Integrand size = 25, antiderivative size = 298 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\frac {\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{3/2} f}+\frac {(3 a-5 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a f}+\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}-\frac {\cos (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f} \]
1/16*(5*a^3-45*a^2*b+15*a*b^2+b^3)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f* x+e)^2)^(1/2))/a^(3/2)/f+1/2*(3*a-5*b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*t an(f*x+e)^2)^(1/2))*b^(1/2)/f-1/16*(5*a^2-26*a*b+b^2)*(a+b+b*tan(f*x+e)^2) ^(1/2)*tan(f*x+e)/a/f+1/48*(5*a^2-40*a*b+3*b^2)*sin(f*x+e)^2*(a+b+b*tan(f* x+e)^2)^(1/2)*tan(f*x+e)/a/f+1/24*(5*a-3*b)*sin(f*x+e)^4*(a+b+b*tan(f*x+e) ^2)^(1/2)*tan(f*x+e)/f-1/6*cos(f*x+e)*sin(f*x+e)^5*(a+b+b*tan(f*x+e)^2)^(3 /2)/f
\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx \]
Time = 0.58 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4620, 369, 439, 440, 27, 444, 27, 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.
| \(\displaystyle \int \sin ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 369 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {\tan ^4(e+f x) \sqrt {b \tan ^2(e+f x)+a+b} \left (8 b \tan ^2(e+f x)+5 (a+b)\right )}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 439 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}-\frac {1}{4} \int \frac {\tan ^4(e+f x) \left (2 (5 a-19 b) b \tan ^2(e+f x)+5 (a-7 b) (a+b)\right )}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int \frac {3 \tan ^2(e+f x) \left (2 b \left (5 a^2-26 b a+b^2\right ) \tan ^2(e+f x)+(a+b) \left (5 a^2-40 b a+3 b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \int \frac {\tan ^2(e+f x) \left (2 b \left (5 a^2-26 b a+b^2\right ) \tan ^2(e+f x)+(a+b) \left (5 a^2-40 b a+3 b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 444 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {\int \frac {2 b \left (8 a (3 a-5 b) b \tan ^2(e+f x)+(a+b) \left (5 a^2-26 b a+b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\int \frac {8 a (3 a-5 b) b \tan ^2(e+f x)+(a+b) \left (5 a^2-26 b a+b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (-\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-8 a b (3 a-5 b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (-\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-8 a b (3 a-5 b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (-\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-8 a \sqrt {b} (3 a-5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (-\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-8 a \sqrt {b} (3 a-5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}-8 a \sqrt {b} (3 a-5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
(-1/6*(Tan[e + f*x]^5*(a + b + b*Tan[e + f*x]^2)^(3/2))/(1 + Tan[e + f*x]^ 2)^3 + (((5*a - 3*b)*Tan[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*(1 + Tan[e + f*x]^2)^2) + (((5*a^2 - 40*a*b + 3*b^2)*Tan[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*a*(1 + Tan[e + f*x]^2)) - (3*(-(((5*a^3 - 45*a^2 *b + 15*a*b^2 + b^3)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/Sqrt[a]) - 8*a*(3*a - 5*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x] )/Sqrt[a + b + b*Tan[e + f*x]^2]] + (5*a^2 - 26*a*b + b^2)*Tan[e + f*x]*Sq rt[a + b + b*Tan[e + f*x]^2]))/(2*a))/4)/6)/f
3.1.86.3.1 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[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 ))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G tQ[q, 0] && !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
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[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ (p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ (b*d*(m + 2*(p + q + 1) + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) ^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && 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. \(1601\) vs. \(2(266)=532\).
Time = 16.25 (sec) , antiderivative size = 1602, normalized size of antiderivative = 5.38
-1/48/f/b/(-a)^(1/2)/a*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos (f*x+e))^2)^(1/2)/(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))*(8*cos(f*x+e)^8*sin(f* x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(-a)^(1/2)*a^2*b+8*((b+a* cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(-a)^(1/2)*a^2*b*cos(f*x+e)^7*sin(f* x+e)-26*cos(f*x+e)^6*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 )*(-a)^(1/2)*a^2*b+14*cos(f*x+e)^6*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f *x+e))^2)^(1/2)*(-a)^(1/2)*a*b^2+60*cos(f*x+e)^3*(-a)^(1/2)*b^(5/2)*ln(4*( ((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b +a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1)) *a+60*cos(f*x+e)^3*(-a)^(1/2)*b^(5/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f* x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e ))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a-26*((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)*(-a)^(1/2)*a^2*b*cos(f*x+e)^5*sin(f*x+e)+14*((b+a*cos (f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(-a)^(1/2)*a*b^2*cos(f*x+e)^5*sin(f*x+e )-36*cos(f*x+e)^3*(-a)^(1/2)*b^(3/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)) ^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a^2-36*cos(f*x+e)^3*(-a)^(1/2) *b^(3/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f* x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b) /(sin(f*x+e)-1))*a^2+33*cos(f*x+e)^4*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+...
Time = 10.53 (sec) , antiderivative size = 1855, normalized size of antiderivative = 6.22 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\text {Too large to display} \]
[-1/384*(3*(5*a^3 - 45*a^2*b + 15*a*b^2 + b^3)*sqrt(-a)*cos(f*x + e)*log(1 28*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))*sq rt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 48*(3*a ^3 - 5*a^2*b)*sqrt(b)*cos(f*x + e)*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) + 8*(8*a^3*cos(f*x + e)^6 - 2*(13*a^3 - 7*a^2*b)*cos (f*x + e)^4 - 24*a^2*b + (33*a^3 - 68*a^2*b + 3*a*b^2)*cos(f*x + e)^2)*sqr t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^2*f*cos(f*x + e) ), 1/384*(96*(3*a^3 - 5*a^2*b)*sqrt(-b)*arctan(-1/2*((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) /((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*cos(f*x + e) - 3*(5*a^3 - 45*a ^2*b + 15*a*b^2 + b^3)*sqrt(-a)*cos(f*x + e)*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*...
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\text {Timed out} \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{6} \,d x } \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]