3.9.86 \(\int \frac {\sec ^4(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [886]

3.9.86.1 Optimal result
3.9.86.2 Mathematica [A] (verified)
3.9.86.3 Rubi [A] (verified)
3.9.86.4 Maple [A] (verified)
3.9.86.5 Fricas [A] (verification not implemented)
3.9.86.6 Sympy [F(-1)]
3.9.86.7 Maxima [A] (verification not implemented)
3.9.86.8 Giac [A] (verification not implemented)
3.9.86.9 Mupad [B] (verification not implemented)

3.9.86.1 Optimal result

Integrand size = 29, antiderivative size = 150 \[ \int \frac {\sec ^4(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))}{128 a d}-\frac {\sec ^6(c+d x)}{6 a d}+\frac {\sec ^8(c+d x)}{8 a d}-\frac {3 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{64 a d}+\frac {\sec ^5(c+d x) \tan (c+d x)}{16 a d}-\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{8 a d} \]

output
-3/128*arctanh(sin(d*x+c))/a/d-1/6*sec(d*x+c)^6/a/d+1/8*sec(d*x+c)^8/a/d-3 
/128*sec(d*x+c)*tan(d*x+c)/a/d-1/64*sec(d*x+c)^3*tan(d*x+c)/a/d+1/16*sec(d 
*x+c)^5*tan(d*x+c)/a/d-1/8*sec(d*x+c)^5*tan(d*x+c)^3/a/d
 
3.9.86.2 Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.61 \[ \int \frac {\sec ^4(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {9 \text {arctanh}(\sin (c+d x))+\frac {4}{(-1+\sin (c+d x))^3}+\frac {3}{(-1+\sin (c+d x))^2}-\frac {9}{-1+\sin (c+d x)}-\frac {6}{(1+\sin (c+d x))^4}+\frac {8}{(1+\sin (c+d x))^3}+\frac {6}{(1+\sin (c+d x))^2}}{384 a d} \]

input
Integrate[(Sec[c + d*x]^4*Tan[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 
output
-1/384*(9*ArcTanh[Sin[c + d*x]] + 4/(-1 + Sin[c + d*x])^3 + 3/(-1 + Sin[c 
+ d*x])^2 - 9/(-1 + Sin[c + d*x]) - 6/(1 + Sin[c + d*x])^4 + 8/(1 + Sin[c 
+ d*x])^3 + 6/(1 + Sin[c + d*x])^2)/(a*d)
 
3.9.86.3 Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {3042, 3314, 3042, 3086, 25, 244, 2009, 3091, 3042, 3091, 3042, 4255, 3042, 4255, 3042, 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.

\(\displaystyle \int \frac {\tan ^3(c+d x) \sec ^4(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^3}{\cos (c+d x)^7 (a \sin (c+d x)+a)}dx\)

\(\Big \downarrow \) 3314

\(\displaystyle \frac {\int \sec ^6(c+d x) \tan ^3(c+d x)dx}{a}-\frac {\int \sec ^5(c+d x) \tan ^4(c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sec (c+d x)^6 \tan (c+d x)^3dx}{a}-\frac {\int \sec (c+d x)^5 \tan (c+d x)^4dx}{a}\)

\(\Big \downarrow \) 3086

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 244

\(\displaystyle -\frac {\int \left (\sec ^5(c+d x)-\sec ^7(c+d x)\right )d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^5 \tan (c+d x)^4dx}{a}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\int \sec (c+d x)^5 \tan (c+d x)^4dx}{a}\)

\(\Big \downarrow \) 3091

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \int \sec ^5(c+d x) \tan ^2(c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \int \sec (c+d x)^5 \tan (c+d x)^2dx}{a}\)

\(\Big \downarrow \) 3091

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \left (\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}-\frac {1}{6} \int \sec ^5(c+d x)dx\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \left (\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}-\frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\right )}{a}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \int \sec ^3(c+d x)dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )}{a}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )}{a}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )}{a}\)

input
Int[(Sec[c + d*x]^4*Tan[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 
output
(-1/6*Sec[c + d*x]^6 + Sec[c + d*x]^8/8)/(a*d) - ((Sec[c + d*x]^5*Tan[c + 
d*x]^3)/(8*d) - (3*((Sec[c + d*x]^5*Tan[c + d*x])/(6*d) + (-1/4*(Sec[c + d 
*x]^3*Tan[c + d*x])/d - (3*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Ta 
n[c + d*x])/(2*d)))/4)/6))/8)/a
 

