\(\int \frac {\sin ^5(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [54]

Optimal result

Integrand size = 23, antiderivative size = 200 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\sqrt {b} \left (15 a^2+70 a b+63 b^2\right ) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{11/2} f}-\frac {\left (a^2+6 a b+6 b^2\right ) \cos (e+f x)}{a^5 f}+\frac {(2 a+3 b) \cos ^3(e+f x)}{3 a^4 f}-\frac {\cos ^5(e+f x)}{5 a^3 f}+\frac {b^2 (a+b)^2 \cos (e+f x)}{4 a^5 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {b (a+b) (9 a+17 b) \cos (e+f x)}{8 a^5 f \left (b+a \cos ^2(e+f x)\right )} \] Output:

1/8*b^(1/2)*(15*a^2+70*a*b+63*b^2)*arctan(a^(1/2)*cos(f*x+e)/b^(1/2))/a^(1 
1/2)/f-(a^2+6*a*b+6*b^2)*cos(f*x+e)/a^5/f+1/3*(2*a+3*b)*cos(f*x+e)^3/a^4/f 
-1/5*cos(f*x+e)^5/a^3/f+1/4*b^2*(a+b)^2*cos(f*x+e)/a^5/f/(b+a*cos(f*x+e)^2 
)^2-1/8*b*(a+b)*(9*a+17*b)*cos(f*x+e)/a^5/f/(b+a*cos(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 9.00 (sec) , antiderivative size = 1641, normalized size of antiderivative = 8.20 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(-900*a^(11/2)*b^(3/2)*Cos[ 
e + f*x] - 109000*a^(9/2)*b^(5/2)*Cos[e + f*x] - 936000*a^(7/2)*b^(7/2)*Co 
s[e + f*x] - 2803072*a^(5/2)*b^(9/2)*Cos[e + f*x] - 3763200*a^(3/2)*b^(11/ 
2)*Cos[e + f*x] - 1935360*Sqrt[a]*b^(13/2)*Cos[e + f*x] - 900*a^(11/2)*b^( 
3/2)*Cos[e + f*x]*Cos[2*(e + f*x)] + 900*a^(9/2)*b^(3/2)*Cos[e + f*x]*(a + 
 2*b + a*Cos[2*(e + f*x)]) + 24000*a^(7/2)*b^(5/2)*Cos[e + f*x]*(a + 2*b + 
 a*Cos[2*(e + f*x)]) + 43200*a^(5/2)*b^(7/2)*Cos[e + f*x]*(a + 2*b + a*Cos 
[2*(e + f*x)]) + 225*a^5*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - 
I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Co 
s[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)]) 
^2 + 115200*a^2*b^3*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin 
[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] 
- I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 
537600*a*b^4*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2] 
)*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin 
[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 483840* 
b^5*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]* 
Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*T 
an[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 225*a^5*ArcTan[( 
(-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)...
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4621, 366, 25, 360, 2341, 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.

\(\displaystyle \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^5}{\left (a+b \sec (e+f x)^2\right )^3}dx\)

\(\Big \downarrow \) 4621

\(\displaystyle -\frac {\int \frac {\cos ^6(e+f x) \left (1-\cos ^2(e+f x)\right )^2}{\left (a \cos ^2(e+f x)+b\right )^3}d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 366

\(\displaystyle -\frac {\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\int -\frac {\cos ^6(e+f x) \left (4 a^2+4 b \cos ^2(e+f x) a-7 (a+b)^2\right )}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{4 a^2 b}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\int \frac {\cos ^6(e+f x) \left (4 a^2+4 b \cos ^2(e+f x) a-7 (a+b)^2\right )}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{4 a^2 b}+\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 360

\(\displaystyle -\frac {\frac {\frac {b^2 (a+b) (3 a+11 b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\int \frac {-8 a^4 b \cos ^6(e+f x)+2 a^3 (a+b) (3 a+11 b) \cos ^4(e+f x)-2 a^2 b (a+b) (3 a+11 b) \cos ^2(e+f x)+a b^2 (a+b) (3 a+11 b)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a^4}}{4 a^2 b}+\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 2341

