3.3.88 \(\int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx\) [288]

3.3.88.1 Optimal result
3.3.88.2 Mathematica [A] (verified)
3.3.88.3 Rubi [A] (verified)
3.3.88.4 Maple [C] (verified)
3.3.88.5 Fricas [A] (verification not implemented)
3.3.88.6 Sympy [B] (verification not implemented)
3.3.88.7 Maxima [B] (verification not implemented)
3.3.88.8 Giac [A] (verification not implemented)
3.3.88.9 Mupad [B] (verification not implemented)

3.3.88.1 Optimal result

Integrand size = 26, antiderivative size = 131 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx=\frac {\sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac {\sec ^5(e+f x)}{9 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac {2 \tan (e+f x)}{3 a^3 c^5 f}+\frac {4 \tan ^3(e+f x)}{9 a^3 c^5 f}+\frac {2 \tan ^5(e+f x)}{15 a^3 c^5 f} \]

output
1/9*sec(f*x+e)^5/a^3/c^3/f/(c-c*sin(f*x+e))^2+1/9*sec(f*x+e)^5/a^3/f/(c^5- 
c^5*sin(f*x+e))+2/3*tan(f*x+e)/a^3/c^5/f+4/9*tan(f*x+e)^3/a^3/c^5/f+2/15*t 
an(f*x+e)^5/a^3/c^5/f
 
3.3.88.2 Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.63 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (19843425 \cos (e+f x)+1310720 \cos (2 (e+f x))+8378335 \cos (3 (e+f x))+1048576 \cos (4 (e+f x))+440965 \cos (5 (e+f x))+262144 \cos (6 (e+f x))-440965 \cos (7 (e+f x))+2949120 \sin (e+f x)-8819300 \sin (2 (e+f x))+1245184 \sin (3 (e+f x))-7055440 \sin (4 (e+f x))+65536 \sin (5 (e+f x))-1763860 \sin (6 (e+f x))-65536 \sin (7 (e+f x)))}{11796480 a^3 c^5 f (-1+\sin (e+f x))^5 (1+\sin (e+f x))^3} \]

input
Integrate[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^5),x]
 
output
-1/11796480*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin 
[(e + f*x)/2])*(19843425*Cos[e + f*x] + 1310720*Cos[2*(e + f*x)] + 8378335 
*Cos[3*(e + f*x)] + 1048576*Cos[4*(e + f*x)] + 440965*Cos[5*(e + f*x)] + 2 
62144*Cos[6*(e + f*x)] - 440965*Cos[7*(e + f*x)] + 2949120*Sin[e + f*x] - 
8819300*Sin[2*(e + f*x)] + 1245184*Sin[3*(e + f*x)] - 7055440*Sin[4*(e + f 
*x)] + 65536*Sin[5*(e + f*x)] - 1763860*Sin[6*(e + f*x)] - 65536*Sin[7*(e 
+ f*x)]))/(a^3*c^5*f*(-1 + Sin[e + f*x])^5*(1 + Sin[e + f*x])^3)
 
3.3.88.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.88, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3215, 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.

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^5} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^5}dx\)

