3.1.82 \(\int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [82]

3.1.82.1 Optimal result
3.1.82.2 Mathematica [A] (verified)
3.1.82.3 Rubi [A] (verified)
3.1.82.4 Maple [A] (verified)
3.1.82.5 Fricas [A] (verification not implemented)
3.1.82.6 Sympy [F]
3.1.82.7 Maxima [A] (verification not implemented)
3.1.82.8 Giac [A] (verification not implemented)
3.1.82.9 Mupad [B] (verification not implemented)

3.1.82.1 Optimal result

Integrand size = 21, antiderivative size = 195 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{128 a^4 d}+\frac {a^2}{48 d (a+a \sin (c+d x))^6}-\frac {7 a}{80 d (a+a \sin (c+d x))^5}+\frac {1}{8 d (a+a \sin (c+d x))^4}-\frac {5}{96 a d (a+a \sin (c+d x))^3}+\frac {1}{256 d \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {5}{256 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {3}{256 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {1}{256 d \left (a^4+a^4 \sin (c+d x)\right )} \]

output
-1/128*arctanh(sin(d*x+c))/a^4/d+1/48*a^2/d/(a+a*sin(d*x+c))^6-7/80*a/d/(a 
+a*sin(d*x+c))^5+1/8/d/(a+a*sin(d*x+c))^4-5/96/a/d/(a+a*sin(d*x+c))^3+1/25 
6/d/(a^2-a^2*sin(d*x+c))^2-5/256/d/(a^2+a^2*sin(d*x+c))^2-3/256/d/(a^4-a^4 
*sin(d*x+c))-1/256/d/(a^4+a^4*sin(d*x+c))
 
3.1.82.2 Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.57 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {30 \text {arctanh}(\sin (c+d x))-\frac {2 \left (-48-177 \sin (c+d x)-132 \sin ^2(c+d x)+257 \sin ^3(c+d x)+440 \sin ^4(c+d x)+65 \sin ^5(c+d x)+60 \sin ^6(c+d x)+15 \sin ^7(c+d x)\right )}{(-1+\sin (c+d x))^2 (1+\sin (c+d x))^6}}{3840 a^4 d} \]

input
Integrate[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^4,x]
 
output
-1/3840*(30*ArcTanh[Sin[c + d*x]] - (2*(-48 - 177*Sin[c + d*x] - 132*Sin[c 
 + d*x]^2 + 257*Sin[c + d*x]^3 + 440*Sin[c + d*x]^4 + 65*Sin[c + d*x]^5 + 
60*Sin[c + d*x]^6 + 15*Sin[c + d*x]^7))/((-1 + Sin[c + d*x])^2*(1 + Sin[c 
+ d*x])^6))/(a^4*d)
 
3.1.82.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 99, 2009}

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.

\(\displaystyle \int \frac {\tan ^5(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^5}{(a \sin (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 3186

\(\displaystyle \frac {\int \frac {a^5 \sin ^5(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^7}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (-\frac {a^2}{8 (\sin (c+d x) a+a)^7}+\frac {7 a}{16 (\sin (c+d x) a+a)^6}-\frac {1}{2 (\sin (c+d x) a+a)^5}+\frac {5}{32 (\sin (c+d x) a+a)^4 a}+\frac {1}{128 (a-a \sin (c+d x))^3 a^2}+\frac {5}{128 (\sin (c+d x) a+a)^3 a^2}-\frac {1}{128 \left (a^2-a^2 \sin ^2(c+d x)\right ) a^3}-\frac {3}{256 (a-a \sin (c+d x))^2 a^3}+\frac {1}{256 (\sin (c+d x) a+a)^2 a^3}\right )d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {\text {arctanh}(\sin (c+d x))}{128 a^4}-\frac {3}{256 a^3 (a-a \sin (c+d x))}-\frac {1}{256 a^3 (a \sin (c+d x)+a)}+\frac {a^2}{48 (a \sin (c+d x)+a)^6}+\frac {1}{256 a^2 (a-a \sin (c+d x))^2}-\frac {5}{256 a^2 (a \sin (c+d x)+a)^2}-\frac {7 a}{80 (a \sin (c+d x)+a)^5}+\frac {1}{8 (a \sin (c+d x)+a)^4}-\frac {5}{96 a (a \sin (c+d x)+a)^3}}{d}\)

input
Int[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^4,x]
 
output
(-1/128*ArcTanh[Sin[c + d*x]]/a^4 + 1/(256*a^2*(a - a*Sin[c + d*x])^2) - 3 
/(256*a^3*(a - a*Sin[c + d*x])) + a^2/(48*(a + a*Sin[c + d*x])^6) - (7*a)/ 
(80*(a + a*Sin[c + d*x])^5) + 1/(8*(a + a*Sin[c + d*x])^4) - 5/(96*a*(a + 
a*Sin[c + d*x])^3) - 5/(256*a^2*(a + a*Sin[c + d*x])^2) - 1/(256*a^3*(a + 
a*Sin[c + d*x])))/d
 

3.1.82.3.1 Defintions of rubi rules used

rule 99
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

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 3186
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) 
^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E 
qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
 
