Integrand size = 23, antiderivative size = 148 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {a (3 a-4 b) \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 (a+b)^{9/2} f}-\frac {a (a-2 b) \cot (e+f x)}{(a+b)^4 f}-\frac {2 a \cot ^3(e+f x)}{3 (a+b)^3 f}-\frac {\cot ^5(e+f x)}{5 (a+b)^2 f}-\frac {a^2 b \tan (e+f x)}{2 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:
-1/2*a*(3*a-4*b)*b^(1/2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/(a+b)^(9/2 )/f-a*(a-2*b)*cot(f*x+e)/(a+b)^4/f-2/3*a*cot(f*x+e)^3/(a+b)^3/f-1/5*cot(f* x+e)^5/(a+b)^2/f-1/2*a^2*b*tan(f*x+e)/(a+b)^4/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 3.18 (sec) , antiderivative size = 777, normalized size of antiderivative = 5.25 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^4(e+f x) \left (\frac {960 a (3 a-4 b) b \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x))) (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}-\csc (e) \csc ^5(e+f x) \sec (2 e) \left (10 a \left (16 a^2+34 a b+123 b^2\right ) \sin (f x)-a \left (16 a^2-223 a b+1336 b^2\right ) \sin (3 f x)+240 a^3 \sin (2 e-f x)+640 a^2 b \sin (2 e-f x)-1460 a b^2 \sin (2 e-f x)+240 b^3 \sin (2 e-f x)-240 a^3 \sin (2 e+f x)-715 a^2 b \sin (2 e+f x)+860 a b^2 \sin (2 e+f x)-240 b^3 \sin (2 e+f x)+160 a^3 \sin (4 e+f x)+415 a^2 b \sin (4 e+f x)+1830 a b^2 \sin (4 e+f x)+165 a^2 b \sin (2 e+3 f x)-30 a b^2 \sin (2 e+3 f x)+120 b^3 \sin (2 e+3 f x)-16 a^3 \sin (4 e+3 f x)+208 a^2 b \sin (4 e+3 f x)-1036 a b^2 \sin (4 e+3 f x)+180 a^2 b \sin (6 e+3 f x)-330 a b^2 \sin (6 e+3 f x)+120 b^3 \sin (6 e+3 f x)+48 a^3 \sin (2 e+5 f x)-268 a^2 b \sin (2 e+5 f x)+290 a b^2 \sin (2 e+5 f x)-24 b^3 \sin (2 e+5 f x)+48 a^3 \sin (6 e+5 f x)-223 a^2 b \sin (6 e+5 f x)+230 a b^2 \sin (6 e+5 f x)-24 b^3 \sin (6 e+5 f x)-45 a^2 b \sin (8 e+5 f x)+60 a b^2 \sin (8 e+5 f x)-16 a^3 \sin (4 e+7 f x)+83 a^2 b \sin (4 e+7 f x)-6 a b^2 \sin (4 e+7 f x)-15 a^2 b \sin (6 e+7 f x)-16 a^3 \sin (8 e+7 f x)+68 a^2 b \sin (8 e+7 f x)-6 a b^2 \sin (8 e+7 f x)\right )\right )}{7680 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^2} \] Input:
Integrate[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*((960*a*(3*a - 4*b)*b*ArcTa n[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f *x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e]) ^4]) - Csc[e]*Csc[e + f*x]^5*Sec[2*e]*(10*a*(16*a^2 + 34*a*b + 123*b^2)*Si n[f*x] - a*(16*a^2 - 223*a*b + 1336*b^2)*Sin[3*f*x] + 240*a^3*Sin[2*e - f* x] + 640*a^2*b*Sin[2*e - f*x] - 1460*a*b^2*Sin[2*e - f*x] + 240*b^3*Sin[2* e - f*x] - 240*a^3*Sin[2*e + f*x] - 715*a^2*b*Sin[2*e + f*x] + 860*a*b^2*S in[2*e + f*x] - 240*b^3*Sin[2*e + f*x] + 160*a^3*Sin[4*e + f*x] + 415*a^2* b*Sin[4*e + f*x] + 1830*a*b^2*Sin[4*e + f*x] + 165*a^2*b*Sin[2*e + 3*f*x] - 30*a*b^2*Sin[2*e + 3*f*x] + 120*b^3*Sin[2*e + 3*f*x] - 16*a^3*Sin[4*e + 3*f*x] + 208*a^2*b*Sin[4*e + 3*f*x] - 1036*a*b^2*Sin[4*e + 3*f*x] + 180*a^ 2*b*Sin[6*e + 3*f*x] - 330*a*b^2*Sin[6*e + 3*f*x] + 120*b^3*Sin[6*e + 3*f* x] + 48*a^3*Sin[2*e + 5*f*x] - 268*a^2*b*Sin[2*e + 5*f*x] + 290*a*b^2*Sin[ 2*e + 5*f*x] - 24*b^3*Sin[2*e + 5*f*x] + 48*a^3*Sin[6*e + 5*f*x] - 223*a^2 *b*Sin[6*e + 5*f*x] + 230*a*b^2*Sin[6*e + 5*f*x] - 24*b^3*Sin[6*e + 5*f*x] - 45*a^2*b*Sin[8*e + 5*f*x] + 60*a*b^2*Sin[8*e + 5*f*x] - 16*a^3*Sin[4*e + 7*f*x] + 83*a^2*b*Sin[4*e + 7*f*x] - 6*a*b^2*Sin[4*e + 7*f*x] - 15*a^2*b *Sin[6*e + 7*f*x] - 16*a^3*Sin[8*e + 7*f*x] + 68*a^2*b*Sin[8*e + 7*f*x] - 6*a*b^2*Sin[8*e + 7*f*x])))/(7680*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^2)
Time = 0.48 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4620, 365, 361, 25, 1584, 2009}
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.
