\(\int \frac {\sin ^5(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [119]

Optimal result

Integrand size = 25, antiderivative size = 204 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {b (a+b)^2 \sec (e+f x)}{3 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {b (a+b) (5 a+11 b) \sec (e+f x)}{3 a^5 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\left (15 a^2+80 a b+73 b^2\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{15 a^5 f}+\frac {2 (5 a+7 b) \cos ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{15 a^4 f}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{5 a^3 f} \] Output:

-1/3*b*(a+b)^2*sec(f*x+e)/a^4/f/(a+b*sec(f*x+e)^2)^(3/2)-1/3*b*(a+b)*(5*a+ 
11*b)*sec(f*x+e)/a^5/f/(a+b*sec(f*x+e)^2)^(1/2)-1/15*(15*a^2+80*a*b+73*b^2 
)*cos(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/a^5/f+2/15*(5*a+7*b)*cos(f*x+e)^3*(a 
+b*sec(f*x+e)^2)^(1/2)/a^4/f-1/5*cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2)/a^3 
/f
 

Mathematica [A] (verified)

Time = 3.06 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (425 a^4+6400 a^3 b+22784 a^2 b^2+32768 a b^3+16384 b^4+48 a \left (11 a^3+150 a^2 b+384 a b^2+256 b^3\right ) \cos (2 (e+f x))+12 a^2 \left (7 a^2+64 a b+64 b^2\right ) \cos (4 (e+f x))-16 a^4 \cos (6 (e+f x))-32 a^3 b \cos (6 (e+f x))+3 a^4 \cos (8 (e+f x))\right ) \sec ^5(e+f x)}{3840 a^5 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

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

Output:

-1/3840*((a + 2*b + a*Cos[2*(e + f*x)])*(425*a^4 + 6400*a^3*b + 22784*a^2* 
b^2 + 32768*a*b^3 + 16384*b^4 + 48*a*(11*a^3 + 150*a^2*b + 384*a*b^2 + 256 
*b^3)*Cos[2*(e + f*x)] + 12*a^2*(7*a^2 + 64*a*b + 64*b^2)*Cos[4*(e + f*x)] 
 - 16*a^4*Cos[6*(e + f*x)] - 32*a^3*b*Cos[6*(e + f*x)] + 3*a^4*Cos[8*(e + 
f*x)])*Sec[e + f*x]^5)/(a^5*f*(a + b*Sec[e + f*x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4622, 365, 25, 359, 245, 209, 208}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

\(\displaystyle \frac {\int \frac {\cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 365

\(\displaystyle \frac {\frac {\int -\frac {\cos ^4(e+f x) \left (2 (5 a+4 b)-5 a \sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{5 a}-\frac {\cos ^5(e+f x)}{5 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \frac {\cos ^4(e+f x) \left (2 (5 a+4 b)-5 a \sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{5 a}-\frac {\cos ^5(e+f x)}{5 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 359

\(\displaystyle \frac {-\frac {-\frac {\left (5 a^2+20 a b+16 b^2\right ) \int \frac {\cos ^2(e+f x)}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{a}-\frac {2 (5 a+4 b) \cos ^3(e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{5 a}-\frac {\cos ^5(e+f x)}{5 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 245

\(\displaystyle \frac {-\frac {-\frac {\left (5 a^2+20 a b+16 b^2\right ) \left (-\frac {4 b \int \frac {1}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {2 (5 a+4 b) \cos ^3(e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{5 a}-\frac {\cos ^5(e+f x)}{5 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 209

\(\displaystyle \frac {-\frac {-\frac {\left (5 a^2+20 a b+16 b^2\right ) \left (-\frac {4 b \left (\frac {2 \int \frac {1}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{3 a}+\frac {\sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {2 (5 a+4 b) \cos ^3(e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{5 a}-\frac {\cos ^5(e+f x)}{5 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 208

\(\displaystyle \frac {-\frac {-\frac {\left (5 a^2+20 a b+16 b^2\right ) \left (-\frac {4 b \left (\frac {2 \sec (e+f x)}{3 a^2 \sqrt {a+b \sec ^2(e+f x)}}+\frac {\sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {2 (5 a+4 b) \cos ^3(e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{5 a}-\frac {\cos ^5(e+f x)}{5 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

Input:

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

Output:

(-1/5*Cos[e + f*x]^5/(a*(a + b*Sec[e + f*x]^2)^(3/2)) - ((-2*(5*a + 4*b)*C 
os[e + f*x]^3)/(3*a*(a + b*Sec[e + f*x]^2)^(3/2)) - ((5*a^2 + 20*a*b + 16* 
b^2)*(-(Cos[e + f*x]/(a*(a + b*Sec[e + f*x]^2)^(3/2))) - (4*b*(Sec[e + f*x 
]/(3*a*(a + b*Sec[e + f*x]^2)^(3/2)) + (2*Sec[e + f*x])/(3*a^2*Sqrt[a + b* 
Sec[e + f*x]^2])))/a))/a)/(5*a))/f
 

Defintions of rubi rules used

rule 25

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

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 209

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
 

rule 245

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + 
 b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) 
   Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si 
mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
 

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 365

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

rule 3042

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

rule 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
 

Maple [A] (verified)

Time = 1.58 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00

\[\frac {a^{2} \left (192 a \,b^{3} \cos \left (f x +e \right )^{2}+\left (-8 \cos \left (f x +e \right )^{6}+60 \cos \left (f x +e \right )^{4}+60 \cos \left (f x +e \right )^{2}\right ) b \,a^{3}+160 a \,b^{3}+40 a^{2} b^{2}+\left (48 \cos \left (f x +e \right )^{4}+240 \cos \left (f x +e \right )^{2}\right ) a^{2} b^{2}+128 b^{4}+\left (3 \cos \left (f x +e \right )^{8}-10 \cos \left (f x +e \right )^{6}+15 \cos \left (f x +e \right )^{4}\right ) a^{4}\right ) \left (a +b \right )^{7} \left (b +a \cos \left (f x +e \right )^{2}\right ) \sec \left (f x +e \right )^{5}}{15 f \left (\sqrt {-a b}+a \right )^{7} \left (\sqrt {-a b}-a \right )^{7} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\]

Input:

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

Output:

1/15/f*a^2/((-a*b)^(1/2)+a)^7/((-a*b)^(1/2)-a)^7*(192*a*b^3*cos(f*x+e)^2+( 
-8*cos(f*x+e)^6+60*cos(f*x+e)^4+60*cos(f*x+e)^2)*b*a^3+160*a*b^3+40*a^2*b^ 
2+(48*cos(f*x+e)^4+240*cos(f*x+e)^2)*a^2*b^2+128*b^4+(3*cos(f*x+e)^8-10*co 
s(f*x+e)^6+15*cos(f*x+e)^4)*a^4)*(a+b)^7*(b+a*cos(f*x+e)^2)/(a+b*sec(f*x+e 
)^2)^(5/2)*sec(f*x+e)^5
 

Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (3 \, a^{4} \cos \left (f x + e\right )^{9} - 2 \, {\left (5 \, a^{4} + 4 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 3 \, {\left (5 \, a^{4} + 20 \, a^{3} b + 16 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 12 \, {\left (5 \, a^{3} b + 20 \, a^{2} b^{2} + 16 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (5 \, a^{2} b^{2} + 20 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}} \] Input:

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

Output:

-1/15*(3*a^4*cos(f*x + e)^9 - 2*(5*a^4 + 4*a^3*b)*cos(f*x + e)^7 + 3*(5*a^ 
4 + 20*a^3*b + 16*a^2*b^2)*cos(f*x + e)^5 + 12*(5*a^3*b + 20*a^2*b^2 + 16* 
a*b^3)*cos(f*x + e)^3 + 8*(5*a^2*b^2 + 20*a*b^3 + 16*b^4)*cos(f*x + e))*sq 
rt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(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. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.64 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3}} - \frac {10 \, {\left ({\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{4}} + \frac {3 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 20 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 90 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{5}} + \frac {5 \, {\left (6 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{3} \cos \left (f x + e\right )^{3}} + \frac {10 \, {\left (9 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{4} \cos \left (f x + e\right )^{3}} + \frac {5 \, {\left (12 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{3} \cos \left (f x + e\right )^{2} - b^{4}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{5} \cos \left (f x + e\right )^{3}}}{15 \, f} \] Input:

