3.19 \(\int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx\)

Optimal. Leaf size=63 \[ \frac {b \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2 \sqrt {a^2+b^2}}-\frac {\tanh ^{-1}(\cos (x))}{a^2}+\frac {1}{a (a \cos (x)+b \sin (x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3093, 3770, 3074, 206} \[ \frac {b \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2 \sqrt {a^2+b^2}}-\frac {\tanh ^{-1}(\cos (x))}{a^2}+\frac {1}{a (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

-(ArcTanh[Cos[x]]/a^2) + (b*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(a^2*Sqrt[a^2 + b^2]) + 1/(a*(a*Co
s[x] + b*Sin[x]))

Rule 206

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

Rule 3074

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 3093

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.33, size = 72, normalized size = 1.14 \[ \frac {-\frac {2 b \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {a \csc (x)}{a \cot (x)+b}+\log \left (\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.56, size = 220, normalized size = 3.49 \[ \frac {2 \, a^{3} + 2 \, a b^{2} + {\left (a b \cos \relax (x) + b^{2} \sin \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) - {\left ({\left (a^{3} + a b^{2}\right )} \cos \relax (x) + {\left (a^{2} b + b^{3}\right )} \sin \relax (x)\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left ({\left (a^{3} + a b^{2}\right )} \cos \relax (x) + {\left (a^{2} b + b^{3}\right )} \sin \relax (x)\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{5} + a^{3} b^{2}\right )} \cos \relax (x) + {\left (a^{4} b + a^{2} b^{3}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a*cos(x)+b*sin(x))^2,x, algorithm="fricas")

[Out]

1/2*(2*a^3 + 2*a*b^2 + (a*b*cos(x) + b^2*sin(x))*sqrt(a^2 + b^2)*log((2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)
^2 - 2*a^2 - b^2 - 2*sqrt(a^2 + b^2)*(b*cos(x) - a*sin(x)))/(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 + b^2)
) - ((a^3 + a*b^2)*cos(x) + (a^2*b + b^3)*sin(x))*log(1/2*cos(x) + 1/2) + ((a^3 + a*b^2)*cos(x) + (a^2*b + b^3
)*sin(x))*log(-1/2*cos(x) + 1/2))/((a^5 + a^3*b^2)*cos(x) + (a^4*b + a^2*b^3)*sin(x))

________________________________________________________________________________________

giac [A]  time = 4.46, size = 109, normalized size = 1.73 \[ \frac {b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, x\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a*cos(x)+b*sin(x))^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.81, size = 106, normalized size = 1.68 \[ \frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}-\frac {2 \tan \left (\frac {x}{2}\right ) b}{a^{2} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )}-\frac {2}{a \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )}-\frac {2 b \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)/(a*cos(x)+b*sin(x))^2,x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 128, normalized size = 2.03 \[ \frac {2 \, {\left (a + \frac {b \sin \relax (x)}{\cos \relax (x) + 1}\right )}}{a^{3} + \frac {2 \, a^{2} b \sin \relax (x)}{\cos \relax (x) + 1} - \frac {a^{3} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}} + \frac {b \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {\log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a*cos(x)+b*sin(x))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.81, size = 492, normalized size = 7.81 \[ \frac {\frac {2}{a}+\frac {2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2}}{-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^2}+\frac {b\,\mathrm {atan}\left (\frac {\frac {b\,\sqrt {a^2+b^2}\,\left (4\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+4\,b^2\right )}{a}+\frac {b\,\left (2\,a^2\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^4+4\,a^2\,b^2\right )}{a}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}\right )\,1{}\mathrm {i}}{a^4+a^2\,b^2}+\frac {b\,\sqrt {a^2+b^2}\,\left (4\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+4\,b^2\right )}{a}-\frac {b\,\left (2\,a^2\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^4+4\,a^2\,b^2\right )}{a}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}\right )\,1{}\mathrm {i}}{a^4+a^2\,b^2}}{\frac {4\,b}{a^2}+\frac {b\,\sqrt {a^2+b^2}\,\left (4\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+4\,b^2\right )}{a}+\frac {b\,\left (2\,a^2\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^4+4\,a^2\,b^2\right )}{a}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}\right )}{a^4+a^2\,b^2}-\frac {b\,\sqrt {a^2+b^2}\,\left (4\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+4\,b^2\right )}{a}-\frac {b\,\left (2\,a^2\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^4+4\,a^2\,b^2\right )}{a}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}\right )}{a^4+a^2\,b^2}}\right )\,\sqrt {a^2+b^2}\,2{}\mathrm {i}}{a^4+a^2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)*(a*cos(x) + b*sin(x))^2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc {\relax (x )}}{\left (a \cos {\relax (x )} + b \sin {\relax (x )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a*cos(x)+b*sin(x))**2,x)

[Out]

Integral(csc(x)/(a*cos(x) + b*sin(x))**2, x)

________________________________________________________________________________________