3.1.87 \(\int \frac {(d+e x^2) (a+b \csc ^{-1}(c x))}{x^3} \, dx\) [87]

3.1.87.1 Optimal result

Integrand size = 19, antiderivative size = 137 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{x^3} \, dx=-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}+\frac {1}{4} b c^2 d \csc ^{-1}(c x)+\frac {1}{2} i b e \csc ^{-1}(c x)^2-\frac {d \left (a+b \csc ^{-1}(c x)\right )}{2 x^2}-b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )+b e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-e \left (a+b \csc ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )+\frac {1}{2} i b e \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right ) \]

output

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

3.1.87.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{x^3} \, dx=-\frac {a d}{2 x^2}-\frac {b c d \sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}}{4 x}-\frac {b d \csc ^{-1}(c x)}{2 x^2}+\frac {1}{4} b c^2 d \arcsin \left (\frac {1}{c x}\right )-b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )+a e \log (x)+\frac {1}{2} i b e \left (\csc ^{-1}(c x)^2+\operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )\right ) \]

input

Integrate[((d + e*x^2)*(a + b*ArcCsc[c*x]))/x^3,x]
 

output

-1/2*(a*d)/x^2 - (b*c*d*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)])/(4*x) - (b*d*ArcCs 
c[c*x])/(2*x^2) + (b*c^2*d*ArcSin[1/(c*x)])/4 - b*e*ArcCsc[c*x]*Log[1 - E^ 
((2*I)*ArcCsc[c*x])] + a*e*Log[x] + (I/2)*b*e*(ArcCsc[c*x]^2 + PolyLog[2, 
E^((2*I)*ArcCsc[c*x])])
 

3.1.87.3 Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5764, 5230, 27, 7293, 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.

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

\(\Big \downarrow \) 5764

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

\(\Big \downarrow \) 5230

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

Int[((d + e*x^2)*(a + b*ArcCsc[c*x]))/x^3,x]
 

output

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

3.1.87.3.1 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 2009

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

rule 5230

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_ 
)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp 
[(a + b*ArcSin[c*x])   u, x] - Simp[b*c   Int[SimplifyIntegrand[u/Sqrt[1 - 
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 
0] && IntegerQ[p] && (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))
 

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]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.1.87.4 Maple [A] (verified)

Time = 2.76 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.39

method	result	size
parts	\(-\frac {a d}{2 x^{2}}+a e \ln \left (x \right )+b \,c^{2} \left (\frac {i \operatorname {arccsc}\left (c x \right )^{2} e}{2 c^{2}}-\frac {e \,\operatorname {arccsc}\left (c x \right ) \ln \left (1+\frac {i}{c x}+\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}{c^{2}}+\frac {i e \operatorname {polylog}\left (2, -\frac {i}{c x}-\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}{c^{2}}-\frac {e \,\operatorname {arccsc}\left (c x \right ) \ln \left (1-\frac {i}{c x}-\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}{c^{2}}+\frac {i e \operatorname {polylog}\left (2, \frac {i}{c x}+\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}{c^{2}}+\frac {\operatorname {arccsc}\left (c x \right ) d \cos \left (2 \,\operatorname {arccsc}\left (c x \right )\right )}{4}-\frac {d \sin \left (2 \,\operatorname {arccsc}\left (c x \right )\right )}{8}\right )\)	\(191\)
derivativedivides	\(c^{2} \left (-\frac {a d}{2 c^{2} x^{2}}+\frac {a e \ln \left (c x \right )}{c^{2}}+\frac {b \left (\frac {i \operatorname {arccsc}\left (c x \right )^{2} e}{2}-e \,\operatorname {arccsc}\left (c x \right ) \ln \left (1-\frac {i}{c x}-\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+i e \operatorname {polylog}\left (2, \frac {i}{c x}+\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )-e \,\operatorname {arccsc}\left (c x \right ) \ln \left (1+\frac {i}{c x}+\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+i e \operatorname {polylog}\left (2, -\frac {i}{c x}-\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\frac {c^{2} d \,\operatorname {arccsc}\left (c x \right ) \cos \left (2 \,\operatorname {arccsc}\left (c x \right )\right )}{4}-\frac {d \,c^{2} \sin \left (2 \,\operatorname {arccsc}\left (c x \right )\right )}{8}\right )}{c^{2}}\right )\)	\(194\)
default	\(c^{2} \left (-\frac {a d}{2 c^{2} x^{2}}+\frac {a e \ln \left (c x \right )}{c^{2}}+\frac {b \left (\frac {i \operatorname {arccsc}\left (c x \right )^{2} e}{2}-e \,\operatorname {arccsc}\left (c x \right ) \ln \left (1-\frac {i}{c x}-\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+i e \operatorname {polylog}\left (2, \frac {i}{c x}+\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )-e \,\operatorname {arccsc}\left (c x \right ) \ln \left (1+\frac {i}{c x}+\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+i e \operatorname {polylog}\left (2, -\frac {i}{c x}-\sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\frac {c^{2} d \,\operatorname {arccsc}\left (c x \right ) \cos \left (2 \,\operatorname {arccsc}\left (c x \right )\right )}{4}-\frac {d \,c^{2} \sin \left (2 \,\operatorname {arccsc}\left (c x \right )\right )}{8}\right )}{c^{2}}\right )\)	\(194\)

