Integrand size = 19, antiderivative size = 52 \[ \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{a^4+b^4 \csc ^2(x)}}{a}\right )}{a}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a^4+b^4 \csc ^2(x)}}{a}\right )}{a} \] Output:
-arctan((a^4+b^4*csc(x)^2)^(1/4)/a)/a+arctanh((a^4+b^4*csc(x)^2)^(1/4)/a)/ a
Leaf count is larger than twice the leaf count of optimal. \(256\) vs. \(2(52)=104\).
Time = 0.20 (sec) , antiderivative size = 256, normalized size of antiderivative = 4.92 \[ \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx=\frac {\sqrt [4]{-a^4-2 b^4+a^4 \cos (2 x)} \left (-2 \arctan \left (1-\frac {\sqrt {2} a \sqrt {\sin (x)}}{\sqrt [4]{-b^4-a^4 \sin ^2(x)}}\right )+2 \arctan \left (1+\frac {\sqrt {2} a \sqrt {\sin (x)}}{\sqrt [4]{-b^4-a^4 \sin ^2(x)}}\right )-\log \left (1+\frac {a^2 \sin (x)}{\sqrt {-b^4-a^4 \sin ^2(x)}}-\frac {\sqrt {2} a \sqrt {\sin (x)}}{\sqrt [4]{-b^4-a^4 \sin ^2(x)}}\right )+\log \left (1+\frac {a^2 \sin (x)}{\sqrt {-b^4-a^4 \sin ^2(x)}}+\frac {\sqrt {2} a \sqrt {\sin (x)}}{\sqrt [4]{-b^4-a^4 \sin ^2(x)}}\right )\right )}{2\ 2^{3/4} a \sqrt [4]{a^4+b^4 \csc ^2(x)} \sqrt {\sin (x)}} \] Input:
Integrate[Cot[x]/(a^4 + b^4*Csc[x]^2)^(1/4),x]
Output:
((-a^4 - 2*b^4 + a^4*Cos[2*x])^(1/4)*(-2*ArcTan[1 - (Sqrt[2]*a*Sqrt[Sin[x] ])/(-b^4 - a^4*Sin[x]^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*a*Sqrt[Sin[x]])/(- b^4 - a^4*Sin[x]^2)^(1/4)] - Log[1 + (a^2*Sin[x])/Sqrt[-b^4 - a^4*Sin[x]^2 ] - (Sqrt[2]*a*Sqrt[Sin[x]])/(-b^4 - a^4*Sin[x]^2)^(1/4)] + Log[1 + (a^2*S in[x])/Sqrt[-b^4 - a^4*Sin[x]^2] + (Sqrt[2]*a*Sqrt[Sin[x]])/(-b^4 - a^4*Si n[x]^2)^(1/4)]))/(2*2^(3/4)*a*(a^4 + b^4*Csc[x]^2)^(1/4)*Sqrt[Sin[x]])
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 25, 4627, 243, 73, 25, 27, 827, 216, 219}
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.
\(\displaystyle \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\tan \left (x+\frac {\pi }{2}\right )}{\sqrt [4]{a^4+b^4 \sec \left (x+\frac {\pi }{2}\right )^2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\tan \left (x+\frac {\pi }{2}\right )}{\sqrt [4]{a^4+b^4 \sec \left (x+\frac {\pi }{2}\right )^2}}dx\) |
\(\Big \downarrow \) 4627 |
\(\displaystyle -\int \frac {\sin (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}}d\csc (x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {1}{2} \int \frac {\sin (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}}d\csc ^2(x)\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 \int -\frac {b^4 \csc ^4(x)}{a^4-\csc ^8(x)}d\sqrt [4]{a^4+b^4 \csc ^2(x)}}{b^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \int \frac {b^4 \csc ^4(x)}{a^4-\csc ^8(x)}d\sqrt [4]{a^4+b^4 \csc ^2(x)}}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {\csc ^4(x)}{a^4-\csc ^8(x)}d\sqrt [4]{a^4+b^4 \csc ^2(x)}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {1}{a^2-\csc ^4(x)}d\sqrt [4]{a^4+b^4 \csc ^2(x)}-\frac {1}{2} \int \frac {1}{\csc ^4(x)+a^2}d\sqrt [4]{a^4+b^4 \csc ^2(x)}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {1}{a^2-\csc ^4(x)}d\sqrt [4]{a^4+b^4 \csc ^2(x)}-\frac {\arctan \left (\frac {\sqrt [4]{a^4+b^4 \csc ^2(x)}}{a}\right )}{2 a}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a^4+b^4 \csc ^2(x)}}{a}\right )}{2 a}-\frac {\arctan \left (\frac {\sqrt [4]{a^4+b^4 \csc ^2(x)}}{a}\right )}{2 a}\right )\) |
