3.1.26 \(\int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx\) [26]

Optimal. Leaf size=59 \[ \frac {a x}{b^2}-\frac {2 \sqrt {a-b} \sqrt {a+b} \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^2}-\frac {\sin (x)}{b} \]

[Out]

a*x/b^2-sin(x)/b-2*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))*(a-b)^(1/2)*(a+b)^(1/2)/b^2

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2774, 2814, 2738, 211} \begin {gather*} -\frac {2 \sqrt {a-b} \sqrt {a+b} \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^2}+\frac {a x}{b^2}-\frac {\sin (x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Cos[x]),x]

[Out]

(a*x)/b^2 - (2*Sqrt[a - b]*Sqrt[a + b]*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/b^2 - Sin[x]/b

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[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]

Rule 2774

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[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] /; FreeQ[{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]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(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]

Rubi steps

\begin {align*} \int \frac {\sin ^2(x)}{a+b \cos (x)} \, dx &=-\frac {\sin (x)}{b}-\frac {\int \frac {-b-a \cos (x)}{a+b \cos (x)} \, dx}{b}\\ &=\frac {a x}{b^2}-\frac {\sin (x)}{b}+\left (1-\frac {a^2}{b^2}\right ) \int \frac {1}{a+b \cos (x)} \, dx\\ &=\frac {a x}{b^2}-\frac {\sin (x)}{b}+\left (2 \left (1-\frac {a^2}{b^2}\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=\frac {a x}{b^2}-\frac {2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^2}-\frac {\sin (x)}{b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 54, normalized size = 0.92 \begin {gather*} \frac {a x-2 \sqrt {-a^2+b^2} \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )-b \sin (x)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + b*Cos[x]),x]

[Out]

(a*x - 2*Sqrt[-a^2 + b^2]*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]] - b*Sin[x])/b^2

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 78, normalized size = 1.32

method	result	size
default	\(\frac {-\frac {2 b \tan \left (\frac {x}{2}\right )}{\tan ^{2}\left (\frac {x}{2}\right )+1}+2 a \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{2}}-\frac {2 \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}\)	\(78\)
risch	\(\frac {a x}{b^{2}}+\frac {i {\mathrm e}^{i x}}{2 b}-\frac {i {\mathrm e}^{-i x}}{2 b}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{b^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{b^{2}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^2*(-b*tan(1/2*x)/(tan(1/2*x)^2+1)+a*arctan(tan(1/2*x)))-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))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

________________________________________________________________________________________

Fricas [A]
time = 0.41, size = 154, normalized size = 2.61 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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]

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 991 vs. \(2 (49) = 98\).
time = 53.72, size = 991, normalized size = 16.80 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} - 1 \right )} \tan ^{2}{\left (\frac {x}{2} \right )}}{\tan ^{2}{\left (\frac {x}{2} \right )} + 1} - \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} - 1 \right )}}{\tan ^{2}{\left (\frac {x}{2} \right )} + 1} + \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac {x}{2} \right )}}{\tan ^{2}{\left (\frac {x}{2} \right )} + 1} + \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )}}{\tan ^{2}{\left (\frac {x}{2} \right )} + 1} - \frac {2 \tan {\left (\frac {x}{2} \right )}}{\tan ^{2}{\left (\frac {x}{2} \right )} + 1}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {x \tan ^{2}{\left (\frac {x}{2} \right )}}{b \tan ^{2}{\left (\frac {x}{2} \right )} + b} + \frac {x}{b \tan ^{2}{\left (\frac {x}{2} \right )} + b} - \frac {2 \tan {\left (\frac {x}{2} \right )}}{b \tan ^{2}{\left (\frac {x}{2} \right )} + b} & \text {for}\: a = b \\- \frac {x \tan ^{2}{\left (\frac {x}{2} \right )}}{b \tan ^{2}{\left (\frac {x}{2} \right )} + b} - \frac {x}{b \tan ^{2}{\left (\frac {x}{2} \right )} + b} - \frac {2 \tan {\left (\frac {x}{2} \right )}}{b \tan ^{2}{\left (\frac {x}{2} \right )} + b} & \text {for}\: a = - b \\\frac {\frac {x \sin ^{2}{\left (x \right )}}{2} + \frac {x \cos ^{2}{\left (x \right )}}{2} - \frac {\sin {\left (x \right )} \cos {\left (x \right )}}{2}}{a} & \text {for}\: b = 0 \\\frac {a x \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {a x \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {a \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )} \tan ^{2}{\left (\frac {x}{2} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {a \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {a \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )} \tan ^{2}{\left (\frac {x}{2} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {a \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {2 b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan {\left (\frac {x}{2} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {b \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )} \tan ^{2}{\left (\frac {x}{2} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {b \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {b \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )} \tan ^{2}{\left (\frac {x}{2} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {b \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )}}{b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \tan ^{2}{\left (\frac {x}{2} \right )} + b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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(t
an(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*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*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*sqr
t(-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*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))) + a*log(sqrt(-a/(a - b) - b/(a - b)) + tan(x/2))*t
an(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 - 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))) - b*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))) + b*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))) + b*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))), True))

________________________________________________________________________________________

Giac [A]
time = 0.45, size = 90, normalized size = 1.53 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

________________________________________________________________________________________

Mupad [B]
time = 0.49, size = 74, normalized size = 1.25 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a + b*cos(x)),x)

[Out]

(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

________________________________________________________________________________________