3.1.82.4 Maple [A] (verified)

Time = 8.04 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.65

method result size
derivativedivides \(\frac {\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {7}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{4}}\) \(127\)
default \(\frac {\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {7}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{4}}\) \(127\)
risch \(\frac {i \left (120 i {\mathrm e}^{2 i \left (d x +c \right )}-5727 \,{\mathrm e}^{5 i \left (d x +c \right )}-365 \,{\mathrm e}^{13 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}+5727 \,{\mathrm e}^{11 i \left (d x +c \right )}+4133 \,{\mathrm e}^{7 i \left (d x +c \right )}+365 \,{\mathrm e}^{3 i \left (d x +c \right )}-4133 \,{\mathrm e}^{9 i \left (d x +c \right )}+15 \,{\mathrm e}^{15 i \left (d x +c \right )}-4240 i {\mathrm e}^{4 i \left (d x +c \right )}+11656 i {\mathrm e}^{6 i \left (d x +c \right )}-8928 i {\mathrm e}^{8 i \left (d x +c \right )}+11656 i {\mathrm e}^{10 i \left (d x +c \right )}-4240 i {\mathrm e}^{12 i \left (d x +c \right )}+120 i {\mathrm e}^{14 i \left (d x +c \right )}\right )}{960 \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{12} d \,a^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a^{4} d}+\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{128 a^{4} d}\) \(254\)

input
int(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
 
output
1/d/a^4*(1/48/(1+sin(d*x+c))^6-7/80/(1+sin(d*x+c))^5+1/8/(1+sin(d*x+c))^4- 
5/96/(1+sin(d*x+c))^3-5/256/(1+sin(d*x+c))^2-1/256/(1+sin(d*x+c))-1/256*ln 
(1+sin(d*x+c))+1/256/(sin(d*x+c)-1)^2+3/256/(sin(d*x+c)-1)+1/256*ln(sin(d* 
x+c)-1))
 
3.1.82.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {120 \, \cos \left (d x + c\right )^{6} - 1240 \, \cos \left (d x + c\right )^{4} + 1856 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{8} - 8 \, \cos \left (d x + c\right )^{6} + 8 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{8} - 8 \, \cos \left (d x + c\right )^{6} + 8 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, \cos \left (d x + c\right )^{6} - 110 \, \cos \left (d x + c\right )^{4} + 432 \, \cos \left (d x + c\right )^{2} - 160\right )} \sin \left (d x + c\right ) - 640}{3840 \, {\left (a^{4} d \cos \left (d x + c\right )^{8} - 8 \, a^{4} d \cos \left (d x + c\right )^{6} + 8 \, a^{4} d \cos \left (d x + c\right )^{4} - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 2 \, a^{4} d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )}} \]

input
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="fricas")
 
output
-1/3840*(120*cos(d*x + c)^6 - 1240*cos(d*x + c)^4 + 1856*cos(d*x + c)^2 + 
15*(cos(d*x + c)^8 - 8*cos(d*x + c)^6 + 8*cos(d*x + c)^4 - 4*(cos(d*x + c) 
^6 - 2*cos(d*x + c)^4)*sin(d*x + c))*log(sin(d*x + c) + 1) - 15*(cos(d*x + 
 c)^8 - 8*cos(d*x + c)^6 + 8*cos(d*x + c)^4 - 4*(cos(d*x + c)^6 - 2*cos(d* 
x + c)^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 2*(15*cos(d*x + c)^6 - 11 
0*cos(d*x + c)^4 + 432*cos(d*x + c)^2 - 160)*sin(d*x + c) - 640)/(a^4*d*co 
s(d*x + c)^8 - 8*a^4*d*cos(d*x + c)^6 + 8*a^4*d*cos(d*x + c)^4 - 4*(a^4*d* 
cos(d*x + c)^6 - 2*a^4*d*cos(d*x + c)^4)*sin(d*x + c))
 
3.1.82.6 Sympy [F]

\[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\tan ^{5}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

input
integrate(tan(d*x+c)**5/(a+a*sin(d*x+c))**4,x)
 
output
Integral(tan(c + d*x)**5/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + 
d*x)**2 + 4*sin(c + d*x) + 1), x)/a**4
 
3.1.82.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{7} + 60 \, \sin \left (d x + c\right )^{6} + 65 \, \sin \left (d x + c\right )^{5} + 440 \, \sin \left (d x + c\right )^{4} + 257 \, \sin \left (d x + c\right )^{3} - 132 \, \sin \left (d x + c\right )^{2} - 177 \, \sin \left (d x + c\right ) - 48\right )}}{a^{4} \sin \left (d x + c\right )^{8} + 4 \, a^{4} \sin \left (d x + c\right )^{7} + 4 \, a^{4} \sin \left (d x + c\right )^{6} - 4 \, a^{4} \sin \left (d x + c\right )^{5} - 10 \, a^{4} \sin \left (d x + c\right )^{4} - 4 \, a^{4} \sin \left (d x + c\right )^{3} + 4 \, a^{4} \sin \left (d x + c\right )^{2} + 4 \, a^{4} \sin \left (d x + c\right ) + a^{4}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4}}}{3840 \, d} \]

