Integrand size = 19, antiderivative size = 205 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \text {arctanh}\left (\frac {b+2 c \cos (x)}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a^2-b^2+2 a c+c^2\right )^2}+\frac {(b-(a+c) \cos (x)) \csc ^2(x)}{2 (a-b+c) (a+b+c)}+\frac {(a+2 b+3 c) \log (1-\cos (x))}{4 (a+b+c)^2}-\frac {(a-2 b+3 c) \log (1+\cos (x))}{4 (a-b+c)^2}-\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 \left (a^2-b^2+2 a c+c^2\right )^2} \]
1/2*(b-(a+c)*cos(x))*csc(x)^2/(a-b+c)/(a+b+c)+1/4*(a+2*b+3*c)*ln(1-cos(x)) /(a+b+c)^2-1/4*(a-2*b+3*c)*ln(1+cos(x))/(a-b+c)^2-1/2*b*(b^2-2*c*(a+c))*ln (a+b*cos(x)+c*cos(x)^2)/(a^2+2*a*c-b^2+c^2)^2+(b^4+2*c^2*(a+c)^2-2*b^2*c*( 2*a+c))*arctanh((b+2*c*cos(x))/(-4*a*c+b^2)^(1/2))/(a^2+2*a*c-b^2+c^2)^2/( -4*a*c+b^2)^(1/2)
Result contains complex when optimal does not.
Time = 1.76 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.91 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {1}{8} \left (\frac {16 i \left (b^3-2 b c (a+c)\right ) x}{(a-b+c)^2 (a+b+c)^2}+\frac {4 i (a-2 b+3 c) \arctan (\tan (x))}{(a-b+c)^2}-\frac {4 i (a+2 b+3 c) \arctan (\tan (x))}{(a+b+c)^2}-\frac {\csc ^2\left (\frac {x}{2}\right )}{a+b+c}-\frac {2 (a-2 b+3 c) \log \left (\cos ^2\left (\frac {x}{2}\right )\right )}{(a-b+c)^2}-\frac {4 \left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)+b^3 \sqrt {b^2-4 a c}-2 b c (a+c) \sqrt {b^2-4 a c}\right ) \log \left (-b+\sqrt {b^2-4 a c}-2 c \cos (x)\right )}{\sqrt {b^2-4 a c} \left (a^2-b^2+2 a c+c^2\right )^2}-\frac {4 \left (-b^4-2 c^2 (a+c)^2+2 b^2 c (2 a+c)+b^3 \sqrt {b^2-4 a c}-2 b c (a+c) \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c \cos (x)\right )}{\sqrt {b^2-4 a c} \left (a^2-b^2+2 a c+c^2\right )^2}+\frac {2 (a+2 b+3 c) \log \left (\sin ^2\left (\frac {x}{2}\right )\right )}{(a+b+c)^2}+\frac {\sec ^2\left (\frac {x}{2}\right )}{a-b+c}\right ) \]
(((16*I)*(b^3 - 2*b*c*(a + c))*x)/((a - b + c)^2*(a + b + c)^2) + ((4*I)*( a - 2*b + 3*c)*ArcTan[Tan[x]])/(a - b + c)^2 - ((4*I)*(a + 2*b + 3*c)*ArcT an[Tan[x]])/(a + b + c)^2 - Csc[x/2]^2/(a + b + c) - (2*(a - 2*b + 3*c)*Lo g[Cos[x/2]^2])/(a - b + c)^2 - (4*(b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c) + b^3*Sqrt[b^2 - 4*a*c] - 2*b*c*(a + c)*Sqrt[b^2 - 4*a*c])*Log[-b + Sqr t[b^2 - 4*a*c] - 2*c*Cos[x]])/(Sqrt[b^2 - 4*a*c]*(a^2 - b^2 + 2*a*c + c^2) ^2) - (4*(-b^4 - 2*c^2*(a + c)^2 + 2*b^2*c*(2*a + c) + b^3*Sqrt[b^2 - 4*a* c] - 2*b*c*(a + c)*Sqrt[b^2 - 4*a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*Cos[ x]])/(Sqrt[b^2 - 4*a*c]*(a^2 - b^2 + 2*a*c + c^2)^2) + (2*(a + 2*b + 3*c)* Log[Sin[x/2]^2])/(a + b + c)^2 + Sec[x/2]^2/(a - b + c))/8
Time = 0.57 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3740, 1301, 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.
\(\displaystyle \int \frac {\csc ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (x)^3 \left (a+b \cos (x)+c \cos (x)^2\right )}dx\) |
\(\Big \downarrow \) 3740 |
\(\displaystyle -\int \frac {1}{\left (1-\cos ^2(x)\right )^2 \left (c \cos ^2(x)+b \cos (x)+a\right )}d\cos (x)\) |
\(\Big \downarrow \) 1301 |
\(\displaystyle -\int \left (\frac {a-2 b+3 c}{4 (a-b+c)^2 (\cos (x)+1)}+\frac {a+2 b+3 c}{4 (a+b+c)^2 (1-\cos (x))}+\frac {b^4-c (3 a+2 c) b^2+c \left (b^2-2 c (a+c)\right ) \cos (x) b+c^2 (a+c)^2}{(a-b+c)^2 (a+b+c)^2 \left (c \cos ^2(x)+b \cos (x)+a\right )}+\frac {1}{4 (a+b+c) (1-\cos (x))^2}+\frac {1}{4 (a-b+c) (\cos (x)+1)^2}\right )d\cos (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-2 b^2 c (2 a+c)+2 c^2 (a+c)^2+b^4\right ) \text {arctanh}\left (\frac {b+2 c \cos (x)}{\sqrt {b^2-4 a c}}\right )}{(a-b+c)^2 (a+b+c)^2 \sqrt {b^2-4 a c}}-\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 (a-b+c)^2 (a+b+c)^2}-\frac {1}{4 (1-\cos (x)) (a+b+c)}+\frac {1}{4 (\cos (x)+1) (a-b+c)}+\frac {(a+2 b+3 c) \log (1-\cos (x))}{4 (a+b+c)^2}-\frac {(a-2 b+3 c) \log (\cos (x)+1)}{4 (a-b+c)^2}\) |
((b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c))*ArcTanh[(b + 2*c*Cos[x])/Sqrt [b^2 - 4*a*c]])/((a - b + c)^2*(a + b + c)^2*Sqrt[b^2 - 4*a*c]) - 1/(4*(a + b + c)*(1 - Cos[x])) + 1/(4*(a - b + c)*(1 + Cos[x])) + ((a + 2*b + 3*c) *Log[1 - Cos[x]])/(4*(a + b + c)^2) - ((a - 2*b + 3*c)*Log[1 + Cos[x]])/(4 *(a - b + c)^2) - (b*(b^2 - 2*c*(a + c))*Log[a + b*Cos[x] + c*Cos[x]^2])/( 2*(a - b + c)^2*(a + b + c)^2)
3.1.5.3.1 Defintions of rubi rules used
Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x _Symbol] :> With[{r = Rt[(-a)*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[(-r + c*x)^p*(r + c*x)^p*(d + e*x + f*x^2)^q, x], x], x] /; EqQ[p, -1] || !Fra ctionalPowerFactorQ[r]] /; FreeQ[{a, c, d, e, f}, x] && ILtQ[p, 0] && Integ erQ[q] && NiceSqrtQ[(-a)*c]
Int[((a_.) + (b_.)*(cos[(d_.) + (e_.)*(x_)]*(f_.))^(n_.) + (c_.)*(cos[(d_.) + (e_.)*(x_)]*(f_.))^(n2_.))^(p_.)*sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol ] :> Module[{g = FreeFactors[Cos[d + e*x], x]}, Simp[-g/e Subst[Int[(1 - g^2*x^2)^((m - 1)/2)*(a + b*(f*g*x)^n + c*(f*g*x)^(2*n))^p, x], x, Cos[d + e*x]/g], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[n2, 2*n] && Integ erQ[(m - 1)/2]
