\(\int \frac {\tan ^3(d+e x)}{(a+b \cot ^2(d+e x)+c \cot ^4(d+e x))^{3/2}} \, dx\) [31]

Optimal result

Integrand size = 35, antiderivative size = 478 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \cot ^2(d+e x)+c \cot ^4(d+e x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a+b \cot ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{2 a^{3/2} e}-\frac {3 b \text {arctanh}\left (\frac {2 a+b \cot ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{4 a^{5/2} e}+\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \cot ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{2 (a-b+c)^{3/2} e}+\frac {b^2-2 a c+b c \cot ^2(d+e x)}{a \left (b^2-4 a c\right ) e \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}-\frac {b^2-2 a c-b c+(b-2 c) c \cot ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}-\frac {\left (b^2-2 a c+b c \cot ^2(d+e x)\right ) \tan ^2(d+e x)}{a \left (b^2-4 a c\right ) e \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)} \tan ^2(d+e x)}{2 a^2 \left (b^2-4 a c\right ) e} \] Output:

-1/2*arctanh(1/2*(2*a+b*cot(e*x+d)^2)/a^(1/2)/(a+b*cot(e*x+d)^2+c*cot(e*x+ 
d)^4)^(1/2))/a^(3/2)/e-3/4*b*arctanh(1/2*(2*a+b*cot(e*x+d)^2)/a^(1/2)/(a+b 
*cot(e*x+d)^2+c*cot(e*x+d)^4)^(1/2))/a^(5/2)/e+1/2*arctanh(1/2*(2*a-b+(b-2 
*c)*cot(e*x+d)^2)/(a-b+c)^(1/2)/(a+b*cot(e*x+d)^2+c*cot(e*x+d)^4)^(1/2))/( 
a-b+c)^(3/2)/e+(b^2-2*a*c+b*c*cot(e*x+d)^2)/a/(-4*a*c+b^2)/e/(a+b*cot(e*x+ 
d)^2+c*cot(e*x+d)^4)^(1/2)-(b^2-2*a*c-b*c+(b-2*c)*c*cot(e*x+d)^2)/(a-b+c)/ 
(-4*a*c+b^2)/e/(a+b*cot(e*x+d)^2+c*cot(e*x+d)^4)^(1/2)-(b^2-2*a*c+b*c*cot( 
e*x+d)^2)*tan(e*x+d)^2/a/(-4*a*c+b^2)/e/(a+b*cot(e*x+d)^2+c*cot(e*x+d)^4)^ 
(1/2)+1/2*(-8*a*c+3*b^2)*(a+b*cot(e*x+d)^2+c*cot(e*x+d)^4)^(1/2)*tan(e*x+d 
)^2/a^2/(-4*a*c+b^2)/e
 

Mathematica [A] (warning: unable to verify)

Time = 11.52 (sec) , antiderivative size = 562, normalized size of antiderivative = 1.18 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \cot ^2(d+e x)+c \cot ^4(d+e x)\right )^{3/2}} \, dx=\frac {\sqrt {2} \sqrt {(3 a+b+3 c-4 (a-c) \cos (2 (d+e x))+(a-b+c) \cos (4 (d+e x))) \csc ^4(d+e x)} \left (\frac {3 b^2 (b-c)^2-4 a^3 c+a^2 \left (b^2+8 b c-4 c^2\right )-2 a \left (b^3+5 b^2 c-10 b c^2+4 c^3\right )}{(a-b+c)^2 \left (-b^2+4 a c\right )}+\frac {8 \left (-b^5+b^4 c+2 a c^3 (a+c)-b^2 c^2 (4 a+c)+b^3 c (5 a+c)-a b c^2 (5 a+3 c)+\left (b^5-3 b^4 c+a b (5 a-9 c) c^2+b^2 (12 a-c) c^2+2 a c^3 (-3 a+c)+b^3 c (-5 a+3 c)\right ) \cos (2 (d+e x))\right )}{(a-b+c)^2 \left (-b^2+4 a c\right ) (3 a+b+3 c-4 (a-c) \cos (2 (d+e x))+(a-b+c) \cos (4 (d+e x)))}+\sec ^2(d+e x)\right )-\frac {2 \left (\frac {(2 a+3 b) (a-b+c) \text {arctanh}\left (\frac {b+2 a \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {c+b \tan ^2(d+e x)+a \tan ^4(d+e x)}}\right )}{\sqrt {a}}-\frac {2 a^2 \text {arctanh}\left (\frac {b-2 c+(2 a-b) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {c+b \tan ^2(d+e x)+a \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}\right ) \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)} \tan ^2(d+e x)}{(a-b+c) \sqrt {c+b \tan ^2(d+e x)+a \tan ^4(d+e x)}}}{8 a^2 e} \] Input:

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

Output:

