Integrand size = 25, antiderivative size = 171 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (15 a^2-40 a b+24 b^2\right ) \cot (e+f x)}{15 a^3 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {2 (5 a-3 b) \cot ^3(e+f x)}{15 a^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\cot ^5(e+f x)}{5 a f \sqrt {a+b \tan ^2(e+f x)}}-\frac {2 b \left (15 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{15 a^4 f \sqrt {a+b \tan ^2(e+f x)}} \]
-1/15*(15*a^2-40*a*b+24*b^2)*cot(f*x+e)/a^3/f/(a+b*tan(f*x+e)^2)^(1/2)-2/1 5*(5*a-3*b)*cot(f*x+e)^3/a^2/f/(a+b*tan(f*x+e)^2)^(1/2)-1/5*cot(f*x+e)^5/a /f/(a+b*tan(f*x+e)^2)^(1/2)-2/15*b*(15*a^2-40*a*b+24*b^2)*tan(f*x+e)/a^4/f /(a+b*tan(f*x+e)^2)^(1/2)
Time = 2.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.79 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (\cot (e+f x) \left (8 a^2-41 a b+33 b^2+a (4 a-9 b) \csc ^2(e+f x)+3 a^2 \csc ^4(e+f x)\right )+\frac {15 (a-b)^2 b \sin (2 (e+f x))}{a+b+(a-b) \cos (2 (e+f x))}\right )}{15 \sqrt {2} a^4 f} \]
-1/15*(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(Cot[e + f* x]*(8*a^2 - 41*a*b + 33*b^2 + a*(4*a - 9*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4) + (15*(a - b)^2*b*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f *x)])))/(Sqrt[2]*a^4*f)
Time = 0.35 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4146, 365, 359, 245, 208}
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.
| \(\displaystyle \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^6 \left (a+b \tan (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int \frac {\cot ^4(e+f x) \left (5 a \tan ^2(e+f x)+2 (5 a-3 b)\right )}{\left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{5 a}-\frac {\cot ^5(e+f x)}{5 a \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {\frac {\left (15 a^2-8 b (5 a-3 b)\right ) \int \frac {\cot ^2(e+f x)}{\left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{3 a}-\frac {2 (5 a-3 b) \cot ^3(e+f x)}{3 a \sqrt {a+b \tan ^2(e+f x)}}}{5 a}-\frac {\cot ^5(e+f x)}{5 a \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 245 |
| \(\displaystyle \frac {\frac {\frac {\left (15 a^2-8 b (5 a-3 b)\right ) \left (-\frac {2 b \int \frac {1}{\left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{a}-\frac {\cot (e+f x)}{a \sqrt {a+b \tan ^2(e+f x)}}\right )}{3 a}-\frac {2 (5 a-3 b) \cot ^3(e+f x)}{3 a \sqrt {a+b \tan ^2(e+f x)}}}{5 a}-\frac {\cot ^5(e+f x)}{5 a \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {\frac {\frac {\left (15 a^2-8 b (5 a-3 b)\right ) \left (-\frac {2 b \tan (e+f x)}{a^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\cot (e+f x)}{a \sqrt {a+b \tan ^2(e+f x)}}\right )}{3 a}-\frac {2 (5 a-3 b) \cot ^3(e+f x)}{3 a \sqrt {a+b \tan ^2(e+f x)}}}{5 a}-\frac {\cot ^5(e+f x)}{5 a \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
(-1/5*Cot[e + f*x]^5/(a*Sqrt[a + b*Tan[e + f*x]^2]) + ((-2*(5*a - 3*b)*Cot [e + f*x]^3)/(3*a*Sqrt[a + b*Tan[e + f*x]^2]) + ((15*a^2 - 8*(5*a - 3*b)*b )*(-(Cot[e + f*x]/(a*Sqrt[a + b*Tan[e + f*x]^2])) - (2*b*Tan[e + f*x])/(a^ 2*Sqrt[a + b*Tan[e + f*x]^2])))/(3*a))/(5*a))/f
3.2.39.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
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] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -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[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Time = 5.47 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.21
| method | result | size |
