\(\int \cot ^5(x) (a+b \tan ^4(x))^{3/2} \, dx\) [397]

Optimal result

Integrand size = 17, antiderivative size = 151 \[ \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\frac {1}{2} b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )+\frac {1}{2} (a+b)^{3/2} \text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )-\frac {1}{4} \sqrt {a} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \tan ^4(x)}}{\sqrt {a}}\right )+\frac {1}{2} a \cot ^2(x) \sqrt {a+b \tan ^4(x)}-\frac {1}{4} a \cot ^4(x) \sqrt {a+b \tan ^4(x)} \] Output:

1/2*b^(3/2)*arctanh(b^(1/2)*tan(x)^2/(a+b*tan(x)^4)^(1/2))+1/2*(a+b)^(3/2) 
*arctanh((a-b*tan(x)^2)/(a+b)^(1/2)/(a+b*tan(x)^4)^(1/2))-1/4*a^(1/2)*(2*a 
+3*b)*arctanh((a+b*tan(x)^4)^(1/2)/a^(1/2))+1/2*a*cot(x)^2*(a+b*tan(x)^4)^ 
(1/2)-1/4*a*cot(x)^4*(a+b*tan(x)^4)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 5.64 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.10 \[ \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\frac {b (4 (a-b) \cos (2 x)+(a+b) (3+\cos (4 x)))^2 \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},1+\frac {b \tan ^4(x)}{a}\right ) \sec ^8(x) \sqrt {a+b \tan ^4(x)}}{640 a^2}+\frac {\sqrt {a} \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a}}\right ) \sqrt {a+b \tan ^4(x)}}{4 \sqrt {1+\frac {b \tan ^4(x)}{a}}}+\frac {a \cot ^2(x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},-\frac {b \tan ^4(x)}{a}\right ) \sqrt {a+b \tan ^4(x)}}{2 \sqrt {1+\frac {b \tan ^4(x)}{a}}}+\frac {1}{4} \left (2 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )+2 (a+b)^{3/2} \text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^4(x)}}{\sqrt {a}}\right )-2 b \sqrt {a+b \tan ^4(x)}+b \tan ^2(x) \sqrt {a+b \tan ^4(x)}\right ) \] Input:

Integrate[Cot[x]^5*(a + b*Tan[x]^4)^(3/2),x]
 

Output:

(b*(4*(a - b)*Cos[2*x] + (a + b)*(3 + Cos[4*x]))^2*Hypergeometric2F1[2, 5/ 
2, 7/2, 1 + (b*Tan[x]^4)/a]*Sec[x]^8*Sqrt[a + b*Tan[x]^4])/(640*a^2) + (Sq 
rt[a]*Sqrt[b]*ArcSinh[(Sqrt[b]*Tan[x]^2)/Sqrt[a]]*Sqrt[a + b*Tan[x]^4])/(4 
*Sqrt[1 + (b*Tan[x]^4)/a]) + (a*Cot[x]^2*Hypergeometric2F1[-3/2, -1/2, 1/2 
, -((b*Tan[x]^4)/a)]*Sqrt[a + b*Tan[x]^4])/(2*Sqrt[1 + (b*Tan[x]^4)/a]) + 
(2*Sqrt[b]*(a + b)*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]] + 2*(a 
 + b)^(3/2)*ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])] - 
 2*a^(3/2)*ArcTanh[Sqrt[a + b*Tan[x]^4]/Sqrt[a]] - 2*b*Sqrt[a + b*Tan[x]^4 
] + b*Tan[x]^2*Sqrt[a + b*Tan[x]^4])/4
 

Rubi [A] (warning: unable to verify)

Time = 0.61 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 4153, 1579, 617, 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.

\(\displaystyle \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \tan (x)^4\right )^{3/2}}{\tan (x)^5}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \int \frac {\cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2}}{\tan ^2(x)+1}d\tan (x)\)

\(\Big \downarrow \) 1579

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

\(\Big \downarrow \) 617

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^4(x)}}{\sqrt {a}}\right )-\frac {3}{2} \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+b \tan ^4(x)}}{\sqrt {a}}\right )+\frac {1}{2} \sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )-\frac {3}{2} a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )+(a+b)^{3/2} \text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )+a \sqrt {a+b \tan ^4(x)}+\frac {3}{2} b \sqrt {a+b \tan ^4(x)}-\frac {3}{2} b \tan ^2(x) \sqrt {a+b \tan ^4(x)}-\frac {1}{2} \left (2 (a+b)-b \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}-\frac {1}{2} \cot ^2(x) \left (a+b \tan ^4(x)\right )^{3/2}+\cot (x) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

Input:

Int[Cot[x]^5*(a + b*Tan[x]^4)^(3/2),x]
 

Output:

((-3*a*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]])/2 + (Sqrt 
[b]*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]])/2 + (a + 
 b)^(3/2)*ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])] - a 
^(3/2)*ArcTanh[Sqrt[a + b*Tan[x]^4]/Sqrt[a]] - (3*Sqrt[a]*b*ArcTanh[Sqrt[a 
 + b*Tan[x]^4]/Sqrt[a]])/2 + a*Sqrt[a + b*Tan[x]^4] + (3*b*Sqrt[a + b*Tan[ 
x]^4])/2 - (3*b*Tan[x]^2*Sqrt[a + b*Tan[x]^4])/2 - ((2*(a + b) - b*Tan[x]^ 
2)*Sqrt[a + b*Tan[x]^4])/2 + Cot[x]*(a + b*Tan[x]^4)^(3/2) - (Cot[x]^2*(a 
+ b*Tan[x]^4)^(3/2))/2)/2
 

Defintions of rubi rules used

rule 617

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Int[ExpandIntegrand[(a + b*x^2)^p, x^m*(c + d*x)^n, x], x] /; FreeQ[{ 
a, b, c, d, p}, x] && ILtQ[n, 0] && IntegerQ[m] && IntegerQ[2*p]
 

rule 1579

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], 
 x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(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 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [F]

\[\int \cot \left (x \right )^{5} \left (a +b \tan \left (x \right )^{4}\right )^{\frac {3}{2}}d x\]

Input:

int(cot(x)^5*(a+b*tan(x)^4)^(3/2),x)
 

Output:

int(cot(x)^5*(a+b*tan(x)^4)^(3/2),x)
 

Fricas [A] (verification not implemented)

Time = 21.45 (sec) , antiderivative size = 1464, normalized size of antiderivative = 9.70 \[ \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

integrate(cot(x)^5*(a+b*tan(x)^4)^(3/2),x, algorithm="fricas")
 

Output:

[1/8*(2*b^(3/2)*log(2*b*tan(x)^4 + 2*sqrt(b*tan(x)^4 + a)*sqrt(b)*tan(x)^2 
 + a)*tan(x)^4 + 2*(a + b)^(3/2)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x 
)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/( 
tan(x)^4 + 2*tan(x)^2 + 1))*tan(x)^4 + (2*a + 3*b)*sqrt(a)*log((b*tan(x)^4 
 - 2*sqrt(b*tan(x)^4 + a)*sqrt(a) + 2*a)/tan(x)^4)*tan(x)^4 + 2*sqrt(b*tan 
(x)^4 + a)*(2*a*tan(x)^2 - a))/tan(x)^4, -1/8*(4*sqrt(-b)*b*arctan(sqrt(-b 
)*tan(x)^2/sqrt(b*tan(x)^4 + a))*tan(x)^4 - 2*(a + b)^(3/2)*log(((a*b + 2* 
b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*s 
qrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1))*tan(x)^4 - (2*a + 3 
*b)*sqrt(a)*log((b*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*sqrt(a) + 2*a)/tan(x) 
^4)*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*(2*a*tan(x)^2 - a))/tan(x)^4, 1/4*(s 
qrt(-a)*(2*a + 3*b)*arctan(sqrt(-a)/sqrt(b*tan(x)^4 + a))*tan(x)^4 + b^(3/ 
2)*log(2*b*tan(x)^4 + 2*sqrt(b*tan(x)^4 + a)*sqrt(b)*tan(x)^2 + a)*tan(x)^ 
4 + (a + b)^(3/2)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b* 
tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*ta 
n(x)^2 + 1))*tan(x)^4 + sqrt(b*tan(x)^4 + a)*(2*a*tan(x)^2 - a))/tan(x)^4, 
 -1/4*(2*sqrt(-b)*b*arctan(sqrt(-b)*tan(x)^2/sqrt(b*tan(x)^4 + a))*tan(x)^ 
4 - sqrt(-a)*(2*a + 3*b)*arctan(sqrt(-a)/sqrt(b*tan(x)^4 + a))*tan(x)^4 - 
(a + b)^(3/2)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan( 
x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan...
 

Sympy [F]

\[ \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\int \left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac {3}{2}} \cot ^{5}{\left (x \right )}\, dx \] Input:

integrate(cot(x)**5*(a+b*tan(x)**4)**(3/2),x)
 

Output:

Integral((a + b*tan(x)**4)**(3/2)*cot(x)**5, x)
 

Maxima [F]

\[ \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\int { {\left (b \tan \left (x\right )^{4} + a\right )}^{\frac {3}{2}} \cot \left (x\right )^{5} \,d x } \] Input:

integrate(cot(x)^5*(a+b*tan(x)^4)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(x)^4 + a)^(3/2)*cot(x)^5, x)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.13 \[ \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\frac {{\left (2 \, a^{2} + 3 \, a b\right )} \arctan \left (-\frac {\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a}} + \frac {{\left (\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a}\right )}^{3} a b - 2 \, {\left (\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a}\right )}^{2} a^{2} \sqrt {b} + {\left (\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a}\right )} a^{2} b + 2 \, a^{3} \sqrt {b}}{2 \, {\left ({\left (\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a}\right )}^{2} - a\right )}^{2}} \] Input:

integrate(cot(x)^5*(a+b*tan(x)^4)^(3/2),x, algorithm="giac")
 

Output:

1/2*(2*a^2 + 3*a*b)*arctan(-(sqrt(b)*tan(x)^2 - sqrt(b*tan(x)^4 + a))/sqrt 
(-a))/sqrt(-a) + 1/2*((sqrt(b)*tan(x)^2 - sqrt(b*tan(x)^4 + a))^3*a*b - 2* 
(sqrt(b)*tan(x)^2 - sqrt(b*tan(x)^4 + a))^2*a^2*sqrt(b) + (sqrt(b)*tan(x)^ 
2 - sqrt(b*tan(x)^4 + a))*a^2*b + 2*a^3*sqrt(b))/((sqrt(b)*tan(x)^2 - sqrt 
(b*tan(x)^4 + a))^2 - a)^2
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (x\right )}^5\,{\left (b\,{\mathrm {tan}\left (x\right )}^4+a\right )}^{3/2} \,d x \] Input:

int(cot(x)^5*(a + b*tan(x)^4)^(3/2),x)
 

Output:

int(cot(x)^5*(a + b*tan(x)^4)^(3/2), x)
 

Reduce [F]

\[ \int \cot ^5(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\int \cot \left (x \right )^{5} \left (\tan \left (x \right )^{4} b +a \right )^{\frac {3}{2}}d x \] Input:

int(cot(x)^5*(a+b*tan(x)^4)^(3/2),x)
 

Output:

int(cot(x)^5*(a+b*tan(x)^4)^(3/2),x)