\(\int \frac {\cot (d+e x)}{(a+b \tan ^2(d+e x)+c \tan ^4(d+e x))^{3/2}} \, dx\) [47]

Optimal result

Integrand size = 33, antiderivative size = 280 \[ \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 a^{3/2} e}+\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 (a-b+c)^{3/2} e}+\frac {b^2-2 a c+b c \tan ^2(d+e x)}{a \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {b^2-2 a c-b c+(b-2 c) c \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \] Output:

-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))/a^(3/2)/e+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))/(a-b+c)^(3/2)/e+(b^2-2*a* 
c+b*c*tan(e*x+d)^2)/a/(-4*a*c+b^2)/e/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/ 
2)-(b^2-2*a*c-b*c+(b-2*c)*c*tan(e*x+d)^2)/(a-b+c)/(-4*a*c+b^2)/e/(a+b*tan( 
e*x+d)^2+c*tan(e*x+d)^4)^(1/2)
 

Mathematica [A] (verified)

Time = 2.34 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.99 \[ \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\frac {\frac {\left (-\frac {b^2}{2}+2 a c\right ) \text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{a^{3/2}}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {-2 a+b-(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 (a-b+c)^{3/2}}+\frac {b^2-2 a c+b c \tan ^2(d+e x)}{a \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {b^2-2 a c-b c+(b-2 c) c \tan ^2(d+e x)}{(a-b+c) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}}{\left (b^2-4 a c\right ) e} \] Input:

Integrate[Cot[d + e*x]/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),x]
 

Output:

(((-1/2*b^2 + 2*a*c)*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])])/a^(3/2) - ((b^2 - 4*a*c)*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])])/(2*(a - b + c)^(3/2)) + (b^2 - 2*a*c + b*c* 
Tan[d + e*x]^2)/(a*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) - (b^2 - 
 2*a*c - b*c + (b - 2*c)*c*Tan[d + e*x]^2)/((a - b + c)*Sqrt[a + b*Tan[d + 
 e*x]^2 + c*Tan[d + e*x]^4]))/((b^2 - 4*a*c)*e)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, 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.

\(\displaystyle \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (d+e x) \left (a+b \tan (d+e x)^2+c \tan (d+e x)^4\right )^{3/2}}dx\)

\(\Big \downarrow \) 4183

\(\displaystyle \frac {\int \frac {\cot (d+e x)}{\left (\tan ^2(d+e x)+1\right ) \left (c \tan ^4(d+e x)+b \tan ^2(d+e x)+a\right )^{3/2}}d\tan (d+e x)}{e}\)

\(\Big \downarrow \) 1578

\(\displaystyle \frac {\int \frac {\cot (d+e x)}{\left (\tan ^2(d+e x)+1\right ) \left (c \tan ^4(d+e x)+b \tan ^2(d+e x)+a\right )^{3/2}}d\tan ^2(d+e x)}{2 e}\)

\(\Big \downarrow \) 1289

\(\displaystyle \frac {\int \left (\frac {\cot (d+e x)}{\left (c \tan ^4(d+e x)+b \tan ^2(d+e x)+a\right )^{3/2}}+\frac {1}{\left (-\tan ^2(d+e x)-1\right ) \left (c \tan ^4(d+e x)+b \tan ^2(d+e x)+a\right )^{3/2}}\right )d\tan ^2(d+e x)}{2 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {\text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{a^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{(a-b+c)^{3/2}}+\frac {2 \left (-2 a c+b^2+b c \tan ^2(d+e x)\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {2 \left (-2 a c+b^2+c (b-2 c) \tan ^2(d+e x)-b c\right )}{(a-b+c) \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}}{2 e}\)

Input:

Int[Cot[d + e*x]/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),x]
 

Output:

(-(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])]/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])]/(a - 
 b + c)^(3/2) + (2*(b^2 - 2*a*c + b*c*Tan[d + e*x]^2))/(a*(b^2 - 4*a*c)*Sq 
rt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) - (2*(b^2 - 2*a*c - b*c + (b 
- 2*c)*c*Tan[d + e*x]^2))/((a - b + c)*(b^2 - 4*a*c)*Sqrt[a + b*Tan[d + e* 
x]^2 + c*Tan[d + e*x]^4]))/(2*e)
 

Defintions of rubi rules used

rule 1289

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]))
 

rule 1578

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4183

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]
 

Maple [F]

\[\int \frac {\cot \left (e x +d \right )}{\left (a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}\right )^{\frac {3}{2}}}d x\]

Input:

int(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x)
 

Output:

int(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 963 vs. \(2 (256) = 512\).

Time = 2.35 (sec) , antiderivative size = 3951, normalized size of antiderivative = 14.11 \[ \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorithm= 
"fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int \frac {\cot {\left (d + e x \right )}}{\left (a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cot(e*x+d)/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(3/2),x)
 

Output:

Integral(cot(d + e*x)/(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4)**(3/2), 
x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorithm= 
"maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-1)]

Timed out. \[ \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorithm= 
"giac")
                                                                                    
                                                                                    
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int \frac {\mathrm {cot}\left (d+e\,x\right )}{{\left (c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:

int(cot(d + e*x)/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(3/2),x)
 

Output:

int(cot(d + e*x)/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\cot (d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\tan \left (e x +d \right )^{4} c +\tan \left (e x +d \right )^{2} b +a}\, \cot \left (e x +d \right )}{\tan \left (e x +d \right )^{8} c^{2}+2 \tan \left (e x +d \right )^{6} b c +2 \tan \left (e x +d \right )^{4} a c +\tan \left (e x +d \right )^{4} b^{2}+2 \tan \left (e x +d \right )^{2} a b +a^{2}}d x \] Input:

int(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x)
 

Output:

int((sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*cot(d + e*x))/(tan(d 
+ e*x)**8*c**2 + 2*tan(d + e*x)**6*b*c + 2*tan(d + e*x)**4*a*c + tan(d + e 
*x)**4*b**2 + 2*tan(d + e*x)**2*a*b + a**2),x)