Integrand size = 22, antiderivative size = 51 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=x-\frac {x}{\sqrt {2}}-\frac {\arctan \left (\frac {\cos (c+d x) \sin (c+d x)}{1+\sqrt {2}+\sin ^2(c+d x)}\right )}{\sqrt {2} d} \] Output:
x-1/2*x*2^(1/2)-1/2*arctan(cos(d*x+c)*sin(d*x+c)/(1+2^(1/2)+sin(d*x+c)^2)) *2^(1/2)/d
Time = 0.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.59 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {c}{d}+x-\frac {\arctan \left (\sqrt {2} \tan (c+d x)\right )}{\sqrt {2} d} \] Input:
Integrate[Sin[c + d*x]/(Csc[c + d*x] + Sin[c + d*x]),x]
Output:
c/d + x - ArcTan[Sqrt[2]*Tan[c + d*x]]/(Sqrt[2]*d)
Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.63, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4889, 1450, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sin (c+d x)}{\sin (c+d x)+\csc (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{\sin (c+d x)+\csc (c+d x)}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \frac {\int \frac {\tan ^2(c+d x)}{2 \tan ^4(c+d x)+3 \tan ^2(c+d x)+1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 1450 |
\(\displaystyle \frac {2 \int \frac {1}{2 \tan ^2(c+d x)+2}d\tan (c+d x)-\int \frac {1}{2 \tan ^2(c+d x)+1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\arctan (\tan (c+d x))-\frac {\arctan \left (\sqrt {2} \tan (c+d x)\right )}{\sqrt {2}}}{d}\) |
Input:
Int[Sin[c + d*x]/(Csc[c + d*x] + Sin[c + d*x]),x]
Output:
(ArcTan[Tan[c + d*x]] - ArcTan[Sqrt[2]*Tan[c + d*x]]/Sqrt[2])/d
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Wi th[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(d^2/2)*(b/q + 1) Int[(d*x)^(m - 2)/(b/ 2 + q/2 + c*x^2), x], x] - Simp[(d^2/2)*(b/q - 1) Int[(d*x)^(m - 2)/(b/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.57
method | result | size |
derivativedivides | \(\frac {\arctan \left (\tan \left (d x +c \right )\right )-\frac {\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {2}\right )}{2}}{d}\) | \(29\) |
default | \(\frac {\arctan \left (\tan \left (d x +c \right )\right )-\frac {\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {2}\right )}{2}}{d}\) | \(29\) |
risch | \(x -\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-2 \sqrt {2}-3\right )}{4 d}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+2 \sqrt {2}-3\right )}{4 d}\) | \(55\) |
Input:
int(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(arctan(tan(d*x+c))-1/2*2^(1/2)*arctan(tan(d*x+c)*2^(1/2)))
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {4 \, d x + \sqrt {2} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2}}{4 \, \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, d} \] Input:
integrate(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x, algorithm="fricas")
Output:
1/4*(4*d*x + sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(d*x + c)^2 - 2*sqrt(2))/(co s(d*x + c)*sin(d*x + c))))/d
\[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + \csc {\left (c + d x \right )}}\, dx \] Input:
integrate(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x)
Output:
Integral(sin(c + d*x)/(sin(c + d*x) + csc(c + d*x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (45) = 90\).
Time = 0.14 (sec) , antiderivative size = 252, normalized size of antiderivative = 4.94 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {4 \, d x - \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sin \left (d x + c\right )}{2 \, {\left (\sqrt {2} + 1\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sqrt {2} + 3}, \frac {\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{2 \, {\left (\sqrt {2} + 1\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sqrt {2} + 3}\right ) + \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sin \left (d x + c\right )}{2 \, {\left (\sqrt {2} - 1\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sqrt {2} + 3}, \frac {\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 1}{2 \, {\left (\sqrt {2} - 1\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sqrt {2} + 3}\right ) + 4 \, c}{4 \, d} \] Input:
integrate(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x, algorithm="maxima")
Output:
1/4*(4*d*x - sqrt(2)*arctan2(2*sqrt(2)*sin(d*x + c)/(2*(sqrt(2) + 1)*cos(d *x + c) + cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sqrt(2) + 3), (cos(d*x + c)^ 2 + sin(d*x + c)^2 + 2*cos(d*x + c) - 1)/(2*(sqrt(2) + 1)*cos(d*x + c) + c os(d*x + c)^2 + sin(d*x + c)^2 + 2*sqrt(2) + 3)) + sqrt(2)*arctan2(2*sqrt( 2)*sin(d*x + c)/(2*(sqrt(2) - 1)*cos(d*x + c) + cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sqrt(2) + 3), (cos(d*x + c)^2 + sin(d*x + c)^2 - 2*cos(d*x + c) - 1)/(2*(sqrt(2) - 1)*cos(d*x + c) + cos(d*x + c)^2 + sin(d*x + c)^2 - 2*s qrt(2) + 3)) + 4*c)/d
Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.61 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {2 \, d x - \sqrt {2} {\left (d x + c + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) - 2 \, \sin \left (2 \, d x + 2 \, c\right )}{\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 2}\right )\right )} + 2 \, c}{2 \, d} \] Input:
integrate(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x, algorithm="giac")
Output:
1/2*(2*d*x - sqrt(2)*(d*x + c + arctan(-(sqrt(2)*sin(2*d*x + 2*c) - 2*sin( 2*d*x + 2*c))/(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2) - 2*cos(2*d*x + 2*c) + 2 ))) + 2*c)/d
Time = 17.45 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.22 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=x-\frac {\sqrt {2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {7\,\sqrt {2}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )\right )}{4\,d} \] Input:
int(sin(c + d*x)/(sin(c + d*x) + 1/sin(c + d*x)),x)
Output:
x - (2^(1/2)*(2*atan((7*2^(1/2)*tan(c/2 + (d*x)/2))/4 + (2^(1/2)*tan(c/2 + (d*x)/2)^3)/4) + 2*atan((2^(1/2)*tan(c/2 + (d*x)/2))/4)))/(4*d)
Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.47 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {-2 \sqrt {2}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {2}+1}\right )+\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i \right ) i -\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i \right ) i +4 d x}{4 d} \] Input:
int(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x)
Output:
( - 2*sqrt(2)*atan(tan((c + d*x)/2)/(sqrt(2) + 1)) + sqrt(2)*log( - sqrt(2 )*i + tan((c + d*x)/2) + i)*i - sqrt(2)*log(sqrt(2)*i + tan((c + d*x)/2) - i)*i + 4*d*x)/(4*d)