Integrand size = 19, antiderivative size = 76 \[ \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=-\frac {\left (b^2-2 c (a+c)\right ) \text {arctanh}\left (\frac {b+2 c \cos (x)}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {\cos (x)}{c}-\frac {b \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 c^2} \] Output:
-(b^2-2*c*(a+c))*arctanh((b+2*c*cos(x))/(-4*a*c+b^2)^(1/2))/c^2/(-4*a*c+b^ 2)^(1/2)+cos(x)/c-1/2*b*ln(a+b*cos(x)+c*cos(x)^2)/c^2
Time = 0.23 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.72 \[ \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {2 c \sqrt {b^2-4 a c} \cos (x)+\left (b^2-2 c (a+c)-b \sqrt {b^2-4 a c}\right ) \log \left (-b+\sqrt {b^2-4 a c}-2 c \cos (x)\right )-\left (b^2-2 c (a+c)+b \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c \cos (x)\right )}{2 c^2 \sqrt {b^2-4 a c}} \] Input:
Integrate[Sin[x]^3/(a + b*Cos[x] + c*Cos[x]^2),x]
Output:
(2*c*Sqrt[b^2 - 4*a*c]*Cos[x] + (b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c])* Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*Cos[x]] - (b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*Cos[x]])/(2*c^2*Sqrt[b^2 - 4*a* c])
Time = 0.34 (sec) , antiderivative size = 76, 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.
\(\displaystyle \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)^3}{a+b \cos (x)+c \cos (x)^2}dx\) |
\(\Big \downarrow \) 3740 |
\(\displaystyle -\int \frac {1-\cos ^2(x)}{c \cos ^2(x)+b \cos (x)+a}d\cos (x)\) |
\(\Big \downarrow \) 2188 |
\(\displaystyle -\int \left (\frac {a+c+b \cos (x)}{c \left (c \cos ^2(x)+b \cos (x)+a\right )}-\frac {1}{c}\right )d\cos (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (b^2-2 c (a+c)\right ) \text {arctanh}\left (\frac {b+2 c \cos (x)}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 c^2}+\frac {\cos (x)}{c}\) |
Input:
Int[Sin[x]^3/(a + b*Cos[x] + c*Cos[x]^2),x]
Output:
-(((b^2 - 2*c*(a + c))*ArcTanh[(b + 2*c*Cos[x])/Sqrt[b^2 - 4*a*c]])/(c^2*S qrt[b^2 - 4*a*c])) + Cos[x]/c - (b*Log[a + b*Cos[x] + c*Cos[x]^2])/(2*c^2)
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 = 0.59 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {\cos \left (x \right )}{c}+\frac {-\frac {b \ln \left (a +b \cos \left (x \right )+c \cos \left (x \right )^{2}\right )}{2 c}+\frac {2 \left (-a -c +\frac {b^{2}}{2 c}\right ) \arctan \left (\frac {b +2 c \cos \left (x \right )}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c}\) | \(82\) |
default | \(\frac {\cos \left (x \right )}{c}+\frac {-\frac {b \ln \left (a +b \cos \left (x \right )+c \cos \left (x \right )^{2}\right )}{2 c}+\frac {2 \left (-a -c +\frac {b^{2}}{2 c}\right ) \arctan \left (\frac {b +2 c \cos \left (x \right )}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c}\) | \(82\) |
risch | \(\text {Expression too large to display}\) | \(1097\) |
Input:
int(sin(x)^3/(a+b*cos(x)+c*cos(x)^2),x,method=_RETURNVERBOSE)
Output:
cos(x)/c+1/c*(-1/2*b/c*ln(a+b*cos(x)+c*cos(x)^2)+2*(-a-c+1/2*b^2/c)/(4*a*c -b^2)^(1/2)*arctan((b+2*c*cos(x))/(4*a*c-b^2)^(1/2)))
Time = 0.13 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.39 \[ \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\left [-\frac {{\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (-\frac {2 \, c^{2} \cos \left (x\right )^{2} + 2 \, b c \cos \left (x\right ) + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c \cos \left (x\right ) + b\right )}}{c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a}\right ) - 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} \cos \left (x\right ) + {\left (b^{3} - 4 \, a b c\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, -\frac {2 \, {\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c \cos \left (x\right ) + b\right )}}{b^{2} - 4 \, a c}\right ) - 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} \cos \left (x\right ) + {\left (b^{3} - 4 \, a b c\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \] Input:
