3.9.45 \(\int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [845]

3.9.45.1 Optimal result
3.9.45.2 Mathematica [A] (verified)
3.9.45.3 Rubi [A] (verified)
3.9.45.4 Maple [C] (verified)
3.9.45.5 Fricas [A] (verification not implemented)
3.9.45.6 Sympy [F]
3.9.45.7 Maxima [B] (verification not implemented)
3.9.45.8 Giac [A] (verification not implemented)
3.9.45.9 Mupad [B] (verification not implemented)

3.9.45.1 Optimal result

Integrand size = 27, antiderivative size = 123 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec ^3(c+d x)}{9 d (a+a \sin (c+d x))^3}-\frac {\sec ^3(c+d x)}{21 a d (a+a \sin (c+d x))^2}-\frac {\sec ^3(c+d x)}{21 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {4 \tan (c+d x)}{21 a^3 d}+\frac {4 \tan ^3(c+d x)}{63 a^3 d} \]

output
1/9*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^3-1/21*sec(d*x+c)^3/a/d/(a+a*sin(d*x+c 
))^2-1/21*sec(d*x+c)^3/d/(a^3+a^3*sin(d*x+c))+4/21*tan(d*x+c)/a^3/d+4/63*t 
an(d*x+c)^3/a^3/d
 
3.9.45.2 Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {10752+900 \cos (c+d x)-6912 \cos (2 (c+d x))+50 \cos (3 (c+d x))-3072 \cos (4 (c+d x))-150 \cos (5 (c+d x))+256 \cos (6 (c+d x))+9216 \sin (c+d x)+675 \sin (2 (c+d x))+512 \sin (3 (c+d x))+300 \sin (4 (c+d x))-1536 \sin (5 (c+d x))-25 \sin (6 (c+d x))}{64512 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a+a \sin (c+d x))^3} \]

input
Integrate[(Sec[c + d*x]^3*Tan[c + d*x])/(a + a*Sin[c + d*x])^3,x]
 
output
(10752 + 900*Cos[c + d*x] - 6912*Cos[2*(c + d*x)] + 50*Cos[3*(c + d*x)] - 
3072*Cos[4*(c + d*x)] - 150*Cos[5*(c + d*x)] + 256*Cos[6*(c + d*x)] + 9216 
*Sin[c + d*x] + 675*Sin[2*(c + d*x)] + 512*Sin[3*(c + d*x)] + 300*Sin[4*(c 
 + d*x)] - 1536*Sin[5*(c + d*x)] - 25*Sin[6*(c + d*x)])/(64512*d*(Cos[(c + 
 d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a 
+ a*Sin[c + d*x])^3)
 
3.9.45.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3338, 3042, 3151, 3042, 3151, 3042, 4254, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3338

\(\displaystyle \frac {\int \frac {\sec ^4(c+d x)}{(\sin (c+d x) a+a)^2}dx}{3 a}+\frac {\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\cos (c+d x)^4 (\sin (c+d x) a+a)^2}dx}{3 a}+\frac {\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {\frac {5 \int \frac {\sec ^4(c+d x)}{\sin (c+d x) a+a}dx}{7 a}-\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}}{3 a}+\frac {\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \int \frac {1}{\cos (c+d x)^4 (\sin (c+d x) a+a)}dx}{7 a}-\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}}{3 a}+\frac {\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {\frac {5 \left (\frac {4 \int \sec ^4(c+d x)dx}{5 a}-\frac {\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)}\right )}{7 a}-\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}}{3 a}+\frac {\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \left (\frac {4 \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{5 a}-\frac {\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)}\right )}{7 a}-\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}}{3 a}+\frac {\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {5 \left (-\frac {4 \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{5 a d}-\frac {\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)}\right )}{7 a}-\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}}{3 a}+\frac {\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3}+\frac {\frac {5 \left (-\frac {4 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{5 a d}-\frac {\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)}\right )}{7 a}-\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}}{3 a}\)

input
Int[(Sec[c + d*x]^3*Tan[c + d*x])/(a + a*Sin[c + d*x])^3,x]
 
output
Sec[c + d*x]^3/(9*d*(a + a*Sin[c + d*x])^3) + (-1/7*Sec[c + d*x]^3/(d*(a + 
 a*Sin[c + d*x])^2) + (5*(-1/5*Sec[c + d*x]^3/(d*(a + a*Sin[c + d*x])) - ( 
4*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/(5*a*d)))/(7*a))/(3*a)
 

3.9.45.3.1 Defintions of rubi rules used

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 3151
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl 
ify[2*m + p + 1])   Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x] 
, x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simpli 
fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]
 

rule 3338
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - 
 a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) 
)), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1))   Int[(g*Cos[e 
+ f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 
]) && NeQ[2*m + p + 1, 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]
 
