Integrand size = 13, antiderivative size = 59 \[ \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx=\frac {a x}{b^2}-\frac {2 \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^2}-\frac {\sin (x)}{b} \] Output:
a*x/b^2-2*(a-b)^(1/2)*(a+b)^(1/2)*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2 ))/b^2-sin(x)/b
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx=\frac {a x-2 \sqrt {-a^2+b^2} \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )-b \sin (x)}{b^2} \] Input:
Integrate[Sin[x]^2/(a + b*Cos[x]),x]
Output:
(a*x - 2*Sqrt[-a^2 + b^2]*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]] - b *Sin[x])/b^2
Time = 0.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.24, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 3174, 25, 3042, 3214, 3042, 3138, 218}
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.
\(\displaystyle \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (x-\frac {\pi }{2}\right )^2}{a-b \sin \left (x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3174 |
\(\displaystyle -\frac {\int -\frac {b+a \cos (x)}{a+b \cos (x)}dx}{b}-\frac {\sin (x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {b+a \cos (x)}{a+b \cos (x)}dx}{b}-\frac {\sin (x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {b+a \sin \left (x+\frac {\pi }{2}\right )}{a+b \sin \left (x+\frac {\pi }{2}\right )}dx}{b}-\frac {\sin (x)}{b}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {a x}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{a+b \cos (x)}dx}{b}}{b}-\frac {\sin (x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a x}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{a+b \sin \left (x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {\sin (x)}{b}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {a x}{b}-\frac {2 \left (a^2-b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {x}{2}\right )+a+b}d\tan \left (\frac {x}{2}\right )}{b}}{b}-\frac {\sin (x)}{b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {a x}{b}-\frac {2 \left (a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {\sin (x)}{b}\) |
Input:
Int[Sin[x]^2/(a + b*Cos[x]),x]
Output:
((a*x)/b - (2*(a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqr t[a - b]*b*Sqrt[a + b]))/b - Sin[x]/b
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(b*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; F reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 0.46 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.32
method | result | size |
default | \(-\frac {2 \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {-\frac {2 b \tan \left (\frac {x}{2}\right )}{1+\tan \left (\frac {x}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{2}}\) | \(78\) |
risch | \(\frac {a x}{b^{2}}+\frac {i {\mathrm e}^{i x}}{2 b}-\frac {i {\mathrm e}^{-i x}}{2 b}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{b^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{b^{2}}\) | \(118\) |
Input:
int(sin(x)^2/(a+b*cos(x)),x,method=_RETURNVERBOSE)
Output:
-2*(a+b)*(a-b)/b^2/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*x)/((a-b)*(a+b ))^(1/2))+2/b^2*(-b*tan(1/2*x)/(1+tan(1/2*x)^2)+a*arctan(tan(1/2*x)))
Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.61 \[ \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx=\left [\frac {2 \, a x - 2 \, b \sin \left (x\right ) + \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right )}{2 \, b^{2}}, \frac {a x - b \sin \left (x\right ) - \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right )}{b^{2}}\right ] \] Input:
integrate(sin(x)^2/(a+b*cos(x)),x, algorithm="fricas")
Output:
[1/2*(2*a*x - 2*b*sin(x) + sqrt(-a^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - b ^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)/(b^ 2*cos(x)^2 + 2*a*b*cos(x) + a^2)))/b^2, (a*x - b*sin(x) - sqrt(a^2 - b^2)* arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x))))/b^2]
Leaf count of result is larger than twice the leaf count of optimal. 991 vs. \(2 (49) = 98\).
