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

Optimal. Leaf size=100 \[ \frac {2 A \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (1-\cos (x))}{2 (a+b)}+\frac {B \log (1+\cos (x))}{2 (a-b)}-\frac {a B \log (a+b \cos (x))}{a^2-b^2} \]

[Out]

1/2*B*ln(1-cos(x))/(a+b)+1/2*B*ln(cos(x)+1)/(a-b)-a*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 = 0.12, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4486, 2738, 211, 2800, 815} \begin {gather*} -\frac {a B \log (a+b \cos (x))}{a^2-b^2}+\frac {2 A \text {ArcTan}\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Cot[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)) - (a*B*Log[a + b*Cos[x]])/(a^2 - b^2)

Rule 211

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

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2738

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

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Mathematica [A]
time = 0.29, size = 134, normalized size = 1.34 \begin {gather*} \frac {(A+B \cot (x)) \left (-2 A \left (a^2-b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )+\sqrt {-a^2+b^2} B \left ((a+b) \log \left (\cos \left (\frac {x}{2}\right )\right )-a \log (a+b \cos (x))+(a-b) \log \left (\sin \left (\frac {x}{2}\right )\right )\right )\right ) \sin (x)}{(a-b) (a+b) \sqrt {-a^2+b^2} (B \cos (x)+A \sin (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A + B*Cot[x])*(-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)*L
og[Cos[x/2]] - a*Log[a + b*Cos[x]] + (a - b)*Log[Sin[x/2]]))*Sin[x])/((a - b)*(a + b)*Sqrt[-a^2 + b^2]*(B*Cos[
x] + A*Sin[x]))

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Maple [A]
time = 0.20, size = 96, normalized size = 0.96

method	result	size
default	\(\frac {-\frac {a B \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+a +b \right )}{a -b}+\frac {\left (2 a A +2 A b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{a +b}+\frac {B \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a +b}\)	\(96\)
risch	\(\frac {2 i x B \,a^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i x B \,b^{2} a}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {i x B}{a -b}-\frac {i x B}{a +b}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {a A -i \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) a B}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {a A -i \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{\left (a +b \right ) \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {a A +i \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) a B}{\left (a +b \right ) \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {a A +i \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{\left (a +b \right ) \left (a -b \right )}+\frac {B \ln \left ({\mathrm e}^{i x}+1\right )}{a -b}+\frac {B \ln \left ({\mathrm e}^{i x}-1\right )}{a +b}\)	\(358\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cot(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 de

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Fricas [A]
time = 3.22, size = 266, normalized size = 2.66 \begin {gather*} \left [-\frac {B a \log \left (b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}\right ) + \sqrt {-a^{2} + b^{2}} A \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) - {\left (B a + B b\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (B a - B b\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}}, -\frac {B a \log \left (b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}\right ) - 2 \, \sqrt {a^{2} - b^{2}} A \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) - {\left (B a + B b\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (B a - B b\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(B*a*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*a*log(b^2*cos(x)^2 + 2*a*b
*cos(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*co
s(x) + 1/2) - (B*a - B*b)*log(-1/2*cos(x) + 1/2))/(a^2 - b^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B \cot {\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.52, size = 116, normalized size = 1.16 \begin {gather*} -\frac {B a \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*a*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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Mupad [B]
time = 3.58, size = 419, normalized size = 4.19 \begin {gather*} \frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a+b}+\frac {\ln \left (3\,B\,a^2\,b^2-B\,b^4-2\,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 )+2\,B\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-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\,a^3+B\,a\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {\ln \left (2\,B\,a^4+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 )+2\,B\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}\right )\,\left (B\,a^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a\,b^2\right )}{a^4-2\,a^2\,b^2+b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cot(x))/(a + b*cos(x)),x)

[Out]

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

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