3.1.50 \(\int \frac {\cot ^3(d+e x)}{(a+b \tan ^2(d+e x)+c \tan ^4(d+e x))^{3/2}} \, dx\) [50]

3.1.50.1 Optimal result

Integrand size = 35, antiderivative size = 477 \[ \int \frac {\cot ^3(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 a^{3/2} e}+\frac {3 b \text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 a^{5/2} e}-\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 (a-b+c)^{3/2} e}-\frac {b^2-2 a c+b c \tan ^2(d+e x)}{a \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac {\cot ^2(d+e x) \left (b^2-2 a c+b c \tan ^2(d+e x)\right )}{a \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac {b^2-2 a c-b c+(b-2 c) c \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\left (3 b^2-8 a c\right ) \cot ^2(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 a^2 \left (b^2-4 a c\right ) e} \]

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

3.1.50.2 Mathematica [A] (verified)

Time = 6.10 (sec) , antiderivative size = 555, normalized size of antiderivative = 1.16 \[ \int \frac {\cot ^3(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\frac {-\frac {2 \left (-\frac {b^2}{2}+2 a c\right ) \text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{a^{3/2} \left (b^2-4 a c\right )}+\frac {8 \left (-\frac {b^2}{2}+2 a c\right ) \text {arctanh}\left (\frac {2 a-b-(-b+2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c} (4 a-4 b+4 c) \left (b^2-4 a c\right )}+\frac {2 \left (-b^2+2 a c-b c \tan ^2(d+e x)\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {2 \cot ^2(d+e x) \left (-b^2+2 a c-b c \tan ^2(d+e x)\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {2 \left (-b^2+2 a c+b c+c (-b+2 c) \tan ^2(d+e x)\right )}{(a-b+c) \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {2 \left (\frac {\left (2 a b c+\frac {1}{2} b \left (-3 b^2+8 a c\right )\right ) \text {arctanh}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 a^{3/2}}+\frac {\left (3 b^2-8 a c\right ) \cot ^2(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 a}\right )}{a \left (b^2-4 a c\right )}}{2 e} \]

input

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

output

((-2*(-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)) + (8*(- 
1/2*b^2 + 2*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])])/(Sqrt[a - b + c]* 
(4*a - 4*b + 4*c)*(b^2 - 4*a*c)) + (2*(-b^2 + 2*a*c - b*c*Tan[d + e*x]^2)) 
/(a*(b^2 - 4*a*c)*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) - (2*Cot[ 
d + e*x]^2*(-b^2 + 2*a*c - b*c*Tan[d + e*x]^2))/(a*(b^2 - 4*a*c)*Sqrt[a + 
b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) - (2*(-b^2 + 2*a*c + b*c + c*(-b + 2 
*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*(((2*a*b*c + (b*(-3*b^2 + 8*a*c))/2)*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)) + ((3*b^2 - 8*a*c)*Cot[d + e*x]^2*Sqrt[a + b*Tan[d + e 
*x]^2 + c*Tan[d + e*x]^4])/(2*a)))/(a*(b^2 - 4*a*c)))/(2*e)
 

3.1.50.3 Rubi [A] (warning: unable to verify)

Time = 0.68 (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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4183

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

\(\Big \downarrow \) 1578

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

\(\Big \downarrow \) 1289

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

\(\Big \downarrow \) 2009

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

input

Int[Cot[d + e*x]^3/(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) + (3*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^(5/2)) - 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*T 
an[d + e*x]^2))/(a*(b^2 - 4*a*c)*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x 
]^4]) + (2*Cot[d + e*x]*(b^2 - 2*a*c + b*c*Tan[d + e*x]^2))/(a*(b^2 - 4*a* 
c)*Sqrt[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]) - ((3*b^2 - 8*a*c)*Cot[d + e*x]*Sqrt[a + b* 
Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(a^2*(b^2 - 4*a*c)))/(2*e)
 

3.1.50.3.1 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]
 

3.1.50.4 Maple [F]

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

input

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

output

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

3.1.50.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1273 vs. \(2 (437) = 874\).

Time = 3.27 (sec) , antiderivative size = 5189, normalized size of antiderivative = 10.88 \[ \int \frac {\cot ^3(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.1.50.6 Sympy [F]

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

input

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

output

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

3.1.50.7 Maxima [F(-1)]

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

input

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

output

Timed out
 

3.1.50.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.1.50.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}