3.5 \(\int \frac{1}{a+b \csc ^2(c+d x)} \, dx\)

Optimal. Leaf size=46 \[ \frac{x}{a}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{b}}\right )}{a d \sqrt{a+b}} \]

[Out]

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

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Rubi [A]  time = 0.0470416, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4127, 3181, 205} \[ \frac{x}{a}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{b}}\right )}{a d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x]^2)^(-1),x]

[Out]

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

Rule 4127

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> Simp[x/a, x] - Dist[b/a, Int[1/(b + a*Cos[e +
f*x]^2), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.204044, size = 46, normalized size = 1. \[ \frac{-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{b}}\right )}{\sqrt{a+b}}+c+d x}{a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csc[c + d*x]^2)^(-1),x]

[Out]

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

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Maple [A]  time = 0.072, size = 50, normalized size = 1.1 \begin{align*}{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{ad}}-{\frac{b}{ad}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*csc(d*x+c)^2),x)

[Out]

1/d/a*arctan(tan(d*x+c))-1/d*b/a/((a+b)*b)^(1/2)*arctan((a+b)*tan(d*x+c)/((a+b)*b)^(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)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.540766, size = 617, normalized size = 13.41 \begin{align*} \left [\frac{4 \, d x + \sqrt{-\frac{b}{a + b}} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} + 5 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt{-\frac{b}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{a^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{4 \, a d}, \frac{2 \, d x + \sqrt{\frac{b}{a + b}} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{\frac{b}{a + b}}}{2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{2 \, a d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/4*(4*d*x + sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(a^2 + 5*a*b + 4*b^2)*cos(d*x + c
)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-b/(a + b))*sin(d*x + c
) + a^2 + 2*a*b + b^2)/(a^2*cos(d*x + c)^4 - 2*(a^2 + a*b)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)))/(a*d), 1/2*(2
*d*x + sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(d*x + c)^2 - a - b)*sqrt(b/(a + b))/(b*cos(d*x + c)*sin(d*x +
c))))/(a*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \csc ^{2}{\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)**2),x)

[Out]

Integral(1/(a + b*csc(c + d*x)**2), x)

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Giac [B]  time = 1.48081, size = 109, normalized size = 2.37 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a b + b^{2}}}\right )\right )} b}{\sqrt{a b + 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)^2),x, algorithm="giac")

[Out]

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