\(\int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx\) [19]

Optimal result

Integrand size = 14, antiderivative size = 63 \[ \int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {\text {arctanh}(\cos (x))}{a^2}+\frac {b \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2 \sqrt {a^2+b^2}}+\frac {1}{a (a \cos (x)+b \sin (x))} \]

[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] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3172, 3855, 3153, 212} \[ \int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {b \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2 \sqrt {a^2+b^2}}-\frac {\text {arctanh}(\cos (x))}{a^2}+\frac {1}{a (a \cos (x)+b \sin (x))} \]

[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 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 3172

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \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 {\text {arctanh}(\cos (x))}{a^2}+\frac {1}{a (a \cos (x)+b \sin (x))}+\frac {b \text {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{a^2} \\ & = -\frac {\text {arctanh}(\cos (x))}{a^2}+\frac {b \text {arctanh}\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] (verified)

Time = 0.61 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.14 \[ \int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {-\frac {2 b \text {arctanh}\left (\frac {-b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {a \csc (x)}{b+a \cot (x)}-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )}{a^2} \]

[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

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {\frac {4 \left (-\frac {b \tan \left (\frac {x}{2}\right )}{2}-\frac {a}{2}\right )}{\tan \left (\frac {x}{2}\right )^{2} a -2 b \tan \left (\frac {x}{2}\right )-a}-\frac {2 b \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{a^{2}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\)	\(85\)
risch	\(\frac {2 \,{\mathrm e}^{i x}}{a \left (-i b \,{\mathrm e}^{2 i x}+a \,{\mathrm e}^{2 i x}+i b +a \right )}-\frac {i b \ln \left ({\mathrm e}^{i x}-\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, a^{2}}+\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, a^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{a^{2}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{a^{2}}\)	\(156\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (59) = 118\).

Time = 0.29 (sec) , antiderivative size = 220, normalized size of antiderivative = 3.49 \[ \int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 \, a^{3} + 2 \, a b^{2} + {\left (a b \cos \left (x\right ) + b^{2} \sin \left (x\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right ) - {\left ({\left (a^{3} + a b^{2}\right )} \cos \left (x\right ) + {\left (a^{2} b + b^{3}\right )} \sin \left (x\right )\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{3} + a b^{2}\right )} \cos \left (x\right ) + {\left (a^{2} b + b^{3}\right )} \sin \left (x\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{5} + a^{3} b^{2}\right )} \cos \left (x\right ) + {\left (a^{4} b + a^{2} b^{3}\right )} \sin \left (x\right )\right )}} \]

[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))

Sympy [F]

\[ \int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\int \frac {\csc {\left (x \right )}}{\left (a \cos {\left (x \right )} + b \sin {\left (x \right )}\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (59) = 118\).

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.03 \[ \int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 \, {\left (a + \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}}{a^{3} + \frac {2 \, a^{2} b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {a^{3} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}} + \frac {b \log \left (\frac {b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {\log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.73 \[ \int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, x\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )} a^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 21.95 (sec) , antiderivative size = 492, normalized size of antiderivative = 7.81 \[ \int \frac {\csc (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {\frac {2}{a}+\frac {2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2}}{-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^2}+\frac {b\,\mathrm {atan}\left (\frac {\frac {b\,\sqrt {a^2+b^2}\,\left (4\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+4\,b^2\right )}{a}+\frac {b\,\left (2\,a^2\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^4+4\,a^2\,b^2\right )}{a}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}\right )\,1{}\mathrm {i}}{a^4+a^2\,b^2}+\frac {b\,\sqrt {a^2+b^2}\,\left (4\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+4\,b^2\right )}{a}-\frac {b\,\left (2\,a^2\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^4+4\,a^2\,b^2\right )}{a}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}\right )\,1{}\mathrm {i}}{a^4+a^2\,b^2}}{\frac {4\,b}{a^2}+\frac {b\,\sqrt {a^2+b^2}\,\left (4\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+4\,b^2\right )}{a}+\frac {b\,\left (2\,a^2\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^4+4\,a^2\,b^2\right )}{a}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}\right )}{a^4+a^2\,b^2}-\frac {b\,\sqrt {a^2+b^2}\,\left (4\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+4\,b^2\right )}{a}-\frac {b\,\left (2\,a^2\,b+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^4+4\,a^2\,b^2\right )}{a}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}\right )}{a^4+a^2\,b^2}}\right )\,\sqrt {a^2+b^2}\,2{}\mathrm {i}}{a^4+a^2\,b^2} \]

[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)