Integrand size = 35, antiderivative size = 249 \[ \int \frac {\cot ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\frac {\text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a} e}+\frac {b \text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 a^{3/2} e}-\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a-b+c} e}-\frac {\cot ^2(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 a e} \]
1/4*b*arctanh(1/2*(2*a+b*tan(e*x+d)^2)/a^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x +d)^4)^(1/2))/a^(3/2)/e+1/2*arctanh(1/2*(2*a+b*tan(e*x+d)^2)/a^(1/2)/(a+b* tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/e/a^(1/2)-1/2*arctanh(1/2*(2*a-b+(b-2* c)*tan(e*x+d)^2)/(a-b+c)^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/e/ (a-b+c)^(1/2)-1/2*cot(e*x+d)^2*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)/a/e
Time = 4.42 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.76 \[ \int \frac {\cot ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\frac {(2 a+b) \text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+2 \sqrt {a} \left (-\frac {a \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}-\cot ^2(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}\right )}{4 a^{3/2} e} \]
((2*a + b)*ArcTanh[(2*a + b*Tan[d + e*x]^2)/(2*Sqrt[a]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])] + 2*Sqrt[a]*(-((a*ArcTanh[(2*a - b + (b - 2*c )*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])])/Sqrt[a - b + c]) - Cot[d + e*x]^2*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]))/(4*a^(3/2)*e)
Time = 0.48 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4183, 1578, 1289, 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.
\(\displaystyle \int \frac {\cot ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (d+e x)^3 \sqrt {a+b \tan (d+e x)^2+c \tan (d+e x)^4}}dx\) |
\(\Big \downarrow \) 4183 |
\(\displaystyle \frac {\int \frac {\cot ^3(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{e}\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {\int \frac {\cot ^2(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 e}\) |
\(\Big \downarrow \) 1289 |
\(\displaystyle \frac {\int \left (\frac {\cot ^2(d+e x)}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}-\frac {\cot (d+e x)}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+\frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}\right )d\tan ^2(d+e x)}{2 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b \text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 a^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a}}-\frac {\text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}-\frac {\cot (d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{a}}{2 e}\) |
(ArcTanh[(2*a + b*Tan[d + e*x]^2)/(2*Sqrt[a]*Sqrt[a + b*Tan[d + e*x]^2 + c *Tan[d + e*x]^4])]/Sqrt[a] + (b*ArcTanh[(2*a + b*Tan[d + e*x]^2)/(2*Sqrt[a ]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])])/(2*a^(3/2)) - ArcTanh[( 2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])]/Sqrt[a - b + c] - (Cot[d + e*x]*Sqrt[a + b*Ta n[d + e*x]^2 + c*Tan[d + e*x]^4])/a)/(2*e)
3.1.40.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && ( IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0]))
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*( x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Simp[f/e Subst[Int[(x/f)^m*((a + b*x^n + c*x^(2*n))^p/(f^2 + x^2)), x ], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n 2, 2*n] && NeQ[b^2 - 4*a*c, 0]
\[\int \frac {\cot \left (e x +d \right )^{3}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}d x\]
Time = 1.29 (sec) , antiderivative size = 1350, normalized size of antiderivative = 5.42 \[ \int \frac {\cot ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Too large to display} \]
[1/8*(2*sqrt(a - b + c)*a^2*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d )^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 - 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1))*tan(e*x + d)^2 + (2*a^2 - a*b - b^2 + (2*a + b)*c)*sqrt(a)*log(((b^2 + 4*a*c)*tan(e*x + d)^4 + 8*a*b*tan(e*x + d)^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(b*tan(e*x + d)^2 + 2*a)*sqrt(a) + 8*a^2)/tan(e*x + d)^4)*tan(e*x + d)^2 - 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*( a^2 - a*b + a*c))/((a^3 - a^2*b + a^2*c)*e*tan(e*x + d)^2), 1/4*(sqrt(a - b + c)*a^2*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 - 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1))*tan(e*x + d )^2 - (2*a^2 - a*b - b^2 + (2*a + b)*c)*sqrt(-a)*arctan(1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(b*tan(e*x + d)^2 + 2*a)*sqrt(-a)/(a*c*tan (e*x + d)^4 + a*b*tan(e*x + d)^2 + a^2))*tan(e*x + d)^2 - 2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(a^2 - a*b + a*c))/((a^3 - a^2*b + a^2*c)* e*tan(e*x + d)^2), -1/8*(4*a^2*sqrt(-a + b - c)*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt( -a + b - c)/(((a - b)*c + c^2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*tan(e...
\[ \int \frac {\cot ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {\cot ^{3}{\left (d + e x \right )}}{\sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \]
Timed out. \[ \int \frac {\cot ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cot ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int { \frac {\cot \left (e x + d\right )^{3}}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\cot ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {{\mathrm {cot}\left (d+e\,x\right )}^3}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \]