3.9.9 \(\int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\) [809]

3.9.9.1 Optimal result
3.9.9.2 Mathematica [A] (verified)
3.9.9.3 Rubi [A] (verified)
3.9.9.4 Maple [A] (verified)
3.9.9.5 Fricas [B] (verification not implemented)
3.9.9.6 Sympy [F(-1)]
3.9.9.7 Maxima [A] (verification not implemented)
3.9.9.8 Giac [A] (verification not implemented)
3.9.9.9 Mupad [B] (verification not implemented)

3.9.9.1 Optimal result

Integrand size = 29, antiderivative size = 87 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {2 a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {10 a^2 \cot (c+d x)}{3 d}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2} \]

output
-2*a^2*arctanh(cos(d*x+c))/d-10/3*a^2*cot(d*x+c)/d+2*a^2*cot(d*x+c)/d/(1-s 
in(d*x+c))+1/3*a^4*cot(d*x+c)/d/(a-a*sin(d*x+c))^2
 
3.9.9.2 Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.55 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-3 \cot \left (\frac {1}{2} (c+d x)\right )-12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right ) (-8+7 \sin (c+d x))}{\left (-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+3 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \]

input
Integrate[Csc[c + d*x]^2*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]
 
output
(a^2*(-3*Cot[(c + d*x)/2] - 12*Log[Cos[(c + d*x)/2]] + 12*Log[Sin[(c + d*x 
)/2]] + 2/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*Sin[(c + d*x)/2]*(- 
8 + 7*Sin[c + d*x]))/(-Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + 3*Tan[(c + 
 d*x)/2]))/(6*d)
 
3.9.9.3 Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3348, 3042, 3245, 27, 3042, 3457, 3042, 3227, 3042, 4254, 24, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \csc ^2(c+d x) \sec ^4(c+d x) (a \sin (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^2}{\sin (c+d x)^2 \cos (c+d x)^4}dx\)

\(\Big \downarrow \) 3348

\(\displaystyle a^4 \int \frac {\csc ^2(c+d x)}{(a-a \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 \int \frac {1}{\sin (c+d x)^2 (a-a \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3245

\(\displaystyle a^4 \left (\frac {\int \frac {2 \csc ^2(c+d x) (\sin (c+d x) a+2 a)}{a-a \sin (c+d x)}dx}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle a^4 \left (\frac {2 \int \frac {\csc ^2(c+d x) (\sin (c+d x) a+2 a)}{a-a \sin (c+d x)}dx}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 \left (\frac {2 \int \frac {\sin (c+d x) a+2 a}{\sin (c+d x)^2 (a-a \sin (c+d x))}dx}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 3457

\(\displaystyle a^4 \left (\frac {2 \left (\frac {\int \csc ^2(c+d x) \left (3 \sin (c+d x) a^2+5 a^2\right )dx}{a^2}+\frac {3 \cot (c+d x)}{d (1-\sin (c+d x))}\right )}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 \left (\frac {2 \left (\frac {\int \frac {3 \sin (c+d x) a^2+5 a^2}{\sin (c+d x)^2}dx}{a^2}+\frac {3 \cot (c+d x)}{d (1-\sin (c+d x))}\right )}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle a^4 \left (\frac {2 \left (\frac {5 a^2 \int \csc ^2(c+d x)dx+3 a^2 \int \csc (c+d x)dx}{a^2}+\frac {3 \cot (c+d x)}{d (1-\sin (c+d x))}\right )}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 \left (\frac {2 \left (\frac {3 a^2 \int \csc (c+d x)dx+5 a^2 \int \csc (c+d x)^2dx}{a^2}+\frac {3 \cot (c+d x)}{d (1-\sin (c+d x))}\right )}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 4254

\(\displaystyle a^4 \left (\frac {2 \left (\frac {3 a^2 \int \csc (c+d x)dx-\frac {5 a^2 \int 1d\cot (c+d x)}{d}}{a^2}+\frac {3 \cot (c+d x)}{d (1-\sin (c+d x))}\right )}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle a^4 \left (\frac {2 \left (\frac {3 a^2 \int \csc (c+d x)dx-\frac {5 a^2 \cot (c+d x)}{d}}{a^2}+\frac {3 \cot (c+d x)}{d (1-\sin (c+d x))}\right )}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

\(\Big \downarrow \) 4257

\(\displaystyle a^4 \left (\frac {2 \left (\frac {-\frac {3 a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {5 a^2 \cot (c+d x)}{d}}{a^2}+\frac {3 \cot (c+d x)}{d (1-\sin (c+d x))}\right )}{3 a^2}+\frac {\cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\)

input
Int[Csc[c + d*x]^2*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]
 
output
a^4*((2*(((-3*a^2*ArcTanh[Cos[c + d*x]])/d - (5*a^2*Cot[c + d*x])/d)/a^2 + 
 (3*Cot[c + d*x])/(d*(1 - Sin[c + d*x]))))/(3*a^2) + Cot[c + d*x]/(3*d*(a 
- a*Sin[c + d*x])^2))
 