input
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="maxima")
 
output
1/3840*(2*(15*sin(d*x + c)^7 + 60*sin(d*x + c)^6 + 65*sin(d*x + c)^5 + 440 
*sin(d*x + c)^4 + 257*sin(d*x + c)^3 - 132*sin(d*x + c)^2 - 177*sin(d*x + 
c) - 48)/(a^4*sin(d*x + c)^8 + 4*a^4*sin(d*x + c)^7 + 4*a^4*sin(d*x + c)^6 
 - 4*a^4*sin(d*x + c)^5 - 10*a^4*sin(d*x + c)^4 - 4*a^4*sin(d*x + c)^3 + 4 
*a^4*sin(d*x + c)^2 + 4*a^4*sin(d*x + c) + a^4) - 15*log(sin(d*x + c) + 1) 
/a^4 + 15*log(sin(d*x + c) - 1)/a^4)/d
 
3.1.82.8 Giac [A] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4}} + \frac {30 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) + 7\right )}}{a^{4} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {147 \, \sin \left (d x + c\right )^{6} + 822 \, \sin \left (d x + c\right )^{5} + 1605 \, \sin \left (d x + c\right )^{4} + 340 \, \sin \left (d x + c\right )^{3} - 675 \, \sin \left (d x + c\right )^{2} - 522 \, \sin \left (d x + c\right ) - 117}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{6}}}{15360 \, d} \]

input
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="giac")
 
output
-1/15360*(60*log(abs(sin(d*x + c) + 1))/a^4 - 60*log(abs(sin(d*x + c) - 1) 
)/a^4 + 30*(3*sin(d*x + c)^2 - 12*sin(d*x + c) + 7)/(a^4*(sin(d*x + c) - 1 
)^2) - (147*sin(d*x + c)^6 + 822*sin(d*x + c)^5 + 1605*sin(d*x + c)^4 + 34 
0*sin(d*x + c)^3 - 675*sin(d*x + c)^2 - 522*sin(d*x + c) - 117)/(a^4*(sin( 
d*x + c) + 1)^6))/d
 
3.1.82.9 Mupad [B] (verification not implemented)

Time = 12.14 (sec) , antiderivative size = 476, normalized size of antiderivative = 2.44 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{8}+\frac {73\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{192}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{12}-\frac {139\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{320}+\frac {1073\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{120}+\frac {10277\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{960}+\frac {237\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{10}+\frac {10277\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{960}+\frac {1073\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{120}-\frac {139\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{320}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}+\frac {73\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+8\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-36\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-120\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-88\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+88\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+198\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+88\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-88\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-120\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-36\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^4\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a^4\,d} \]

input
int(tan(c + d*x)^5/(a + a*sin(c + d*x))^4,x)
 
output
(tan(c/2 + (d*x)/2)/64 + tan(c/2 + (d*x)/2)^2/8 + (73*tan(c/2 + (d*x)/2)^3 
)/192 + (5*tan(c/2 + (d*x)/2)^4)/12 - (139*tan(c/2 + (d*x)/2)^5)/320 + (10 
73*tan(c/2 + (d*x)/2)^6)/120 + (10277*tan(c/2 + (d*x)/2)^7)/960 + (237*tan 
(c/2 + (d*x)/2)^8)/10 + (10277*tan(c/2 + (d*x)/2)^9)/960 + (1073*tan(c/2 + 
 (d*x)/2)^10)/120 - (139*tan(c/2 + (d*x)/2)^11)/320 + (5*tan(c/2 + (d*x)/2 
)^12)/12 + (73*tan(c/2 + (d*x)/2)^13)/192 + tan(c/2 + (d*x)/2)^14/8 + tan( 
c/2 + (d*x)/2)^15/64)/(d*(24*a^4*tan(c/2 + (d*x)/2)^2 + 24*a^4*tan(c/2 + ( 
d*x)/2)^3 - 36*a^4*tan(c/2 + (d*x)/2)^4 - 120*a^4*tan(c/2 + (d*x)/2)^5 - 8 
8*a^4*tan(c/2 + (d*x)/2)^6 + 88*a^4*tan(c/2 + (d*x)/2)^7 + 198*a^4*tan(c/2 
 + (d*x)/2)^8 + 88*a^4*tan(c/2 + (d*x)/2)^9 - 88*a^4*tan(c/2 + (d*x)/2)^10 
 - 120*a^4*tan(c/2 + (d*x)/2)^11 - 36*a^4*tan(c/2 + (d*x)/2)^12 + 24*a^4*t 
an(c/2 + (d*x)/2)^13 + 24*a^4*tan(c/2 + (d*x)/2)^14 + 8*a^4*tan(c/2 + (d*x 
)/2)^15 + a^4*tan(c/2 + (d*x)/2)^16 + a^4 + 8*a^4*tan(c/2 + (d*x)/2))) - a 
tanh(tan(c/2 + (d*x)/2))/(64*a^4*d)