(Sqrt[2]*Sqrt[(3*a + b + 3*c - 4*(a - c)*Cos[2*(d + e*x)] + (a - b + c)*Co 
s[4*(d + e*x)])*Csc[d + e*x]^4]*((3*b^2*(b - c)^2 - 4*a^3*c + a^2*(b^2 + 8 
*b*c - 4*c^2) - 2*a*(b^3 + 5*b^2*c - 10*b*c^2 + 4*c^3))/((a - b + c)^2*(-b 
^2 + 4*a*c)) + (8*(-b^5 + b^4*c + 2*a*c^3*(a + c) - b^2*c^2*(4*a + c) + b^ 
3*c*(5*a + c) - a*b*c^2*(5*a + 3*c) + (b^5 - 3*b^4*c + a*b*(5*a - 9*c)*c^2 
 + b^2*(12*a - c)*c^2 + 2*a*c^3*(-3*a + c) + b^3*c*(-5*a + 3*c))*Cos[2*(d 
+ e*x)]))/((a - b + c)^2*(-b^2 + 4*a*c)*(3*a + b + 3*c - 4*(a - c)*Cos[2*( 
d + e*x)] + (a - b + c)*Cos[4*(d + e*x)])) + Sec[d + e*x]^2) - (2*(((2*a + 
 3*b)*(a - b + c)*ArcTanh[(b + 2*a*Tan[d + e*x]^2)/(2*Sqrt[a]*Sqrt[c + b*T 
an[d + e*x]^2 + a*Tan[d + e*x]^4])])/Sqrt[a] - (2*a^2*ArcTanh[(b - 2*c + ( 
2*a - b)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[c + b*Tan[d + e*x]^2 + a* 
Tan[d + e*x]^4])])/Sqrt[a - b + c])*Sqrt[a + b*Cot[d + e*x]^2 + c*Cot[d + 
e*x]^4]*Tan[d + e*x]^2)/((a - b + c)*Sqrt[c + b*Tan[d + e*x]^2 + a*Tan[d + 
 e*x]^4]))/(8*a^2*e)
 

Rubi [A] (warning: unable to verify)

Time = 0.70 (sec) , antiderivative size = 454, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4184, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4184

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

\(\Big \downarrow \) 1578

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

\(\Big \downarrow \) 1289

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {3 b \text {arctanh}\left (\frac {2 a+b \cot ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{2 a^{5/2}}+\frac {\text {arctanh}\left (\frac {2 a+b \cot ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{a^{3/2}}-\frac {\left (3 b^2-8 a c\right ) \tan (d+e x) \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}{a^2 \left (b^2-4 a c\right )}-\frac {\text {arctanh}\left (\frac {2 a+(b-2 c) \cot ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{(a-b+c)^{3/2}}-\frac {2 \left (-2 a c+b^2+b c \cot ^2(d+e x)\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}+\frac {2 \left (-2 a c+b^2+c (b-2 c) \cot ^2(d+e x)-b c\right )}{(a-b+c) \left (b^2-4 a c\right ) \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}+\frac {2 \tan (d+e x) \left (-2 a c+b^2+b c \cot ^2(d+e x)\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}}{2 e}\)

Input:

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

Output:

-1/2*(ArcTanh[(2*a + b*Cot[d + e*x]^2)/(2*Sqrt[a]*Sqrt[a + b*Cot[d + e*x]^ 
2 + c*Cot[d + e*x]^4])]/a^(3/2) + (3*b*ArcTanh[(2*a + b*Cot[d + e*x]^2)/(2 
*Sqrt[a]*Sqrt[a + b*Cot[d + e*x]^2 + c*Cot[d + e*x]^4])])/(2*a^(5/2)) - Ar 
cTanh[(2*a - b + (b - 2*c)*Cot[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*C 
ot[d + e*x]^2 + c*Cot[d + e*x]^4])]/(a - b + c)^(3/2) - (2*(b^2 - 2*a*c + 
b*c*Cot[d + e*x]^2))/(a*(b^2 - 4*a*c)*Sqrt[a + b*Cot[d + e*x]^2 + c*Cot[d 
+ e*x]^4]) + (2*(b^2 - 2*a*c - b*c + (b - 2*c)*c*Cot[d + e*x]^2))/((a - b 
+ c)*(b^2 - 4*a*c)*Sqrt[a + b*Cot[d + e*x]^2 + c*Cot[d + e*x]^4]) + (2*(b^ 
2 - 2*a*c + b*c*Cot[d + e*x]^2)*Tan[d + e*x])/(a*(b^2 - 4*a*c)*Sqrt[a + b* 
Cot[d + e*x]^2 + c*Cot[d + e*x]^4]) - ((3*b^2 - 8*a*c)*Sqrt[a + b*Cot[d + 
e*x]^2 + c*Cot[d + e*x]^4]*Tan[d + e*x])/(a^2*(b^2 - 4*a*c)))/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 4184

Int[cot[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*(cot[(d_.) + (e_.)*(x_)]*( 
f_.))^(n_.) + (c_.)*(cot[(d_.) + (e_.)*(x_)]*(f_.))^(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*Cot[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[ 
n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
 

Maple [F]

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

Input:

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

Output:

int(tan(e*x+d)^3/(a+b*cot(e*x+d)^2+c*cot(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. 1300 vs. \(2 (438) = 876\).

Time = 2.97 (sec) , antiderivative size = 5274, normalized size of antiderivative = 11.03 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \cot ^2(d+e x)+c \cot ^4(d+e x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

int((sqrt(cot(d + e*x)**4*c + cot(d + e*x)**2*b + a)*tan(d + e*x)**3)/(cot 
(d + e*x)**8*c**2 + 2*cot(d + e*x)**6*b*c + 2*cot(d + e*x)**4*a*c + cot(d 
+ e*x)**4*b**2 + 2*cot(d + e*x)**2*a*b + a**2),x)