\(\Big \downarrow \) 3215

\(\displaystyle \frac {\int \frac {\sec ^6(e+f x)}{(c-c \sin (e+f x))^2}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\cos (e+f x)^6 (c-c \sin (e+f x))^2}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {\frac {7 \int \frac {\sec ^6(e+f x)}{c-c \sin (e+f x)}dx}{9 c}+\frac {\sec ^5(e+f x)}{9 f (c-c \sin (e+f x))^2}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7 \int \frac {1}{\cos (e+f x)^6 (c-c \sin (e+f x))}dx}{9 c}+\frac {\sec ^5(e+f x)}{9 f (c-c \sin (e+f x))^2}}{a^3 c^3}\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {\frac {7 \left (\frac {6 \int \sec ^6(e+f x)dx}{7 c}+\frac {\sec ^5(e+f x)}{7 f (c-c \sin (e+f x))}\right )}{9 c}+\frac {\sec ^5(e+f x)}{9 f (c-c \sin (e+f x))^2}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7 \left (\frac {6 \int \csc \left (e+f x+\frac {\pi }{2}\right )^6dx}{7 c}+\frac {\sec ^5(e+f x)}{7 f (c-c \sin (e+f x))}\right )}{9 c}+\frac {\sec ^5(e+f x)}{9 f (c-c \sin (e+f x))^2}}{a^3 c^3}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {7 \left (\frac {\sec ^5(e+f x)}{7 f (c-c \sin (e+f x))}-\frac {6 \int \left (\tan ^4(e+f x)+2 \tan ^2(e+f x)+1\right )d(-\tan (e+f x))}{7 c f}\right )}{9 c}+\frac {\sec ^5(e+f x)}{9 f (c-c \sin (e+f x))^2}}{a^3 c^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\sec ^5(e+f x)}{9 f (c-c \sin (e+f x))^2}+\frac {7 \left (\frac {\sec ^5(e+f x)}{7 f (c-c \sin (e+f x))}-\frac {6 \left (-\frac {1}{5} \tan ^5(e+f x)-\frac {2}{3} \tan ^3(e+f x)-\tan (e+f x)\right )}{7 c f}\right )}{9 c}}{a^3 c^3}\)

input
Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^5),x]
 
output
(Sec[e + f*x]^5/(9*f*(c - c*Sin[e + f*x])^2) + (7*(Sec[e + f*x]^5/(7*f*(c 
- c*Sin[e + f*x])) - (6*(-Tan[e + f*x] - (2*Tan[e + f*x]^3)/3 - Tan[e + f* 
x]^5/5))/(7*c*f)))/(9*c))/(a^3*c^3)
 

3.3.88.3.1 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 3215
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + 
 d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 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]
 
3.3.88.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.84

method result size
risch \(-\frac {32 i \left (-20 i {\mathrm e}^{5 i \left (f x +e \right )}+45 \,{\mathrm e}^{6 i \left (f x +e \right )}-16 i {\mathrm e}^{3 i \left (f x +e \right )}+19 \,{\mathrm e}^{4 i \left (f x +e \right )}-4 i {\mathrm e}^{i \left (f x +e \right )}+{\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{45 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{9} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} f \,c^{5} a^{3}}\) \(110\)
parallelrisch \(\frac {-\frac {4}{9}-\frac {32 \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\frac {538 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{45}+4 \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {4 \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\frac {46 \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15}+\frac {236 \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15}-\frac {20 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{9}-\frac {64 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15}-\frac {344 \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15}-\frac {76 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{9}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{9}+\frac {32 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{9}-2 \left (\tan ^{13}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f \,c^{5} a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}\) \(207\)
derivativedivides \(\frac {-\frac {4}{9 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}-\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{8}}-\frac {5}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {49}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {49}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {35}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {49}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {51}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {99}{64 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {1}{20 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {13}{48 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {9}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {29}{64 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} c^{5} f}\) \(223\)
default \(\frac {-\frac {4}{9 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}-\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{8}}-\frac {5}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {49}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {49}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {35}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {49}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {51}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {99}{64 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {1}{20 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {13}{48 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {9}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {29}{64 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} c^{5} f}\) \(223\)
norman \(\frac {\frac {4 \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {344 \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 a c f}-\frac {4}{9 a c f}-\frac {2 \left (\tan ^{13}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}+\frac {32 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{9 a c f}-\frac {4 \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{9 a c f}-\frac {76 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{9 a c f}-\frac {32 \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}+\frac {46 \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 a c f}-\frac {20 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{9 a c f}+\frac {538 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{45 a c f}-\frac {64 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 a c f}+\frac {236 \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 a c f}}{a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}\) \(330\)

input
int(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x,method=_RETURNVERBOSE)
 
output
-32/45*I*(-20*I*exp(5*I*(f*x+e))+45*exp(6*I*(f*x+e))-16*I*exp(3*I*(f*x+e)) 
+19*exp(4*I*(f*x+e))-4*I*exp(I*(f*x+e))+exp(2*I*(f*x+e))-1)/(exp(I*(f*x+e) 
)-I)^9/(exp(I*(f*x+e))+I)^5/f/c^5/a^3
 
3.3.88.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx=-\frac {32 \, \cos \left (f x + e\right )^{6} - 16 \, \cos \left (f x + e\right )^{4} - 4 \, \cos \left (f x + e\right )^{2} - {\left (16 \, \cos \left (f x + e\right )^{6} - 24 \, \cos \left (f x + e\right )^{4} - 10 \, \cos \left (f x + e\right )^{2} - 7\right )} \sin \left (f x + e\right ) - 2}{45 \, {\left (a^{3} c^{5} f \cos \left (f x + e\right )^{7} + 2 \, a^{3} c^{5} f \cos \left (f x + e\right )^{5} \sin \left (f x + e\right ) - 2 \, a^{3} c^{5} f \cos \left (f x + e\right )^{5}\right )}} \]

input
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="fricas")
 
output
-1/45*(32*cos(f*x + e)^6 - 16*cos(f*x + e)^4 - 4*cos(f*x + e)^2 - (16*cos( 
f*x + e)^6 - 24*cos(f*x + e)^4 - 10*cos(f*x + e)^2 - 7)*sin(f*x + e) - 2)/ 
(a^3*c^5*f*cos(f*x + e)^7 + 2*a^3*c^5*f*cos(f*x + e)^5*sin(f*x + e) - 2*a^ 
3*c^5*f*cos(f*x + e)^5)
 
3.3.88.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4335 vs. \(2 (117) = 234\).

Time = 36.76 (sec) , antiderivative size = 4335, normalized size of antiderivative = 33.09 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx=\text {Too large to display} \]

input
integrate(1/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**5,x)
 
output
Piecewise((-90*tan(e/2 + f*x/2)**13/(45*a**3*c**5*f*tan(e/2 + f*x/2)**14 - 
 180*a**3*c**5*f*tan(e/2 + f*x/2)**13 + 45*a**3*c**5*f*tan(e/2 + f*x/2)**1 
2 + 720*a**3*c**5*f*tan(e/2 + f*x/2)**11 - 855*a**3*c**5*f*tan(e/2 + f*x/2 
)**10 - 900*a**3*c**5*f*tan(e/2 + f*x/2)**9 + 2025*a**3*c**5*f*tan(e/2 + f 
*x/2)**8 - 2025*a**3*c**5*f*tan(e/2 + f*x/2)**6 + 900*a**3*c**5*f*tan(e/2 
+ f*x/2)**5 + 855*a**3*c**5*f*tan(e/2 + f*x/2)**4 - 720*a**3*c**5*f*tan(e/ 
2 + f*x/2)**3 - 45*a**3*c**5*f*tan(e/2 + f*x/2)**2 + 180*a**3*c**5*f*tan(e 
/2 + f*x/2) - 45*a**3*c**5*f) + 180*tan(e/2 + f*x/2)**12/(45*a**3*c**5*f*t 
an(e/2 + f*x/2)**14 - 180*a**3*c**5*f*tan(e/2 + f*x/2)**13 + 45*a**3*c**5* 
f*tan(e/2 + f*x/2)**12 + 720*a**3*c**5*f*tan(e/2 + f*x/2)**11 - 855*a**3*c 
**5*f*tan(e/2 + f*x/2)**10 - 900*a**3*c**5*f*tan(e/2 + f*x/2)**9 + 2025*a* 
*3*c**5*f*tan(e/2 + f*x/2)**8 - 2025*a**3*c**5*f*tan(e/2 + f*x/2)**6 + 900 
*a**3*c**5*f*tan(e/2 + f*x/2)**5 + 855*a**3*c**5*f*tan(e/2 + f*x/2)**4 - 7 
20*a**3*c**5*f*tan(e/2 + f*x/2)**3 - 45*a**3*c**5*f*tan(e/2 + f*x/2)**2 + 
180*a**3*c**5*f*tan(e/2 + f*x/2) - 45*a**3*c**5*f) - 60*tan(e/2 + f*x/2)** 
11/(45*a**3*c**5*f*tan(e/2 + f*x/2)**14 - 180*a**3*c**5*f*tan(e/2 + f*x/2) 
**13 + 45*a**3*c**5*f*tan(e/2 + f*x/2)**12 + 720*a**3*c**5*f*tan(e/2 + f*x 
/2)**11 - 855*a**3*c**5*f*tan(e/2 + f*x/2)**10 - 900*a**3*c**5*f*tan(e/2 + 
 f*x/2)**9 + 2025*a**3*c**5*f*tan(e/2 + f*x/2)**8 - 2025*a**3*c**5*f*tan(e 
/2 + f*x/2)**6 + 900*a**3*c**5*f*tan(e/2 + f*x/2)**5 + 855*a**3*c**5*f*...
 
