Integrand size = 26, antiderivative size = 68 \[ \int \frac {\tan ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a^2}{7 f \left (a \cos ^2(e+f x)\right )^{7/2}}-\frac {2 a}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}+\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]
1/7*a^2/f/(a*cos(f*x+e)^2)^(7/2)-2/5*a/f/(a*cos(f*x+e)^2)^(5/2)+1/3/f/(a*c os(f*x+e)^2)^(3/2)
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\left (15-42 \cos ^2(e+f x)+35 \cos ^4(e+f x)\right ) \sec ^4(e+f x)}{105 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]
((15 - 42*Cos[e + f*x]^2 + 35*Cos[e + f*x]^4)*Sec[e + f*x]^4)/(105*f*(a*Co s[e + f*x]^2)^(3/2))
Time = 0.37 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3655, 3042, 25, 3684, 8, 53, 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.
| \(\displaystyle \int \frac {\tan ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^5}{\left (a-a \sin (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3655 |
| \(\displaystyle \int \frac {\tan ^5(e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\tan \left (e+f x+\frac {\pi }{2}\right )^5 \left (a \sin \left (e+f x+\frac {\pi }{2}\right )^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\left (a \sin \left (\frac {1}{2} (2 e+\pi )+f x\right )^2\right )^{3/2} \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^5}dx\) |
| \(\Big \downarrow \) 3684 |
| \(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right )^2 \sec ^6(e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}}d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 8 |
| \(\displaystyle -\frac {a^3 \int \frac {\left (1-\cos ^2(e+f x)\right )^2}{\left (a \cos ^2(e+f x)\right )^{9/2}}d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle -\frac {a^3 \int \left (\frac {1}{\left (a \cos ^2(e+f x)\right )^{9/2}}-\frac {2}{\left (a \cos ^2(e+f x)\right )^{7/2} a}+\frac {1}{\left (a \cos ^2(e+f x)\right )^{5/2} a^2}\right )d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 \left (-\frac {2}{3 a^3 \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac {4}{5 a^2 \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {2}{7 a \left (a \cos ^2(e+f x)\right )^{7/2}}\right )}{2 f}\) |
-1/2*(a^3*(-2/(7*a*(a*Cos[e + f*x]^2)^(7/2)) + 4/(5*a^2*(a*Cos[e + f*x]^2) ^(5/2)) - 2/(3*a^3*(a*Cos[e + f*x]^2)^(3/2))))/f
3.5.78.3.1 Defintions of rubi rules used
Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m Int[u*(a* x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*cos[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a + b, 0]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_. ), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1 )/2)/(2*f) Subst[Int[x^((m - 1)/2)*((b*ff^(n/2)*x^(n/2))^p/(1 - ff*x)^((m + 1)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && Inte gerQ[(m - 1)/2] && IntegerQ[n/2]
Time = 1.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(\frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (35 \left (\cos ^{4}\left (f x +e \right )\right )-42 \left (\cos ^{2}\left (f x +e \right )\right )+15\right )}{105 a^{2} \cos \left (f x +e \right )^{8} f}\) | \(51\) |
| risch | \(\frac {\frac {8 \,{\mathrm e}^{10 i \left (f x +e \right )}}{3}-\frac {32 \,{\mathrm e}^{8 i \left (f x +e \right )}}{15}+\frac {304 \,{\mathrm e}^{6 i \left (f x +e \right )}}{35}-\frac {32 \,{\mathrm e}^{4 i \left (f x +e \right )}}{15}+\frac {8 \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{6} a}\) | \(104\) |
Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {{\left (35 \, \cos \left (f x + e\right )^{4} - 42 \, \cos \left (f x + e\right )^{2} + 15\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{105 \, a^{2} f \cos \left (f x + e\right )^{8}} \]
1/105*(35*cos(f*x + e)^4 - 42*cos(f*x + e)^2 + 15)*sqrt(a*cos(f*x + e)^2)/ (a^2*f*cos(f*x + e)^8)
\[ \int \frac {\tan ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {35 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )}^{2} a^{3} + 42 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + 15 \, a^{5}}{105 \, {\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {7}{2}} a^{3} f} \]
1/105*(35*(a*sin(f*x + e)^2 - a)^2*a^3 + 42*(a*sin(f*x + e)^2 - a)*a^4 + 1 5*a^5)/((-a*sin(f*x + e)^2 + a)^(7/2)*a^3*f)
Time = 2.97 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.37 \[ \int \frac {\tan ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {16 \, {\left (70 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} + 35 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 21 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 7 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}}{105 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{7} a^{\frac {3}{2}} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \]
16/105*(70*tan(1/2*f*x + 1/2*e)^8 + 35*tan(1/2*f*x + 1/2*e)^6 + 21*tan(1/2 *f*x + 1/2*e)^4 - 7*tan(1/2*f*x + 1/2*e)^2 + 1)/((tan(1/2*f*x + 1/2*e)^2 - 1)^7*a^(3/2)*f*sgn(tan(1/2*f*x + 1/2*e)^4 - 1))
Time = 29.97 (sec) , antiderivative size = 583, normalized size of antiderivative = 8.57 \[ \int \frac {\tan ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {16\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {464\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{15\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {3072\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{35\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {4736\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{35\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^5\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {768\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{7\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^6\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {256\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{7\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^7\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \]
(16*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f *x*1i)*1i)/2)^2)^(1/2))/(3*a^2*f*(exp(e*2i + f*x*2i) + 1)^2*(exp(e*1i + f* x*1i) + exp(e*3i + f*x*3i))) - (464*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(15*a^2*f*(exp(e*2 i + f*x*2i) + 1)^3*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) + (3072*exp( e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1 i)/2)^2)^(1/2))/(35*a^2*f*(exp(e*2i + f*x*2i) + 1)^4*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (4736*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x *1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(35*a^2*f*(exp(e*2i + f* x*2i) + 1)^5*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) + (768*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2 )^(1/2))/(7*a^2*f*(exp(e*2i + f*x*2i) + 1)^6*(exp(e*1i + f*x*1i) + exp(e*3 i + f*x*3i))) - (256*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/ 2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(7*a^2*f*(exp(e*2i + f*x*2i) + 1) ^7*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))