Input:
Int[Cot[x]/(a^4 + b^4*Csc[x]^2)^(1/4),x]
Output:
2*(-1/2*ArcTan[(a^4 + b^4*Csc[x]^2)^(1/4)/a]/a + ArcTanh[(a^4 + b^4*Csc[x] ^2)^(1/4)/a]/(2*a))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si mp[1/f Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] , x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers Q[2*n, p])
\[\int \frac {\cot \left (x \right )}{\left (a^{4}+b^{4} \csc \left (x \right )^{2}\right )^{\frac {1}{4}}}d x\]
Input:
int(cot(x)/(a^4+b^4*csc(x)^2)^(1/4),x)
Output:
int(cot(x)/(a^4+b^4*csc(x)^2)^(1/4),x)
Timed out. \[ \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx=\text {Timed out} \] Input:
integrate(cot(x)/(a^4+b^4*csc(x)^2)^(1/4),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx=\int \frac {\cot {\left (x \right )}}{\sqrt [4]{a^{4} + b^{4} \csc ^{2}{\left (x \right )}}}\, dx \] Input:
integrate(cot(x)/(a**4+b**4*csc(x)**2)**(1/4),x)
Output:
Integral(cot(x)/(a**4 + b**4*csc(x)**2)**(1/4), x)
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.37 \[ \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx=-\frac {\arctan \left (\frac {{\left (a^{4} + \frac {b^{4}}{\sin \left (x\right )^{2}}\right )}^{\frac {1}{4}}}{a}\right )}{a} + \frac {\log \left (a + {\left (a^{4} + \frac {b^{4}}{\sin \left (x\right )^{2}}\right )}^{\frac {1}{4}}\right )}{2 \, a} - \frac {\log \left (-a + {\left (a^{4} + \frac {b^{4}}{\sin \left (x\right )^{2}}\right )}^{\frac {1}{4}}\right )}{2 \, a} \] Input:
integrate(cot(x)/(a^4+b^4*csc(x)^2)^(1/4),x, algorithm="maxima")
Output:
-arctan((a^4 + b^4/sin(x)^2)^(1/4)/a)/a + 1/2*log(a + (a^4 + b^4/sin(x)^2) ^(1/4))/a - 1/2*log(-a + (a^4 + b^4/sin(x)^2)^(1/4))/a
Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.40 \[ \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx=-\frac {\arctan \left (\frac {{\left (a^{4} + \frac {b^{4}}{\sin \left (x\right )^{2}}\right )}^{\frac {1}{4}}}{a}\right )}{a} + \frac {\log \left ({\left | a + {\left (a^{4} + \frac {b^{4}}{\sin \left (x\right )^{2}}\right )}^{\frac {1}{4}} \right |}\right )}{2 \, a} - \frac {\log \left ({\left | -a + {\left (a^{4} + \frac {b^{4}}{\sin \left (x\right )^{2}}\right )}^{\frac {1}{4}} \right |}\right )}{2 \, a} \] Input:
integrate(cot(x)/(a^4+b^4*csc(x)^2)^(1/4),x, algorithm="giac")
Output:
-arctan((a^4 + b^4/sin(x)^2)^(1/4)/a)/a + 1/2*log(abs(a + (a^4 + b^4/sin(x )^2)^(1/4)))/a - 1/2*log(abs(-a + (a^4 + b^4/sin(x)^2)^(1/4)))/a
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88 \[ \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx=-\frac {\mathrm {atan}\left (\frac {{\left (\frac {b^4}{{\sin \left (x\right )}^2}+a^4\right )}^{1/4}}{a}\right )-\mathrm {atanh}\left (\frac {{\left (\frac {b^4}{{\sin \left (x\right )}^2}+a^4\right )}^{1/4}}{a}\right )}{a} \] Input:
int(cot(x)/(b^4/sin(x)^2 + a^4)^(1/4),x)
Output:
-(atan((b^4/sin(x)^2 + a^4)^(1/4)/a) - atanh((b^4/sin(x)^2 + a^4)^(1/4)/a) )/a
\[ \int \frac {\cot (x)}{\sqrt [4]{a^4+b^4 \csc ^2(x)}} \, dx=\int \frac {\cot \left (x \right )}{\left (\csc \left (x \right )^{2} b^{4}+a^{4}\right )^{\frac {1}{4}}}d x \] Input:
int(cot(x)/(a^4+b^4*csc(x)^2)^(1/4),x)
Output:
int(cot(x)/(csc(x)**2*b**4 + a**4)**(1/4),x)