| \(\displaystyle \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int \frac {\cot ^4(e+f x) \left (5 (a+b) \tan ^2(e+f x)+10 a+3 b\right )}{\left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle \frac {\frac {-\frac {1}{2} b \int -\frac {\cot ^4(e+f x) \left (-\frac {\left (5 a^2+2 b^2\right ) \tan ^4(e+f x)}{(a+b)^3}+\frac {2 \left (5 a^2+2 b^2\right ) \tan ^2(e+f x)}{b (a+b)^2}+\frac {2 (10 a+3 b)}{b (a+b)}\right )}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)-\frac {b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{2 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {1}{2} b \int \frac {\cot ^4(e+f x) \left (-\frac {\left (5 a^2+2 b^2\right ) \tan ^4(e+f x)}{(a+b)^3}+\frac {2 \left (5 a^2+2 b^2\right ) \tan ^2(e+f x)}{b (a+b)^2}+\frac {2 (10 a+3 b)}{b (a+b)}\right )}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)-\frac {b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{2 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {\frac {\frac {1}{2} b \int \left (\frac {2 (10 a+3 b) \cot ^4(e+f x)}{b (a+b)^2}-\frac {2 \left (-5 a^2+10 b a+b^2\right ) \cot ^2(e+f x)}{b (a+b)^3}-\frac {5 a (3 a-4 b)}{(a+b)^3 \left (b \tan ^2(e+f x)+a+b\right )}\right )d\tan (e+f x)-\frac {b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{2 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {1}{2} b \left (-\frac {2 \left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{b (a+b)^3}-\frac {5 a (3 a-4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {b} (a+b)^{7/2}}-\frac {2 (10 a+3 b) \cot ^3(e+f x)}{3 b (a+b)^2}\right )-\frac {b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{2 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
Input:
Int[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]
Output:
(-1/5*Cot[e + f*x]^5/((a + b)*(a + b + b*Tan[e + f*x]^2)) + ((b*((-5*a*(3* a - 4*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(Sqrt[b]*(a + b)^(7/2 )) - (2*(5*a^2 - 10*a*b - b^2)*Cot[e + f*x])/(b*(a + b)^3) - (2*(10*a + 3* b)*Cot[e + f*x]^3)/(3*b*(a + b)^2)))/2 - (b*(5*a^2 + 2*b^2)*Tan[e + f*x])/ (2*(a + b)^3*(a + b + b*Tan[e + f*x]^2)))/(5*(a + b)))/f
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
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[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
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]
Time = 1.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.84
| method | result | size |
| derivativedivides | \(\frac {-\frac {b a \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (3 a -4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{4}}-\frac {1}{5 \left (a +b \right )^{2} \tan \left (f x +e \right )^{5}}-\frac {a \left (a -2 b \right )}{\left (a +b \right )^{4} \tan \left (f x +e \right )}-\frac {2 a}{3 \left (a +b \right )^{3} \tan \left (f x +e \right )^{3}}}{f}\) | \(124\) |
| default | \(\frac {-\frac {b a \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (3 a -4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{4}}-\frac {1}{5 \left (a +b \right )^{2} \tan \left (f x +e \right )^{5}}-\frac {a \left (a -2 b \right )}{\left (a +b \right )^{4} \tan \left (f x +e \right )}-\frac {2 a}{3 \left (a +b \right )^{3} \tan \left (f x +e \right )^{3}}}{f}\) | \(124\) |
| risch | \(-\frac {i \left (-45 a^{2} b \,{\mathrm e}^{12 i \left (f x +e \right )}+60 a \,b^{2} {\mathrm e}^{12 i \left (f x +e \right )}+180 a^{2} b \,{\mathrm e}^{10 i \left (f x +e \right )}-330 a \,b^{2} {\mathrm e}^{10 i \left (f x +e \right )}+120 b^{3} {\mathrm e}^{10 i \left (f x +e \right )}+160 a^{3} {\mathrm e}^{8 i \left (f x +e \right )}+415 a^{2} b \,{\mathrm e}^{8 i \left (f x +e \right )}+1830 a \,b^{2} {\mathrm e}^{8 i \left (f x +e \right )}+240 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+640 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}-1460 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+240 b^{3} {\mathrm e}^{6 i \left (f x +e \right )}+16 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}-223 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+1336 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-48 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+268 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{2} b -290 \,{\mathrm e}^{2 i \left (f x +e \right )} a \,b^{2}+24 b^{3} {\mathrm e}^{2 i \left (f x +e \right )}+16 a^{3}-83 a^{2} b +6 a \,b^{2}\right )}{15 f \left (a +b \right )^{4} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}-\frac {3 \sqrt {-\left (a +b \right ) b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 \left (a +b \right )^{5} f}+\frac {\sqrt {-\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b}{\left (a +b \right )^{5} f}+\frac {3 \sqrt {-\left (a +b \right ) b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 \left (a +b \right )^{5} f}-\frac {\sqrt {-\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b}{\left (a +b \right )^{5} f}\) | \(572\) |
Input:
int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(-b*a/(a+b)^4*(1/2*a*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+1/2*(3*a-4*b)/((a +b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))-1/5/(a+b)^2/tan(f*x+e)^ 5-a*(a-2*b)/(a+b)^4/tan(f*x+e)-2/3*a/(a+b)^3/tan(f*x+e)^3)
Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (132) = 264\).
Time = 0.15 (sec) , antiderivative size = 987, normalized size of antiderivative = 6.67 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
[-1/120*(4*(16*a^3 - 83*a^2*b + 6*a*b^2)*cos(f*x + e)^7 - 4*(40*a^3 - 201* a^2*b + 68*a*b^2 - 6*b^3)*cos(f*x + e)^5 + 20*(6*a^3 - 29*a^2*b + 28*a*b^2 )*cos(f*x + e)^3 + 15*((3*a^3 - 4*a^2*b)*cos(f*x + e)^6 - (6*a^3 - 11*a^2* b + 4*a*b^2)*cos(f*x + e)^4 + 3*a^2*b - 4*a*b^2 + (3*a^3 - 10*a^2*b + 8*a* b^2)*cos(f*x + e)^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f* x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/ (a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) + 60*(3*a^ 2*b - 4*a*b^2)*cos(f*x + e))/(((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a* b^4)*f*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 - b^5)*f*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^ 4 - 2*b^5)*f*cos(f*x + e)^2 + (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b ^5)*f)*sin(f*x + e)), -1/60*(2*(16*a^3 - 83*a^2*b + 6*a*b^2)*cos(f*x + e)^ 7 - 2*(40*a^3 - 201*a^2*b + 68*a*b^2 - 6*b^3)*cos(f*x + e)^5 + 10*(6*a^3 - 29*a^2*b + 28*a*b^2)*cos(f*x + e)^3 - 15*((3*a^3 - 4*a^2*b)*cos(f*x + e)^ 6 - (6*a^3 - 11*a^2*b + 4*a*b^2)*cos(f*x + e)^4 + 3*a^2*b - 4*a*b^2 + (3*a ^3 - 10*a^2*b + 8*a*b^2)*cos(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*si n(f*x + e) + 30*(3*a^2*b - 4*a*b^2)*cos(f*x + e))/(((a^5 + 4*a^4*b + 6*a^3 *b^2 + 4*a^2*b^3 + a*b^4)*f*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b...
Timed out. \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(csc(f*x+e)**6/(a+b*sec(f*x+e)**2)**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (132) = 264\).
