Integrand size = 25, antiderivative size = 149 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f} \]
arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-cot(f*x+e )*(a+b+b*tan(f*x+e)^2)^(1/2)/f-2/15*(5*a+4*b)*cot(f*x+e)^3*(a+b+b*tan(f*x+ e)^2)^(3/2)/(a+b)^2/f-1/5*cot(f*x+e)^5*(a+b+b*tan(f*x+e)^2)^(3/2)/(a+b)/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 8.73 (sec) , antiderivative size = 941, normalized size of antiderivative = 6.32 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {2} \csc ^5(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (-3 (a+b) \cos ^2(e+f x)-4 a \cos ^2(e+f x) \sin ^2(e+f x)-16 b \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)+8 b \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)-\frac {8 a^2 \cos ^2(e+f x) \sin ^4(e+f x)}{a+b}-\frac {8 a b \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x)}{a+b}-\frac {16 a b \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x)}{a+b}+\frac {24 a^2 b \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x)}{(a+b)^2}+\frac {8 a^2 b \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x)}{(a+b)^2}-\frac {16 b^2 \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a+b}+\frac {8 b^2 \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a+b}-\frac {8 a b^2 \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x) \tan ^2(e+f x)}{(a+b)^2}-\frac {16 a b^2 \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x) \tan ^2(e+f x)}{(a+b)^2}+\frac {24 a^2 b^2 \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x) \tan ^2(e+f x)}{(a+b)^3}+\frac {8 a^2 b^2 \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x) \tan ^2(e+f x)}{(a+b)^3}-\frac {4 a \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \cos ^2(e+f x) \sin ^2(e+f x) \sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}}{\sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}}}+\frac {3 b \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \sin ^2(e+f x)}{\sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}}+\frac {8 a^2 b \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \sin ^6(e+f x)}{(a+b)^2 \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}}\right )}{15 f \sqrt {a+2 b+a \cos (2 e+2 f x)} \sqrt {a+b-a \sin ^2(e+f x)}} \]
(Sqrt[2]*Csc[e + f*x]^5*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*(1 - (a*Si n[e + f*x]^2)/(a + b))*(-3*(a + b)*Cos[e + f*x]^2 - 4*a*Cos[e + f*x]^2*Sin [e + f*x]^2 - 16*b*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2 + 8*b*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[ e + f*x]^2)/(a + b))]*Sin[e + f*x]^2 - (8*a^2*Cos[e + f*x]^2*Sin[e + f*x]^ 4)/(a + b) - (8*a*b*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^4)/(a + b) - (16*a*b*HypergeometricPFQ[{2, 2, 2}, {1, 3 /2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^4)/(a + b) + (24*a^2*b*Hy pergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^6)/ (a + b)^2 + (8*a^2*b*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f *x]^2)/(a + b))]*Sin[e + f*x]^6)/(a + b)^2 - (16*b^2*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(a + b) + (8*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/( a + b))]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(a + b) - (8*a*b^2*Hypergeometric2 F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^4*Tan[e + f*x]^2 )/(a + b)^2 - (16*a*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^4*Tan[e + f*x]^2)/(a + b)^2 + (24*a^2*b^2 *Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^ 6*Tan[e + f*x]^2)/(a + b)^3 + (8*a^2*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^6*Tan[e + f*x]^2)/(a ...
Time = 0.33 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4620, 365, 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.
| \(\displaystyle \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \sec (e+f x)^2}}{\sin (e+f x)^6}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int \cot ^4(e+f x) \sqrt {b \tan ^2(e+f x)+a+b} \left (5 (a+b) \tan ^2(e+f x)+2 (5 a+4 b)\right )d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle \frac {\frac {5 (a+b) \int \cot ^2(e+f x) \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {\frac {5 (a+b) \left (b \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {5 (a+b) \left (b \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {5 (a+b) \left (\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\) |
(-1/5*(Cot[e + f*x]^5*(a + b + b*Tan[e + f*x]^2)^(3/2))/(a + b) + ((-2*(5* a + 4*b)*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(3/2))/(3*(a + b)) + 5* (a + b)*(Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x ]^2]] - Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2]))/(5*(a + b)))/f
3.1.79.3.1 Defintions of rubi rules used
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[((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]
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]
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]
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. \(1679\) vs. \(2(133)=266\).
Time = 7.18 (sec) , antiderivative size = 1680, normalized size of antiderivative = 11.28
-1/30/f/(a+b)^2*(15*sin(f*x+e)^3*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))*cos(f*x+e)+15*sin(f*x+e)^3 *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))*cos(f*x+e)+30*sin(f*x+e)^3*b^(3/2)*a*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))*cos(f*x+e)+30 *sin(f*x+e)^3*b^(3/2)*a*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)-si n(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+e)-15*sin(f*x+e)^3*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))- 15*sin(f*x+e)^3*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)-si n(f*x+e)*a+a+b)/(sin(f*x+e)-1))+15*sin(f*x+e)^3*b^(1/2)*a^2*ln(4*(((b+a*co s(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))*cos(f*x +e)+15*sin(f*x+e)^3*b^(1/2)*a^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))^...
Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (133) = 266\).
Time = 1.39 (sec) , antiderivative size = 656, normalized size of antiderivative = 4.40 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (8 \, a^{2} + 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (20 \, a^{2} + 59 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 40 \, a b + 23 \, 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}}}}{60 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}, \frac {15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left ({\left (8 \, a^{2} + 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (20 \, a^{2} + 59 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 40 \, a b + 23 \, 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 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]
[1/60*(15*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos( f*x + e)^2 + a^2 + 2*a*b + b^2)*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)*sin(f*x + e) - 4*((8*a^2 + 25*a*b + 15*b^2)*cos (f*x + e)^5 - (20*a^2 + 59*a*b + 35*b^2)*cos(f*x + e)^3 + (15*a^2 + 40*a*b + 23*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a ^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^ 2 + (a^2 + 2*a*b + b^2)*f)*sin(f*x + e)), 1/30*(15*((a^2 + 2*a*b + b^2)*co s(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*s qrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*s qrt((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)))*sin(f*x + e) - 2*((8*a^2 + 25*a*b + 15*b^2)*cos(f*x + e)^5 - ( 20*a^2 + 59*a*b + 35*b^2)*cos(f*x + e)^3 + (15*a^2 + 40*a*b + 23*b^2)*cos( f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^2 + 2*a*b + b^ 2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a* b + b^2)*f)*sin(f*x + e))]
Timed out. \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {15 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - \frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{\tan \left (f x + e\right )} - \frac {10 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}} + \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}{{\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} - \frac {3 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]
1/15*(15*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) - 15*sqrt(b*tan(f *x + e)^2 + a + b)/tan(f*x + e) - 10*(b*tan(f*x + e)^2 + a + b)^(3/2)/((a + b)*tan(f*x + e)^3) + 2*(b*tan(f*x + e)^2 + a + b)^(3/2)*b/((a + b)^2*tan (f*x + e)^3) - 3*(b*tan(f*x + e)^2 + a + b)^(3/2)/((a + b)*tan(f*x + e)^5) )/f
\[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{{\sin \left (e+f\,x\right )}^6} \,d x \]