Integrand size = 25, antiderivative size = 146 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a-2 b) \cot (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 (a-2 b) b \tan (e+f x)}{3 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 (a-2 b) b \tan (e+f x)}{3 a^4 f \sqrt {a+b \tan ^2(e+f x)}} \]
-8/3*(a-2*b)*b*tan(f*x+e)/a^4/f/(a+b*tan(f*x+e)^2)^(1/2)-(a-2*b)*cot(f*x+e )/a^2/f/(a+b*tan(f*x+e)^2)^(3/2)-1/3*cot(f*x+e)^3/a/f/(a+b*tan(f*x+e)^2)^( 3/2)-4/3*(a-2*b)*b*tan(f*x+e)/a^3/f/(a+b*tan(f*x+e)^2)^(3/2)
Time = 2.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.96 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-\cot (e+f x) \left (2 a-8 b+a \csc ^2(e+f x)\right )+\frac {2 b \left (-3 a^2+2 a b+4 b^2+\left (-3 a^2+7 a b-4 b^2\right ) \cos (2 (e+f x))\right ) \sin (2 (e+f x))}{(a+b+(a-b) \cos (2 (e+f x)))^2}\right )}{3 \sqrt {2} a^4 f} \]
(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-(Cot[e + f*x]*( 2*a - 8*b + a*Csc[e + f*x]^2)) + (2*b*(-3*a^2 + 2*a*b + 4*b^2 + (-3*a^2 + 7*a*b - 4*b^2)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2* (e + f*x)])^2))/(3*Sqrt[2]*a^4*f)
Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4146, 359, 245, 209, 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.
| \(\displaystyle \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^4 \left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\cot ^4(e+f x) \left (\tan ^2(e+f x)+1\right )}{\left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {(a-2 b) \int \frac {\cot ^2(e+f x)}{\left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan (e+f x)}{a}-\frac {\cot ^3(e+f x)}{3 a \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 245 |
| \(\displaystyle \frac {\frac {(a-2 b) \left (-\frac {4 b \int \frac {1}{\left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan (e+f x)}{a}-\frac {\cot (e+f x)}{a \left (a+b \tan ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cot ^3(e+f x)}{3 a \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {\frac {(a-2 b) \left (-\frac {4 b \left (\frac {2 \int \frac {1}{\left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{3 a}+\frac {\tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cot (e+f x)}{a \left (a+b \tan ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cot ^3(e+f x)}{3 a \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {\frac {(a-2 b) \left (-\frac {4 b \left (\frac {2 \tan (e+f x)}{3 a^2 \sqrt {a+b \tan ^2(e+f x)}}+\frac {\tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cot (e+f x)}{a \left (a+b \tan ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cot ^3(e+f x)}{3 a \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
(-1/3*Cot[e + f*x]^3/(a*(a + b*Tan[e + f*x]^2)^(3/2)) + ((a - 2*b)*(-(Cot[ e + f*x]/(a*(a + b*Tan[e + f*x]^2)^(3/2))) - (4*b*(Tan[e + f*x]/(3*a*(a + b*Tan[e + f*x]^2)^(3/2)) + (2*Tan[e + f*x])/(3*a^2*Sqrt[a + b*Tan[e + f*x] ^2])))/a))/a)/f
3.2.50.3.1 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[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
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[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 = 6.29 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.23
| method | result | size |
| default | \(\frac {\left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (2 a^{3} \cos \left (f x +e \right )^{6}+18 \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )^{2} a^{2} b +32 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4} a \,b^{2}+16 \sin \left (f x +e \right )^{6} b^{3}-3 a^{3} \cos \left (f x +e \right )^{4}-12 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a^{2} b -8 a \,b^{2} \sin \left (f x +e \right )^{4}\right ) \sec \left (f x +e \right )^{5} \csc \left (f x +e \right )^{3}}{3 f \,a^{4} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\) | \(180\) |
1/3/f/a^4*(a*cos(f*x+e)^2+b*sin(f*x+e)^2)*(2*a^3*cos(f*x+e)^6+18*cos(f*x+e )^4*sin(f*x+e)^2*a^2*b+32*cos(f*x+e)^2*sin(f*x+e)^4*a*b^2+16*sin(f*x+e)^6* b^3-3*a^3*cos(f*x+e)^4-12*cos(f*x+e)^2*sin(f*x+e)^2*a^2*b-8*a*b^2*sin(f*x+ e)^4)/(a+b*tan(f*x+e)^2)^(5/2)*sec(f*x+e)^5*csc(f*x+e)^3
Time = 47.75 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.64 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (2 \, {\left (a^{3} - 9 \, a^{2} b + 16 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (f x + e\right )^{7} - 3 \, {\left (a^{3} - 10 \, a^{2} b + 24 \, a b^{2} - 16 \, b^{3}\right )} \cos \left (f x + e\right )^{5} - 12 \, {\left (a^{2} b - 4 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left (a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{6} - a^{4} b^{2} f - {\left (a^{6} - 4 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{4} - {\left (2 \, a^{5} b - 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )} \]
-1/3*(2*(a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*x + e)^7 - 3*(a^3 - 10*a^ 2*b + 24*a*b^2 - 16*b^3)*cos(f*x + e)^5 - 12*(a^2*b - 4*a*b^2 + 4*b^3)*cos (f*x + e)^3 - 8*(a*b^2 - 2*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^6 - 2*a^5*b + a^4*b^2)*f*cos(f*x + e)^6 - a^4*b ^2*f - (a^6 - 4*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^4 - (2*a^5*b - 3*a^4*b^2 )*f*cos(f*x + e)^2)*sin(f*x + e))
\[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\csc ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.34 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {8 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {4 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {16 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{4}} - \frac {8 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )} - \frac {6 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \tan \left (f x + e\right )} + \frac {1}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )^{3}}}{3 \, f} \]
-1/3*(8*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^3) + 4*b*tan(f*x + e) /((b*tan(f*x + e)^2 + a)^(3/2)*a^2) - 16*b^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^4) - 8*b^2*tan(f*x + e)/((b*tan(f*x + e)^2 + a)^(3/2)*a^3) + 3/((b*tan(f*x + e)^2 + a)^(3/2)*a*tan(f*x + e)) - 6*b/((b*tan(f*x + e)^2 + a)^(3/2)*a^2*tan(f*x + e)) + 1/((b*tan(f*x + e)^2 + a)^(3/2)*a*tan(f*x + e)^3))/f
\[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{4}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \]