Integrand size = 25, antiderivative size = 477 \[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\cos ^2(c+d x) \cot (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt {a+b} \cos (c+d x) \sin (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)} \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )}-\frac {\sqrt [4]{a+b} \cos ^2(c+d x) E\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt [4]{a+b} \cos ^2(c+d x) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}} \]
-(a+b)^(1/4)*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))^2 )^(1/2)/cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticE(sin(2*arct an((a+b)^(1/4)*tan(d*x+c)/a^(1/4))),1/2*(2-2*a^(1/2)/(a+b)^(1/2))^(1/2))*( (a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2) ^2)^(1/2)*(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)/a^(3/4)/d/(a+b*sin(d*x+c)^4)^ (1/2)+1/2*(a+b)^(1/4)*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^ (1/4)))^2)^(1/2)/cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticF(s in(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4))),1/2*(2-2*a^(1/2)/(a+b)^(1/2)) ^(1/2))*((a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*tan( d*x+c)^2)^2)^(1/2)*(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)/a^(3/4)/d/(a+b*sin(d *x+c)^4)^(1/2)-cos(d*x+c)^2*cot(d*x+c)*(a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c )^4)/a/d/(a+b*sin(d*x+c)^4)^(1/2)+cos(d*x+c)*sin(d*x+c)*(a+b)^(1/2)*(a+2*a *tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/a/d/(a+b*sin(d*x+c)^4)^(1/2)/(a^(1/2)+(a +b)^(1/2)*tan(d*x+c)^2)
Result contains complex when optimal does not.
Time = 9.88 (sec) , antiderivative size = 378, normalized size of antiderivative = 0.79 \[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))} \cot (c+d x)}{2 \sqrt {2} a d}-\frac {\cos ^4(c+d x) \left (a \sec ^4(c+d x) \tan (c+d x)+b \tan ^5(c+d x)+\frac {\left (i a+\sqrt {a} \sqrt {b}\right ) \left (E\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )\right ) \sec ^2(c+d x) \sqrt {1+\left (1-\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)}}{\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{a d \sqrt {\cos ^4(c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}} \]
-1/2*(Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]]*Cot[c + d*x])/(Sqrt[2]*a*d) - (Cos[c + d*x]^4*(a*Sec[c + d*x]^4*Tan[c + d*x] + b*T an[c + d*x]^5 + ((I*a + Sqrt[a]*Sqrt[b])*(EllipticE[I*ArcSinh[Sqrt[1 - (I* Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], (Sqrt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b ])] - EllipticF[I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], (Sq rt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])])*Sec[c + d*x]^2*Sqrt[1 + (1 - (I *Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2]*Sqrt[1 + (1 + (I*Sqrt[b])/Sqrt[a])*Tan[ c + d*x]^2])/Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]))/(a*d*Sqrt[Cos[c + d*x]^4*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)])
Time = 0.59 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3711, 1443, 27, 1459, 27, 1416, 1509}
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.
| \(\displaystyle \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^2 \sqrt {a+b \sin (c+d x)^4}}dx\) |
| \(\Big \downarrow \) 3711 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \int \frac {\cot ^2(c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 1443 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {\int \frac {(a+b) \tan ^2(c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{a}-\frac {\cot (c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{a}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {(a+b) \int \frac {\tan ^2(c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{a}-\frac {\cot (c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{a}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 1459 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {(a+b) \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{\sqrt {a+b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {a+b} \tan ^2(c+d x)}{\sqrt {a} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{\sqrt {a+b}}\right )}{a}-\frac {\cot (c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{a}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {(a+b) \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{\sqrt {a+b}}-\frac {\int \frac {\sqrt {a}-\sqrt {a+b} \tan ^2(c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{\sqrt {a+b}}\right )}{a}-\frac {\cot (c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{a}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {(a+b) \left (\frac {\sqrt [4]{a} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 (a+b)^{3/4} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}-\frac {\int \frac {\sqrt {a}-\sqrt {a+b} \tan ^2(c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{\sqrt {a+b}}\right )}{a}-\frac {\cot (c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{a}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 1509 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {(a+b) \left (\frac {\sqrt [4]{a} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 (a+b)^{3/4} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{\sqrt [4]{a+b} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}-\frac {\tan (c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}}}{\sqrt {a+b}}\right )}{a}-\frac {\cot (c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{a}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
(Cos[c + d*x]^2*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4]*(-(( Cot[c + d*x]*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4])/a) + ( (a + b)*((a^(1/4)*EllipticF[2*ArcTan[((a + b)^(1/4)*Tan[c + d*x])/a^(1/4)] , (1 - Sqrt[a]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt [(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]* Tan[c + d*x]^2)^2])/(2*(a + b)^(3/4)*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b) *Tan[c + d*x]^4]) - (-((Tan[c + d*x]*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b) *Tan[c + d*x]^4])/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)) + (a^(1/4)*Ellip ticE[2*ArcTan[((a + b)^(1/4)*Tan[c + d*x])/a^(1/4)], (1 - Sqrt[a]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt[(a + 2*a*Tan[c + d*x]^ 2 + (a + b)*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)^2])/((a + b)^(1/4)*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4]))/Sqrt[a + b]))/a))/(d*Sqrt[a + b*Sin[c + d*x]^4])
3.6.63.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c /a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ (2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) ], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1)/(a*d*(m + 1))), x] - Sim p[1/(a*d^2*(m + 1)) Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p + 5)* x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[1/q Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q ^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 /(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_)*((d_.)*tan[(e_.) + (f_.)* (x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff *(a + b*Sin[e + f*x]^4)^p*((Sec[e + f*x]^2)^(2*p)/(f*Apart[a*(1 + Tan[e + f *x]^2)^2 + b*Tan[e + f*x]^4]^p)) Subst[Int[(d*ff*x)^m*(ExpandToSum[a*(1 + ff^2*x^2)^2 + b*ff^4*x^4, x]^p/(1 + ff^2*x^2)^(2*p + 1)), x], x, Tan[e + f *x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ[p - 1/2]
\[\int \frac {\cot ^{2}\left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}d x\]
\[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\cot \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]
\[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\cot ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\cot \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]
Timed out. \[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^2}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]