Integrand size = 13, antiderivative size = 70 \[ \int \frac {1}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \, dx=-\frac {4 \cos ^5(x) \sin (x)}{9 \sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}}-\frac {8 \cos ^3(x) \sin ^3(x)}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}}+\frac {4 \cos (x) \sin ^5(x)}{7 \sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \] Output:
-4/9*cos(x)^5*sin(x)/(cos(x)^11*sin(x)^13)^(1/4)-8*cos(x)^3*sin(x)^3/(cos( x)^11*sin(x)^13)^(1/4)+4/7*cos(x)*sin(x)^5/(cos(x)^11*sin(x)^13)^(1/4)
Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \, dx=-\frac {4 \cos (x) (15+8 \cos (2 x)-16 \cos (4 x)) \sin (x)}{63 \sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \] Input:
Integrate[(Cos[x]^11*Sin[x]^13)^(-1/4),x]
Output:
(-4*Cos[x]*(15 + 8*Cos[2*x] - 16*Cos[4*x])*Sin[x])/(63*(Cos[x]^11*Sin[x]^1 3)^(1/4))
Time = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4889, 7270, 244, 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.
\(\displaystyle \int \frac {1}{\sqrt [4]{\sin ^{13}(x) \cos ^{11}(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt [4]{\sin (x)^{13} \cos (x)^{11}}}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {1}{\sqrt [4]{\frac {\tan ^{13}(x)}{\left (\tan ^2(x)+1\right )^{12}}} \left (\tan ^2(x)+1\right )}d\tan (x)\) |
\(\Big \downarrow \) 7270 |
\(\displaystyle \frac {\tan ^{\frac {13}{4}}(x) \int \frac {\left (\tan ^2(x)+1\right )^2}{\tan ^{\frac {13}{4}}(x)}d\tan (x)}{\sqrt [4]{\frac {\tan ^{13}(x)}{\left (\tan ^2(x)+1\right )^{12}}} \left (\tan ^2(x)+1\right )^3}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\tan ^{\frac {13}{4}}(x) \int \left (\tan ^{\frac {3}{4}}(x)+\frac {2}{\tan ^{\frac {5}{4}}(x)}+\frac {1}{\tan ^{\frac {13}{4}}(x)}\right )d\tan (x)}{\sqrt [4]{\frac {\tan ^{13}(x)}{\left (\tan ^2(x)+1\right )^{12}}} \left (\tan ^2(x)+1\right )^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\tan ^{\frac {13}{4}}(x) \left (\frac {4}{7} \tan ^{\frac {7}{4}}(x)-\frac {4}{9 \tan ^{\frac {9}{4}}(x)}-\frac {8}{\sqrt [4]{\tan (x)}}\right )}{\sqrt [4]{\frac {\tan ^{13}(x)}{\left (\tan ^2(x)+1\right )^{12}}} \left (\tan ^2(x)+1\right )^3}\) |
Input:
Int[(Cos[x]^11*Sin[x]^13)^(-1/4),x]
Output:
(Tan[x]^(13/4)*(-4/(9*Tan[x]^(9/4)) - 8/Tan[x]^(1/4) + (4*Tan[x]^(7/4))/7) )/((Tan[x]^13/(1 + Tan[x]^2)^12)^(1/4)*(1 + Tan[x]^2)^3)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p ]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p]))) Int[u*v ^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] && !IntegerQ[p] && !Free Q[v, x] && !FreeQ[w, x]
\[\int \frac {1}{\left (\cos \left (x \right )^{11} \sin \left (x \right )^{13}\right )^{\frac {1}{4}}}d x\]
Input:
int(1/(cos(x)^11*sin(x)^13)^(1/4),x)
Output:
int(1/(cos(x)^11*sin(x)^13)^(1/4),x)
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \, dx=\frac {4 \, {\left (128 \, \cos \left (x\right )^{4} - 144 \, \cos \left (x\right )^{2} + 9\right )} \left ({\left (\cos \left (x\right )^{23} - 6 \, \cos \left (x\right )^{21} + 15 \, \cos \left (x\right )^{19} - 20 \, \cos \left (x\right )^{17} + 15 \, \cos \left (x\right )^{15} - 6 \, \cos \left (x\right )^{13} + \cos \left (x\right )^{11}\right )} \sin \left (x\right )\right )^{\frac {3}{4}}}{63 \, {\left (\cos \left (x\right )^{22} - 6 \, \cos \left (x\right )^{20} + 15 \, \cos \left (x\right )^{18} - 20 \, \cos \left (x\right )^{16} + 15 \, \cos \left (x\right )^{14} - 6 \, \cos \left (x\right )^{12} + \cos \left (x\right )^{10}\right )}} \] Input:
integrate(1/(cos(x)^11*sin(x)^13)^(1/4),x, algorithm="fricas")
Output:
4/63*(128*cos(x)^4 - 144*cos(x)^2 + 9)*((cos(x)^23 - 6*cos(x)^21 + 15*cos( x)^19 - 20*cos(x)^17 + 15*cos(x)^15 - 6*cos(x)^13 + cos(x)^11)*sin(x))^(3/ 4)/(cos(x)^22 - 6*cos(x)^20 + 15*cos(x)^18 - 20*cos(x)^16 + 15*cos(x)^14 - 6*cos(x)^12 + cos(x)^10)
Timed out. \[ \int \frac {1}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \, dx=\text {Timed out} \] Input:
integrate(1/(cos(x)**11*sin(x)**13)**(1/4),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \, dx=\frac {4}{23} \, \tan \left (x\right )^{\frac {23}{4}} + \frac {8}{15} \, \tan \left (x\right )^{\frac {15}{4}} + \frac {4}{7} \, \tan \left (x\right )^{\frac {7}{4}} - \frac {4 \, {\left (35 \, \tan \left (x\right )^{7} + 161 \, \tan \left (x\right )^{5} + 345 \, \tan \left (x\right )^{3} - 805 \, \tan \left (x\right )\right )}}{805 \, \tan \left (x\right )^{\frac {5}{4}}} + \frac {4 \, {\left (21 \, \tan \left (x\right )^{7} + 135 \, \tan \left (x\right )^{5} - 945 \, \tan \left (x\right )^{3} - 35 \, \tan \left (x\right )\right )}}{315 \, \tan \left (x\right )^{\frac {13}{4}}} \] Input:
integrate(1/(cos(x)^11*sin(x)^13)^(1/4),x, algorithm="maxima")
Output:
4/23*tan(x)^(23/4) + 8/15*tan(x)^(15/4) + 4/7*tan(x)^(7/4) - 4/805*(35*tan (x)^7 + 161*tan(x)^5 + 345*tan(x)^3 - 805*tan(x))/tan(x)^(5/4) + 4/315*(21 *tan(x)^7 + 135*tan(x)^5 - 945*tan(x)^3 - 35*tan(x))/tan(x)^(13/4)
\[ \int \frac {1}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \, dx=\int { \frac {1}{\left (\cos \left (x\right )^{11} \sin \left (x\right )^{13}\right )^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(cos(x)^11*sin(x)^13)^(1/4),x, algorithm="giac")
Output:
integrate((cos(x)^11*sin(x)^13)^(-1/4), x)
Time = 1.97 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \, dx=-\frac {2^{3/4}\,\left (-32\,{\cos \left (2\,x\right )}^2+8\,\cos \left (2\,x\right )+31\right )\,{\left (924\,\sin \left (2\,x\right )-132\,\sin \left (4\,x\right )-660\,\sin \left (6\,x\right )+165\,\sin \left (8\,x\right )+330\,\sin \left (10\,x\right )-110\,\sin \left (12\,x\right )-110\,\sin \left (14\,x\right )+44\,\sin \left (16\,x\right )+22\,\sin \left (18\,x\right )-10\,\sin \left (20\,x\right )-2\,\sin \left (22\,x\right )+\sin \left (24\,x\right )\right )}^{3/4}}{2016\,{\left (\cos \left (2\,x\right )-1\right )}^6\,{\left (\cos \left (2\,x\right )+1\right )}^5} \] Input:
int(1/(cos(x)^11*sin(x)^13)^(1/4),x)
Output:
-(2^(3/4)*(8*cos(2*x) - 32*cos(2*x)^2 + 31)*(924*sin(2*x) - 132*sin(4*x) - 660*sin(6*x) + 165*sin(8*x) + 330*sin(10*x) - 110*sin(12*x) - 110*sin(14* x) + 44*sin(16*x) + 22*sin(18*x) - 10*sin(20*x) - 2*sin(22*x) + sin(24*x)) ^(3/4))/(2016*(cos(2*x) - 1)^6*(cos(2*x) + 1)^5)
\[ \int \frac {1}{\sqrt [4]{\cos ^{11}(x) \sin ^{13}(x)}} \, dx=\int \frac {1}{\sin \left (x \right )^{\frac {13}{4}} \cos \left (x \right )^{\frac {11}{4}}}d x \] Input:
int(1/(cos(x)^11*sin(x)^13)^(1/4),x)
Output:
int(1/(sin(x)**(1/4)*cos(x)**(3/4)*cos(x)**2*sin(x)**3),x)