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

Optimal. Leaf size=92 \[ \frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 d (a+b)^{3/2}}+\frac{x}{a^2}+\frac{b \cot (c+d x)}{2 a d (a+b) \left (a+b \cot ^2(c+d x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.107348, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4128, 414, 522, 203, 205} \[ \frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 d (a+b)^{3/2}}+\frac{x}{a^2}+\frac{b \cot (c+d x)}{2 a d (a+b) \left (a+b \cot ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x]^2)^(-2),x]

[Out]

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

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 414

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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])

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}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{b \cot (c+d x)}{2 a (a+b) d \left (a+b+b \cot ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{2 a+b-b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\cot (c+d x)\right )}{2 a (a+b) d}\\ &=\frac{b \cot (c+d x)}{2 a (a+b) d \left (a+b+b \cot ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{a^2 d}+\frac{(b (3 a+2 b)) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\cot (c+d x)\right )}{2 a^2 (a+b) d}\\ &=\frac{x}{a^2}+\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 (a+b)^{3/2} d}+\frac{b \cot (c+d x)}{2 a (a+b) d \left (a+b+b \cot ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.800738, size = 166, normalized size = 1.8 \[ -\frac{\csc ^4(c+d x) (a \cos (2 (c+d x))-a-2 b) \left (\sqrt{a+b} \left (2 \left (a^2+3 a b+2 b^2\right ) (c+d x)+a b \sin (2 (c+d x))-2 a (a+b) (c+d x) \cos (2 (c+d x))\right )-\sqrt{b} (3 a+2 b) (a (-\cos (2 (c+d x)))+a+2 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{b}}\right )\right )}{8 a^2 d (a+b)^{3/2} \left (a+b \csc ^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-a - 2*b + a*Cos[2*(c + d*x)])*Csc[c + d*x]^4*(-(Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt
[b]]*(a + 2*b - a*Cos[2*(c + d*x)])) + Sqrt[a + b]*(2*(a^2 + 3*a*b + 2*b^2)*(c + d*x) - 2*a*(a + b)*(c + d*x)*
Cos[2*(c + d*x)] + a*b*Sin[2*(c + d*x)])))/(8*a^2*(a + b)^(3/2)*d*(a + b*Csc[c + d*x]^2)^2)

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Maple [A]  time = 0.086, size = 140, normalized size = 1.5 \begin{align*}{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d{a}^{2}}}+{\frac{b\tan \left ( dx+c \right ) }{2\,ad \left ( a+b \right ) \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}+b \right ) }}-{\frac{3\,b}{2\,ad \left ( a+b \right ) }\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}}}}-{\frac{{b}^{2}}{d{a}^{2} \left ( a+b \right ) }\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)^2,x)

[Out]

1/d/a^2*arctan(tan(d*x+c))+1/2/d*b/a/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+b*tan(d*x+c)^2+b)-3/2/d*b/a/(a+b)/((a+b)
*b)^(1/2)*arctan((a+b)*tan(d*x+c)/((a+b)*b)^(1/2))-1/d*b^2/a^2/(a+b)/((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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.596462, size = 1165, normalized size = 12.66 \begin{align*} \left [\frac{8 \,{\left (a^{2} + a b\right )} d x \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 8 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x +{\left ({\left (3 \, a^{2} + 2 \, a b\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \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 )}{8 \,{\left ({\left (a^{4} + a^{3} b\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}, \frac{4 \,{\left (a^{2} + a b\right )} d x \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x +{\left ({\left (3 \, a^{2} + 2 \, a b\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \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 )}{4 \,{\left ({\left (a^{4} + a^{3} b\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.47066, size = 189, normalized size = 2.05 \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 )}{\left (3 \, a b + 2 \, b^{2}\right )}}{{\left (a^{3} + a^{2} b\right )} \sqrt{a b + b^{2}}} - \frac{b \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + b\right )}{\left (a^{2} + a b\right )}} - \frac{2 \,{\left (d x + c\right )}}{a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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