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

Optimal. Leaf size=99 \[ \frac {b B \log (a+b \cos (x))}{a^2-b^2}+\frac {2 A \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (1-\cos (x))}{2 (a+b)}-\frac {B \log (\cos (x)+1)}{2 (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {4225, 4401, 2659, 205, 2668, 706, 31, 633} \[ \frac {b B \log (a+b \cos (x))}{a^2-b^2}+\frac {2 A \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (1-\cos (x))}{2 (a+b)}-\frac {B \log (\cos (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Csc[x])/(a + b*Cos[x]),x]

[Out]

(2*A*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) + (B*Log[1 - Cos[x]])/(2*(a + b)) -
(B*Log[1 + Cos[x]])/(2*(a - b)) + (b*B*Log[a + b*Cos[x]])/(a^2 - b^2)

Rule 31

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

Rule 205

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

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 2659

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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rule 4401

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

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 116, normalized size = 1.17 \[ \frac {-2 A \left (a^2-b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )-B \sqrt {b^2-a^2} \left ((b-a) \log \left (\sin \left (\frac {x}{2}\right )\right )+(a+b) \log \left (\cos \left (\frac {x}{2}\right )\right )-b \log (a+b \cos (x))\right )}{(a-b) (a+b) \sqrt {b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 8.19, size = 265, normalized size = 2.68 \[ \left [\frac {B b \log \left (b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}\right ) - \sqrt {-a^{2} + b^{2}} A \log \left (\frac {2 \, a b \cos \relax (x) + {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \relax (x) + b\right )} \sin \relax (x) - a^{2} + 2 \, b^{2}}{b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}}\right ) - {\left (B a + B b\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (B a - B b\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}}, \frac {B b \log \left (b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}\right ) + 2 \, \sqrt {a^{2} - b^{2}} A \arctan \left (-\frac {a \cos \relax (x) + b}{\sqrt {a^{2} - b^{2}} \sin \relax (x)}\right ) - {\left (B a + B b\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (B a - B b\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.41, size = 115, normalized size = 1.16 \[ \frac {B b \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} - a - b\right )}{a^{2} - b^{2}} - \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} A}{\sqrt {a^{2} - b^{2}}} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 134, normalized size = 1.35 \[ \frac {B b \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b +a +b \right )}{\left (a +b \right ) \left (a -b \right )}+\frac {2 \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a A}{\left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A b}{\left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {B \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a +b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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mupad [B]  time = 3.54, size = 417, normalized size = 4.21 \[ \frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a+b}+\frac {\ln \left (3\,B\,a^2\,b^2-2\,B\,b^4-B\,a^4+A\,a\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}+A\,b\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}+A\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+A\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )-B\,a\,b^3+B\,a^3\,b-2\,A\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )+B\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-2\,B\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}\right )\,\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {\ln \left (B\,a^4+2\,B\,b^4-3\,B\,a^2\,b^2+A\,a\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}+A\,b\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-A\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )-A\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )+B\,a\,b^3-B\,a^3\,b+2\,A\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )+B\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-2\,B\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}\right )\,\left (B\,b^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/sin(x))/(a + b*cos(x)),x)

[Out]

(B*log(tan(x/2)))/(a + b) + (log(3*B*a^2*b^2 - 2*B*b^4 - B*a^4 + A*a*(-(a + b)^3*(a - b)^3)^(1/2) + A*b*(-(a +
b)^3*(a - b)^3)^(1/2) + A*a^4*tan(x/2) + A*b^4*tan(x/2) - B*a*b^3 + B*a^3*b - 2*A*a^2*b^2*tan(x/2) + B*a*tan(
x/2)*(-(a + b)^3*(a - b)^3)^(1/2) - 2*B*b*tan(x/2)*(-(a + b)^3*(a - b)^3)^(1/2))*(A*(-(a + b)^3*(a - b)^3)^(1/
2) - B*b^3 + B*a^2*b))/(a^4 + b^4 - 2*a^2*b^2) - (log(B*a^4 + 2*B*b^4 - 3*B*a^2*b^2 + A*a*(-(a + b)^3*(a - b)^
3)^(1/2) + A*b*(-(a + b)^3*(a - b)^3)^(1/2) - A*a^4*tan(x/2) - A*b^4*tan(x/2) + B*a*b^3 - B*a^3*b + 2*A*a^2*b^
2*tan(x/2) + B*a*tan(x/2)*(-(a + b)^3*(a - b)^3)^(1/2) - 2*B*b*tan(x/2)*(-(a + b)^3*(a - b)^3)^(1/2))*(B*b^3 +
A*(-(a + b)^3*(a - b)^3)^(1/2) - B*a^2*b))/(a^4 + b^4 - 2*a^2*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*csc(x))/(a+b*cos(x)),x)

[Out]

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

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