| default | \(-\frac {\left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (8 a^{3} \cos \left (f x +e \right )^{6}+64 \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )^{2} a^{2} b +104 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4} a \,b^{2}+48 \sin \left (f x +e \right )^{6} b^{3}-20 a^{3} \cos \left (f x +e \right )^{4}-100 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a^{2} b -80 a \,b^{2} \sin \left (f x +e \right )^{4}+15 a^{3} \cos \left (f x +e \right )^{2}+30 a^{2} b \sin \left (f x +e \right )^{2}\right ) \sec \left (f x +e \right )^{3} \csc \left (f x +e \right )^{5}}{15 f \,a^{4} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\) | \(207\) |
-1/15/f/a^4*(a*cos(f*x+e)^2+b*sin(f*x+e)^2)*(8*a^3*cos(f*x+e)^6+64*cos(f*x +e)^4*sin(f*x+e)^2*a^2*b+104*cos(f*x+e)^2*sin(f*x+e)^4*a*b^2+48*sin(f*x+e) ^6*b^3-20*a^3*cos(f*x+e)^4-100*cos(f*x+e)^2*sin(f*x+e)^2*a^2*b-80*a*b^2*si n(f*x+e)^4+15*a^3*cos(f*x+e)^2+30*a^2*b*sin(f*x+e)^2)/(a+b*tan(f*x+e)^2)^( 3/2)*sec(f*x+e)^3*csc(f*x+e)^5
Time = 18.75 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.36 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (8 \, {\left (a^{3} - 8 \, a^{2} b + 13 \, a b^{2} - 6 \, b^{3}\right )} \cos \left (f x + e\right )^{7} - 4 \, {\left (5 \, a^{3} - 41 \, a^{2} b + 72 \, a b^{2} - 36 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{3} - 130 \, a^{2} b + 264 \, a b^{2} - 144 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (15 \, a^{2} b - 40 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{5} - a^{4} b\right )} f \cos \left (f x + e\right )^{6} + a^{4} b f - {\left (2 \, a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )} \]
-1/15*(8*(a^3 - 8*a^2*b + 13*a*b^2 - 6*b^3)*cos(f*x + e)^7 - 4*(5*a^3 - 41 *a^2*b + 72*a*b^2 - 36*b^3)*cos(f*x + e)^5 + (15*a^3 - 130*a^2*b + 264*a*b ^2 - 144*b^3)*cos(f*x + e)^3 + 2*(15*a^2*b - 40*a*b^2 + 24*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^5 - a^4*b)*f*co s(f*x + e)^6 + a^4*b*f - (2*a^5 - 3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4 *b)*f*cos(f*x + e)^2)*sin(f*x + e))
\[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\csc ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.49 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {30 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{2}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {48 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{4}} + \frac {15}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a \tan \left (f x + e\right )} - \frac {40 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{2} \tan \left (f x + e\right )} + \frac {24 \, b^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3} \tan \left (f x + e\right )} + \frac {10}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a \tan \left (f x + e\right )^{3}} - \frac {6 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a \tan \left (f x + e\right )^{5}}}{15 \, f} \]
-1/15*(30*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^2) - 80*b^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^3) + 48*b^3*tan(f*x + e)/(sqrt(b*tan(f *x + e)^2 + a)*a^4) + 15/(sqrt(b*tan(f*x + e)^2 + a)*a*tan(f*x + e)) - 40* b/(sqrt(b*tan(f*x + e)^2 + a)*a^2*tan(f*x + e)) + 24*b^2/(sqrt(b*tan(f*x + e)^2 + a)*a^3*tan(f*x + e)) + 10/(sqrt(b*tan(f*x + e)^2 + a)*a*tan(f*x + e)^3) - 6*b/(sqrt(b*tan(f*x + e)^2 + a)*a^2*tan(f*x + e)^3) + 3/(sqrt(b*ta n(f*x + e)^2 + a)*a*tan(f*x + e)^5))/f
\[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{6}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Hanged} \]