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

Optimal. Leaf size=61 \[ \frac{2 (a A-b B) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2}}-\frac{B \tanh ^{-1}(\cos (x))}{a} \]

[Out]

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

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Rubi [A]  time = 0.130843, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {2828, 3001, 3770, 2660, 618, 204} \[ \frac{2 (a A-b B) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2}}-\frac{B \tanh ^{-1}(\cos (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Csc[x])/(a + b*Sin[x]),x]

[Out]

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

Rule 2828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[((a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n)/Sin[e + f*x]^n, x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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])

Rubi steps

\begin{align*} \int \frac{A+B \csc (x)}{a+b \sin (x)} \, dx &=\int \frac{\csc (x) (B+A \sin (x))}{a+b \sin (x)} \, dx\\ &=\frac{B \int \csc (x) \, dx}{a}+\frac{(a A-b B) \int \frac{1}{a+b \sin (x)} \, dx}{a}\\ &=-\frac{B \tanh ^{-1}(\cos (x))}{a}+\frac{(2 (a A-b B)) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{a}\\ &=-\frac{B \tanh ^{-1}(\cos (x))}{a}-\frac{(4 (a A-b B)) \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 )}{a}\\ &=\frac{2 (a A-b B) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2}}-\frac{B \tanh ^{-1}(\cos (x))}{a}\\ \end{align*}

Mathematica [A]  time = 0.122134, size = 72, normalized size = 1.18 \[ \frac{\frac{2 (a A-b B) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+B \left (\log \left (\sin \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )\right )\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.089, size = 94, normalized size = 1.5 \begin{align*} 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 ) }-2\,{\frac{Bb}{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*csc(x))/(a+b*sin(x)),x)

[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))-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*b+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*csc(x))/(a+b*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.2715, size = 630, normalized size = 10.33 \begin{align*} \left [\frac{{\left (A a - B b\right )} \sqrt{-a^{2} + b^{2}} \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 (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (B a^{2} - B b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{3} - a b^{2}\right )}}, -\frac{2 \,{\left (A a - B b\right )} \sqrt{a^{2} - b^{2}} \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 (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) -{\left (B a^{2} - B b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\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*csc(x))/(a+b*sin(x)),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.10979, size = 96, normalized size = 1.57 \begin{align*} \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{a} + \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 )}{\left (A a - B b\right )}}{\sqrt{a^{2} - b^{2}} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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