Integrand size = 19, antiderivative size = 136 \[ \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \text {arctanh}\left (\frac {b+2 c \cos (x)}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}-\frac {\left (b^2-c (a+2 c)\right ) \cos (x)}{c^3}+\frac {b \cos ^2(x)}{2 c^2}-\frac {\cos ^3(x)}{3 c}+\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 c^4} \] Output:
(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))/c^4/(-4*a*c+b^2)^(1/2)-(b^2-c*(a+2*c))*cos(x)/c^3+1/2*b*cos(x)^2/c^2- 1/3*cos(x)^3/c+1/2*b*(b^2-2*c*(a+c))*ln(a+b*cos(x)+c*cos(x)^2)/c^4
Time = 0.48 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.76 \[ \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {3 c \left (-4 b^2+c (4 a+7 c)\right ) \cos (x)+3 b c^2 \cos (2 x)-c^3 \cos (3 x)+\frac {6 \left (-b^4-2 c^2 (a+c)^2+2 b^2 c (2 a+c)+b^3 \sqrt {b^2-4 a c}-2 b c (a+c) \sqrt {b^2-4 a c}\right ) \log \left (-b+\sqrt {b^2-4 a c}-2 c \cos (x)\right )}{\sqrt {b^2-4 a c}}+\frac {6 \left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)+b^3 \sqrt {b^2-4 a c}-2 b c (a+c) \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c \cos (x)\right )}{\sqrt {b^2-4 a c}}}{12 c^4} \] Input:
Integrate[Sin[x]^5/(a + b*Cos[x] + c*Cos[x]^2),x]
Output:
(3*c*(-4*b^2 + c*(4*a + 7*c))*Cos[x] + 3*b*c^2*Cos[2*x] - c^3*Cos[3*x] + ( 6*(-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]])/S qrt[b^2 - 4*a*c] + (6*(b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c) + b^3*Sqr t[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])/(12*c^4)
Time = 0.45 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3740, 2188, 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.
\(\displaystyle \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)^5}{a+b \cos (x)+c \cos (x)^2}dx\) |
\(\Big \downarrow \) 3740 |
\(\displaystyle -\int \frac {\left (1-\cos ^2(x)\right )^2}{c \cos ^2(x)+b \cos (x)+a}d\cos (x)\) |
\(\Big \downarrow \) 2188 |
\(\displaystyle -\int \left (\frac {\cos ^2(x)}{c}-\frac {b \cos (x)}{c^2}+\frac {b^2-c (a+2 c)}{c^3}-\frac {-c^3-a^2 c+a \left (b^2-2 c^2\right )+b \left (b^2-2 c (a+c)\right ) \cos (x)}{c^3 \left (c \cos ^2(x)+b \cos (x)+a\right )}\right )d\cos (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-2 b^2 c (2 a+c)+2 c^2 (a+c)^2+b^4\right ) \text {arctanh}\left (\frac {b+2 c \cos (x)}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 c^4}-\frac {\cos (x) \left (b^2-c (a+2 c)\right )}{c^3}+\frac {b \cos ^2(x)}{2 c^2}-\frac {\cos ^3(x)}{3 c}\) |
Input:
Int[Sin[x]^5/(a + b*Cos[x] + c*Cos[x]^2),x]
Output:
((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]])/(c^4*Sqrt[b^2 - 4*a*c]) - ((b^2 - c*(a + 2*c))*Cos[x])/c^3 + (b*Cos[x]^2)/(2*c^2) - Cos[x]^3/(3*c) + (b*(b^2 - 2*c*(a + c))*Log[a + b*Cos[x] + c*Cos[x]^2])/(2*c^4)
Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq , x] && IGtQ[p, -2]
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 = 1.72 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.18
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (x \right )^{3} c^{2}}{3}+\frac {\cos \left (x \right )^{2} b c}{2}+\cos \left (x \right ) a c -\cos \left (x \right ) b^{2}+2 \cos \left (x \right ) c^{2}}{c^{3}}+\frac {\frac {\left (-2 c a b +b^{3}-2 b \,c^{2}\right ) \ln \left (a +b \cos \left (x \right )+c \cos \left (x \right )^{2}\right )}{2 c}+\frac {2 \left (-a^{2} c +a \,b^{2}-2 a \,c^{2}-c^{3}-\frac {\left (-2 c a b +b^{3}-2 b \,c^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {b +2 c \cos \left (x \right )}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c^{3}}\) | \(160\) |
