\(\int \frac {\csc (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) [6]

Optimal result

Integrand size = 17, antiderivative size = 244 \[ \int \frac {\csc (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {\sqrt {2} c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{a \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}-\frac {\sqrt {2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{a \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}-\frac {\text {arctanh}(\cos (x))}{a} \] Output:

-2^(1/2)*c*(1+b/(-4*a*c+b^2)^(1/2))*arctan(1/2*(2*c+(b-(-4*a*c+b^2)^(1/2)) 
*tan(1/2*x))*2^(1/2)/(b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1/2))^(1/2))/a/(b^2-2* 
c*(a+c)-b*(-4*a*c+b^2)^(1/2))^(1/2)-2^(1/2)*c*(1-b/(-4*a*c+b^2)^(1/2))*arc 
tan(1/2*(2*c+(b+(-4*a*c+b^2)^(1/2))*tan(1/2*x))*2^(1/2)/(b^2-2*c*(a+c)+b*( 
-4*a*c+b^2)^(1/2))^(1/2))/a/(b^2-2*c*(a+c)+b*(-4*a*c+b^2)^(1/2))^(1/2)-arc 
tanh(cos(x))/a
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.68 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.25 \[ \int \frac {\csc (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {\frac {c \left (-i b+\sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}+\frac {c \left (i b+\sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}+\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )}{a} \] Input:

Integrate[Csc[x]/(a + b*Sin[x] + c*Sin[x]^2),x]
 

Output:

-(((c*((-I)*b + Sqrt[-b^2 + 4*a*c])*ArcTan[(2*c + (b - I*Sqrt[-b^2 + 4*a*c 
])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]])])/ 
(Sqrt[-1/2*b^2 + 2*a*c]*Sqrt[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]]) 
+ (c*(I*b + Sqrt[-b^2 + 4*a*c])*ArcTan[(2*c + (b + I*Sqrt[-b^2 + 4*a*c])*T 
an[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]])])/(Sqr 
t[-1/2*b^2 + 2*a*c]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]]) + Lo 
g[Cos[x/2]] - Log[Sin[x/2]])/a)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3042, 3737, 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.

\(\displaystyle \int \frac {\csc (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (x) \left (a+b \sin (x)+c \sin (x)^2\right )}dx\)

\(\Big \downarrow \) 3737

\(\displaystyle \int \left (\frac {-b-c \sin (x)}{a \left (a+b \sin (x)+c \sin ^2(x)\right )}+\frac {\csc (x)}{a}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sqrt {2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}-\frac {\sqrt {2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}-\frac {\text {arctanh}(\cos (x))}{a}\)

Input:

Int[Csc[x]/(a + b*Sin[x] + c*Sin[x]^2),x]
 

Output:

-((Sqrt[2]*c*(1 + b/Sqrt[b^2 - 4*a*c])*ArcTan[(2*c + (b - Sqrt[b^2 - 4*a*c 
])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]])])/(a* 
Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]])) - (Sqrt[2]*c*(1 - b/Sqrt[b 
^2 - 4*a*c])*ArcTan[(2*c + (b + Sqrt[b^2 - 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt 
[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]])])/(a*Sqrt[b^2 - 2*c*(a + c) + b 
*Sqrt[b^2 - 4*a*c]]) - ArcTanh[Cos[x]]/a
 

Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3737

Int[sin[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n 
_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]^(n2_.))^(p_), x_Symbol] :> Int[ExpandTr 
ig[sin[d + e*x]^m*(a + b*sin[d + e*x]^n + c*sin[d + e*x]^(2*n))^p, x], x] / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && Integ 
ersQ[m, n, p]
 

Maple [A] (verified)

Time = 1.86 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.16

method	result	size
default	\(-\frac {2 \left (2 \sqrt {-4 a c +b^{2}}\, a c -\sqrt {-4 a c +b^{2}}\, b^{2}-4 a b c +b^{3}\right ) \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {2 \left (-2 \sqrt {-4 a c +b^{2}}\, a c +\sqrt {-4 a c +b^{2}}\, b^{2}-4 a b c +b^{3}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{a}\)	\(284\)
risch	\(\text {Expression too large to display}\)	\(1303\)

Input:

int(csc(x)/(a+b*sin(x)+c*sin(x)^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5296 vs. \(2 (208) = 416\).

