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

Optimal. Leaf size=108 \[ \frac{2 b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac{b^2 \cot (c+d x)}{a d \left (a^2-b^2\right ) (a+b \csc (c+d x))}+\frac{x}{a^2} \]

[Out]

x/a^2 + (2*b*(2*a^2 - b^2)*ArcTanh[(a + b*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(3/2)*d) - (b^2
*Cot[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Csc[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.171399, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3785, 3919, 3831, 2660, 618, 206} \[ \frac{2 b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac{b^2 \cot (c+d x)}{a d \left (a^2-b^2\right ) (a+b \csc (c+d x))}+\frac{x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x])^(-2),x]

[Out]

x/a^2 + (2*b*(2*a^2 - b^2)*ArcTanh[(a + b*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(3/2)*d) - (b^2
*Cot[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Csc[c + d*x]))

Rule 3785

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, 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))^2} \, dx &=-\frac{b^2 \cot (c+d x)}{a \left (a^2-b^2\right ) d (a+b \csc (c+d x))}-\frac{\int \frac{-a^2+b^2+a b \csc (c+d x)}{a+b \csc (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{x}{a^2}-\frac{b^2 \cot (c+d x)}{a \left (a^2-b^2\right ) d (a+b \csc (c+d x))}-\frac{\left (b \left (2 a^2-b^2\right )\right ) \int \frac{\csc (c+d x)}{a+b \csc (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac{x}{a^2}-\frac{b^2 \cot (c+d x)}{a \left (a^2-b^2\right ) d (a+b \csc (c+d x))}-\frac{\left (2 a^2-b^2\right ) \int \frac{1}{1+\frac{a \sin (c+d x)}{b}} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac{x}{a^2}-\frac{b^2 \cot (c+d x)}{a \left (a^2-b^2\right ) d (a+b \csc (c+d x))}-\frac{\left (2 \left (2 a^2-b^2\right )\right ) \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^2 \left (a^2-b^2\right ) d}\\ &=\frac{x}{a^2}-\frac{b^2 \cot (c+d x)}{a \left (a^2-b^2\right ) d (a+b \csc (c+d x))}+\frac{\left (4 \left (2 a^2-b^2\right )\right ) \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^2 \left (a^2-b^2\right ) d}\\ &=\frac{x}{a^2}+\frac{2 b \left (2 a^2-b^2\right ) \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^2 \left (a^2-b^2\right )^{3/2} d}-\frac{b^2 \cot (c+d x)}{a \left (a^2-b^2\right ) d (a+b \csc (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.455729, size = 139, normalized size = 1.29 \[ \frac{\csc (c+d x) (a \sin (c+d x)+b) \left (-\frac{2 b \left (b^2-2 a^2\right ) \tan ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right ) (a+b \csc (c+d x))}{\left (b^2-a^2\right )^{3/2}}+\frac{a b^2 \cot (c+d x)}{(b-a) (a+b)}+(c+d x) (a+b \csc (c+d x))\right )}{a^2 d (a+b \csc (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*((a*b^2*Cot[c + d*x])/((-a + b)*(a + b)) + (c + d*x)*(a + b*Csc[c + d*x]) - (2*b*(-2*a^2 + b^2)*
ArcTan[(a + b*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]*(a + b*Csc[c + d*x]))/(-a^2 + b^2)^(3/2))*(b + a*Sin[c + d*x
]))/(a^2*d*(a + b*Csc[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.093, size = 247, normalized size = 2.3 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-2\,{\frac{b\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+2\,a\tan \left ( 1/2\,dx+c/2 \right ) +b \right ) \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{{b}^{2}}{ad \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+2\,a\tan \left ( 1/2\,dx+c/2 \right ) +b \right ) \left ({a}^{2}-{b}^{2} \right ) }}-4\,{\frac{b}{d \left ({a}^{2}-{b}^{2} \right ) \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 ) }+2\,{\frac{{b}^{3}}{d{a}^{2} \left ({a}^{2}-{b}^{2} \right ) \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))^2,x)

[Out]

2/d/a^2*arctan(tan(1/2*d*x+1/2*c))-2/d*b/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)/(a^2-b^2)*tan(1/2*d
*x+1/2*c)-2/d*b^2/a/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)/(a^2-b^2)-4/d*b/(a^2-b^2)/(-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))+2/d*b^3/a^2/(a^2-b^2)/(-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))

________________________________________________________________________________________

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

________________________________________________________________________________________

Fricas [B]  time = 0.563269, size = 1076, normalized size = 9.96 \begin{align*} \left [\frac{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \sin \left (d x + c\right ) + 2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x +{\left (2 \, a^{2} b^{2} - b^{4} +{\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \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} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d x + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \sin \left (d x + c\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x +{\left (2 \, a^{2} b^{2} - b^{4} +{\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \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} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d x + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} 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]

[1/2*(2*(a^5 - 2*a^3*b^2 + a*b^4)*d*x*sin(d*x + c) + 2*(a^4*b - 2*a^2*b^3 + b^5)*d*x + (2*a^2*b^2 - b^4 + (2*a
^3*b - a*b^3)*sin(d*x + c))*sqrt(a^2 - b^2)*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)) - 2*(a^3*b^2 - a*b^4)*cos(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*sin(d*x + c) + (a^6*b - 2*a^4
*b^3 + a^2*b^5)*d), ((a^5 - 2*a^3*b^2 + a*b^4)*d*x*sin(d*x + c) + (a^4*b - 2*a^2*b^3 + b^5)*d*x + (2*a^2*b^2 -
b^4 + (2*a^3*b - a*b^3)*sin(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*sin(d*x + c) + a)/((a^2 -
b^2)*cos(d*x + c))) - (a^3*b^2 - a*b^4)*cos(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*sin(d*x + c) + (a^6*b - 2
*a^4*b^3 + a^2*b^5)*d)]

________________________________________________________________________________________

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

________________________________________________________________________________________

Giac [A]  time = 1.38376, size = 213, normalized size = 1.97 \begin{align*} -\frac{\frac{2 \,{\left (2 \, a^{2} b - b^{3}\right )}{\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 )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \,{\left (a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{2}\right )}}{{\left (a^{3} - a b^{2}\right )}{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b\right )}} - \frac{d x + c}{a^{2}}}{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]

-(2*(2*a^2*b - b^3)*(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)))/((a^4 - a^2*b^2)*sqrt(-a^2 + b^2)) + 2*(a*b*tan(1/2*d*x + 1/2*c) + b^2)/((a^3 - a*b^2)*(b*tan(1/2*d*x
+ 1/2*c)^2 + 2*a*tan(1/2*d*x + 1/2*c) + b)) - (d*x + c)/a^2)/d