Time = 0.12 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.81 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {15 \, {\left (3 \, a^{2} b - 4 \, a b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, {\left (3 \, a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{6} + 10 \, {\left (3 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{4} + 6 \, a^{3} + 18 \, a^{2} b + 18 \, a b^{2} + 6 \, b^{3} + 2 \, {\left (10 \, a^{3} + 23 \, a^{2} b + 16 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{7} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5}}}{30 \, f} \] Input:
integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
-1/30*(15*(3*a^2*b - 4*a*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt((a + b)*b)) + (15*(3*a^2*b - 4*a*b^2)*tan(f*x + e)^6 + 10*(3*a^3 - a^2*b - 4*a*b^2)*tan(f*x + e)^4 + 6* a^3 + 18*a^2*b + 18*a*b^2 + 6*b^3 + 2*(10*a^3 + 23*a^2*b + 16*a*b^2 + 3*b^ 3)*tan(f*x + e)^2)/((a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*tan(f* x + e)^7 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*tan(f *x + e)^5))/f
Time = 0.26 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.72 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {15 \, a^{2} b \tan \left (f x + e\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac {15 \, {\left (3 \, a^{2} b - 4 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {a b + b^{2}}} + \frac {2 \, {\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 30 \, a b \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 10 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{5}}}{30 \, f} \] Input:
integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
-1/30*(15*a^2*b*tan(f*x + e)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)* (b*tan(f*x + e)^2 + a + b)) + 15*(3*a^2*b - 4*a*b^2)*(pi*floor((f*x + e)/p i + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt(a*b + b^2)) + 2*(15*a^2*tan(f*x + e)^4 - 30*a*b*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 + 10*a*b*tan(f*x + e)^2 + 3 *a^2 + 6*a*b + 3*b^2)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*tan(f*x + e)^5))/f
Time = 14.37 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.34 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {a\,\sqrt {b}\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (3\,a-4\,b\right )\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )}{{\left (a+b\right )}^{9/2}\,\left (4\,a\,b-3\,a^2\right )}\right )\,\left (3\,a-4\,b\right )}{2\,f\,{\left (a+b\right )}^{9/2}}-\frac {\frac {1}{5\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (4\,a\,b-3\,a^2\right )}{3\,{\left (a+b\right )}^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,a+3\,b\right )}{15\,{\left (a+b\right )}^2}-\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (4\,a\,b-3\,a^2\right )}{2\,{\left (a+b\right )}^4}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^7+\left (a+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )} \] Input:
int(1/(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^2),x)
Output:
(a*b^(1/2)*atan((a*b^(1/2)*tan(e + f*x)*(3*a - 4*b)*(4*a*b^3 + 4*a^3*b + a ^4 + b^4 + 6*a^2*b^2))/((a + b)^(9/2)*(4*a*b - 3*a^2)))*(3*a - 4*b))/(2*f* (a + b)^(9/2)) - (1/(5*(a + b)) - (tan(e + f*x)^4*(4*a*b - 3*a^2))/(3*(a + b)^3) + (tan(e + f*x)^2*(10*a + 3*b))/(15*(a + b)^2) - (b*tan(e + f*x)^6* (4*a*b - 3*a^2))/(2*(a + b)^4))/(f*(tan(e + f*x)^5*(a + b) + b*tan(e + f*x )^7))
Time = 0.25 (sec) , antiderivative size = 927, normalized size of antiderivative = 6.26 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x)
Output:
( - 45*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s qrt(b))*sin(e + f*x)**7*a**3 + 60*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*ta n((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**7*a**2*b + 45*sqrt(b)*sqr t(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f* x)**5*a**3 - 15*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - s qrt(a))/sqrt(b))*sin(e + f*x)**5*a**2*b - 60*sqrt(b)*sqrt(a + b)*atan((sqr t(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**5*a*b**2 - 45* sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b)) *sin(e + f*x)**7*a**3 + 60*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**7*a**2*b + 45*sqrt(b)*sqrt(a + b )*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**5*a **3 - 15*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a)) /sqrt(b))*sin(e + f*x)**5*a**2*b - 60*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b )*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**5*a*b**2 - 16*cos(e + f*x)*sin(e + f*x)**6*a**4 + 67*cos(e + f*x)*sin(e + f*x)**6*a**3*b + 77*c os(e + f*x)*sin(e + f*x)**6*a**2*b**2 - 6*cos(e + f*x)*sin(e + f*x)**6*a*b **3 + 8*cos(e + f*x)*sin(e + f*x)**4*a**4 - 40*cos(e + f*x)*sin(e + f*x)** 4*a**3*b - 98*cos(e + f*x)*sin(e + f*x)**4*a**2*b**2 - 44*cos(e + f*x)*sin (e + f*x)**4*a*b**3 + 6*cos(e + f*x)*sin(e + f*x)**4*b**4 + 2*cos(e + f*x) *sin(e + f*x)**2*a**4 - 6*cos(e + f*x)*sin(e + f*x)**2*a**3*b - 30*cos(...