\(\displaystyle -\frac {\frac {\frac {b^2 (a+b) (3 a+11 b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\int \left (-8 a^3 b \cos ^4(e+f x)+2 a^2 (a+3 b) (3 a+5 b) \cos ^2(e+f x)-4 a b \left (3 a^2+14 b a+13 b^2\right )+\frac {63 a b^4+70 a^2 b^3+15 a^3 b^2}{a \cos ^2(e+f x)+b}\right )d\cos (e+f x)}{2 a^4}}{4 a^2 b}+\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}+\frac {\frac {b^2 (a+b) (3 a+11 b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {-\frac {8}{5} a^3 b \cos ^5(e+f x)+\sqrt {a} b^{3/2} \left (15 a^2+70 a b+63 b^2\right ) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )-4 a b \left (3 a^2+14 a b+13 b^2\right ) \cos (e+f x)+\frac {2}{3} a^2 (a+3 b) (3 a+5 b) \cos ^3(e+f x)}{2 a^4}}{4 a^2 b}}{f}\)

Input:

Int[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

-((((a + b)^2*Cos[e + f*x]^7)/(4*a^2*b*(b + a*Cos[e + f*x]^2)^2) + ((b^2*( 
a + b)*(3*a + 11*b)*Cos[e + f*x])/(2*a^3*(b + a*Cos[e + f*x]^2)) - (Sqrt[a 
]*b^(3/2)*(15*a^2 + 70*a*b + 63*b^2)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b] 
] - 4*a*b*(3*a^2 + 14*a*b + 13*b^2)*Cos[e + f*x] + (2*a^2*(a + 3*b)*(3*a + 
 5*b)*Cos[e + f*x]^3)/3 - (8*a^3*b*Cos[e + f*x]^5)/5)/(2*a^4))/(4*a^2*b))/ 
f)
 

Defintions of rubi rules used

rule 25

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

rule 360

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 366

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, 
x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* 
b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1))   Int[(e*x)^m*(a + b*x^2)^(p 
 + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 
2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p 
, -1]
 

rule 2009

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

rule 2341

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 3042

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

rule 4621

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
   Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), 
x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 
2] && IntegerQ[n] && IntegerQ[p]
 

Maple [A] (verified)

