\(\int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx\) [884]

Optimal result

Integrand size = 29, antiderivative size = 152 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}+\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}-\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}+\frac {\tan ^6(c+d x)}{6 a d}+\frac {\tan ^8(c+d x)}{8 a d} \]

[Out]

5/128*arctanh(sin(d*x+c))/a/d+5/128*sec(d*x+c)*tan(d*x+c)/a/d-5/64*sec(d*x+c)^3*tan(d*x+c)/a/d+5/48*sec(d*x+c)
^3*tan(d*x+c)^3/a/d-1/8*sec(d*x+c)^3*tan(d*x+c)^5/a/d+1/6*tan(d*x+c)^6/a/d+1/8*tan(d*x+c)^8/a/d

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2914, 2687, 14, 2691, 3853, 3855} \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {\tan ^8(c+d x)}{8 a d}+\frac {\tan ^6(c+d x)}{6 a d}-\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 a d}+\frac {5 \tan ^3(c+d x) \sec ^3(c+d x)}{48 a d}-\frac {5 \tan (c+d x) \sec ^3(c+d x)}{64 a d}+\frac {5 \tan (c+d x) \sec (c+d x)}{128 a d} \]

[In]

Int[(Sec[c + d*x]^2*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

(5*ArcTanh[Sin[c + d*x]])/(128*a*d) + (5*Sec[c + d*x]*Tan[c + d*x])/(128*a*d) - (5*Sec[c + d*x]^3*Tan[c + d*x]
)/(64*a*d) + (5*Sec[c + d*x]^3*Tan[c + d*x]^3)/(48*a*d) - (Sec[c + d*x]^3*Tan[c + d*x]^5)/(8*a*d) + Tan[c + d*
x]^6/(6*a*d) + Tan[c + d*x]^8/(8*a*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 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 2691

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] - Dist[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] && Integers
Q[2*m, 2*n]

Rule 2914

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[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 3853

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] + Dist[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 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^4(c+d x) \tan ^5(c+d x) \, dx}{a}-\frac {\int \sec ^3(c+d x) \tan ^6(c+d x) \, dx}{a} \\ & = -\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}+\frac {5 \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx}{8 a}+\frac {\text {Subst}\left (\int x^5 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a d} \\ & = \frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}-\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}-\frac {5 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{16 a}+\frac {\text {Subst}\left (\int \left (x^5+x^7\right ) \, dx,x,\tan (c+d x)\right )}{a d} \\ & = -\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}+\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}-\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}+\frac {\tan ^6(c+d x)}{6 a d}+\frac {\tan ^8(c+d x)}{8 a d}+\frac {5 \int \sec ^3(c+d x) \, dx}{64 a} \\ & = \frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}+\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}-\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}+\frac {\tan ^6(c+d x)}{6 a d}+\frac {\tan ^8(c+d x)}{8 a d}+\frac {5 \int \sec (c+d x) \, dx}{128 a} \\ & = \frac {5 \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}+\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}-\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}+\frac {\tan ^6(c+d x)}{6 a d}+\frac {\tan ^8(c+d x)}{8 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.61 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15 \text {arctanh}(\sin (c+d x))-\frac {4}{(-1+\sin (c+d x))^3}-\frac {15}{(-1+\sin (c+d x))^2}-\frac {15}{-1+\sin (c+d x)}+\frac {6}{(1+\sin (c+d x))^4}-\frac {24}{(1+\sin (c+d x))^3}+\frac {30}{(1+\sin (c+d x))^2}}{384 a d} \]

[In]

Integrate[(Sec[c + d*x]^2*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

(15*ArcTanh[Sin[c + d*x]] - 4/(-1 + Sin[c + d*x])^3 - 15/(-1 + Sin[c + d*x])^2 - 15/(-1 + Sin[c + d*x]) + 6/(1
+ Sin[c + d*x])^4 - 24/(1 + Sin[c + d*x])^3 + 30/(1 + Sin[c + d*x])^2)/(384*a*d)

