\(\int \frac {\sin ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [99]

Optimal result

Integrand size = 25, antiderivative size = 193 \[ \int \frac {\sin ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {5 (a+b)^3 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{7/2} f}-\frac {\left (33 a^2+40 a b+15 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a^3 f}+\frac {(9 a+5 b) \cos ^3(e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 a f} \] Output:

5/16*(a+b)^3*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(7/2) 
/f-1/48*(33*a^2+40*a*b+15*b^2)*cos(f*x+e)*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^ 
(1/2)/a^3/f+1/24*(9*a+5*b)*cos(f*x+e)^3*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1 
/2)/a^2/f+1/6*cos(f*x+e)^3*sin(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/a/f
 

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\sqrt {a+2 b+a \cos (2 (e+f x))} \sec (e+f x) \left (15 (a+b)^3 \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )-\sqrt {a} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)} \left (15 (a+b)^2+10 a (a+b) \sin ^2(e+f x)+8 a^2 \sin ^4(e+f x)\right )\right )}{48 \sqrt {2} a^{7/2} f \sqrt {a+b \sec ^2(e+f x)}} \] Input:

Integrate[Sin[e + f*x]^6/Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*Sec[e + f*x]*(15*(a + b)^3*ArcTan[(Sqr 
t[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]] - Sqrt[a]*Sin[e + f*x]* 
Sqrt[a + b - a*Sin[e + f*x]^2]*(15*(a + b)^2 + 10*a*(a + b)*Sin[e + f*x]^2 
 + 8*a^2*Sin[e + f*x]^4)))/(48*Sqrt[2]*a^(7/2)*f*Sqrt[a + b*Sec[e + f*x]^2 
])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4620, 372, 440, 402, 27, 291, 216}

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.

\(\displaystyle \int \frac {\sin ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^6}{\sqrt {a+b \sec (e+f x)^2}}dx\)

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \frac {\tan ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right )^4 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 440

\(\displaystyle \frac {\frac {\tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\int \frac {(a+b) (9 a+5 b)-2 \left (12 a^2+13 b a+5 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}-\frac {(9 a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (33 a^2+40 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int \frac {15 (a+b)^3}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}}{4 a}-\frac {(9 a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (33 a^2+40 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {15 (a+b)^3 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}}{4 a}-\frac {(9 a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (33 a^2+40 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {15 (a+b)^3 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 a}}{4 a}-\frac {(9 a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (33 a^2+40 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {15 (a+b)^3 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 a^{3/2}}}{4 a}-\frac {(9 a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}}{f}\)

Input:

Int[Sin[e + f*x]^6/Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

((Tan[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(6*a*(1 + Tan[e + f*x]^2) 
^3) - (-1/4*((9*a + 5*b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(a*( 
1 + Tan[e + f*x]^2)^2) + ((-15*(a + b)^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqr 
t[a + b + b*Tan[e + f*x]^2]])/(2*a^(3/2)) + ((33*a^2 + 40*a*b + 15*b^2)*Ta 
n[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*a*(1 + Tan[e + f*x]^2)))/(4* 
a))/(6*a))/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 291

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

rule 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 402

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

rule 440

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

rule 3042

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

rule 4620

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(632\) vs. \(2(173)=346\).

Time = 19.71 (sec) , antiderivative size = 633, normalized size of antiderivative = 3.28

method	result	size
default	\(\frac {15 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (\sec \left (f x +e \right )+1\right )+45 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (\sec \left (f x +e \right )+1\right )+45 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a \,b^{2} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (\sec \left (f x +e \right )+1\right )+15 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (\sec \left (f x +e \right )+1\right )+\sin \left (f x +e \right ) \cos \left (f x +e \right ) \left (-8 \cos \left (f x +e \right )^{4}+26 \cos \left (f x +e \right )^{2}-33\right ) a^{3} \sqrt {-a}+\left (2 \cos \left (f x +e \right )^{4}-14 \cos \left (f x +e \right )^{2}-33\right ) \sqrt {-a}\, a^{2} b \tan \left (f x +e \right )+5 \left (-8-\cos \left (f x +e \right )^{2}\right ) \sqrt {-a}\, a \,b^{2} \tan \left (f x +e \right )-15 \sqrt {-a}\, b^{3} \tan \left (f x +e \right )}{48 f \,a^{3} \sqrt {-a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(633\)

Input:

int(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/48/f/a^3/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2)*(15*((b+a*cos(f*x+e)^2)/(1+ 
cos(f*x+e))^2)^(1/2)*a^3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e) 
)^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 
1/2)-4*sin(f*x+e)*a)*(sec(f*x+e)+1)+45*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 
2)^(1/2)*a^2*b*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) 
*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin 
(f*x+e)*a)*(sec(f*x+e)+1)+45*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a 
*b^2*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e 
)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a) 
*(sec(f*x+e)+1)+15*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3*ln(4*(- 
a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/ 
2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(sec(f*x+e) 
+1)+sin(f*x+e)*cos(f*x+e)*(-8*cos(f*x+e)^4+26*cos(f*x+e)^2-33)*a^3*(-a)^(1 
/2)+(2*cos(f*x+e)^4-14*cos(f*x+e)^2-33)*(-a)^(1/2)*a^2*b*tan(f*x+e)+5*(-8- 
cos(f*x+e)^2)*(-a)^(1/2)*a*b^2*tan(f*x+e)-15*(-a)^(1/2)*b^3*tan(f*x+e))
 

Fricas [A] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 639, normalized size of antiderivative = 3.31 \[ \int \frac {\sin ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/384*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(-a)*log(128*a^4*cos(f*x + 
 e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^ 
2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 
 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 
- 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x 
 + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*c 
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(8*a^3*cos(f*x + e)^5 
 - 2*(13*a^3 + 5*a^2*b)*cos(f*x + e)^3 + (33*a^3 + 40*a^2*b + 15*a*b^2)*co 
s(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^4 
*f), -1/192*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a)*arctan(1/4*(8*a^2* 
cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f* 
x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f* 
x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) 
+ 4*(8*a^3*cos(f*x + e)^5 - 2*(13*a^3 + 5*a^2*b)*cos(f*x + e)^3 + (33*a^3 
+ 40*a^2*b + 15*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + 
 e)^2)*sin(f*x + e))/(a^4*f)]
 

Sympy [F]

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

integrate(sin(f*x+e)**6/(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sin(e + f*x)**6/sqrt(a + b*sec(e + f*x)**2), x)
 

Maxima [F]

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

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sin(f*x + e)^6/sqrt(b*sec(f*x + e)^2 + a), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sin ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^6}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \] Input:

int(sin(e + f*x)^6/(a + b/cos(e + f*x)^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(sin(e + f*x)^6/(a + b/cos(e + f*x)^2)^(1/2), x)
 

Reduce [F]

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

int(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

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