\(\int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [785]

Optimal result

Integrand size = 29, antiderivative size = 130 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {4 \tan (c+d x)}{a^2 d}+\frac {5 \tan ^3(c+d x)}{3 a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d} \]

[Out]

2*arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a^2/d-2*sec(d*x+c)/a^2/d-2/3*sec(d*x+c)^3/a^2/d-2/5*sec(d*x+c)^5/a^2/d+
4*tan(d*x+c)/a^2/d+5/3*tan(d*x+c)^3/a^2/d+2/5*tan(d*x+c)^5/a^2/d

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2954, 2952, 3852, 2702, 308, 213, 2700, 276} \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d}+\frac {5 \tan ^3(c+d x)}{3 a^2 d}+\frac {4 \tan (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec (c+d x)}{a^2 d} \]

[In]

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

[Out]

(2*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a^2*d) - (2*Sec[c + d*x])/(a^2*d) - (2*Sec[c + d*x]^3)/(3*a^
2*d) - (2*Sec[c + d*x]^5)/(5*a^2*d) + (4*Tan[c + d*x])/(a^2*d) + (5*Tan[c + d*x]^3)/(3*a^2*d) + (2*Tan[c + d*x
]^5)/(5*a^2*d)

Rule 213

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

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2700

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

Rule 2702

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

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2954

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e +
f*x])^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc ^2(c+d x) \sec ^6(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4} \\ & = \frac {\int \left (a^2 \sec ^6(c+d x)-2 a^2 \csc (c+d x) \sec ^6(c+d x)+a^2 \csc ^2(c+d x) \sec ^6(c+d x)\right ) \, dx}{a^4} \\ & = \frac {\int \sec ^6(c+d x) \, dx}{a^2}+\frac {\int \csc ^2(c+d x) \sec ^6(c+d x) \, dx}{a^2}-\frac {2 \int \csc (c+d x) \sec ^6(c+d x) \, dx}{a^2} \\ & = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {\text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}-\frac {2 \text {Subst}\left (\int \frac {x^6}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d} \\ & = \frac {\tan (c+d x)}{a^2 d}+\frac {2 \tan ^3(c+d x)}{3 a^2 d}+\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {\text {Subst}\left (\int \left (3+\frac {1}{x^2}+3 x^2+x^4\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {2 \text {Subst}\left (\int \left (1+x^2+x^4+\frac {1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d} \\ & = -\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {4 \tan (c+d x)}{a^2 d}+\frac {5 \tan ^3(c+d x)}{3 a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d} \\ & = \frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {4 \tan (c+d x)}{a^2 d}+\frac {5 \tan ^3(c+d x)}{3 a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(130)=260\).

Time = 1.01 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.22 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^3(c+d x) \left (40+48 \cos (2 (c+d x))+112 \cos (3 (c+d x))-28 \cos (4 (c+d x))+60 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 \cos (c+d x) \left (28+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-60 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+58 \sin (c+d x)-168 \sin (2 (c+d x))-90 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+90 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+82 \sin (3 (c+d x))+28 \sin (4 (c+d x))+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))\right )}{15 a^2 d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\sin (c+d x))^2} \]

[In]

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

[Out]

-1/15*(Csc[c + d*x]^3*(40 + 48*Cos[2*(c + d*x)] + 112*Cos[3*(c + d*x)] - 28*Cos[4*(c + d*x)] + 60*Cos[3*(c + d
*x)]*Log[Cos[(c + d*x)/2]] - 4*Cos[c + d*x]*(28 + 15*Log[Cos[(c + d*x)/2]] - 15*Log[Sin[(c + d*x)/2]]) - 60*Co
s[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + 58*Sin[c + d*x] - 168*Sin[2*(c + d*x)] - 90*Log[Cos[(c + d*x)/2]]*Sin[2
*(c + d*x)] + 90*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] + 82*Sin[3*(c + d*x)] + 28*Sin[4*(c + d*x)] + 15*Log[C
os[(c + d*x)/2]]*Sin[4*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)]))/(a^2*d*(Csc[(c + d*x)/2]^2 - S
ec[(c + d*x)/2]^2)*(1 + Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {26}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {31}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{2 d \,a^{2}}\)	\(134\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {26}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {31}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{2 d \,a^{2}}\)	\(134\)
parallelrisch	\(\frac {-60 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-420 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-960 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-550 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+560 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+796 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+304}{30 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(160\)
risch	\(-\frac {4 \left (-85 \,{\mathrm e}^{5 i \left (d x +c \right )}+60 i {\mathrm e}^{6 i \left (d x +c \right )}-40 i {\mathrm e}^{4 i \left (d x +c \right )}+97 \,{\mathrm e}^{i \left (d x +c \right )}-108 i {\mathrm e}^{2 i \left (d x +c \right )}+28 i-27 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d \,a^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}\)	\(174\)
norman	\(\frac {\frac {1}{2 a d}-\frac {55 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {7 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {29 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {199 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{30 d a}+\frac {188 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(201\)

[In]

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

[Out]

1/2/d/a^2*(tan(1/2*d*x+1/2*c)-1/2/(tan(1/2*d*x+1/2*c)-1)-1/tan(1/2*d*x+1/2*c)-4*ln(tan(1/2*d*x+1/2*c))-8/5/(ta
n(1/2*d*x+1/2*c)+1)^5+4/(tan(1/2*d*x+1/2*c)+1)^4-26/3/(tan(1/2*d*x+1/2*c)+1)^3+9/(tan(1/2*d*x+1/2*c)+1)^2-31/2
/(tan(1/2*d*x+1/2*c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.68 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {56 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (41 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) + 9}{15 \, {\left (2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) + {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/15*(56*cos(d*x + c)^4 - 80*cos(d*x + c)^2 - 15*(2*cos(d*x + c)^3 + (cos(d*x + c)^3 - 2*cos(d*x + c))*sin(d*
x + c) - 2*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 15*(2*cos(d*x + c)^3 + (cos(d*x + c)^3 - 2*cos(d*x + c)
)*sin(d*x + c) - 2*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(41*cos(d*x + c)^2 - 3)*sin(d*x + c) + 9)/(2
*a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c) + (a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c))*sin(d*x + c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (122) = 244\).

Time = 0.23 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.38 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {244 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {571 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {320 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {475 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {660 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {255 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 15}{\frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {15 \, \sin \left (d x + c\right )}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{30 \, d} \]

[In]

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

[Out]

-1/30*((244*sin(d*x + c)/(cos(d*x + c) + 1) + 571*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 320*sin(d*x + c)^3/(co
s(d*x + c) + 1)^3 - 475*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 660*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 255*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15)/(a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^2*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 5*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 5*a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 4*a^2*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6 - a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) + 60*log(sin(d*x + c)/(cos(d*x + c
) + 1))/a^2 - 15*sin(d*x + c)/(a^2*(cos(d*x + c) + 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.24 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {15 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{2}} + \frac {465 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1590 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 383}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \]

[In]

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

[Out]

-1/60*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 30*tan(1/2*d*x + 1/2*c)/a^2 - 15*(4*tan(1/2*d*x + 1/2*c)^2 - 7
*tan(1/2*d*x + 1/2*c) + 2)/((tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c))*a^2) + (465*tan(1/2*d*x + 1/2*c)^4
+ 1590*tan(1/2*d*x + 1/2*c)^3 + 2240*tan(1/2*d*x + 1/2*c)^2 + 1450*tan(1/2*d*x + 1/2*c) + 383)/(a^2*(tan(1/2*
d*x + 1/2*c) + 1)^5))/d

Mupad [B] (verification not implemented)

Time = 14.12 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.66 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {-17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-44\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {95\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {571\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {244\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+1}{d\,\left (-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

[In]

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

[Out]

tan(c/2 + (d*x)/2)/(2*a^2*d) - (2*log(tan(c/2 + (d*x)/2)))/(a^2*d) - ((244*tan(c/2 + (d*x)/2))/15 + (571*tan(c
/2 + (d*x)/2)^2)/15 + (64*tan(c/2 + (d*x)/2)^3)/3 - (95*tan(c/2 + (d*x)/2)^4)/3 - 44*tan(c/2 + (d*x)/2)^5 - 17
*tan(c/2 + (d*x)/2)^6 + 1)/(d*(8*a^2*tan(c/2 + (d*x)/2)^2 + 10*a^2*tan(c/2 + (d*x)/2)^3 - 10*a^2*tan(c/2 + (d*
x)/2)^5 - 8*a^2*tan(c/2 + (d*x)/2)^6 - 2*a^2*tan(c/2 + (d*x)/2)^7 + 2*a^2*tan(c/2 + (d*x)/2)))