) , antiderivative size = 965, normalized size of antiderivative = 1.20

method	result	size
parts	\(\text {Expression too large to display}\)	\(965\)
derivativedivides	\(\text {Expression too large to display}\)	\(987\)
default	\(\text {Expression too large to display}\)	\(987\)

Input:

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

Output:

a*(1/e^2*x-1/e^2*d*(-1/2*x/(e*x^2+d)+3/2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2 
))))+b/c^5*(1/2*x*c^5*arccsc(c*x)*(2*c^2*e*x^2+3*c^2*d)/(c^2*e*x^2+c^2*d)/ 
e^2-1/2*(-(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*((e*(c^2*d+e))^(1/2)* 
c^2*d+2*c^2*d*e+2*(e*(c^2*d+e))^(1/2)*e+2*e^2)*c*arctan(c*d*(I/c/x+(1-1/c^ 
2/x^2)^(1/2))/((-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e)*d)^(1/2))/e^2/(c^2*d+e)/ 
d^2-1/2*((c^2*d+2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*(-(e*(c^2*d+e))^(1/2)* 
c^2*d+2*c^2*d*e-2*(e*(c^2*d+e))^(1/2)*e+2*e^2)*c*arctanh(c*d*(I/c/x+(1-1/c 
^2/x^2)^(1/2))/((c^2*d+2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2))/e^2/(c^2*d+e)/ 
d^2+1/e^2*c^4*ln(1+I/c/x+(1-1/c^2/x^2)^(1/2))-1/e^2*c^4*ln(-1+I/c/x+(1-1/c 
^2/x^2)^(1/2))-3/16/e^3*c^6*d*sum((_R1^2*c^2*d-c^2*d-4*e)/_R1/(_R1^2*c^2*d 
-c^2*d-2*e)*(I*arccsc(c*x)*ln((_R1-I/c/x-(1-1/c^2/x^2)^(1/2))/_R1)+dilog(( 
_R1-I/c/x-(1-1/c^2/x^2)^(1/2))/_R1)),_R1=RootOf(c^2*d*_Z^4+(-2*c^2*d-4*e)* 
_Z^2+c^2*d))+3/16/e^3*c^6*d*sum((_R1^2*c^2*d+4*_R1^2*e-c^2*d)/_R1/(_R1^2*c 
^2*d-c^2*d-2*e)*(I*arccsc(c*x)*ln((_R1-I/c/x-(1-1/c^2/x^2)^(1/2))/_R1)+dil 
og((_R1-I/c/x-(1-1/c^2/x^2)^(1/2))/_R1)),_R1=RootOf(c^2*d*_Z^4+(-2*c^2*d-4 
*e)*_Z^2+c^2*d))+1/2*(-(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*(c^2*d+2 
*(e*(c^2*d+e))^(1/2)+2*e)*c*arctan(c*d*(I/c/x+(1-1/c^2/x^2)^(1/2))/((-c^2* 
d+2*(e*(c^2*d+e))^(1/2)-2*e)*d)^(1/2))/d^2/e^2+1/2*((c^2*d+2*(e*(c^2*d+e)) 
^(1/2)+2*e)*d)^(1/2)*(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*c*arctanh(c*d*(I/c/ 
x+(1-1/c^2/x^2)^(1/2))/((c^2*d+2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2))/d^2...
 

Fricas [F]

\[ \int \frac {x^4 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {x^4 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^{4} \left (a + b \operatorname {acsc}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^4*(a+b*arccsc(c*x))/(e*x^2+d)^2,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(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^4 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^2} \,d x \] Input:

int((x^4*(a + b*asin(1/(c*x))))/(d + e*x^2)^2,x)
 

Output:

int((x^4*(a + b*asin(1/(c*x))))/(d + e*x^2)^2, x)
 

Reduce [F]

\[ \int \frac {x^4 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\frac {-3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a d -3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a e \,x^{2}+2 \left (\int \frac {\mathit {acsc} \left (c x \right ) x^{4}}{e^{2} x^{4}+2 d e \,x^{2}+d^{2}}d x \right ) b d \,e^{3}+2 \left (\int \frac {\mathit {acsc} \left (c x \right ) x^{4}}{e^{2} x^{4}+2 d e \,x^{2}+d^{2}}d x \right ) b \,e^{4} x^{2}+3 a d e x +2 a \,e^{2} x^{3}}{2 e^{3} \left (e \,x^{2}+d \right )} \] Input:

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

Output:

( - 3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d - 3*sqrt(e)*sqrt(d 
)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*e*x**2 + 2*int((acsc(c*x)*x**4)/(d**2 + 
2*d*e*x**2 + e**2*x**4),x)*b*d*e**3 + 2*int((acsc(c*x)*x**4)/(d**2 + 2*d*e 
*x**2 + e**2*x**4),x)*b*e**4*x**2 + 3*a*d*e*x + 2*a*e**2*x**3)/(2*e**3*(d 
+ e*x**2))