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

Optimal. Leaf size=63 \[ \frac{2 A \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{B \log (a+b \sin (x))}{a}+\frac{B \log (\sin (x))}{a} \]

[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

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Rubi [A]  time = 0.147516, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {4401, 2660, 618, 204, 2721, 36, 29, 31} \[ \frac{2 A \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{B \log (a+b \sin (x))}{a}+\frac{B \log (\sin (x))}{a} \]

Antiderivative was successfully verified.

[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 4401

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

Rule 2660

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 618

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 204

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

Rule 2721

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 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 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]

Rubi steps

\begin{align*} \int \frac{A+B \cot (x)}{a+b \sin (x)} \, dx &=\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) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+B \operatorname{Subst}\left (\int \frac{1}{x (a+x)} \, dx,x,b \sin (x)\right )\\ &=-\left ((4 A) \operatorname{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 \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,b \sin (x)\right )}{a}-\frac{B \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \sin (x)\right )}{a}\\ &=\frac{2 A \tan ^{-1}\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]  time = 0.166278, size = 60, normalized size = 0.95 \[ \frac{2 A \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{B (\log (\sin (x))-\log (a+b \sin (x)))}{a} \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.091, size = 75, normalized size = 1.2 \begin{align*} -{\frac{B}{a}\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) }+2\,{\frac{A}{\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+{\frac{B}{a}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.44425, size = 660, normalized size = 10.48 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \cot{\left (x \right )}}{a + b \sin{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11491, size = 115, normalized size = 1.83 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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