3.9.9.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3245
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( 
a*(2*m + 1)*(b*c - a*d))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + 
f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (Intege 
rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3348
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) 
 + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m)   Int[(d* 
Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, 
 x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
 

rule 3457
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 4254
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
3.9.9.4 Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07

method result size
parallelrisch \(\frac {\left (4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )-19 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {50}{3}\right ) a^{2}}{2 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\) \(93\)
derivativedivides \(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a^{2} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )}{d}\) \(118\)
default \(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a^{2} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )}{d}\) \(118\)
risch \(\frac {-\frac {44 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{3}-12 i {\mathrm e}^{3 i \left (d x +c \right )} a^{2}+\frac {20 a^{2}}{3}+16 i a^{2} {\mathrm e}^{i \left (d x +c \right )}+4 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\) \(138\)
norman \(\frac {\frac {a^{2}}{2 d}-\frac {7 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {35 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {10 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {35 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {7 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {8 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {16 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) \(278\)

input
int(csc(d*x+c)^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 
output
1/2*(4*(tan(1/2*d*x+1/2*c)-1)^3*ln(tan(1/2*d*x+1/2*c))+tan(1/2*d*x+1/2*c)^ 
4-19*tan(1/2*d*x+1/2*c)^2+cot(1/2*d*x+1/2*c)+31*tan(1/2*d*x+1/2*c)-50/3)*a 
^2/d/(tan(1/2*d*x+1/2*c)-1)^3
 
3.9.9.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (82) = 164\).

Time = 0.27 (sec) , antiderivative size = 329, normalized size of antiderivative = 3.78 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {10 \, a^{2} \cos \left (d x + c\right )^{3} - 4 \, a^{2} \cos \left (d x + c\right )^{2} - 13 \, a^{2} \cos \left (d x + c\right ) + a^{2} - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (10 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

input
integrate(csc(d*x+c)^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="frica 
s")
 
output
1/3*(10*a^2*cos(d*x + c)^3 - 4*a^2*cos(d*x + c)^2 - 13*a^2*cos(d*x + c) + 
a^2 - 3*(a^2*cos(d*x + c)^3 + 2*a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2* 
a^2 - (a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2)*sin(d*x + c))*log(1/ 
2*cos(d*x + c) + 1/2) + 3*(a^2*cos(d*x + c)^3 + 2*a^2*cos(d*x + c)^2 - a^2 
*cos(d*x + c) - 2*a^2 - (a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2)*si 
n(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + (10*a^2*cos(d*x + c)^2 + 14*a^2 
*cos(d*x + c) + a^2)*sin(d*x + c))/(d*cos(d*x + c)^3 + 2*d*cos(d*x + c)^2 
- d*cos(d*x + c) - (d*cos(d*x + c)^2 - d*cos(d*x + c) - 2*d)*sin(d*x + c) 
- 2*d)
 
3.9.9.6 Sympy [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

input
integrate(csc(d*x+c)**2*sec(d*x+c)**4*(a+a*sin(d*x+c))**2,x)
 
output
Timed out
 
3.9.9.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.23 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {{\left (\tan \left (d x + c\right )^{3} - \frac {3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a^{2} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{3 \, d} \]

input
integrate(csc(d*x+c)^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxim 
a")
 
output
1/3*((tan(d*x + c)^3 - 3/tan(d*x + c) + 6*tan(d*x + c))*a^2 + (tan(d*x + c 
)^3 + 3*tan(d*x + c))*a^2 + a^2*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 - 
 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)))/d
 
3.9.9.8 Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.36 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {12 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {4 \, {\left (9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

input
integrate(csc(d*x+c)^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac" 
)
 
output
1/6*(12*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a^2*tan(1/2*d*x + 1/2*c) - 
3*(4*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c) - 4*(9*a^2*tan(1 
/2*d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x + 1/2*c) + 8*a^2)/(tan(1/2*d*x + 1/ 
2*c) - 1)^3)/d
 
3.9.9.9 Mupad [B] (verification not implemented)

Time = 10.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.66 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+23\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {41\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+a^2}{d\,\left (-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

input
int((a + a*sin(c + d*x))^2/(cos(c + d*x)^4*sin(c + d*x)^2),x)
 
output
(2*a^2*log(tan(c/2 + (d*x)/2)))/d - (23*a^2*tan(c/2 + (d*x)/2)^2 - 13*a^2* 
tan(c/2 + (d*x)/2)^3 + a^2 - (41*a^2*tan(c/2 + (d*x)/2))/3)/(d*(2*tan(c/2 
+ (d*x)/2) - 6*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^3 - 2*tan(c/2 + 
 (d*x)/2)^4)) + (a^2*tan(c/2 + (d*x)/2))/(2*d)