3.3.88.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 610 vs. \(2 (123) = 246\).

Time = 0.21 (sec) , antiderivative size = 610, normalized size of antiderivative = 4.66 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx=\frac {2 \, {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {80 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {190 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {50 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {269 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {96 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {516 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {354 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - \frac {69 \, \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac {240 \, \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}} + \frac {30 \, \sin \left (f x + e\right )^{11}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{11}} - \frac {90 \, \sin \left (f x + e\right )^{12}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{12}} + \frac {45 \, \sin \left (f x + e\right )^{13}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{13}} + 10\right )}}{45 \, {\left (a^{3} c^{5} - \frac {4 \, a^{3} c^{5} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {a^{3} c^{5} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {16 \, a^{3} c^{5} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {19 \, a^{3} c^{5} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {20 \, a^{3} c^{5} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {45 \, a^{3} c^{5} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {45 \, a^{3} c^{5} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac {20 \, a^{3} c^{5} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac {19 \, a^{3} c^{5} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}} - \frac {16 \, a^{3} c^{5} \sin \left (f x + e\right )^{11}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{11}} - \frac {a^{3} c^{5} \sin \left (f x + e\right )^{12}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{12}} + \frac {4 \, a^{3} c^{5} \sin \left (f x + e\right )^{13}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{13}} - \frac {a^{3} c^{5} \sin \left (f x + e\right )^{14}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{14}}\right )} f} \]

input
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="maxima")
 
output
2/45*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 80*sin(f*x + e)^2/(cos(f*x + e) 
+ 1)^2 + 190*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 50*sin(f*x + e)^4/(cos( 
f*x + e) + 1)^4 - 269*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 96*sin(f*x + e 
)^6/(cos(f*x + e) + 1)^6 + 516*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 354*s 
in(f*x + e)^8/(cos(f*x + e) + 1)^8 - 69*sin(f*x + e)^9/(cos(f*x + e) + 1)^ 
9 + 240*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 30*sin(f*x + e)^11/(cos(f* 
x + e) + 1)^11 - 90*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 45*sin(f*x + e 
)^13/(cos(f*x + e) + 1)^13 + 10)/((a^3*c^5 - 4*a^3*c^5*sin(f*x + e)/(cos(f 
*x + e) + 1) + a^3*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 16*a^3*c^5*si 
n(f*x + e)^3/(cos(f*x + e) + 1)^3 - 19*a^3*c^5*sin(f*x + e)^4/(cos(f*x + e 
) + 1)^4 - 20*a^3*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 45*a^3*c^5*sin 
(f*x + e)^6/(cos(f*x + e) + 1)^6 - 45*a^3*c^5*sin(f*x + e)^8/(cos(f*x + e) 
 + 1)^8 + 20*a^3*c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 19*a^3*c^5*sin( 
f*x + e)^10/(cos(f*x + e) + 1)^10 - 16*a^3*c^5*sin(f*x + e)^11/(cos(f*x + 
e) + 1)^11 - a^3*c^5*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 4*a^3*c^5*sin 
(f*x + e)^13/(cos(f*x + e) + 1)^13 - a^3*c^5*sin(f*x + e)^14/(cos(f*x + e) 
 + 1)^14)*f)
 
