\(\int \sqrt {a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx\) [67]

Optimal result

Integrand size = 25, antiderivative size = 139 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f} \] Output:

b^(1/2)*arctanh(b^(1/2)*sec(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2))/f-cos(f*x+e)* 
(a+b*sec(f*x+e)^2)^(1/2)/f+2/15*(5*a+b)*cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3 
/2)/a^2/f-1/5*cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2)/a/f
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.09 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx=-\frac {\cos (e+f x) \left (-2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b+a \cos ^2(e+f x)}}{\sqrt {b}}\right )+2 \sqrt {b+a \cos ^2(e+f x)}-\frac {2 (2 a+b) \left (b+a \cos ^2(e+f x)\right )^{3/2}}{3 a^2}+\frac {2 \left (b+a \cos ^2(e+f x)\right )^{5/2}}{5 a^2}\right ) \sqrt {a+b \sec ^2(e+f x)}}{\sqrt {2} f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \] Input:

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

Output:

-((Cos[e + f*x]*(-2*Sqrt[b]*ArcTanh[Sqrt[b + a*Cos[e + f*x]^2]/Sqrt[b]] + 
2*Sqrt[b + a*Cos[e + f*x]^2] - (2*(2*a + b)*(b + a*Cos[e + f*x]^2)^(3/2))/ 
(3*a^2) + (2*(b + a*Cos[e + f*x]^2)^(5/2))/(5*a^2))*Sqrt[a + b*Sec[e + f*x 
]^2])/(Sqrt[2]*f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03, 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, 358, 247, 224, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x)^5 \sqrt {a+b \sec (e+f x)^2}dx\)

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 365

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 358

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-1/5*(Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2)^(3/2))/a - ((-2*(5*a + b)*Cos 
[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(3*a) - 5*a*(Sqrt[b]*ArcTanh[(Sq 
rt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]] - Cos[e + f*x]*Sqrt[a + b* 
Sec[e + f*x]^2]))/(5*a))/f
 

Defintions of rubi rules used

rule 25

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

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 247

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

rule 358

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] + S 
imp[d/e^2   Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e 
, m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, 
 -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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(325\) vs. \(2(123)=246\).

Time = 5.25 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.35

method	result	size
default	\(\frac {\left (15 \sqrt {b}\, \ln \left (-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sec \left (f x +e \right )-4 b \sec \left (f x +e \right )\right ) a^{2}+\left (-3 \cos \left (f x +e \right )^{5}-3 \cos \left (f x +e \right )^{4}+10 \cos \left (f x +e \right )^{3}+10 \cos \left (f x +e \right )^{2}-15 \cos \left (f x +e \right )-15\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}+\left (-\cos \left (f x +e \right )^{3}-\cos \left (f x +e \right )^{2}+10 \cos \left (f x +e \right )+10\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b +\left (2 \cos \left (f x +e \right )+2\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}\right ) \cos \left (f x +e \right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}{15 f \,a^{2} \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(326\)

Input:

int((a+b*sec(f*x+e)^2)^(1/2)*sin(f*x+e)^5,x,method=_RETURNVERBOSE)
 

Output:

1/15/f/a^2*(15*b^(1/2)*ln(-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sec(f*x+e)-4* 
b*sec(f*x+e))*a^2+(-3*cos(f*x+e)^5-3*cos(f*x+e)^4+10*cos(f*x+e)^3+10*cos(f 
*x+e)^2-15*cos(f*x+e)-15)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+ 
(-cos(f*x+e)^3-cos(f*x+e)^2+10*cos(f*x+e)+10)*((b+a*cos(f*x+e)^2)/(1+cos(f 
*x+e))^2)^(1/2)*a*b+(2*cos(f*x+e)+2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)*b^2)*cos(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/(1+cos(f*x+e))/((b+a*cos(f 
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.20 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx=\left [\frac {15 \, a^{2} \sqrt {b} \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - {\left (10 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, a^{2} f}, -\frac {15 \, a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) + {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - {\left (10 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, a^{2} f}\right ] \] Input:

integrate((a+b*sec(f*x+e)^2)^(1/2)*sin(f*x+e)^5,x, algorithm="fricas")
 

Output:

[1/30*(15*a^2*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + 
e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) - 2*(3*a^2*c 
os(f*x + e)^5 - (10*a^2 - a*b)*cos(f*x + e)^3 + (15*a^2 - 10*a*b - 2*b^2)* 
cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^2*f), -1/15* 
(15*a^2*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^ 
2)*cos(f*x + e)/(a*cos(f*x + e)^2 + b)) + (3*a^2*cos(f*x + e)^5 - (10*a^2 
- a*b)*cos(f*x + e)^3 + (15*a^2 - 10*a*b - 2*b^2)*cos(f*x + e))*sqrt((a*co 
s(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^2*f)]
 

Sympy [F(-1)]

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

integrate((a+b*sec(f*x+e)**2)**(1/2)*sin(f*x+e)**5,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.23 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx=\frac {\frac {20 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}{a} - 30 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - 15 \, \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) - \frac {2 \, {\left (3 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 5 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3}\right )}}{a^{2}}}{30 \, f} \] Input:

integrate((a+b*sec(f*x+e)^2)^(1/2)*sin(f*x+e)^5,x, algorithm="maxima")
 

Output:

1/30*(20*(a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3/a - 30*sqrt(a + b/cos 
(f*x + e)^2)*cos(f*x + e) - 15*sqrt(b)*log((sqrt(a + b/cos(f*x + e)^2)*cos 
(f*x + e) - sqrt(b))/(sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b))) 
- 2*(3*(a + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 5*(a + b/cos(f*x + e) 
^2)^(3/2)*b*cos(f*x + e)^3)/a^2)/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1281 vs. \(2 (123) = 246\).

Time = 0.58 (sec) , antiderivative size = 1281, normalized size of antiderivative = 9.22 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sec(f*x+e)^2)^(1/2)*sin(f*x+e)^5,x, algorithm="giac")
 

Output:

-2/15*(15*b*arctan(-1/2*(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*tan(1/2*f*x + 1/2*e)^2 + a + b) - sqrt(a + b))/sqrt(-b))/sqrt(-b) - 
2*(15*(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*tan(1/2*f*x 
+ 1/2*e)^2 + a + b))^9*b + 165*(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*tan(1/2*f*x + 1/2*e)^2 + a + b))^8*sqrt(a + b)*b - 20*(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*tan(1/2*f*x + 1/2*e)^2 + 
a + b))^7*(16*a^2 + 5*a*b - 27*b^2) + 20*(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*tan(1/2*f*x + 1/2*e)^2 + a + b))^6*(32*a^2 - 83*a*b 
 + 33*b^2)*sqrt(a + b) + 2*(416*a^3 + 625*a^2*b - 1230*a*b^2 - 15*b^3)*(sq 
rt(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))^5 - 10*(256*a^3 - 391*a^2*b + 90*a*b^2 + 81*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*tan(1/2*f*x + 1/2*e)^2 + a + b) 
)^4*sqrt(a + b) + 20*(16*a^4 - 161*a^3*b + 157*a^2*b^2 + 45*a*b^3 - 33*...
 

Mupad [F(-1)]

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

int(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)^(1/2),x)
 

Output:

int(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)^(1/2), x)
 

Reduce [F]

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

int((a+b*sec(f*x+e)^2)^(1/2)*sin(f*x+e)^5,x)
 

Output:

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