Integrand size = 9, antiderivative size = 50 \[ \int (\sin (x) \tan (x))^{5/2} \, dx=\frac {64}{15} \cot (x) \sqrt {\sin (x) \tan (x)}+\frac {16}{15} \tan (x) \sqrt {\sin (x) \tan (x)}-\frac {2}{5} \sin ^2(x) \tan (x) \sqrt {\sin (x) \tan (x)} \]
64/15*cot(x)*(sin(x)*tan(x))^(1/2)+16/15*(sin(x)*tan(x))^(1/2)*tan(x)-2/5* sin(x)^2*(sin(x)*tan(x))^(1/2)*tan(x)
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.58 \[ \int (\sin (x) \tan (x))^{5/2} \, dx=\frac {2}{15} \left (5+3 \cos ^2(x)+32 \cot ^2(x)\right ) \tan (x) \sqrt {\sin (x) \tan (x)} \]
Time = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.52, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.889, Rules used = {3042, 4900, 3042, 3078, 3042, 3074, 3042, 3069}
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.
\(\displaystyle \int (\sin (x) \tan (x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (\sin (x) \tan (x))^{5/2}dx\) |
\(\Big \downarrow \) 4900 |
\(\displaystyle \frac {\sqrt {\sin (x) \tan (x)} \int \sin ^{\frac {5}{2}}(x) \tan ^{\frac {5}{2}}(x)dx}{\sqrt {\sin (x)} \sqrt {\tan (x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (x) \tan (x)} \int \sin (x)^{5/2} \tan (x)^{5/2}dx}{\sqrt {\sin (x)} \sqrt {\tan (x)}}\) |
\(\Big \downarrow \) 3078 |
\(\displaystyle \frac {\sqrt {\sin (x) \tan (x)} \left (\frac {8}{5} \int \sqrt {\sin (x)} \tan ^{\frac {5}{2}}(x)dx-\frac {2}{5} \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)\right )}{\sqrt {\sin (x)} \sqrt {\tan (x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (x) \tan (x)} \left (\frac {8}{5} \int \sqrt {\sin (x)} \tan (x)^{5/2}dx-\frac {2}{5} \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)\right )}{\sqrt {\sin (x)} \sqrt {\tan (x)}}\) |
\(\Big \downarrow \) 3074 |
\(\displaystyle \frac {\sqrt {\sin (x) \tan (x)} \left (\frac {8}{5} \left (\frac {2}{3} \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)-\frac {4}{3} \int \sqrt {\sin (x)} \sqrt {\tan (x)}dx\right )-\frac {2}{5} \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)\right )}{\sqrt {\sin (x)} \sqrt {\tan (x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (x) \tan (x)} \left (\frac {8}{5} \left (\frac {2}{3} \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)-\frac {4}{3} \int \sqrt {\sin (x)} \sqrt {\tan (x)}dx\right )-\frac {2}{5} \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)\right )}{\sqrt {\sin (x)} \sqrt {\tan (x)}}\) |
\(\Big \downarrow \) 3069 |
\(\displaystyle \frac {\sqrt {\sin (x) \tan (x)} \left (\frac {8}{5} \left (\frac {2}{3} \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)+\frac {8 \sqrt {\sin (x)}}{3 \sqrt {\tan (x)}}\right )-\frac {2}{5} \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)\right )}{\sqrt {\sin (x)} \sqrt {\tan (x)}}\) |
(Sqrt[Sin[x]*Tan[x]]*((-2*Sin[x]^(5/2)*Tan[x]^(3/2))/5 + (8*((8*Sqrt[Sin[x ]])/(3*Sqrt[Tan[x]]) + (2*Sqrt[Sin[x]]*Tan[x]^(3/2))/3))/5))/(Sqrt[Sin[x]] *Sqrt[Tan[x]])
3.2.49.3.1 Defintions of rubi rules used
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f* m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] - Simp[b^2*((m + n - 1)/(n - 1)) Int[(a*Sin[e + f*x])^m*(b*Ta n[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && In tegersQ[2*m, 2*n] && !(GtQ[m, 1] && !IntegerQ[(m - 1)/2])
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/( f*m)), x] + Simp[a^2*((m + n - 1)/m) Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan[ e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1 ] && EqQ[n, 1/2])) && IntegersQ[2*m, 2*n]
Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTri g[u], vv = ActivateTrig[v], ww = ActivateTrig[w]}, Simp[(vv^m*ww^n)^FracPar t[p]/(vv^(m*FracPart[p])*ww^(n*FracPart[p])) Int[uu*vv^(m*p)*ww^(n*p), x] , x]] /; FreeQ[{m, n, p}, x] && !IntegerQ[p] && ( !InertTrigFreeQ[v] || ! InertTrigFreeQ[w])
Leaf count of result is larger than twice the leaf count of optimal. \(282\) vs. \(2(38)=76\).
Time = 2.68 (sec) , antiderivative size = 283, normalized size of antiderivative = 5.66
method | result | size |
default | \(\frac {\tan \left (x \right ) \left (6 \cos \left (x \right )^{4}-15 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}\, \ln \left (\frac {2 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-\cos \left (x \right )+1}{\cos \left (x \right )+1}\right ) \cos \left (x \right )^{2}+15 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}\, \ln \left (\frac {4 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-2 \cos \left (x \right )+4 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2}{\cos \left (x \right )+1}\right ) \cos \left (x \right )^{2}-15 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}\, \ln \left (\frac {2 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-\cos \left (x \right )+1}{\cos \left (x \right )+1}\right ) \cos \left (x \right )+15 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}\, \ln \left (\frac {4 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-2 \cos \left (x \right )+4 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2}{\cos \left (x \right )+1}\right ) \cos \left (x \right )-60 \cos \left (x \right )^{2}-10\right ) \sqrt {\sin \left (x \right ) \tan \left (x \right )}\, \sqrt {4}}{30 \cos \left (x \right )^{2}-30}\) | \(283\) |
1/30*tan(x)*(6*cos(x)^4-15*(-cos(x)/(cos(x)+1)^2)^(1/2)*ln((2*cos(x)*(-cos (x)/(cos(x)+1)^2)^(1/2)+2*(-cos(x)/(cos(x)+1)^2)^(1/2)-cos(x)+1)/(cos(x)+1 ))*cos(x)^2+15*(-cos(x)/(cos(x)+1)^2)^(1/2)*ln(2*(2*cos(x)*(-cos(x)/(cos(x )+1)^2)^(1/2)+2*(-cos(x)/(cos(x)+1)^2)^(1/2)-cos(x)+1)/(cos(x)+1))*cos(x)^ 2-15*(-cos(x)/(cos(x)+1)^2)^(1/2)*ln((2*cos(x)*(-cos(x)/(cos(x)+1)^2)^(1/2 )+2*(-cos(x)/(cos(x)+1)^2)^(1/2)-cos(x)+1)/(cos(x)+1))*cos(x)+15*(-cos(x)/ (cos(x)+1)^2)^(1/2)*ln(2*(2*cos(x)*(-cos(x)/(cos(x)+1)^2)^(1/2)+2*(-cos(x) /(cos(x)+1)^2)^(1/2)-cos(x)+1)/(cos(x)+1))*cos(x)-60*cos(x)^2-10)*(sin(x)* tan(x))^(1/2)/(cos(x)^2-1)*4^(1/2)
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int (\sin (x) \tan (x))^{5/2} \, dx=-\frac {2 \, {\left (3 \, \cos \left (x\right )^{4} - 30 \, \cos \left (x\right )^{2} - 5\right )} \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}}}{15 \, \cos \left (x\right ) \sin \left (x\right )} \]
Timed out. \[ \int (\sin (x) \tan (x))^{5/2} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (38) = 76\).
Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.64 \[ \int (\sin (x) \tan (x))^{5/2} \, dx=-\frac {32 \, {\left (\frac {5 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + \frac {2 \, \sin \left (x\right )^{10}}{{\left (\cos \left (x\right ) + 1\right )}^{10}} - 2\right )}}{15 \, {\left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \]
-32/15*(5*sin(x)^4/(cos(x) + 1)^4 - 5*sin(x)^6/(cos(x) + 1)^6 + 2*sin(x)^1 0/(cos(x) + 1)^10 - 2)/((sin(x)/(cos(x) + 1) + 1)^(5/2)*(-sin(x)/(cos(x) + 1) + 1)^(5/2)*(sin(x)^2/(cos(x) + 1)^2 + 1)^(5/2))
\[ \int (\sin (x) \tan (x))^{5/2} \, dx=\int { \left (\sin \left (x\right ) \tan \left (x\right )\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (\sin (x) \tan (x))^{5/2} \, dx=\int {\left (\sin \left (x\right )\,\mathrm {tan}\left (x\right )\right )}^{5/2} \,d x \]