3.44 \(\int \frac{1}{a+b \csc (c+d x)} \, dx\)

Optimal. Leaf size=57 \[ \frac{2 b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a d \sqrt{a^2-b^2}}+\frac{x}{a} \]

[Out]

x/a + (2*b*ArcTanh[(a + b*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*Sqrt[a^2 - b^2]*d)

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Rubi [A]  time = 0.064512, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3783, 2660, 618, 206} \[ \frac{2 b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a d \sqrt{a^2-b^2}}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x])^(-1),x]

[Out]

x/a + (2*b*ArcTanh[(a + b*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*Sqrt[a^2 - b^2]*d)

Rule 3783

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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])

Rubi steps

\begin{align*} \int \frac{1}{a+b \csc (c+d x)} \, dx &=\frac{x}{a}-\frac{\int \frac{1}{1+\frac{a \sin (c+d x)}{b}} \, dx}{a}\\ &=\frac{x}{a}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{1+\frac{2 a x}{b}+x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a d}\\ &=\frac{x}{a}+\frac{4 \operatorname{Subst}\left (\int \frac{1}{-4 \left (1-\frac{a^2}{b^2}\right )-x^2} \, dx,x,\frac{2 a}{b}+2 \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a d}\\ &=\frac{x}{a}+\frac{2 b \tanh ^{-1}\left (\frac{b \left (\frac{a}{b}+\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}\\ \end{align*}

Mathematica [A]  time = 0.107481, size = 59, normalized size = 1.04 \[ \frac{-\frac{2 b \tan ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{d \sqrt{b^2-a^2}}+\frac{c}{d}+x}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csc[c + d*x])^(-1),x]

[Out]

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

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Maple [A]  time = 0.056, size = 70, normalized size = 1.2 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{ad}}-2\,{\frac{b}{ad\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( 1/2\,dx+c/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*csc(d*x+c)),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.521268, size = 525, normalized size = 9.21 \begin{align*} \left [\frac{2 \,{\left (a^{2} - b^{2}\right )} d x + \sqrt{a^{2} - b^{2}} b \log \left (\frac{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a b \sin \left (d x + c\right ) + a^{2} + b^{2} + 2 \,{\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{2 \,{\left (a^{3} - a b^{2}\right )} d}, \frac{{\left (a^{2} - b^{2}\right )} d x + \sqrt{-a^{2} + b^{2}} b \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \sin \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )}\right )}{{\left (a^{3} - a b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*(2*(a^2 - b^2)*d*x + sqrt(a^2 - b^2)*b*log(((a^2 - 2*b^2)*cos(d*x + c)^2 + 2*a*b*sin(d*x + c) + a^2 + b^2
+ 2*(b*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c))*sqrt(a^2 - b^2))/(a^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c)
- a^2 - b^2)))/((a^3 - a*b^2)*d), ((a^2 - b^2)*d*x + sqrt(-a^2 + b^2)*b*arctan(-sqrt(-a^2 + b^2)*(b*sin(d*x +
c) + a)/((a^2 - b^2)*cos(d*x + c))))/((a^3 - a*b^2)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \csc{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(d*x+c)),x)

[Out]

Integral(1/(a + b*csc(c + d*x)), x)

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Giac [A]  time = 1.45478, size = 104, normalized size = 1.82 \begin{align*} -\frac{\frac{2 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a}{\sqrt{-a^{2} + b^{2}}}\right )\right )} b}{\sqrt{-a^{2} + b^{2}} a} - \frac{d x + c}{a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(d*x+c)),x, algorithm="giac")

[Out]

-(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*d*x + 1/2*c) + a)/sqrt(-a^2 + b^2)))*b/(sqrt(
-a^2 + b^2)*a) - (d*x + c)/a)/d