Time = 7.33 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\cos \left (f x +e \right )^{5} a^{2}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}-a \cos \left (f x +e \right )^{3} b +a^{2} \cos \left (f x +e \right )+6 a b \cos \left (f x +e \right )+6 b^{2} \cos \left (f x +e \right )}{a^{5}}+\frac {b \left (\frac {\left (-\frac {9}{8} a^{3}-\frac {13}{4} a^{2} b -\frac {17}{8} a \,b^{2}\right ) \cos \left (f x +e \right )^{3}-\frac {b \left (7 a^{2}+22 a b +15 b^{2}\right ) \cos \left (f x +e \right )}{8}}{\left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+70 a b +63 b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}}{f}\)	\(190\)
default	\(\frac {-\frac {\frac {\cos \left (f x +e \right )^{5} a^{2}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}-a \cos \left (f x +e \right )^{3} b +a^{2} \cos \left (f x +e \right )+6 a b \cos \left (f x +e \right )+6 b^{2} \cos \left (f x +e \right )}{a^{5}}+\frac {b \left (\frac {\left (-\frac {9}{8} a^{3}-\frac {13}{4} a^{2} b -\frac {17}{8} a \,b^{2}\right ) \cos \left (f x +e \right )^{3}-\frac {b \left (7 a^{2}+22 a b +15 b^{2}\right ) \cos \left (f x +e \right )}{8}}{\left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+70 a b +63 b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}}{f}\)	\(190\)
risch	\(-\frac {{\mathrm e}^{5 i \left (f x +e \right )}}{160 a^{3} f}+\frac {5 \,{\mathrm e}^{3 i \left (f x +e \right )}}{96 a^{3} f}+\frac {{\mathrm e}^{3 i \left (f x +e \right )} b}{8 a^{4} f}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )}}{16 a^{3} f}-\frac {21 \,{\mathrm e}^{i \left (f x +e \right )} b}{8 a^{4} f}-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )} b^{2}}{a^{5} f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )}}{16 a^{3} f}-\frac {21 \,{\mathrm e}^{-i \left (f x +e \right )} b}{8 a^{4} f}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{a^{5} f}+\frac {5 \,{\mathrm e}^{-3 i \left (f x +e \right )}}{96 a^{3} f}+\frac {{\mathrm e}^{-3 i \left (f x +e \right )} b}{8 a^{4} f}-\frac {{\mathrm e}^{-5 i \left (f x +e \right )}}{160 a^{3} f}-\frac {b \left (9 a^{3} {\mathrm e}^{7 i \left (f x +e \right )}+26 a^{2} b \,{\mathrm e}^{7 i \left (f x +e \right )}+17 a \,b^{2} {\mathrm e}^{7 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{5 i \left (f x +e \right )}+106 a^{2} b \,{\mathrm e}^{5 i \left (f x +e \right )}+139 a \,b^{2} {\mathrm e}^{5 i \left (f x +e \right )}+60 b^{3} {\mathrm e}^{5 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{3 i \left (f x +e \right )}+106 a^{2} b \,{\mathrm e}^{3 i \left (f x +e \right )}+139 a \,b^{2} {\mathrm e}^{3 i \left (f x +e \right )}+60 b^{3} {\mathrm e}^{3 i \left (f x +e \right )}+9 a^{3} {\mathrm e}^{i \left (f x +e \right )}+26 a^{2} b \,{\mathrm e}^{i \left (f x +e \right )}+17 a \,b^{2} {\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{5} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}+\frac {63 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b^{2}}{16 a^{6} f}-\frac {35 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{8 a^{5} f}-\frac {63 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b^{2}}{16 a^{6} f}+\frac {35 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{8 a^{5} f}-\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}+\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}\)	\(753\)

Input:

int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/a^5*(1/5*cos(f*x+e)^5*a^2-2/3*a^2*cos(f*x+e)^3-a*cos(f*x+e)^3*b+a^ 
2*cos(f*x+e)+6*a*b*cos(f*x+e)+6*b^2*cos(f*x+e))+b/a^5*(((-9/8*a^3-13/4*a^2 
*b-17/8*a*b^2)*cos(f*x+e)^3-1/8*b*(7*a^2+22*a*b+15*b^2)*cos(f*x+e))/(b+a*c 
os(f*x+e)^2)^2+1/8*(15*a^2+70*a*b+63*b^2)/(a*b)^(1/2)*arctan(a*cos(f*x+e)/ 
(a*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 579, normalized size of antiderivative = 2.90 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [-\frac {48 \, a^{4} \cos \left (f x + e\right )^{9} - 16 \, {\left (10 \, a^{4} + 9 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 16 \, {\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 50 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4}\right )} \cos \left (f x + e\right )}{240 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}}, -\frac {24 \, a^{4} \cos \left (f x + e\right )^{9} - 8 \, {\left (10 \, a^{4} + 9 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 8 \, {\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 25 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4}\right )} \cos \left (f x + e\right )}{120 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}}\right ] \] Input:

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

[-1/240*(48*a^4*cos(f*x + e)^9 - 16*(10*a^4 + 9*a^3*b)*cos(f*x + e)^7 + 16 
*(15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos(f*x + e)^5 + 50*(15*a^3*b + 70*a^2*b 
^2 + 63*a*b^3)*cos(f*x + e)^3 - 15*((15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos(f 
*x + e)^4 + 15*a^2*b^2 + 70*a*b^3 + 63*b^4 + 2*(15*a^3*b + 70*a^2*b^2 + 63 
*a*b^3)*cos(f*x + e)^2)*sqrt(-b/a)*log(-(a*cos(f*x + e)^2 + 2*a*sqrt(-b/a) 
*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) + 30*(15*a^2*b^2 + 70*a*b^3 + 6 
3*b^4)*cos(f*x + e))/(a^7*f*cos(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^ 
5*b^2*f), -1/120*(24*a^4*cos(f*x + e)^9 - 8*(10*a^4 + 9*a^3*b)*cos(f*x + e 
)^7 + 8*(15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos(f*x + e)^5 + 25*(15*a^3*b + 7 
0*a^2*b^2 + 63*a*b^3)*cos(f*x + e)^3 - 15*((15*a^4 + 70*a^3*b + 63*a^2*b^2 
)*cos(f*x + e)^4 + 15*a^2*b^2 + 70*a*b^3 + 63*b^4 + 2*(15*a^3*b + 70*a^2*b 
^2 + 63*a*b^3)*cos(f*x + e)^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b 
) + 15*(15*a^2*b^2 + 70*a*b^3 + 63*b^4)*cos(f*x + e))/(a^7*f*cos(f*x + e)^ 
4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**5/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, {\left ({\left (9 \, a^{3} b + 26 \, a^{2} b^{2} + 17 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (7 \, a^{2} b^{2} + 22 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (f x + e\right )\right )}}{a^{7} \cos \left (f x + e\right )^{4} + 2 \, a^{6} b \cos \left (f x + e\right )^{2} + a^{5} b^{2}} - \frac {15 \, {\left (15 \, a^{2} b + 70 \, a b^{2} + 63 \, b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} + \frac {8 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )\right )}}{a^{5}}}{120 \, f} \] Input:

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

-1/120*(15*((9*a^3*b + 26*a^2*b^2 + 17*a*b^3)*cos(f*x + e)^3 + (7*a^2*b^2 
+ 22*a*b^3 + 15*b^4)*cos(f*x + e))/(a^7*cos(f*x + e)^4 + 2*a^6*b*cos(f*x + 
 e)^2 + a^5*b^2) - 15*(15*a^2*b + 70*a*b^2 + 63*b^3)*arctan(a*cos(f*x + e) 
/sqrt(a*b))/(sqrt(a*b)*a^5) + 8*(3*a^2*cos(f*x + e)^5 - 5*(2*a^2 + 3*a*b)* 
cos(f*x + e)^3 + 15*(a^2 + 6*a*b + 6*b^2)*cos(f*x + e))/a^5)/f
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {{\left (15 \, a^{2} b + 70 \, a b^{2} + 63 \, b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5} f} - \frac {9 \, a^{3} b \cos \left (f x + e\right )^{3} + 26 \, a^{2} b^{2} \cos \left (f x + e\right )^{3} + 17 \, a b^{3} \cos \left (f x + e\right )^{3} + 7 \, a^{2} b^{2} \cos \left (f x + e\right ) + 22 \, a b^{3} \cos \left (f x + e\right ) + 15 \, b^{4} \cos \left (f x + e\right )}{8 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} a^{5} f} - \frac {3 \, a^{12} f^{4} \cos \left (f x + e\right )^{5} - 10 \, a^{12} f^{4} \cos \left (f x + e\right )^{3} - 15 \, a^{11} b f^{4} \cos \left (f x + e\right )^{3} + 15 \, a^{12} f^{4} \cos \left (f x + e\right ) + 90 \, a^{11} b f^{4} \cos \left (f x + e\right ) + 90 \, a^{10} b^{2} f^{4} \cos \left (f x + e\right )}{15 \, a^{15} f^{5}} \] Input:

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

1/8*(15*a^2*b + 70*a*b^2 + 63*b^3)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt( 
a*b)*a^5*f) - 1/8*(9*a^3*b*cos(f*x + e)^3 + 26*a^2*b^2*cos(f*x + e)^3 + 17 
*a*b^3*cos(f*x + e)^3 + 7*a^2*b^2*cos(f*x + e) + 22*a*b^3*cos(f*x + e) + 1 
5*b^4*cos(f*x + e))/((a*cos(f*x + e)^2 + b)^2*a^5*f) - 1/15*(3*a^12*f^4*co 
s(f*x + e)^5 - 10*a^12*f^4*cos(f*x + e)^3 - 15*a^11*b*f^4*cos(f*x + e)^3 + 
 15*a^12*f^4*cos(f*x + e) + 90*a^11*b*f^4*cos(f*x + e) + 90*a^10*b^2*f^4*c 
os(f*x + e))/(a^15*f^5)
 