default | \(\frac {-\frac {\cos \left (x \right )^{3} c^{2}}{3}+\frac {\cos \left (x \right )^{2} b c}{2}+\cos \left (x \right ) a c -\cos \left (x \right ) b^{2}+2 \cos \left (x \right ) c^{2}}{c^{3}}+\frac {\frac {\left (-2 c a b +b^{3}-2 b \,c^{2}\right ) \ln \left (a +b \cos \left (x \right )+c \cos \left (x \right )^{2}\right )}{2 c}+\frac {2 \left (-a^{2} c +a \,b^{2}-2 a \,c^{2}-c^{3}-\frac {\left (-2 c a b +b^{3}-2 b \,c^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {b +2 c \cos \left (x \right )}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c^{3}}\) | \(160\) |
risch | \(\text {Expression too large to display}\) | \(4217\) |
Input:
int(sin(x)^5/(a+b*cos(x)+c*cos(x)^2),x,method=_RETURNVERBOSE)
Output:
1/c^3*(-1/3*cos(x)^3*c^2+1/2*cos(x)^2*b*c+cos(x)*a*c-cos(x)*b^2+2*cos(x)*c ^2)+1/c^3*(1/2*(-2*a*b*c+b^3-2*b*c^2)/c*ln(a+b*cos(x)+c*cos(x)^2)+2*(-a^2* c+a*b^2-2*a*c^2-c^3-1/2*(-2*a*b*c+b^3-2*b*c^2)*b/c)/(4*a*c-b^2)^(1/2)*arct an((b+2*c*cos(x))/(4*a*c-b^2)^(1/2)))
Time = 0.19 (sec) , antiderivative size = 491, normalized size of antiderivative = 3.61 \[ \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\left [-\frac {2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \cos \left (x\right )^{3} - 3 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} \cos \left (x\right )^{2} - 3 \, {\left (b^{4} - 4 \, a b^{2} c + 4 \, a c^{3} + 2 \, c^{4} + 2 \, {\left (a^{2} - b^{2}\right )} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (-\frac {2 \, c^{2} \cos \left (x\right )^{2} + 2 \, b c \cos \left (x\right ) + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c \cos \left (x\right ) + b\right )}}{c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a}\right ) + 6 \, {\left (b^{4} c - 5 \, a b^{2} c^{2} + 8 \, a c^{4} + 2 \, {\left (2 \, a^{2} - b^{2}\right )} c^{3}\right )} \cos \left (x\right ) - 3 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a b c^{3} + 2 \, {\left (4 \, a^{2} b - b^{3}\right )} c^{2}\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{6 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )}}, -\frac {2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \cos \left (x\right )^{3} - 3 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} \cos \left (x\right )^{2} - 6 \, {\left (b^{4} - 4 \, a b^{2} c + 4 \, a c^{3} + 2 \, c^{4} + 2 \, {\left (a^{2} - b^{2}\right )} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c \cos \left (x\right ) + b\right )}}{b^{2} - 4 \, a c}\right ) + 6 \, {\left (b^{4} c - 5 \, a b^{2} c^{2} + 8 \, a c^{4} + 2 \, {\left (2 \, a^{2} - b^{2}\right )} c^{3}\right )} \cos \left (x\right ) - 3 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a b c^{3} + 2 \, {\left (4 \, a^{2} b - b^{3}\right )} c^{2}\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{6 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )}}\right ] \] Input:
integrate(sin(x)^5/(a+b*cos(x)+c*cos(x)^2),x, algorithm="fricas")
Output:
[-1/6*(2*(b^2*c^3 - 4*a*c^4)*cos(x)^3 - 3*(b^3*c^2 - 4*a*b*c^3)*cos(x)^2 - 3*(b^4 - 4*a*b^2*c + 4*a*c^3 + 2*c^4 + 2*(a^2 - b^2)*c^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)) + 6*(b^4*c - 5*a*b^2*c^2 + 8 *a*c^4 + 2*(2*a^2 - b^2)*c^3)*cos(x) - 3*(b^5 - 6*a*b^3*c + 8*a*b*c^3 + 2* (4*a^2*b - b^3)*c^2)*log(c*cos(x)^2 + b*cos(x) + a))/(b^2*c^4 - 4*a*c^5), -1/6*(2*(b^2*c^3 - 4*a*c^4)*cos(x)^3 - 3*(b^3*c^2 - 4*a*b*c^3)*cos(x)^2 - 6*(b^4 - 4*a*b^2*c + 4*a*c^3 + 2*c^4 + 2*(a^2 - b^2)*c^2)*sqrt(-b^2 + 4*a* c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*cos(x) + b)/(b^2 - 4*a*c)) + 6*(b^4*c - 5*a*b^2*c^2 + 8*a*c^4 + 2*(2*a^2 - b^2)*c^3)*cos(x) - 3*(b^5 - 6*a*b^3*c + 8*a*b*c^3 + 2*(4*a^2*b - b^3)*c^2)*log(c*cos(x)^2 + b*cos(x) + a))/(b^2* c^4 - 4*a*c^5)]
Timed out. \[ \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(sin(x)**5/(a+b*cos(x)+c*cos(x)**2),x)
Output:
Timed out
