Integrand size = 25, antiderivative size = 240 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\frac {\left (5 a^3-15 a^2 b-5 a b^2-b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{5/2} f}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {(a-b) (5 a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a^2 f}-\frac {(5 a-b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a f}-\frac {\cos (e+f x) \sin ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f} \]
1/16*(5*a^3-15*a^2*b-5*a*b^2-b^3)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x +e)^2)^(1/2))/a^(5/2)/f+arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1 /2))*b^(1/2)/f-1/16*(a-b)*(5*a+b)*cos(f*x+e)*sin(f*x+e)*(a+b+b*tan(f*x+e)^ 2)^(1/2)/a^2/f-1/24*(5*a-b)*cos(f*x+e)*sin(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^( 1/2)/a/f-1/6*cos(f*x+e)*sin(f*x+e)^5*(a+b+b*tan(f*x+e)^2)^(1/2)/f
\[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx \]
Time = 0.51 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4620, 369, 440, 27, 440, 398, 224, 219, 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.
| \(\displaystyle \int \sin ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^6 \sqrt {a+b \sec (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \sqrt {b \tan ^2(e+f x)+a+b}}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 369 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {\tan ^4(e+f x) \left (6 b \tan ^2(e+f x)+5 (a+b)\right )}{\left (\tan ^2(e+f x)+1\right )^3 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {\int \frac {3 \tan ^2(e+f x) \left (8 a b \tan ^2(e+f x)+(5 a-b) (a+b)\right )}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}-\frac {(5 a-b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3 \int \frac {\tan ^2(e+f x) \left (8 a b \tan ^2(e+f x)+(5 a-b) (a+b)\right )}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}-\frac {(5 a-b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3 \left (\frac {\int \frac {16 a^2 b \tan ^2(e+f x)+(5 a+b) \left (a^2-b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}-\frac {(a-b) (5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}\right )}{4 a}-\frac {(5 a-b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3 \left (\frac {16 a^2 b \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\left (16 a^2 b-(5 a+b) \left (a^2-b^2\right )\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}-\frac {(a-b) (5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}\right )}{4 a}-\frac {(5 a-b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3 \left (\frac {16 a^2 b \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\left (16 a^2 b-(5 a+b) \left (a^2-b^2\right )\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}-\frac {(a-b) (5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}\right )}{4 a}-\frac {(5 a-b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3 \left (\frac {16 a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )-\left (16 a^2 b-(5 a+b) \left (a^2-b^2\right )\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}-\frac {(a-b) (5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}\right )}{4 a}-\frac {(5 a-b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3 \left (\frac {16 a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )-\left (16 a^2 b-(5 a+b) \left (a^2-b^2\right )\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 a}-\frac {(a-b) (5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}\right )}{4 a}-\frac {(5 a-b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3 \left (\frac {16 a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )-\frac {\left (16 a^2 b-(5 a+b) \left (a^2-b^2\right )\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}}{2 a}-\frac {(a-b) (5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}\right )}{4 a}-\frac {(5 a-b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
(-1/6*(Tan[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(1 + Tan[e + f*x]^2) ^3 + (-1/4*((5*a - b)*Tan[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(a*(1 + Tan[e + f*x]^2)^2) + (3*((-(((16*a^2*b - (5*a + b)*(a^2 - b^2))*ArcTan[ (Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/Sqrt[a]) + 16*a^2* Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2 *a) - ((a - b)*(5*a + b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*a *(1 + Tan[e + f*x]^2))))/(4*a))/6)/f
3.1.73.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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])
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]
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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2* b*(p + 1))), x] - Simp[e^2/(2*b*(p + 1)) Int[(e*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(m - 1) + d*(m + 2*q - 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 0 ] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1179\) vs. \(2(214)=428\).
Time = 12.64 (sec) , antiderivative size = 1180, normalized size of antiderivative = 4.92
-1/48/f/a^2/(-a)^(1/2)*(8*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) ^(1/2)*a^2*cos(f*x+e)^5*sin(f*x+e)+8*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^4*(( b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-26*(-a)^(1/2)*((b+a*cos(f*x+ e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*cos(f*x+e)^3*sin(f*x+e)+2*(-a)^(1/2)*((b +a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b*cos(f*x+e)^3*sin(f*x+e)-26*(- a)^(1/2)*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 2)*a^2+2*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x +e))^2)^(1/2)*a*b+33*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 )*a^2*cos(f*x+e)*sin(f*x+e)-14*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e ))^2)^(1/2)*a*b*cos(f*x+e)*sin(f*x+e)-3*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)*b^2*cos(f*x+e)*sin(f*x+e)+33*(-a)^(1/2)*sin(f*x+e)*(( b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-14*(-a)^(1/2)*sin(f*x+e)*((b +a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-3*(-a)^(1/2)*sin(f*x+e)*((b+a *cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2-24*(-a)^(1/2)*b^(1/2)*ln(4*(-(( b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+sin(f*x+e)*a- b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-a-b)/(sin(f*x+e)-1))*a ^2-24*(-a)^(1/2)*b^(1/2)*ln(-4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 2)*b^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)+a+b)/(sin(f*x+e)+1))*a^2-15*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)/...
Time = 2.89 (sec) , antiderivative size = 1715, normalized size of antiderivative = 7.15 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\text {Too large to display} \]
[1/384*(96*a^3*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b )*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f* x + e)^4) + 3*(5*a^3 - 15*a^2*b - 5*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)*co s(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt ((a*cos(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 - a^2*b)*cos(f*x + e)^3 + (33*a^3 - 14*a^2*b - 3*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 ^3*f), 1/384*(192*a^3*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*c os(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*co s(f*x + e)^2 + b^2)*sin(f*x + e))) + 3*(5*a^3 - 15*a^2*b - 5*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 + 3 2*(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)...
Timed out. \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\text {Timed out} \]
\[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{6} \,d x } \]
\[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^6\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \]