3.216 \(\int \frac{\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx\)

Optimal. Leaf size=51 \[ -\frac{\tan ^{-1}\left (\frac{\sin (c+d x) \cos (c+d x)}{\sin ^2(c+d x)+\sqrt{2}+1}\right )}{\sqrt{2} d}-\frac{x}{\sqrt{2}}+x \]

[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)

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Rubi [A]  time = 0.172122, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1130, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sin (c+d x) \cos (c+d x)}{\sin ^2(c+d x)+\sqrt{2}+1}\right )}{\sqrt{2} d}-\frac{x}{\sqrt{2}}+x \]

Antiderivative was successfully verified.

[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 1130

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*(b/q + 1))/2, Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2*(b/q - 1))/2, 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]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\sin (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{1+3 x^2+2 x^4} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{1+2 x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{2+2 x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=x-\frac{x}{\sqrt{2}}-\frac{\tan ^{-1}\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]  time = 0.0722669, size = 30, normalized size = 0.59 \[ -\frac{\tan ^{-1}\left (\sqrt{2} \tan (c+d x)\right )}{\sqrt{2} d}+\frac{c}{d}+x \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.087, size = 30, normalized size = 0.6 \begin{align*} -{\frac{\sqrt{2}\arctan \left ( \sqrt{2}\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{dx+c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.69157, size = 340, normalized size = 6.67 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.49443, size = 140, normalized size = 2.75 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + \csc{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.19908, size = 111, normalized size = 2.18 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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