3.1.15 \(\int \frac {\sec (x)}{a+b \cot (x)} \, dx\) [15]

Optimal. Leaf size=47 \[ \frac {\tanh ^{-1}(\sin (x))}{a}+\frac {b \tanh ^{-1}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2}} \]

[Out]

arctanh(sin(x))/a+b*arctanh((a*cos(x)-b*sin(x))/(a^2+b^2)^(1/2))/a/(a^2+b^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {3599, 3189, 3855, 3153, 212} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2}}+\frac {\tanh ^{-1}(\sin (x))}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]/(a + b*Cot[x]),x]

[Out]

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

Rule 212

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

Rule 3153

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Dist[-d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3189

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*(sin[c + d*x]^n/(a*cos[c + d*x] + b*sin[c + d*
x])), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3599

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Sin[e + f*x]^
m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/Cos[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3855

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

________________________________________________________________________________________

Mathematica [A]
time = 0.10, size = 76, normalized size = 1.62 \begin {gather*} \frac {-\frac {2 b \tanh ^{-1}\left (\frac {-a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]/(a + b*Cot[x]),x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.37, size = 63, normalized size = 1.34

method	result	size
default	\(-\frac {\ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a}+\frac {2 b \arctanh \left (\frac {-2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}\)	\(63\)
risch	\(\frac {b \ln \left ({\mathrm e}^{i x}-\frac {i b -a}{\sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}\, a}-\frac {b \ln \left ({\mathrm e}^{i x}+\frac {i b -a}{\sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}\, a}+\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{a}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{a}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)/(a+b*cot(x)),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (43) = 86\).
time = 0.50, size = 98, normalized size = 2.09 \begin {gather*} \frac {b \log \left (\frac {a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a} + \frac {\log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(a+b*cot(x)),x, algorithm="maxima")

[Out]

b*log((a - b*sin(x)/(cos(x) + 1) + sqrt(a^2 + b^2))/(a - b*sin(x)/(cos(x) + 1) - sqrt(a^2 + b^2)))/(sqrt(a^2 +
b^2)*a) + log(sin(x)/(cos(x) + 1) + 1)/a - log(sin(x)/(cos(x) + 1) - 1)/a

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (43) = 86\).
time = 3.87, size = 140, normalized size = 2.98 \begin {gather*} \frac {\sqrt {a^{2} + b^{2}} b \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) + {\left (a^{2} + b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a^{2} + b^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{3} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(a+b*cot(x)),x, algorithm="fricas")

[Out]

1/2*(sqrt(a^2 + b^2)*b*log((2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 - a^2 - 2*b^2 - 2*sqrt(a^2 + b^2)*(a*co
s(x) - b*sin(x)))/(2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 + a^2)) + (a^2 + b^2)*log(sin(x) + 1) - (a^2 + b
^2)*log(-sin(x) + 1))/(a^3 + a*b^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(a+b*cot(x)),x)

[Out]

Integral(sec(x)/(a + b*cot(x)), x)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (43) = 86\).
time = 0.44, size = 90, normalized size = 1.91 \begin {gather*} \frac {b \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(a+b*cot(x)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.66, size = 408, normalized size = 8.68 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a}-\frac {2\,b\,\mathrm {atanh}\left (\frac {64\,b^3}{\sqrt {a^2+b^2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}-\frac {64\,b^5}{{\left (a^2+b^2\right )}^{3/2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}+\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}\,\left (\frac {64\,a^2\,b^3}{a^2+b^2}+128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}\right )}-\frac {128\,b^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a^2+b^2\right )}^{3/2}\,\left (\frac {64\,a^2\,b^3}{a^2+b^2}+128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}\right )}+\frac {128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}-\frac {192\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a^2+b^2\right )}^{3/2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}\right )}{a\,\sqrt {a^2+b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)*(a + b*cot(x))),x)

[Out]

(2*atanh(tan(x/2)))/a - (2*b*atanh((64*b^3)/((a^2 + b^2)^(1/2)*(128*b^2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b
^2) + (64*a*b^3)/(a^2 + b^2))) - (64*b^5)/((a^2 + b^2)^(3/2)*(128*b^2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2
) + (64*a*b^3)/(a^2 + b^2))) + (128*b^4*tan(x/2))/((a^2 + b^2)^(1/2)*((64*a^2*b^3)/(a^2 + b^2) + 128*a*b^2*tan
(x/2) - (128*a*b^4*tan(x/2))/(a^2 + b^2))) - (128*b^6*tan(x/2))/((a^2 + b^2)^(3/2)*((64*a^2*b^3)/(a^2 + b^2) +
128*a*b^2*tan(x/2) - (128*a*b^4*tan(x/2))/(a^2 + b^2))) + (128*a*b^2*tan(x/2))/((a^2 + b^2)^(1/2)*(128*b^2*ta
n(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2) + (64*a*b^3)/(a^2 + b^2))) - (192*a*b^4*tan(x/2))/((a^2 + b^2)^(3/2)*(
128*b^2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2) + (64*a*b^3)/(a^2 + b^2)))))/(a*(a^2 + b^2)^(1/2))

________________________________________________________________________________________