3.3.88.8 Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.55 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx=-\frac {\frac {3 \, {\left (435 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 1470 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2060 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1330 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 353\right )}}{a^{3} c^{5} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} + \frac {4455 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 26460 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 78120 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 137340 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 157374 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 118356 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 57744 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 16596 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2339}{a^{3} c^{5} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{9}}}{2880 \, f} \]

input
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="giac")
 
output
-1/2880*(3*(435*tan(1/2*f*x + 1/2*e)^4 + 1470*tan(1/2*f*x + 1/2*e)^3 + 206 
0*tan(1/2*f*x + 1/2*e)^2 + 1330*tan(1/2*f*x + 1/2*e) + 353)/(a^3*c^5*(tan( 
1/2*f*x + 1/2*e) + 1)^5) + (4455*tan(1/2*f*x + 1/2*e)^8 - 26460*tan(1/2*f* 
x + 1/2*e)^7 + 78120*tan(1/2*f*x + 1/2*e)^6 - 137340*tan(1/2*f*x + 1/2*e)^ 
5 + 157374*tan(1/2*f*x + 1/2*e)^4 - 118356*tan(1/2*f*x + 1/2*e)^3 + 57744* 
tan(1/2*f*x + 1/2*e)^2 - 16596*tan(1/2*f*x + 1/2*e) + 2339)/(a^3*c^5*(tan( 
1/2*f*x + 1/2*e) - 1)^9))/f
 
3.3.88.9 Mupad [B] (verification not implemented)

Time = 8.23 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx=-\frac {\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {65\,\cos \left (\frac {5\,e}{2}+\frac {5\,f\,x}{2}\right )}{32}-\frac {225\,\cos \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}{32}-5\,\cos \left (\frac {7\,e}{2}+\frac {7\,f\,x}{2}\right )+\cos \left (\frac {9\,e}{2}+\frac {9\,f\,x}{2}\right )-\frac {37\,\cos \left (\frac {11\,e}{2}+\frac {11\,f\,x}{2}\right )}{32}+\frac {5\,\cos \left (\frac {13\,e}{2}+\frac {13\,f\,x}{2}\right )}{32}-\frac {89\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}+11\,\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )-\frac {63\,\sin \left (\frac {5\,e}{2}+\frac {5\,f\,x}{2}\right )}{8}+\frac {25\,\sin \left (\frac {7\,e}{2}+\frac {7\,f\,x}{2}\right )}{8}-\frac {5\,\sin \left (\frac {9\,e}{2}+\frac {9\,f\,x}{2}\right )}{8}+\frac {3\,\sin \left (\frac {11\,e}{2}+\frac {11\,f\,x}{2}\right )}{8}+\frac {\sin \left (\frac {13\,e}{2}+\frac {13\,f\,x}{2}\right )}{4}\right )}{2880\,a^3\,c^5\,f\,{\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}^5\,{\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}^9} \]

input
int(1/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^5),x)
 
output
-(cos(e/2 + (f*x)/2)*((65*cos((5*e)/2 + (5*f*x)/2))/32 - (225*cos((3*e)/2 
+ (3*f*x)/2))/32 - 5*cos((7*e)/2 + (7*f*x)/2) + cos((9*e)/2 + (9*f*x)/2) - 
 (37*cos((11*e)/2 + (11*f*x)/2))/32 + (5*cos((13*e)/2 + (13*f*x)/2))/32 - 
(89*sin(e/2 + (f*x)/2))/4 + 11*sin((3*e)/2 + (3*f*x)/2) - (63*sin((5*e)/2 
+ (5*f*x)/2))/8 + (25*sin((7*e)/2 + (7*f*x)/2))/8 - (5*sin((9*e)/2 + (9*f* 
x)/2))/8 + (3*sin((11*e)/2 + (11*f*x)/2))/8 + sin((13*e)/2 + (13*f*x)/2)/4 
))/(2880*a^3*c^5*f*cos(e/2 - pi/4 + (f*x)/2)^5*cos(e/2 + pi/4 + (f*x)/2)^9 
)