Time = 70.19 (sec) , antiderivative size = 5296, normalized size of antiderivative = 21.70 \[ \int \frac {\csc (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

integrate(csc(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(csc(x)/(a+b*sin(x)+c*sin(x)**2),x)
 

Output:

Integral(csc(x)/(a + b*sin(x) + c*sin(x)**2), x)
 

Maxima [F]

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

integrate(csc(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="maxima")
 

Output:

-1/2*(2*a*integrate(2*(2*b*c*cos(3*x)^2 + 2*b*c*cos(x)^2 + 2*b*c*sin(3*x)^ 
2 + 2*b*c*sin(x)^2 + 4*(2*a*b + b*c)*cos(2*x)^2 + 2*(2*b^2 + 2*a*c + c^2)* 
cos(x)*sin(2*x) + 4*(2*a*b + b*c)*sin(2*x)^2 + c^2*sin(x) - (2*b*c*cos(2*x 
) + c^2*sin(3*x) - c^2*sin(x))*cos(4*x) - 2*(2*b*c*cos(x) + (2*b^2 + 2*a*c 
 + c^2)*sin(2*x))*cos(3*x) - 2*(b*c + (2*b^2 + 2*a*c + c^2)*sin(x))*cos(2* 
x) + (c^2*cos(3*x) - c^2*cos(x) - 2*b*c*sin(2*x))*sin(4*x) - (4*b*c*sin(x) 
 + c^2 - 2*(2*b^2 + 2*a*c + c^2)*cos(2*x))*sin(3*x))/(a*c^2*cos(4*x)^2 + 4 
*a*b^2*cos(3*x)^2 + 4*a*b^2*cos(x)^2 + a*c^2*sin(4*x)^2 + 4*a*b^2*sin(3*x) 
^2 + 4*a*b^2*sin(x)^2 + 4*a*b*c*sin(x) + a*c^2 + 4*(4*a^3 + 4*a^2*c + a*c^ 
2)*cos(2*x)^2 + 8*(2*a^2*b + a*b*c)*cos(x)*sin(2*x) + 4*(4*a^3 + 4*a^2*c + 
 a*c^2)*sin(2*x)^2 - 2*(2*a*b*c*sin(3*x) - 2*a*b*c*sin(x) - a*c^2 + 2*(2*a 
^2*c + a*c^2)*cos(2*x))*cos(4*x) - 8*(a*b^2*cos(x) + (2*a^2*b + a*b*c)*sin 
(2*x))*cos(3*x) - 4*(2*a^2*c + a*c^2 + 2*(2*a^2*b + a*b*c)*sin(x))*cos(2*x 
) + 4*(a*b*c*cos(3*x) - a*b*c*cos(x) - (2*a^2*c + a*c^2)*sin(2*x))*sin(4*x 
) - 4*(2*a*b^2*sin(x) + a*b*c - 2*(2*a^2*b + a*b*c)*cos(2*x))*sin(3*x)), x 
) + log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1) - log(cos(x)^2 + sin(x)^2 - 2* 
cos(x) + 1))/a
 

Giac [F(-1)]

Timed out. \[ \int \frac {\csc (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \] Input:

integrate(csc(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 45.82 (sec) , antiderivative size = 11540, normalized size of antiderivative = 47.30 \[ \int \frac {\csc (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

int(1/(sin(x)*(a + c*sin(x)^2 + b*sin(x))),x)
 

Output:

atan((((8*a^2*c^4 - b^6 + 8*a^3*c^3 - b^3*(-(4*a*c - b^2)^3)^(1/2) + b^4*c 
^2 - 6*a*b^2*c^3 + b*c^2*(-(4*a*c - b^2)^3)^(1/2) - 18*a^2*b^2*c^2 + 8*a*b 
^4*c + 2*a*b*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(a^4*b^4 - a^2*b^6 + 16*a^4*c^ 
4 + 32*a^5*c^3 + 16*a^6*c^2 + 10*a^3*b^4*c - 8*a^5*b^2*c + a^2*b^4*c^2 - 8 
*a^3*b^2*c^3 - 32*a^4*b^2*c^2)))^(1/2)*(((8*a^2*c^4 - b^6 + 8*a^3*c^3 - b^ 
3*(-(4*a*c - b^2)^3)^(1/2) + b^4*c^2 - 6*a*b^2*c^3 + b*c^2*(-(4*a*c - b^2) 
^3)^(1/2) - 18*a^2*b^2*c^2 + 8*a*b^4*c + 2*a*b*c*(-(4*a*c - b^2)^3)^(1/2)) 
/(2*(a^4*b^4 - a^2*b^6 + 16*a^4*c^4 + 32*a^5*c^3 + 16*a^6*c^2 + 10*a^3*b^4 
*c - 8*a^5*b^2*c + a^2*b^4*c^2 - 8*a^3*b^2*c^3 - 32*a^4*b^2*c^2)))^(1/2)*( 
((8*a^2*c^4 - b^6 + 8*a^3*c^3 - b^3*(-(4*a*c - b^2)^3)^(1/2) + b^4*c^2 - 6 
*a*b^2*c^3 + b*c^2*(-(4*a*c - b^2)^3)^(1/2) - 18*a^2*b^2*c^2 + 8*a*b^4*c + 
 2*a*b*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(a^4*b^4 - a^2*b^6 + 16*a^4*c^4 + 32 
*a^5*c^3 + 16*a^6*c^2 + 10*a^3*b^4*c - 8*a^5*b^2*c + a^2*b^4*c^2 - 8*a^3*b 
^2*c^3 - 32*a^4*b^2*c^2)))^(1/2)*(((8*a^2*c^4 - b^6 + 8*a^3*c^3 - b^3*(-(4 
*a*c - b^2)^3)^(1/2) + b^4*c^2 - 6*a*b^2*c^3 + b*c^2*(-(4*a*c - b^2)^3)^(1 
/2) - 18*a^2*b^2*c^2 + 8*a*b^4*c + 2*a*b*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(a 
^4*b^4 - a^2*b^6 + 16*a^4*c^4 + 32*a^5*c^3 + 16*a^6*c^2 + 10*a^3*b^4*c - 8 
*a^5*b^2*c + a^2*b^4*c^2 - 8*a^3*b^2*c^3 - 32*a^4*b^2*c^2)))^(1/2)*(tan(x/ 
2)*(256*a^6*c - 512*a*b^6 + 544*a^3*b^4 - 64*a^5*b^2 + 6144*a^3*c^4 + 1228 
8*a^4*c^3 + 6400*a^5*c^2 + 512*a*b^4*c^2 + 4608*a^2*b^4*c - 3776*a^4*b^...
 