Time = 50.68 (sec) , antiderivative size = 991, normalized size of antiderivative = 16.80 \[ \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx=\text {Too large to display} \] Input:
integrate(sin(x)**2/(a+b*cos(x)),x)
Output:
Piecewise((zoo*(-log(tan(x/2) - 1)*tan(x/2)**2/(tan(x/2)**2 + 1) - log(tan (x/2) - 1)/(tan(x/2)**2 + 1) + log(tan(x/2) + 1)*tan(x/2)**2/(tan(x/2)**2 + 1) + log(tan(x/2) + 1)/(tan(x/2)**2 + 1) - 2*tan(x/2)/(tan(x/2)**2 + 1)) , Eq(a, 0) & Eq(b, 0)), (x*tan(x/2)**2/(b*tan(x/2)**2 + b) + x/(b*tan(x/2) **2 + b) - 2*tan(x/2)/(b*tan(x/2)**2 + b), Eq(a, b)), (-x*tan(x/2)**2/(b*t an(x/2)**2 + b) - x/(b*tan(x/2)**2 + b) - 2*tan(x/2)/(b*tan(x/2)**2 + b), Eq(a, -b)), ((x*sin(x)**2/2 + x*cos(x)**2/2 - sin(x)*cos(x)/2)/a, Eq(b, 0) ), (a*x*sqrt(-a/(a - b) - b/(a - b))*tan(x/2)**2/(b**2*sqrt(-a/(a - b) - b /(a - b))*tan(x/2)**2 + b**2*sqrt(-a/(a - b) - b/(a - b))) + a*x*sqrt(-a/( a - b) - b/(a - b))/(b**2*sqrt(-a/(a - b) - b/(a - b))*tan(x/2)**2 + b**2* sqrt(-a/(a - b) - b/(a - b))) - a*log(-sqrt(-a/(a - b) - b/(a - b)) + tan( x/2))*tan(x/2)**2/(b**2*sqrt(-a/(a - b) - b/(a - b))*tan(x/2)**2 + b**2*sq rt(-a/(a - b) - b/(a - b))) - a*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(x/ 2))/(b**2*sqrt(-a/(a - b) - b/(a - b))*tan(x/2)**2 + b**2*sqrt(-a/(a - b) - b/(a - b))) + a*log(sqrt(-a/(a - b) - b/(a - b)) + tan(x/2))*tan(x/2)**2 /(b**2*sqrt(-a/(a - b) - b/(a - b))*tan(x/2)**2 + b**2*sqrt(-a/(a - b) - b /(a - b))) + a*log(sqrt(-a/(a - b) - b/(a - b)) + tan(x/2))/(b**2*sqrt(-a/ (a - b) - b/(a - b))*tan(x/2)**2 + b**2*sqrt(-a/(a - b) - b/(a - b))) - 2* b*sqrt(-a/(a - b) - b/(a - b))*tan(x/2)/(b**2*sqrt(-a/(a - b) - b/(a - b)) *tan(x/2)**2 + b**2*sqrt(-a/(a - b) - b/(a - b))) - b*log(-sqrt(-a/(a -...
Exception generated. \[ \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(x)^2/(a+b*cos(x)),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*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.53 \[ \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx=\frac {a x}{b^{2}} + \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} \sqrt {a^{2} - b^{2}}}{b^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, x\right )}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )} b} \] Input:
integrate(sin(x)^2/(a+b*cos(x)),x, algorithm="giac")
Output:
a*x/b^2 + 2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2 *x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*sqrt(a^2 - b^2)/b^2 - 2*tan(1/2*x)/( (tan(1/2*x)^2 + 1)*b)
Time = 44.73 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.25 \[ \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}}{a\,\cos \left (\frac {x}{2}\right )+b\,\cos \left (\frac {x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{b^2}-\frac {\sin \left (x\right )}{b}+\frac {2\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{b^2} \] Input:
int(sin(x)^2/(a + b*cos(x)),x)
Output:
(2*atanh((sin(x/2)*(b^2 - a^2)^(1/2))/(a*cos(x/2) + b*cos(x/2)))*(b^2 - a^ 2)^(1/2))/b^2 - sin(x)/b + (2*a*atan(sin(x/2)/cos(x/2)))/b^2
Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right )-\sin \left (x \right ) b +a x}{b^{2}} \] Input:
int(sin(x)^2/(a+b*cos(x)),x)
Output:
( - 2*sqrt(a**2 - b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt(a**2 - b**2)) - sin(x)*b + a*x)/b**2