\(\int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx\) [7]

Optimal result

Integrand size = 15, antiderivative size = 63 \[ \int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {B \log (\sin (x))}{a}-\frac {B \log (a+b \sin (x))}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {4486, 2739, 632, 210, 2800, 36, 29, 31} \[ \int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {B \log (a+b \sin (x))}{a}+\frac {B \log (\sin (x))}{a} \]

[In]

Int[(A + B*Cot[x])/(a + b*Sin[x]),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*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 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {A}{a+b \sin (x)}+\frac {B \cot (x)}{a+b \sin (x)}\right ) \, dx \\ & = A \int \frac {1}{a+b \sin (x)} \, dx+B \int \frac {\cot (x)}{a+b \sin (x)} \, dx \\ & = (2 A) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+B \text {Subst}\left (\int \frac {1}{x (a+x)} \, dx,x,b \sin (x)\right ) \\ & = -\left ((4 A) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right )\right )+\frac {B \text {Subst}\left (\int \frac {1}{x} \, dx,x,b \sin (x)\right )}{a}-\frac {B \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (x)\right )}{a} \\ & = \frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {B \log (\sin (x))}{a}-\frac {B \log (a+b \sin (x))}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {B (\log (\sin (x))-\log (a+b \sin (x)))}{a} \]

[In]

Integrate[(A + B*Cot[x])/(a + b*Sin[x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98

method	result	size
parts	\(\frac {2 A \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}-\frac {B \ln \left (a +b \sin \left (x \right )\right )}{a}+\frac {B \ln \left (\sin \left (x \right )\right )}{a}\)	\(62\)
default	\(\frac {-B \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )+\frac {2 a A \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a}+\frac {B \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a}\)	\(78\)
risch	\(\frac {2 i x B \,a^{3}}{a^{4}-a^{2} b^{2}}-\frac {2 i x B a \,b^{2}}{a^{4}-a^{2} b^{2}}-\frac {2 i x B}{a}-\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i A \,a^{2}+\sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i A \,a^{2}+\sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B \,b^{2}}{\left (a^{2}-b^{2}\right ) a}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i A \,a^{2}+\sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{\left (a^{2}-b^{2}\right ) a}-\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i A \,a^{2}-\sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i A \,a^{2}-\sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B \,b^{2}}{\left (a^{2}-b^{2}\right ) a}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i A \,a^{2}-\sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{\left (a^{2}-b^{2}\right ) a}+\frac {B \ln \left ({\mathrm e}^{2 i x}-1\right )}{a}\)	\(491\)

[In]

int((A+B*cot(x))/(a+b*sin(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 280, normalized size of antiderivative = 4.44 \[ \int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} A a \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right ) - 2 \, {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (x\right )\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}, -\frac {2 \, \sqrt {a^{2} - b^{2}} A a \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right ) - 2 \, {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (x\right )\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}\right ] \]

[In]

integrate((A+B*cot(x))/(a+b*sin(x)),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(-a^2 + b^2)*A*a*log(((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 + 2*(a*cos(x)*sin(x) + b*co
s(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + (B*a^2 - B*b^2)*log(-b^2*cos(x)^2 + 2*a*b
*sin(x) + a^2 + b^2) - 2*(B*a^2 - B*b^2)*log(1/2*sin(x)))/(a^3 - a*b^2), -1/2*(2*sqrt(a^2 - b^2)*A*a*arctan(-(
a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) + (B*a^2 - B*b^2)*log(-b^2*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2) - 2*(B
*a^2 - B*b^2)*log(1/2*sin(x)))/(a^3 - a*b^2)]

Sympy [F]

\[ \int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx=\int \frac {A + B \cot {\left (x \right )}}{a + b \sin {\left (x \right )}}\, dx \]

[In]

integrate((A+B*cot(x))/(a+b*sin(x)),x)

[Out]

Integral((A + B*cot(x))/(a + b*sin(x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+B*cot(x))/(a+b*sin(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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.35 \[ \int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} A}{\sqrt {a^{2} - b^{2}}} - \frac {B \log \left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}{a} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a} \]

[In]

integrate((A+B*cot(x))/(a+b*sin(x)),x, algorithm="giac")

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*A/sqrt(a^2 - b^2) - B*log(a*t
an(1/2*x)^2 + 2*b*tan(1/2*x) + a)/a + B*log(abs(tan(1/2*x)))/a

Mupad [B] (verification not implemented)

Time = 15.42 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.97 \[ \int \frac {A+B \cot (x)}{a+b \sin (x)} \, dx=\frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a}-\ln \left (b+a\,\mathrm {tan}\left (\frac {x}{2}\right )+\sqrt {b^2-a^2}\right )\,\left (\frac {B}{a}+\frac {A\,a\,\sqrt {b^2-a^2}}{a\,b^2-a^3}\right )-\ln \left (b+a\,\mathrm {tan}\left (\frac {x}{2}\right )-\sqrt {b^2-a^2}\right )\,\left (\frac {B}{a}-\frac {A\,a\,\sqrt {b^2-a^2}}{a\,b^2-a^3}\right ) \]

[In]

int((A + B*cot(x))/(a + b*sin(x)),x)

[Out]

(B*log(tan(x/2)))/a - log(b + a*tan(x/2) + (b^2 - a^2)^(1/2))*(B/a + (A*a*(b^2 - a^2)^(1/2))/(a*b^2 - a^3)) -
log(b + a*tan(x/2) - (b^2 - a^2)^(1/2))*(B/a - (A*a*(b^2 - a^2)^(1/2))/(a*b^2 - a^3))