3.1291 \(\int \frac{\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=114 \[ \frac{2 b \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d}+\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.456389, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2889, 3056, 3055, 3001, 3770, 2660, 618, 204} \[ \frac{2 b \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d}+\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^2*Csc[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

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

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 204

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

Rubi steps

\begin{align*} \int \frac{\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx &=\int \frac{\csc ^3(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx\\ &=-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\int \frac{\csc ^2(c+d x) \left (-2 b-a \sin (c+d x)+b \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a}\\ &=\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\int \frac{\csc (c+d x) \left (-a^2+2 b^2+a b \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^2}\\ &=\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}-\frac{\left (a^2-2 b^2\right ) \int \csc (c+d x) \, dx}{2 a^3}+\frac{\left (b \left (a^2-b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^3}\\ &=\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\left (2 b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 d}\\ &=\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}-\frac{\left (4 b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 d}\\ &=\frac{2 b \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 d}+\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}\\ \end{align*}

Mathematica [A]  time = 0.84715, size = 181, normalized size = 1.59 \[ \frac{16 b \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )+a^2 \left (-\csc ^2\left (\frac{1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )-4 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-4 a b \tan \left (\frac{1}{2} (c+d x)\right )+4 a b \cot \left (\frac{1}{2} (c+d x)\right )+8 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-8 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.108, size = 166, normalized size = 1.5 \begin{align*}{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{b}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{b\sqrt{{a}^{2}-{b}^{2}}}{d{a}^{3}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{1}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

1/8/d/a*tan(1/2*d*x+1/2*c)^2-1/2/d/a^2*tan(1/2*d*x+1/2*c)*b+2/d*b*(a^2-b^2)^(1/2)/a^3*arctan(1/2*(2*a*tan(1/2*
d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-1/8/d/a/tan(1/2*d*x+1/2*c)^2-1/2/d/a*ln(tan(1/2*d*x+1/2*c))+1/d/a^3*ln(tan(1/
2*d*x+1/2*c))*b^2+1/2/d/a^2*b/tan(1/2*d*x+1/2*c)

________________________________________________________________________________________

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(cos(d*x+c)^2*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.94687, size = 1121, normalized size = 9.83 \begin{align*} \left [-\frac{4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right ) - 2 \,{\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{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 (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) -{\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}}, -\frac{4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right ) + 4 \,{\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) -{\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**3/(a + b*sin(c + d*x)), x)

________________________________________________________________________________________

Giac [A]  time = 1.22197, size = 267, normalized size = 2.34 \begin{align*} \frac{\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} - \frac{4 \,{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{16 \,{\left (a^{2} b - b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} a^{3}} + \frac{6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*((a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 - 4*(a^2 - 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))
)/a^3 + 16*(a^2*b - b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a
^2 - b^2)))/(sqrt(a^2 - b^2)*a^3) + (6*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 + 4*a*b*tan(
1/2*d*x + 1/2*c) - a^2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d