Exception generated. \[ \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(x)^5/(a+b*cos(x)+c*cos(x)^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(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12 \[ \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=-\frac {2 \, c^{2} \cos \left (x\right )^{3} - 3 \, b c \cos \left (x\right )^{2} + 6 \, b^{2} \cos \left (x\right ) - 6 \, a c \cos \left (x\right ) - 12 \, c^{2} \cos \left (x\right )}{6 \, c^{3}} + \frac {{\left (b^{3} - 2 \, a b c - 2 \, b c^{2}\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{2 \, c^{4}} - \frac {{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + 2 \, c^{4}\right )} \arctan \left (\frac {2 \, c \cos \left (x\right ) + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{4}} \] Input:
integrate(sin(x)^5/(a+b*cos(x)+c*cos(x)^2),x, algorithm="giac")
Output:
-1/6*(2*c^2*cos(x)^3 - 3*b*c*cos(x)^2 + 6*b^2*cos(x) - 6*a*c*cos(x) - 12*c ^2*cos(x))/c^3 + 1/2*(b^3 - 2*a*b*c - 2*b*c^2)*log(c*cos(x)^2 + b*cos(x) + a)/c^4 - (b^4 - 4*a*b^2*c + 2*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 + 2*c^4)*arct an((2*c*cos(x) + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^4)
Time = 0.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.45 \[ \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\cos \left (x\right )\,\left (\frac {a}{c^2}+\frac {2}{c}-\frac {b^2}{c^3}\right )-\frac {{\cos \left (x\right )}^3}{3\,c}-\frac {\ln \left (c\,{\cos \left (x\right )}^2+b\,\cos \left (x\right )+a\right )\,\left (8\,a^2\,b\,c^2-6\,a\,b^3\,c+8\,a\,b\,c^3+b^5-2\,b^3\,c^2\right )}{2\,\left (4\,a\,c^5-b^2\,c^4\right )}+\frac {b\,{\cos \left (x\right )}^2}{2\,c^2}-\frac {\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,\cos \left (x\right )}{\sqrt {4\,a\,c-b^2}}\right )\,\left (2\,a^2\,c^2-4\,a\,b^2\,c+4\,a\,c^3+b^4-2\,b^2\,c^2+2\,c^4\right )}{c^4\,\sqrt {4\,a\,c-b^2}} \] Input:
int(sin(x)^5/(a + b*cos(x) + c*cos(x)^2),x)
Output:
cos(x)*(a/c^2 + 2/c - b^2/c^3) - cos(x)^3/(3*c) - (log(a + b*cos(x) + c*co s(x)^2)*(b^5 - 2*b^3*c^2 + 8*a^2*b*c^2 + 8*a*b*c^3 - 6*a*b^3*c))/(2*(4*a*c ^5 - b^2*c^4)) + (b*cos(x)^2)/(2*c^2) - (atan(b/(4*a*c - b^2)^(1/2) + (2*c *cos(x))/(4*a*c - b^2)^(1/2))*(4*a*c^3 + b^4 + 2*c^4 + 2*a^2*c^2 - 2*b^2*c ^2 - 4*a*b^2*c))/(c^4*(4*a*c - b^2)^(1/2))
Time = 3.84 (sec) , antiderivative size = 2069, normalized size of antiderivative = 15.21 \[ \int \frac {\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx =\text {Too large to display} \] Input:
int(sin(x)^5/(a+b*cos(x)+c*cos(x)^2),x)
Output:
( - 12*sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(sqrt(a + b + c)* sqrt(a - b + c) - a + c)*atan((sqrt(sqrt(a + b + c)*sqrt(a - b + c) - a + c)*sqrt(2) - 2*sqrt(a - b + c)*tan(x/2))/(sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(2)))*a**2*c**2 + 24*sqrt(sqrt(a + b + c)*sqrt(a - b + c ) + a - c)*sqrt(sqrt(a + b + c)*sqrt(a - b + c) - a + c)*atan((sqrt(sqrt(a + b + c)*sqrt(a - b + c) - a + c)*sqrt(2) - 2*sqrt(a - b + c)*tan(x/2))/( sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(2)))*a*b**2*c - 24*sqrt (sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(sqrt(a + b + c)*sqrt(a - b + c) - a + c)*atan((sqrt(sqrt(a + b + c)*sqrt(a - b + c) - a + c)*sqrt(2) - 2*sqrt(a - b + c)*tan(x/2))/(sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(2)))*a*c**3 - 6*sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt (sqrt(a + b + c)*sqrt(a - b + c) - a + c)*atan((sqrt(sqrt(a + b + c)*sqrt( a - b + c) - a + c)*sqrt(2) - 2*sqrt(a - b + c)*tan(x/2))/(sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(2)))*b**4 + 12*sqrt(sqrt(a + b + c)*sq rt(a - b + c) + a - c)*sqrt(sqrt(a + b + c)*sqrt(a - b + c) - a + c)*atan( (sqrt(sqrt(a + b + c)*sqrt(a - b + c) - a + c)*sqrt(2) - 2*sqrt(a - b + c) *tan(x/2))/(sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(2)))*b**2*c **2 - 12*sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(sqrt(a + b + c )*sqrt(a - b + c) - a + c)*atan((sqrt(sqrt(a + b + c)*sqrt(a - b + c) - a + c)*sqrt(2) - 2*sqrt(a - b + c)*tan(x/2))/(sqrt(sqrt(a + b + c)*sqrt(a...