integrate(sin(x)^3/(a+b*cos(x)+c*cos(x)^2),x, algorithm="fricas")
Output:
[-1/2*((b^2 - 2*a*c - 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)) - 2*(b^2*c - 4*a*c^2)*cos(x) + (b^3 - 4*a*b*c)*log(c*cos(x )^2 + b*cos(x) + a))/(b^2*c^2 - 4*a*c^3), -1/2*(2*(b^2 - 2*a*c - 2*c^2)*sq rt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*cos(x) + b)/(b^2 - 4*a*c) ) - 2*(b^2*c - 4*a*c^2)*cos(x) + (b^3 - 4*a*b*c)*log(c*cos(x)^2 + b*cos(x) + a))/(b^2*c^2 - 4*a*c^3)]
Timed out. \[ \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(sin(x)**3/(a+b*cos(x)+c*cos(x)**2),x)
Output:
Timed out
Exception generated. \[ \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(x)^3/(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 = 76, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {\cos \left (x\right )}{c} - \frac {b \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{2 \, c^{2}} + \frac {{\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \arctan \left (\frac {2 \, c \cos \left (x\right ) + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{2}} \] Input:
integrate(sin(x)^3/(a+b*cos(x)+c*cos(x)^2),x, algorithm="giac")
Output:
cos(x)/c - 1/2*b*log(c*cos(x)^2 + b*cos(x) + a)/c^2 + (b^2 - 2*a*c - 2*c^2 )*arctan((2*c*cos(x) + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^2)
Time = 1.10 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.97 \[ \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {\cos \left (x\right )}{c}-\frac {2\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,\cos \left (x\right )}{\sqrt {4\,a\,c-b^2}}\right )}{\sqrt {4\,a\,c-b^2}}+\frac {b^3\,\ln \left (c\,{\cos \left (x\right )}^2+b\,\cos \left (x\right )+a\right )}{2\,\left (4\,a\,c^3-b^2\,c^2\right )}+\frac {b^2\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,\cos \left (x\right )}{\sqrt {4\,a\,c-b^2}}\right )}{c^2\,\sqrt {4\,a\,c-b^2}}-\frac {2\,a\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,\cos \left (x\right )}{\sqrt {4\,a\,c-b^2}}\right )}{c\,\sqrt {4\,a\,c-b^2}}-\frac {2\,a\,b\,c\,\ln \left (c\,{\cos \left (x\right )}^2+b\,\cos \left (x\right )+a\right )}{4\,a\,c^3-b^2\,c^2} \] Input:
int(sin(x)^3/(a + b*cos(x) + c*cos(x)^2),x)
Output:
cos(x)/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 - b^2)^(1/2) + (b^3*log(a + b*cos(x) + c*cos(x)^2))/(2*(4*a*c^3 - b^2*c^2)) + (b^2*atan(b/(4*a*c - b^2)^(1/2) + (2*c*cos(x))/(4*a*c - b^2 )^(1/2)))/(c^2*(4*a*c - b^2)^(1/2)) - (2*a*atan(b/(4*a*c - b^2)^(1/2) + (2 *c*cos(x))/(4*a*c - b^2)^(1/2)))/(c*(4*a*c - b^2)^(1/2)) - (2*a*b*c*log(a + b*cos(x) + c*cos(x)^2))/(4*a*c^3 - b^2*c^2)
Time = 0.19 (sec) , antiderivative size = 916, normalized size of antiderivative = 12.05 \[ \int \frac {\sin ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx =\text {Too large to display} \] Input:
int(sin(x)^3/(a+b*cos(x)+c*cos(x)^2),x)
Output:
( - 4*sqrt(sqrt(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(sqrt(a + b + c)*s qrt(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 + 2*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(sqr t(a + b + c)*sqrt(a - b + c) + a - c)*sqrt(2)))*b**2 - 4*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)))* c**2 - 4*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 + 2*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**2 - 4*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...