\(\int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx\) [216]

Optimal result

Integrand size = 22, antiderivative size = 51 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=x-\frac {x}{\sqrt {2}}-\frac {\arctan \left (\frac {\cos (c+d x) \sin (c+d x)}{1+\sqrt {2}+\sin ^2(c+d x)}\right )}{\sqrt {2} d} \]

[Out]

x-1/2*x*2^(1/2)-1/2*arctan(cos(d*x+c)*sin(d*x+c)/(1+sin(d*x+c)^2+2^(1/2)))/d*2^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1144, 209} \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=-\frac {\arctan \left (\frac {\sin (c+d x) \cos (c+d x)}{\sin ^2(c+d x)+\sqrt {2}+1}\right )}{\sqrt {2} d}-\frac {x}{\sqrt {2}}+x \]

[In]

Int[Sin[c + d*x]/(Csc[c + d*x] + Sin[c + d*x]),x]

[Out]

x - x/Sqrt[2] - ArcTan[(Cos[c + d*x]*Sin[c + d*x])/(1 + Sqrt[2] + Sin[c + d*x]^2)]/(Sqrt[2]*d)

Rule 209

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

Rule 1144

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2/2)*(b/q + 1), Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2}{1+3 x^2+2 x^4} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac {2 \text {Subst}\left (\int \frac {1}{2+2 x^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.59 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {c}{d}+x-\frac {\arctan \left (\sqrt {2} \tan (c+d x)\right )}{\sqrt {2} d} \]

[In]

Integrate[Sin[c + d*x]/(Csc[c + d*x] + Sin[c + d*x]),x]

[Out]

c/d + x - ArcTan[Sqrt[2]*Tan[c + d*x]]/(Sqrt[2]*d)

Maple [A] (verified)

Time = 0.40 (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	\(x -\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-2 \sqrt {2}-3\right )}{4 d}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+2 \sqrt {2}-3\right )}{4 d}\)	\(55\)

[In]

int(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(arctan(tan(d*x+c))-1/2*2^(1/2)*arctan(tan(d*x+c)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {4 \, d x + \sqrt {2} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2}}{4 \, \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, d} \]

[In]

integrate(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(4*d*x + sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(d*x + c)^2 - 2*sqrt(2))/(cos(d*x + c)*sin(d*x + c))))/d

Sympy [F]

\[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + \csc {\left (c + d x \right )}}\, dx \]

[In]

integrate(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x)

[Out]

Integral(sin(c + d*x)/(sin(c + d*x) + csc(c + d*x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (45) = 90\).

Time = 0.34 (sec) , antiderivative size = 252, normalized size of antiderivative = 4.94 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {4 \, d x - \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sin \left (d x + c\right )}{2 \, {\left (\sqrt {2} + 1\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sqrt {2} + 3}, \frac {\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{2 \, {\left (\sqrt {2} + 1\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sqrt {2} + 3}\right ) + \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sin \left (d x + c\right )}{2 \, {\left (\sqrt {2} - 1\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sqrt {2} + 3}, \frac {\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 1}{2 \, {\left (\sqrt {2} - 1\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sqrt {2} + 3}\right ) + 4 \, c}{4 \, d} \]

[In]

integrate(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x, algorithm="maxima")

[Out]

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) +
cos(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)*c
os(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*sqrt(2) + 3)) + 4*c)/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.61 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=\frac {2 \, d x - \sqrt {2} {\left (d x + c + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) - 2 \, \sin \left (2 \, d x + 2 \, c\right )}{\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 2}\right )\right )} + 2 \, c}{2 \, d} \]

[In]

integrate(sin(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 23.80 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.22 \[ \int \frac {\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx=x-\frac {\sqrt {2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {7\,\sqrt {2}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )\right )}{4\,d} \]

[In]

int(sin(c + d*x)/(sin(c + d*x) + 1/sin(c + d*x)),x)

[Out]

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)*ta
n(c/2 + (d*x)/2))/4)))/(4*d)