input

int((e*x^2+d)*(a+b*arccsc(c*x))/x^3,x,method=_RETURNVERBOSE)
 

output

-1/2*a*d/x^2+a*e*ln(x)+b*c^2*(1/2*I/c^2*arccsc(c*x)^2*e-1/c^2*e*arccsc(c*x 
)*ln(1+I/c/x+(1-1/c^2/x^2)^(1/2))+I/c^2*e*polylog(2,-I/c/x-(1-1/c^2/x^2)^( 
1/2))-1/c^2*e*arccsc(c*x)*ln(1-I/c/x-(1-1/c^2/x^2)^(1/2))+I/c^2*e*polylog( 
2,I/c/x+(1-1/c^2/x^2)^(1/2))+1/4*arccsc(c*x)*d*cos(2*arccsc(c*x))-1/8*d*si 
n(2*arccsc(c*x)))
 

3.1.87.5 Fricas [F]

\[ \int \frac {\left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arccsc}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

input

integrate((e*x^2+d)*(a+b*arccsc(c*x))/x^3,x, algorithm="fricas")
 

output

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arccsc(c*x))/x^3, x)
 

3.1.87.6 Sympy [F]

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

input

integrate((e*x**2+d)*(a+b*acsc(c*x))/x**3,x)
 

output

Integral((a + b*acsc(c*x))*(d + e*x**2)/x**3, x)
 

3.1.87.7 Maxima [F]

\[ \int \frac {\left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arccsc}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

input

integrate((e*x^2+d)*(a+b*arccsc(c*x))/x^3,x, algorithm="maxima")
 

output

(c^2*integrate(sqrt(c*x + 1)*sqrt(c*x - 1)*log(x)/(c^4*x^3 - c^2*x), x) + 
arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1))*log(x))*b*e + 1/4*b*d*((c^4*x*sqrt 
(-1/(c^2*x^2) + 1)/(c^2*x^2*(1/(c^2*x^2) - 1) - 1) - c^3*arctan(c*x*sqrt(- 
1/(c^2*x^2) + 1)))/c - 2*arccsc(c*x)/x^2) + a*e*log(x) - 1/2*a*d/x^2
 

3.1.87.8 Giac [F]

\[ \int \frac {\left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arccsc}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

input

integrate((e*x^2+d)*(a+b*arccsc(c*x))/x^3,x, algorithm="giac")
 

output

integrate((e*x^2 + d)*(b*arccsc(c*x) + a)/x^3, x)
 

3.1.87.9 Mupad [B] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{x^3} \, dx=-a\,e\,\ln \left (\frac {1}{x}\right )-\frac {a\,d}{2\,x^2}-b\,e\,\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{c\,x}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (\frac {1}{c\,x}\right )-\frac {b\,c\,d\,\sqrt {1-\frac {1}{c^2\,x^2}}}{4\,x}-\frac {b\,c^2\,d\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\,\left (\frac {2}{c^2\,x^2}-1\right )}{4}+\frac {b\,e\,\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{c\,x}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}+\frac {b\,e\,{\mathrm {asin}\left (\frac {1}{c\,x}\right )}^2\,1{}\mathrm {i}}{2} \]

input

int(((d + e*x^2)*(a + b*asin(1/(c*x))))/x^3,x)
 

output

(b*e*polylog(2, exp(asin(1/(c*x))*2i))*1i)/2 - a*e*log(1/x) + (b*e*asin(1/ 
(c*x))^2*1i)/2 - (a*d)/(2*x^2) - b*e*log(1 - exp(asin(1/(c*x))*2i))*asin(1 
/(c*x)) - (b*c*d*(1 - 1/(c^2*x^2))^(1/2))/(4*x) - (b*c^2*d*asin(1/(c*x))*( 
2/(c^2*x^2) - 1))/4