Reduce [F]

\[ \int \frac {\csc (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {too large to display} \] Input:

int(csc(x)/(a+b*sin(x)+c*sin(x)^2),x)
 

Output:

(2*int(tan(x/2)/(tan(x/2)**4*a**3 + 6*tan(x/2)**4*a**2*c - 2*tan(x/2)**4*a 
*b**2 + 2*tan(x/2)**3*a**2*b + 12*tan(x/2)**3*a*b*c - 4*tan(x/2)**3*b**3 + 
 2*tan(x/2)**2*a**3 + 16*tan(x/2)**2*a**2*c - 4*tan(x/2)**2*a*b**2 + 24*ta 
n(x/2)**2*a*c**2 - 8*tan(x/2)**2*b**2*c + 2*tan(x/2)*a**2*b + 12*tan(x/2)* 
a*b*c - 4*tan(x/2)*b**3 + a**3 + 6*a**2*c - 2*a*b**2),x)*a**4*c - 2*int(ta 
n(x/2)/(tan(x/2)**4*a**3 + 6*tan(x/2)**4*a**2*c - 2*tan(x/2)**4*a*b**2 + 2 
*tan(x/2)**3*a**2*b + 12*tan(x/2)**3*a*b*c - 4*tan(x/2)**3*b**3 + 2*tan(x/ 
2)**2*a**3 + 16*tan(x/2)**2*a**2*c - 4*tan(x/2)**2*a*b**2 + 24*tan(x/2)**2 
*a*c**2 - 8*tan(x/2)**2*b**2*c + 2*tan(x/2)*a**2*b + 12*tan(x/2)*a*b*c - 4 
*tan(x/2)*b**3 + a**3 + 6*a**2*c - 2*a*b**2),x)*a**3*b**2 - 16*int(tan(x/2 
)/(tan(x/2)**4*a**3 + 6*tan(x/2)**4*a**2*c - 2*tan(x/2)**4*a*b**2 + 2*tan( 
x/2)**3*a**2*b + 12*tan(x/2)**3*a*b*c - 4*tan(x/2)**3*b**3 + 2*tan(x/2)**2 
*a**3 + 16*tan(x/2)**2*a**2*c - 4*tan(x/2)**2*a*b**2 + 24*tan(x/2)**2*a*c* 
*2 - 8*tan(x/2)**2*b**2*c + 2*tan(x/2)*a**2*b + 12*tan(x/2)*a*b*c - 4*tan( 
x/2)*b**3 + a**3 + 6*a**2*c - 2*a*b**2),x)*a**3*c**2 - 8*int(tan(x/2)/(tan 
(x/2)**4*a**3 + 6*tan(x/2)**4*a**2*c - 2*tan(x/2)**4*a*b**2 + 2*tan(x/2)** 
3*a**2*b + 12*tan(x/2)**3*a*b*c - 4*tan(x/2)**3*b**3 + 2*tan(x/2)**2*a**3 
+ 16*tan(x/2)**2*a**2*c - 4*tan(x/2)**2*a*b**2 + 24*tan(x/2)**2*a*c**2 - 8 
*tan(x/2)**2*b**2*c + 2*tan(x/2)*a**2*b + 12*tan(x/2)*a*b*c - 4*tan(x/2)*b 
**3 + a**3 + 6*a**2*c - 2*a*b**2),x)*a**2*b**2*c - 184*int(tan(x/2)/(ta...