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

Output:

-1/15*(15*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^3 - 10*((a + b/cos(f*x 
 + e)^2)^(3/2)*cos(f*x + e)^3 - 9*sqrt(a + b/cos(f*x + e)^2)*b*cos(f*x + e 
))/a^4 + (3*(a + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 20*(a + b/cos(f* 
x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 90*sqrt(a + b/cos(f*x + e)^2)*b^2*cos(f 
*x + e))/a^5 + 5*(6*(a + b/cos(f*x + e)^2)*b*cos(f*x + e)^2 - b^2)/((a + b 
/cos(f*x + e)^2)^(3/2)*a^3*cos(f*x + e)^3) + 10*(9*(a + b/cos(f*x + e)^2)* 
b^2*cos(f*x + e)^2 - b^3)/((a + b/cos(f*x + e)^2)^(3/2)*a^4*cos(f*x + e)^3 
) + 5*(12*(a + b/cos(f*x + e)^2)*b^3*cos(f*x + e)^2 - b^4)/((a + b/cos(f*x 
 + e)^2)^(3/2)*a^5*cos(f*x + e)^3))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1638 vs. \(2 (184) = 368\).

Time = 1.62 (sec) , antiderivative size = 1638, normalized size of antiderivative = 8.03 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/15*(5*((((6*a^16*b^3*sgn(cos(f*x + e)) + 23*a^15*b^4*sgn(cos(f*x + e)) 
+ 28*a^14*b^5*sgn(cos(f*x + e)) + 11*a^13*b^6*sgn(cos(f*x + e)))*tan(1/2*f 
*x + 1/2*e)^2/(a^18*b^2) - 3*(2*a^16*b^3*sgn(cos(f*x + e)) + a^15*b^4*sgn( 
cos(f*x + e)) - 12*a^14*b^5*sgn(cos(f*x + e)) - 11*a^13*b^6*sgn(cos(f*x + 
e)))/(a^18*b^2))*tan(1/2*f*x + 1/2*e)^2 - 3*(2*a^16*b^3*sgn(cos(f*x + e)) 
+ a^15*b^4*sgn(cos(f*x + e)) - 12*a^14*b^5*sgn(cos(f*x + e)) - 11*a^13*b^6 
*sgn(cos(f*x + e)))/(a^18*b^2))*tan(1/2*f*x + 1/2*e)^2 + (6*a^16*b^3*sgn(c 
os(f*x + e)) + 23*a^15*b^4*sgn(cos(f*x + e)) + 28*a^14*b^5*sgn(cos(f*x + e 
)) + 11*a^13*b^6*sgn(cos(f*x + e)))/(a^18*b^2))/(a*tan(1/2*f*x + 1/2*e)^4 
+ b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x 
+ 1/2*e)^2 + a + b)^(3/2) + 4*(15*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sq 
rt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 
 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^9*(2*a*b + 3*b^2) + 15*(s 
qrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan( 
1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) 
^2 + a + b))^8*(22*a*b + 21*b^2)*sqrt(a + b) + 20*(16*a^3 + 38*a^2*b + 63* 
a*b^2 + 45*b^3)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 
 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*ta 
n(1/2*f*x + 1/2*e)^2 + a + b))^7 - 20*(32*a^3 + 118*a^2*b + 51*a*b^2 - 63* 
b^3)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{5}}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

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

Output:

int((sqrt(sec(e + f*x)**2*b + a)*sin(e + f*x)**5)/(sec(e + f*x)**6*b**3 + 
3*sec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)