\(\int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [87]

Optimal result

Integrand size = 21, antiderivative size = 73 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {2 \csc ^3(c+d x)}{3 a^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d} \]

[Out]

-1/3*cot(d*x+c)^3/a^2/d-2/5*cot(d*x+c)^5/a^2/d-2/3*csc(d*x+c)^3/a^2/d+2/5*csc(d*x+c)^5/a^2/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3957, 2790, 2687, 30, 2686, 14} \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d}-\frac {2 \csc ^3(c+d x)}{3 a^2 d} \]

[In]

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

[Out]

-1/3*Cot[c + d*x]^3/(a^2*d) - (2*Cot[c + d*x]^5)/(5*a^2*d) - (2*Csc[c + d*x]^3)/(3*a^2*d) + (2*Csc[c + d*x]^5)
/(5*a^2*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2790

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[a^(2*
m), Int[ExpandIntegrand[(g*Tan[e + f*x])^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x]
/; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cot ^2(c+d x)}{(-a-a \cos (c+d x))^2} \, dx \\ & = \frac {\int \left (a^2 \cot ^4(c+d x) \csc ^2(c+d x)-2 a^2 \cot ^3(c+d x) \csc ^3(c+d x)+a^2 \cot ^2(c+d x) \csc ^4(c+d x)\right ) \, dx}{a^4} \\ & = \frac {\int \cot ^4(c+d x) \csc ^2(c+d x) \, dx}{a^2}+\frac {\int \cot ^2(c+d x) \csc ^4(c+d x) \, dx}{a^2}-\frac {2 \int \cot ^3(c+d x) \csc ^3(c+d x) \, dx}{a^2} \\ & = \frac {\text {Subst}\left (\int x^4 \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {\text {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {2 \text {Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\csc (c+d x)\right )}{a^2 d} \\ & = -\frac {\cot ^5(c+d x)}{5 a^2 d}+\frac {\text {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {2 \text {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\csc (c+d x)\right )}{a^2 d} \\ & = -\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {2 \csc ^3(c+d x)}{3 a^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.44 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\csc (c) \csc (c+d x) \sec ^2(c+d x) (-80 \sin (c)+80 \sin (d x)+55 \sin (c+d x)+44 \sin (2 (c+d x))+11 \sin (3 (c+d x))-60 \sin (2 c+d x)+16 \sin (c+2 d x)+4 \sin (2 c+3 d x))}{240 a^2 d (1+\sec (c+d x))^2} \]

[In]

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

[Out]

(Csc[c]*Csc[c + d*x]*Sec[c + d*x]^2*(-80*Sin[c] + 80*Sin[d*x] + 55*Sin[c + d*x] + 44*Sin[2*(c + d*x)] + 11*Sin
[3*(c + d*x)] - 60*Sin[2*c + d*x] + 16*Sin[c + 2*d*x] + 4*Sin[2*c + 3*d*x]))/(240*a^2*d*(1 + Sec[c + d*x])^2)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79

method	result	size
parallelrisch	\(\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-15 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{120 a^{2} d}\)	\(58\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{8 d \,a^{2}}\)	\(60\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{8 d \,a^{2}}\)	\(60\)
norman	\(\frac {-\frac {1}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{24 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{40 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a}\)	\(82\)
risch	\(-\frac {2 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+20 \,{\mathrm e}^{3 i \left (d x +c \right )}+20 \,{\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{15 a^{2} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}\)	\(82\)

[In]

int(csc(d*x+c)^2/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/120*(3*tan(1/2*d*x+1/2*c)^5-5*tan(1/2*d*x+1/2*c)^3-15*cot(1/2*d*x+1/2*c)-15*tan(1/2*d*x+1/2*c))/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 4}{15 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

-1/15*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + 8*cos(d*x + c) + 4)/((a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) +
a^2*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\csc ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(csc(d*x+c)**2/(a+a*sec(d*x+c))**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.23 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} + \frac {15 \, {\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2} \sin \left (d x + c\right )}}{120 \, d} \]

[In]

integrate(csc(d*x+c)^2/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/120*((15*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 3*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5)/a^2 + 15*(cos(d*x + c) + 1)/(a^2*sin(d*x + c)))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {15}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {3 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{10}}}{120 \, d} \]

[In]

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

[Out]

-1/120*(15/(a^2*tan(1/2*d*x + 1/2*c)) - (3*a^8*tan(1/2*d*x + 1/2*c)^5 - 5*a^8*tan(1/2*d*x + 1/2*c)^3 - 15*a^8*
tan(1/2*d*x + 1/2*c))/a^10)/d

Mupad [B] (verification not implemented)

Time = 13.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+14\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-3}{120\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

[In]

int(1/(sin(c + d*x)^2*(a + a/cos(c + d*x))^2),x)

[Out]

-(14*cos(c/2 + (d*x)/2)^2 - 4*cos(c/2 + (d*x)/2)^4 + 8*cos(c/2 + (d*x)/2)^6 - 3)/(120*a^2*d*cos(c/2 + (d*x)/2)
^5*sin(c/2 + (d*x)/2))