Integrand size = 26, antiderivative size = 65 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {2 a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac {1}{f \sqrt {a \cos ^2(e+f x)}} \]
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {15-10 \sec ^2(e+f x)+3 \sec ^4(e+f x)}{15 f \sqrt {a \cos ^2(e+f x)}} \]
Time = 0.36 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.11, 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.
\(\displaystyle \int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^5}{\sqrt {a-a \sin (e+f x)^2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {\tan ^5(e+f x)}{\sqrt {a \cos ^2(e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\tan \left (e+f x+\frac {\pi }{2}\right )^5 \sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\sqrt {a \sin \left (\frac {1}{2} (2 e+\pi )+f x\right )^2} \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^5}dx\) |
\(\Big \downarrow \) 3684 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right )^2 \sec ^6(e+f x)}{\sqrt {a \cos ^2(e+f x)}}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle -\frac {a^3 \int \frac {\left (1-\cos ^2(e+f x)\right )^2}{\left (a \cos ^2(e+f x)\right )^{7/2}}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {a^3 \int \left (\frac {1}{\left (a \cos ^2(e+f x)\right )^{7/2}}-\frac {2}{\left (a \cos ^2(e+f x)\right )^{5/2} a}+\frac {1}{\left (a \cos ^2(e+f x)\right )^{3/2} a^2}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \left (-\frac {2}{a^3 \sqrt {a \cos ^2(e+f x)}}+\frac {4}{3 a^2 \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac {2}{5 a \left (a \cos ^2(e+f x)\right )^{5/2}}\right )}{2 f}\) |
-1/2*(a^3*(-2/(5*a*(a*Cos[e + f*x]^2)^(5/2)) + 4/(3*a^2*(a*Cos[e + f*x]^2) ^(3/2)) - 2/(a^3*Sqrt[a*Cos[e + f*x]^2])))/f
3.5.68.3.1 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 = 0.64 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74
method | result | size |
default | \(\frac {15 \left (\cos ^{4}\left (f x +e \right )\right )-10 \left (\cos ^{2}\left (f x +e \right )\right )+3}{15 \cos \left (f x +e \right )^{4} \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\) | \(48\) |
risch | \(\frac {2 \,{\mathrm e}^{8 i \left (f x +e \right )}+\frac {8 \,{\mathrm e}^{6 i \left (f x +e \right )}}{3}+\frac {116 \,{\mathrm e}^{4 i \left (f x +e \right )}}{15}+\frac {8 \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}+2}{\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4} f}\) | \(91\) |
Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {{\left (15 \, \cos \left (f x + e\right )^{4} - 10 \, \cos \left (f x + e\right )^{2} + 3\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{15 \, a f \cos \left (f x + e\right )^{6}} \]
1/15*(15*cos(f*x + e)^4 - 10*cos(f*x + e)^2 + 3)*sqrt(a*cos(f*x + e)^2)/(a *f*cos(f*x + e)^6)
\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {15 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )}^{2} a^{3} + 10 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + 3 \, a^{5}}{15 \, {\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} a^{3} f} \]
1/15*(15*(a*sin(f*x + e)^2 - a)^2*a^3 + 10*(a*sin(f*x + e)^2 - a)*a^4 + 3* a^5)/((-a*sin(f*x + e)^2 + a)^(5/2)*a^3*f)
Time = 1.66 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {16 \, {\left (10 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5} \sqrt {a} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \]
16/15*(10*tan(1/2*f*x + 1/2*e)^4 - 5*tan(1/2*f*x + 1/2*e)^2 + 1)/((tan(1/2 *f*x + 1/2*e)^2 - 1)^5*sqrt(a)*f*sgn(tan(1/2*f*x + 1/2*e)^4 - 1))
Time = 21.86 (sec) , antiderivative size = 486, normalized size of antiderivative = 7.48 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {4\,{\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}}{a\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\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 {32\,{\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\,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 {352\,{\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\,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 {128\,{\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}}{5\,a\,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 {64\,{\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}}{5\,a\,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 )} \]
(4*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))/(a*f*(exp(e*2i + f*x*2i) + 1)*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (32*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1 i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(3*a*f*(exp(e*2i + f*x*2i) + 1)^2*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) + (352*exp(e*3i + f*x*3 i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/ 2))/(15*a*f*(exp(e*2i + f*x*2i) + 1)^3*(exp(e*1i + f*x*1i) + exp(e*3i + f* x*3i))) - (128*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (e xp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(5*a*f*(exp(e*2i + f*x*2i) + 1)^4*(exp( e*1i + f*x*1i) + exp(e*3i + f*x*3i))) + (64*exp(e*3i + f*x*3i)*(a - a*((ex p(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(5*a*f*(ex p(e*2i + f*x*2i) + 1)^5*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))