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]

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

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Rubi [A]  time = 0.13, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.12, 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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fricas [A]  time = 1.04, size = 265, normalized size = 4.34 \[ \left [\frac {{\left (A a - B b\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} - 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}, -\frac {2 \, {\left (A a - B b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}\right ] \]

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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giac [A]  time = 0.17, size = 71, normalized size = 1.16 \[ \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a} + \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (A a - B b\right )}}{\sqrt {a^{2} - b^{2}} a} \]

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)

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maple [A]  time = 0.24, size = 94, normalized size = 1.54 \[ \frac {B \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a}+\frac {2 A \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}-\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) B b}{a \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 15.62, size = 134, normalized size = 2.20 \[ \frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {x}{2}\right )+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )}{a\,b^2-a^3}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {x}{2}\right )-\sqrt {b^2-a^2}\right )\,\left (A\,a\,\sqrt {b^2-a^2}-B\,b\,\sqrt {b^2-a^2}\right )}{a\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \csc {\relax (x )}}{a + b \sin {\relax (x )}}\, dx \]

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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