3.9.86.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 244
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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

rule 3086
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_.), x_Symbol] :> Simp[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 3091
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m 
 + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1))   Int[(a*Sec[e + f*x])^m*( 
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & 
& NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
 

rule 3314
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( 
a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[Cos[e + f 
*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d)   Int[Cos[e + f*x]^(p 
 - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & 
& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p 
+ 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
 

rule 4255
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

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

Time = 0.94 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.69

method result size
derivativedivides \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) \(103\)
default \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) \(103\)
risch \(\frac {i \left (-1161 \,{\mathrm e}^{5 i \left (d x +c \right )}+102 i {\mathrm e}^{10 i \left (d x +c \right )}-102 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{13 i \left (d x +c \right )}-172 i {\mathrm e}^{8 i \left (d x +c \right )}-18 i {\mathrm e}^{2 i \left (d x +c \right )}+18 i {\mathrm e}^{12 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+42 \,{\mathrm e}^{11 i \left (d x +c \right )}-1161 \,{\mathrm e}^{9 i \left (d x +c \right )}+42 \,{\mathrm e}^{3 i \left (d x +c \right )}+1196 \,{\mathrm e}^{7 i \left (d x +c \right )}+172 i {\mathrm e}^{6 i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 d a}\) \(231\)
parallelrisch \(\frac {\left (504 \cos \left (2 d x +2 c \right )+252 \cos \left (4 d x +4 c \right )+72 \cos \left (6 d x +6 c \right )+9 \cos \left (8 d x +8 c \right )+315\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-504 \cos \left (2 d x +2 c \right )-252 \cos \left (4 d x +4 c \right )-72 \cos \left (6 d x +6 c \right )-9 \cos \left (8 d x +8 c \right )-315\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+1998 \sin \left (3 d x +3 c \right )-138 \sin \left (5 d x +5 c \right )-18 \sin \left (7 d x +7 c \right )-3200 \cos \left (2 d x +2 c \right )+448 \cos \left (4 d x +4 c \right )+128 \cos \left (6 d x +6 c \right )+16 \cos \left (8 d x +8 c \right )-4026 \sin \left (d x +c \right )+2608}{384 a d \left (\cos \left (8 d x +8 c \right )+8 \cos \left (6 d x +6 c \right )+28 \cos \left (4 d x +4 c \right )+56 \cos \left (2 d x +2 c \right )+35\right )}\) \(260\)
norman \(\frac {\frac {111 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {111 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {3 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {3 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {277 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}+\frac {277 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {7 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {125 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {125 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}-\frac {43 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{128 a d}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{128 a d}\) \(315\)

input
int(sec(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d/a*(-1/96/(sin(d*x+c)-1)^3-1/128/(sin(d*x+c)-1)^2+3/128/(sin(d*x+c)-1)+ 
3/256*ln(sin(d*x+c)-1)+1/64/(1+sin(d*x+c))^4-1/48/(1+sin(d*x+c))^3-1/64/(1 
+sin(d*x+c))^2-3/256*ln(1+sin(d*x+c)))
 
3.9.86.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.11 \[ \int \frac {\sec ^4(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {18 \, \cos \left (d x + c\right )^{6} - 6 \, \cos \left (d x + c\right )^{4} - 156 \, \cos \left (d x + c\right )^{2} - 9 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 9 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (9 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) + 112}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

input
integrate(sec(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 
output
1/768*(18*cos(d*x + c)^6 - 6*cos(d*x + c)^4 - 156*cos(d*x + c)^2 - 9*(cos( 
d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 9*(cos(d 
*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(9*cos 
(d*x + c)^4 + 6*cos(d*x + c)^2 - 8)*sin(d*x + c) + 112)/(a*d*cos(d*x + c)^ 
6*sin(d*x + c) + a*d*cos(d*x + c)^6)
 