Time = 3.21 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.18
method | result | size |
default | \(\frac {\frac {\left (2 a b \,c^{2}-b^{3} c +2 b \,c^{3}\right ) \ln \left (a +\cos \left (x \right ) b +c \left (\cos ^{2}\left (x \right )\right )\right )}{2 c}+\frac {2 \left (-a^{2} c^{2}+3 a \,b^{2} c -2 a \,c^{3}-b^{4}+2 b^{2} c^{2}-c^{4}-\frac {\left (2 a b \,c^{2}-b^{3} c +2 b \,c^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {b +2 c \cos \left (x \right )}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a -b +c \right )^{2} \left (a +b +c \right )^{2}}+\frac {1}{\left (4 a -4 b +4 c \right ) \left (1+\cos \left (x \right )\right )}+\frac {\left (-a +2 b -3 c \right ) \ln \left (1+\cos \left (x \right )\right )}{4 \left (a -b +c \right )^{2}}+\frac {1}{\left (4 a +4 b +4 c \right ) \left (\cos \left (x \right )-1\right )}+\frac {\left (a +2 b +3 c \right ) \ln \left (\cos \left (x \right )-1\right )}{4 \left (a +b +c \right )^{2}}\) | \(241\) |
risch | \(\text {Expression too large to display}\) | \(6192\) |
1/(a-b+c)^2/(a+b+c)^2*(1/2*(2*a*b*c^2-b^3*c+2*b*c^3)/c*ln(a+cos(x)*b+c*cos (x)^2)+2*(-a^2*c^2+3*a*b^2*c-2*a*c^3-b^4+2*b^2*c^2-c^4-1/2*(2*a*b*c^2-b^3* c+2*b*c^3)*b/c)/(4*a*c-b^2)^(1/2)*arctan((b+2*c*cos(x))/(4*a*c-b^2)^(1/2)) )+1/(4*a-4*b+4*c)/(1+cos(x))+1/4/(a-b+c)^2*(-a+2*b-3*c)*ln(1+cos(x))+1/(4* a+4*b+4*c)/(cos(x)-1)+1/4*(a+2*b+3*c)/(a+b+c)^2*ln(cos(x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 983 vs. \(2 (194) = 388\).
Time = 5.43 (sec) , antiderivative size = 1991, normalized size of antiderivative = 9.71 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \]
[1/4*(2*a^2*b^3 - 2*b^5 - 8*a*b*c^3 - 2*(8*a^2*b - b^3)*c^2 + 2*(b^4 - 4*a *b^2*c + 4*a*c^3 + 2*c^4 + 2*(a^2 - b^2)*c^2 - (b^4 - 4*a*b^2*c + 4*a*c^3 + 2*c^4 + 2*(a^2 - b^2)*c^2)*cos(x)^2)*sqrt(b^2 - 4*a*c)*log(-(2*c^2*cos(x )^2 + 2*b*c*cos(x) + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*cos(x) + b))/(c* cos(x)^2 + b*cos(x) + a)) - 4*(2*a^3*b - 3*a*b^3)*c - 2*(a^3*b^2 - a*b^4 - 4*a*c^4 - (12*a^2 - b^2)*c^3 - (12*a^3 - 7*a*b^2)*c^2 - (4*a^4 - 7*a^2*b^ 2 + b^4)*c)*cos(x) - 2*(b^5 - 6*a*b^3*c + 8*a*b*c^3 + 2*(4*a^2*b - b^3)*c^ 2 - (b^5 - 6*a*b^3*c + 8*a*b*c^3 + 2*(4*a^2*b - b^3)*c^2)*cos(x)^2)*log(c* cos(x)^2 + b*cos(x) + a) - (a^3*b^2 - 3*a*b^4 - 2*b^5 - 12*a*c^4 - (28*a^2 + 16*a*b - 3*b^2)*c^3 - (20*a^3 + 16*a^2*b - 11*a*b^2 - 4*b^3)*c^2 - (a^3 *b^2 - 3*a*b^4 - 2*b^5 - 12*a*c^4 - (28*a^2 + 16*a*b - 3*b^2)*c^3 - (20*a^ 3 + 16*a^2*b - 11*a*b^2 - 4*b^3)*c^2 - (4*a^4 - 17*a^2*b^2 - 12*a*b^3 + b^ 4)*c)*cos(x)^2 - (4*a^4 - 17*a^2*b^2 - 12*a*b^3 + b^4)*c)*log(1/2*cos(x) + 1/2) + (a^3*b^2 - 3*a*b^4 + 2*b^5 - 12*a*c^4 - (28*a^2 - 16*a*b - 3*b^2)* c^3 - (20*a^3 - 16*a^2*b - 11*a*b^2 + 4*b^3)*c^2 - (a^3*b^2 - 3*a*b^4 + 2* b^5 - 12*a*c^4 - (28*a^2 - 16*a*b - 3*b^2)*c^3 - (20*a^3 - 16*a^2*b - 11*a *b^2 + 4*b^3)*c^2 - (4*a^4 - 17*a^2*b^2 + 12*a*b^3 + b^4)*c)*cos(x)^2 - (4 *a^4 - 17*a^2*b^2 + 12*a*b^3 + b^4)*c)*log(-1/2*cos(x) + 1/2))/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - (a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5...
\[ \int \frac {\csc ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\int \frac {\csc ^{3}{\left (x \right )}}{a + b \cos {\left (x \right )} + c \cos ^{2}{\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {\csc ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.30 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.84 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=-\frac {{\left (b^{3} - 2 \, a b c - 2 \, b c^{2}\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} + 4 \, a^{3} c - 4 \, a b^{2} c + 6 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + c^{4}\right )}} - \frac {{\left (a - 2 \, b + 3 \, c\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2} + 2 \, a c - 2 \, b c + c^{2}\right )}} + \frac {{\left (a + 2 \, b + 3 \, c\right )} \log \left (-\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2} + 2 \, a c + 2 \, b c + c^{2}\right )}} - \frac {{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + 2 \, c^{4}\right )} \arctan \left (\frac {2 \, c \cos \left (x\right ) + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} + 4 \, a^{3} c - 4 \, a b^{2} c + 6 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + c^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {a^{2} b - b^{3} + 2 \, a b c + b c^{2} - {\left (a^{3} - a b^{2} + 3 \, a^{2} c - b^{2} c + 3 \, a c^{2} + c^{3}\right )} \cos \left (x\right )}{2 \, {\left (a + b + c\right )}^{2} {\left (a - b + c\right )}^{2} {\left (\cos \left (x\right ) + 1\right )} {\left (\cos \left (x\right ) - 1\right )}} \]