Mupad [B] (verification not implemented)

Time = 12.51 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.28 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {b}{a^4}+\frac {2}{3\,a^3}\right )}{f}-\frac {\left (\frac {9\,a^3\,b}{8}+\frac {13\,a^2\,b^2}{4}+\frac {17\,a\,b^3}{8}\right )\,{\cos \left (e+f\,x\right )}^3+\left (\frac {7\,a^2\,b^2}{8}+\frac {11\,a\,b^3}{4}+\frac {15\,b^4}{8}\right )\,\cos \left (e+f\,x\right )}{f\,\left (a^7\,{\cos \left (e+f\,x\right )}^4+2\,a^6\,b\,{\cos \left (e+f\,x\right )}^2+a^5\,b^2\right )}-\frac {{\cos \left (e+f\,x\right )}^5}{5\,a^3\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {1}{a^3}-\frac {3\,b^2}{a^5}+\frac {3\,b\,\left (\frac {3\,b}{a^4}+\frac {2}{a^3}\right )}{a}\right )}{f}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,\left (15\,a^2+70\,a\,b+63\,b^2\right )}{15\,a^2\,b+70\,a\,b^2+63\,b^3}\right )\,\left (15\,a^2+70\,a\,b+63\,b^2\right )}{8\,a^{11/2}\,f} \] Input:

int(sin(e + f*x)^5/(a + b/cos(e + f*x)^2)^3,x)
                                                                                    
                                                                                    
 

Output:

(cos(e + f*x)^3*(b/a^4 + 2/(3*a^3)))/f - (cos(e + f*x)^3*((17*a*b^3)/8 + ( 
9*a^3*b)/8 + (13*a^2*b^2)/4) + cos(e + f*x)*((11*a*b^3)/4 + (15*b^4)/8 + ( 
7*a^2*b^2)/8))/(f*(a^5*b^2 + a^7*cos(e + f*x)^4 + 2*a^6*b*cos(e + f*x)^2)) 
 - cos(e + f*x)^5/(5*a^3*f) - (cos(e + f*x)*(1/a^3 - (3*b^2)/a^5 + (3*b*(( 
3*b)/a^4 + 2/a^3))/a))/f + (b^(1/2)*atan((a^(1/2)*b^(1/2)*cos(e + f*x)*(70 
*a*b + 15*a^2 + 63*b^2))/(70*a*b^2 + 15*a^2*b + 63*b^3))*(70*a*b + 15*a^2 
+ 63*b^2))/(8*a^(11/2)*f)
 

Reduce [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 1391, normalized size of antiderivative = 6.96 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

( - 225*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt 
(b))*sin(e + f*x)**4*a**4 - 1050*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e 
+ f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b - 945*sqrt(b)*sqrt(a) 
*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a* 
*2*b**2 + 450*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a) 
)/sqrt(b))*sin(e + f*x)**2*a**4 + 2550*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*t 
an((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3*b + 3990*sqrt(b)* 
sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x 
)**2*a**2*b**2 + 1890*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - 
 sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b**3 - 225*sqrt(b)*sqrt(a)*atan((sqrt 
(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**4 - 1500*sqrt(b)*sqrt(a)*a 
tan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**3*b - 3270*sqrt(b 
)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**2*b**2 
 - 2940*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt 
(b))*a*b**3 - 945*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqr 
t(a))/sqrt(b))*b**4 + 225*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/ 
2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**4 + 1050*sqrt(b)*sqrt(a)*atan((s 
qrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b + 9 
45*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))* 
sin(e + f*x)**4*a**2*b**2 - 450*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((...