3.9.45.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.89

method result size
risch \(-\frac {16 i \left (2 i {\mathrm e}^{3 i \left (d x +c \right )}-12 \,{\mathrm e}^{2 i \left (d x +c \right )}-6 i {\mathrm e}^{i \left (d x +c \right )}-27 \,{\mathrm e}^{4 i \left (d x +c \right )}+1+36 i {\mathrm e}^{5 i \left (d x +c \right )}+42 \,{\mathrm e}^{6 i \left (d x +c \right )}\right )}{63 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{9} d \,a^{3}}\) \(109\)
parallelrisch \(\frac {-\frac {2}{63}+\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {256 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{63}-\frac {36 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {50 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{21}-6 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}\) \(152\)
derivativedivides \(\frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}+\frac {64}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {40}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {27}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {39}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {59}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {13}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{3}}\) \(190\)
default \(\frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}+\frac {64}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {40}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {27}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {39}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {59}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {13}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{3}}\) \(190\)
norman \(\frac {\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {2}{63 a d}-\frac {4 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {6 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {36 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7 d a}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{21 d a}+\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7 d a}-\frac {256 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{63 d a}-\frac {50 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}\) \(209\)

input
int(sec(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 
output
-16/63*I*(2*I*exp(3*I*(d*x+c))-12*exp(2*I*(d*x+c))-6*I*exp(I*(d*x+c))-27*e 
xp(4*I*(d*x+c))+1+36*I*exp(5*I*(d*x+c))+42*exp(6*I*(d*x+c)))/(exp(I*(d*x+c 
))-I)^3/(exp(I*(d*x+c))+I)^9/d/a^3
 
3.9.45.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {8 \, \cos \left (d x + c\right )^{6} - 36 \, \cos \left (d x + c\right )^{4} + 15 \, \cos \left (d x + c\right )^{2} - {\left (24 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 7\right )} \sin \left (d x + c\right ) + 14}{63 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

input
integrate(sec(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas" 
)
 
output
-1/63*(8*cos(d*x + c)^6 - 36*cos(d*x + c)^4 + 15*cos(d*x + c)^2 - (24*cos( 
d*x + c)^4 - 20*cos(d*x + c)^2 - 7)*sin(d*x + c) + 14)/(3*a^3*d*cos(d*x + 
c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + 
c)^3)*sin(d*x + c))
 
3.9.45.6 Sympy [F]

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

input
integrate(sec(d*x+c)**4*sin(d*x+c)/(a+a*sin(d*x+c))**3,x)
 
output
Integral(sin(c + d*x)*sec(c + d*x)**4/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 
 + 3*sin(c + d*x) + 1), x)/a**3
 
3.9.45.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (113) = 226\).

Time = 0.23 (sec) , antiderivative size = 442, normalized size of antiderivative = 3.59 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {75 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {128 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {162 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {42 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {189 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {126 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {63 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + 1\right )}}{63 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {36 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {27 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {12 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \]

input
integrate(sec(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima" 
)
 
output
2/63*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 75*sin(d*x + c)^2/(cos(d*x + c) 
+ 1)^2 + 128*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 162*sin(d*x + c)^4/(cos 
(d*x + c) + 1)^4 - 36*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 42*sin(d*x + c 
)^6/(cos(d*x + c) + 1)^6 + 189*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 126*s 
in(d*x + c)^9/(cos(d*x + c) + 1)^9 + 63*sin(d*x + c)^10/(cos(d*x + c) + 1) 
^10 + 1)/((a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*x + 
c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 27 
*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*a^3*sin(d*x + c)^5/(cos(d*x 
+ c) + 1)^5 + 36*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 27*a^3*sin(d*x 
+ c)^8/(cos(d*x + c) + 1)^8 - 2*a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 
12*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 6*a^3*sin(d*x + c)^11/(cos( 
d*x + c) + 1)^11 - a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*d)
 
3.9.45.8 Giac [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {21 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 756 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4200 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 11340 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 14994 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 13356 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6768 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2196 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 209}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{2016 \, d} \]

input
integrate(sec(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")
 
output
-1/2016*(21*(15*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 13)/(a^ 
3*(tan(1/2*d*x + 1/2*c) - 1)^3) - (315*tan(1/2*d*x + 1/2*c)^8 - 756*tan(1/ 
2*d*x + 1/2*c)^7 - 4200*tan(1/2*d*x + 1/2*c)^6 - 11340*tan(1/2*d*x + 1/2*c 
)^5 - 14994*tan(1/2*d*x + 1/2*c)^4 - 13356*tan(1/2*d*x + 1/2*c)^3 - 6768*t 
an(1/2*d*x + 1/2*c)^2 - 2196*tan(1/2*d*x + 1/2*c) - 209)/(a^3*(tan(1/2*d*x 
 + 1/2*c) + 1)^9))/d
 
3.9.45.9 Mupad [B] (verification not implemented)

Time = 16.00 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.27 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{63}+\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}+\frac {50\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{21}+\frac {256\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{63}+\frac {36\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}-\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{7}-\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^9} \]

input
int(sin(c + d*x)/(cos(c + d*x)^4*(a + a*sin(c + d*x))^3),x)
 
output
((2*cos(c/2 + (d*x)/2)^12)/63 + (4*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2 
))/21 + 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^10 + 4*cos(c/2 + (d*x)/2 
)^3*sin(c/2 + (d*x)/2)^9 + 6*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 - ( 
4*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^6)/3 - (8*cos(c/2 + (d*x)/2)^7*s 
in(c/2 + (d*x)/2)^5)/7 + (36*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4)/7 
+ (256*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^3)/63 + (50*cos(c/2 + (d*x) 
/2)^10*sin(c/2 + (d*x)/2)^2)/21)/(a^3*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d 
*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^9)