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.68

method	result	size
derivativedivides	\(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\)	\(103\)
default	\(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\)	\(103\)
risch	\(-\frac {i \left (625 \,{\mathrm e}^{5 i \left (d x +c \right )}+15 \,{\mathrm e}^{13 i \left (d x +c \right )}-442 \,{\mathrm e}^{11 i \left (d x +c \right )}+86 i {\mathrm e}^{4 i \left (d x +c \right )}-140 i {\mathrm e}^{6 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}+625 \,{\mathrm e}^{9 i \left (d x +c \right )}+30 i {\mathrm e}^{12 i \left (d x +c \right )}-442 \,{\mathrm e}^{3 i \left (d x +c \right )}+140 i {\mathrm e}^{8 i \left (d x +c \right )}-86 i {\mathrm e}^{10 i \left (d x +c \right )}-1420 \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 d a}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a d}\)	\(231\)
parallelrisch	\(\frac {\left (-840 \cos \left (2 d x +2 c \right )-420 \cos \left (4 d x +4 c \right )-120 \cos \left (6 d x +6 c \right )-15 \cos \left (8 d x +8 c \right )-525\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (840 \cos \left (2 d x +2 c \right )+420 \cos \left (4 d x +4 c \right )+120 \cos \left (6 d x +6 c \right )+15 \cos \left (8 d x +8 c \right )+525\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+1790 \sin \left (3 d x +3 c \right )-794 \sin \left (5 d x +5 c \right )+30 \sin \left (7 d x +7 c \right )-2944 \cos \left (2 d x +2 c \right )+1088 \cos \left (4 d x +4 c \right )-128 \cos \left (6 d x +6 c \right )-16 \cos \left (8 d x +8 c \right )-3530 \sin \left (d x +c \right )+2000}{384 a d \left (\cos \left (8 d x +8 c \right )+8 \cos \left (6 d x +6 c \right )+28 \cos \left (4 d x +4 c \right )+56 \cos \left (2 d x +2 c \right )+35\right )}\)	\(260\)
norman	\(\frac {-\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {5 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {5 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {413 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}+\frac {413 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}+\frac {85 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}+\frac {85 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}+\frac {35 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}+\frac {35 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}+\frac {157 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}-\frac {113 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}-\frac {113 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{128 a d}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{128 a d}\)	\(315\)

[In]

int(sec(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-1/96/(sin(d*x+c)-1)^3-5/128/(sin(d*x+c)-1)^2-5/128/(sin(d*x+c)-1)-5/256*ln(sin(d*x+c)-1)+1/64/(1+sin(d
*x+c))^4-1/16/(1+sin(d*x+c))^3+5/64/(1+sin(d*x+c))^2+5/256*ln(1+sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \cos \left (d x + c\right )^{6} - 266 \, \cos \left (d x + c\right )^{4} + 316 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 22 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 112}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/768*(30*cos(d*x + c)^6 - 266*cos(d*x + c)^4 + 316*cos(d*x + c)^2 - 15*(cos(d*x + c)^6*sin(d*x + c) + cos(d*
x + c)^6)*log(sin(d*x + c) + 1) + 15*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2
*(15*cos(d*x + c)^4 - 22*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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**7*sin(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{6} + 15 \, \sin \left (d x + c\right )^{5} + 88 \, \sin \left (d x + c\right )^{4} - 8 \, \sin \left (d x + c\right )^{3} - 63 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) + 16\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \]

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/768*(2*(15*sin(d*x + c)^6 + 15*sin(d*x + c)^5 + 88*sin(d*x + c)^4 - 8*sin(d*x + c)^3 - 63*sin(d*x + c)^2 +
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) - 15*log(sin(d*x + c) + 1)/a + 15*log(sin(d*x + c) - 1)/a)
/d

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (55 \, \sin \left (d x + c\right )^{3} - 225 \, \sin \left (d x + c\right )^{2} + 225 \, \sin \left (d x + c\right ) - 71\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 500 \, \sin \left (d x + c\right )^{3} + 510 \, \sin \left (d x + c\right )^{2} + 212 \, \sin \left (d x + c\right ) + 29}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/3072*(60*log(abs(sin(d*x + c) + 1))/a - 60*log(abs(sin(d*x + c) - 1))/a + 2*(55*sin(d*x + c)^3 - 225*sin(d*x
+ c)^2 + 225*sin(d*x + c) - 71)/(a*(sin(d*x + c) - 1)^3) - (125*sin(d*x + c)^4 + 500*sin(d*x + c)^3 + 510*sin
(d*x + c)^2 + 212*sin(d*x + c) + 29)/(a*(sin(d*x + c) + 1)^4))/d

Mupad [B] (verification not implemented)

Time = 20.71 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.55 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a\,d}+\frac {-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}+\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}-\frac {113\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {413\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{48}+\frac {157\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}+\frac {413\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48}-\frac {113\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}+\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{96}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}-\frac {5\,\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 )} \]

[In]

int(sin(c + d*x)^5/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)

[Out]

(5*atanh(tan(c/2 + (d*x)/2)))/(64*a*d) + ((35*tan(c/2 + (d*x)/2)^3)/96 - (5*tan(c/2 + (d*x)/2)^2)/32 - (5*tan(
c/2 + (d*x)/2))/64 + (85*tan(c/2 + (d*x)/2)^4)/96 - (113*tan(c/2 + (d*x)/2)^5)/192 + (413*tan(c/2 + (d*x)/2)^6
)/48 + (157*tan(c/2 + (d*x)/2)^7)/48 + (413*tan(c/2 + (d*x)/2)^8)/48 - (113*tan(c/2 + (d*x)/2)^9)/192 + (85*ta
n(c/2 + (d*x)/2)^10)/96 + (35*tan(c/2 + (d*x)/2)^11)/96 - (5*tan(c/2 + (d*x)/2)^12)/32 - (5*tan(c/2 + (d*x)/2)
^13)/64)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 5*a*tan(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))