-1/2*(b^3 - 2*a*b*c - 2*b*c^2)*log(c*cos(x)^2 + b*cos(x) + a)/(a^4 - 2*a^2 *b^2 + b^4 + 4*a^3*c - 4*a*b^2*c + 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 + c^4) - 1/4*(a - 2*b + 3*c)*log(cos(x) + 1)/(a^2 - 2*a*b + b^2 + 2*a*c - 2*b*c + c^2) + 1/4*(a + 2*b + 3*c)*log(-cos(x) + 1)/(a^2 + 2*a*b + b^2 + 2*a*c + 2*b*c + c^2) - (b^4 - 4*a*b^2*c + 2*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 + 2*c^4) *arctan((2*c*cos(x) + b)/sqrt(-b^2 + 4*a*c))/((a^4 - 2*a^2*b^2 + b^4 + 4*a ^3*c - 4*a*b^2*c + 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 + c^4)*sqrt(-b^2 + 4*a* c)) - 1/2*(a^2*b - b^3 + 2*a*b*c + b*c^2 - (a^3 - a*b^2 + 3*a^2*c - b^2*c + 3*a*c^2 + c^3)*cos(x))/((a + b + c)^2*(a - b + c)^2*(cos(x) + 1)*(cos(x) - 1))
Time = 21.56 (sec) , antiderivative size = 2742, normalized size of antiderivative = 13.38 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \]
(b/(2*(2*a*c + a^2 - b^2 + c^2)) - (cos(x)*(a + c))/(2*(2*a*c + a^2 - b^2 + c^2)))/sin(x)^2 - log(cos(x) + 1)*(1/(4*(a - b + c)) - (b/4 - c/2)/(a - b + c)^2) + log(cos(x) - 1)*((b/4 + c/2)/(a + b + c)^2 + 1/(4*(a + b + c)) ) - (log((c^4*(4*a*c + a^2 - 4*b^2 + 3*c^2))/(4*(2*a*c + a^2 - b^2 + c^2)^ 2) - (b*c^5*cos(x))/(2*a*c + a^2 - b^2 + c^2)^2 - (((((c*(a*b^4 + 28*a*c^4 + 4*a^4*c - 5*b^4*c + 8*c^5 - a^3*b^2 + 36*a^2*c^3 + 20*a^3*c^2 + 5*b^2*c ^3 - 3*a*b^2*c^2 - 9*a^2*b^2*c))/(2*(2*a*c + a^2 - b^2 + c^2)) - (2*c*((b^ 4*(b^2 - 4*a*c)^(1/2))/2 - b^5/2 + c^4*(b^2 - 4*a*c)^(1/2) + b^3*c^2 + 2*a *c^3*(b^2 - 4*a*c)^(1/2) - 4*a^2*b*c^2 + a^2*c^2*(b^2 - 4*a*c)^(1/2) - b^2 *c^2*(b^2 - 4*a*c)^(1/2) - 4*a*b*c^3 + 3*a*b^3*c - 2*a*b^2*c*(b^2 - 4*a*c) ^(1/2))*(4*a*b^3 + 2*b*c^3 + 2*b^3*c + 3*b^4*cos(x) + 4*c^4*cos(x) + 4*a*c ^3*cos(x) - 4*a^3*c*cos(x) + a^2*b^2*cos(x) - 4*a^2*c^2*cos(x) - 3*b^2*c^2 *cos(x) - 12*a*b*c^2 - 14*a^2*b*c - 10*a*b^2*c*cos(x)))/((4*a*c - b^2)*(2* a*c + a^2 - b^2 + c^2)^2) + (b*c*cos(x)*(36*a*c^3 + 4*a^3*c + 3*b^4 + 16*c ^4 - a^2*b^2 + 24*a^2*c^2 - 13*b^2*c^2 - 18*a*b^2*c))/(2*a*c + a^2 - b^2 + c^2))*((b^4*(b^2 - 4*a*c)^(1/2))/2 - b^5/2 + c^4*(b^2 - 4*a*c)^(1/2) + b^ 3*c^2 + 2*a*c^3*(b^2 - 4*a*c)^(1/2) - 4*a^2*b*c^2 + a^2*c^2*(b^2 - 4*a*c)^ (1/2) - b^2*c^2*(b^2 - 4*a*c)^(1/2) - 4*a*b*c^3 + 3*a*b^3*c - 2*a*b^2*c*(b ^2 - 4*a*c)^(1/2)))/((4*a*c - b^2)*(2*a*c + a^2 - b^2 + c^2)^2) - (b*c*(2* a*b^4 - 20*a*c^4 + 3*a^4*c - 6*b^4*c + 7*c^5 - a^3*b^2 - 26*a^2*c^3 + 4...