3.9.86.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**7*sin(d*x+c)**3/(a+a*sin(d*x+c)),x)
 
output
Timed out
 
3.9.86.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^4(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (9 \, \sin \left (d x + c\right )^{6} + 9 \, \sin \left (d x + c\right )^{5} - 24 \, \sin \left (d x + c\right )^{4} - 24 \, \sin \left (d x + c\right )^{3} - 57 \, \sin \left (d x + c\right )^{2} + 7 \, \sin \left (d x + c\right ) + 16\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {9 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {9 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \]

input
integrate(sec(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 
output
1/768*(2*(9*sin(d*x + c)^6 + 9*sin(d*x + c)^5 - 24*sin(d*x + c)^4 - 24*sin 
(d*x + c)^3 - 57*sin(d*x + c)^2 + 7*sin(d*x + c) + 16)/(a*sin(d*x + c)^7 + 
 a*sin(d*x + c)^6 - 3*a*sin(d*x + c)^5 - 3*a*sin(d*x + c)^4 + 3*a*sin(d*x 
+ c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a) - 9*log(sin(d*x + c) + 1 
)/a + 9*log(sin(d*x + c) - 1)/a)/d
 
3.9.86.8 Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^4(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {36 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {36 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (33 \, \sin \left (d x + c\right )^{3} - 135 \, \sin \left (d x + c\right )^{2} + 183 \, \sin \left (d x + c\right ) - 65\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {75 \, \sin \left (d x + c\right )^{4} + 300 \, \sin \left (d x + c\right )^{3} + 402 \, \sin \left (d x + c\right )^{2} + 140 \, \sin \left (d x + c\right ) + 11}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

input
integrate(sec(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
 
output
-1/3072*(36*log(abs(sin(d*x + c) + 1))/a - 36*log(abs(sin(d*x + c) - 1))/a 
 + 2*(33*sin(d*x + c)^3 - 135*sin(d*x + c)^2 + 183*sin(d*x + c) - 65)/(a*( 
sin(d*x + c) - 1)^3) - (75*sin(d*x + c)^4 + 300*sin(d*x + c)^3 + 402*sin(d 
*x + c)^2 + 140*sin(d*x + c) + 11)/(a*(sin(d*x + c) + 1)^4))/d
 
3.9.86.9 Mupad [B] (verification not implemented)

Time = 19.19 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.59 \[ \int \frac {\sec ^4(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{32}+\frac {111\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{32}+\frac {125\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}+\frac {277\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{48}-\frac {43\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}+\frac {277\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48}+\frac {125\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{64}+\frac {111\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a\,d} \]

input
int(sin(c + d*x)^3/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)
 
output
((3*tan(c/2 + (d*x)/2))/64 + (3*tan(c/2 + (d*x)/2)^2)/32 - (7*tan(c/2 + (d 
*x)/2)^3)/32 + (111*tan(c/2 + (d*x)/2)^4)/32 + (125*tan(c/2 + (d*x)/2)^5)/ 
64 + (277*tan(c/2 + (d*x)/2)^6)/48 - (43*tan(c/2 + (d*x)/2)^7)/48 + (277*t 
an(c/2 + (d*x)/2)^8)/48 + (125*tan(c/2 + (d*x)/2)^9)/64 + (111*tan(c/2 + ( 
d*x)/2)^10)/32 - (7*tan(c/2 + (d*x)/2)^11)/32 + (3*tan(c/2 + (d*x)/2)^12)/ 
32 + (3*tan(c/2 + (d*x)/2)^13)/64)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 5*a*ta 
n(c/2 + (d*x)/2)^2 - 12*a*tan(c/2 + (d*x)/2)^3 + 9*a*tan(c/2 + (d*x)/2)^4 
+ 30*a*tan(c/2 + (d*x)/2)^5 - 5*a*tan(c/2 + (d*x)/2)^6 - 40*a*tan(c/2 + (d 
*x)/2)^7 - 5*a*tan(c/2 + (d*x)/2)^8 + 30*a*tan(c/2 + (d*x)/2)^9 + 9*a*tan( 
c/2 + (d*x)/2)^10 - 12*a*tan(c/2 + (d*x)/2)^11 - 5*a*tan(c/2 + (d*x)/2)^12 
 + 2*a*tan(c/2 + (d*x)/2)^13 + a*tan(c/2 + (d*x)/2)^14)) - (3*atanh(tan(c/ 
2 + (d*x)/2)))/(64*a*d)