\(\int \frac {\cot (e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [349]

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}-\frac {b}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(2 a-b) b}{a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}} \] Output:

-arctanh((a+b*tan(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2)/f+arctanh((a+b*tan(f*x+ 
e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/f-1/3*b/a/(a-b)/f/(a+b*tan(f*x+e)^2)^ 
(3/2)-(2*a-b)*b/a^2/(a-b)^2/f/(a+b*tan(f*x+e)^2)^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.64 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {-a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan ^2(e+f x)}{a-b}\right )+(a-b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \tan ^2(e+f x)}{a}\right )}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

(-(a*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[e + f*x]^2)/(a - b)]) + ( 
a - b)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*Tan[e + f*x]^2)/a])/(3*a*(a 
 - b)*f*(a + b*Tan[e + f*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4153, 354, 96, 169, 27, 174, 73, 221}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (e+f x) \left (a+b \tan (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 96

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((-2*b)/(3*a*(a - b)*(a + b*Tan[e + f*x]^2)^(3/2)) + (((-2*(a - b)^2*ArcTa 
nh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/Sqrt[a] + (2*a^2*ArcTanh[Sqrt[a + 
b*Tan[e + f*x]^2]/Sqrt[a - b]])/Sqrt[a - b])/(a*(a - b)) - (2*(2*a - b)*b) 
/(a*(a - b)*Sqrt[a + b*Tan[e + f*x]^2]))/(a*(a - b)))/(2*f)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 96

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[f*((e + f*x)^(p + 1)/((p + 1)*(b*e - a*f)*(d*e - c*f))), x] + S 
imp[1/((b*e - a*f)*(d*e - c*f))   Int[(b*d*e - b*c*f - a*d*f - b*d*f*x)*((e 
 + f*x)^(p + 1)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, 
 x] && LtQ[p, -1]
 

rule 169

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

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 [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(39147\) vs. \(2(129)=258\).

Time = 15.18 (sec) , antiderivative size = 39148, normalized size of antiderivative = 266.31

method	result	size
default	\(\text {Expression too large to display}\)	\(39148\)

Input:

int(cot(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (129) = 258\).

Time = 0.14 (sec) , antiderivative size = 1627, normalized size of antiderivative = 11.07 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/6*(3*(a^3*b^2*tan(f*x + e)^4 + 2*a^4*b*tan(f*x + e)^2 + a^5)*sqrt(a - b 
)*log((b*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 2*a - 
 b)/(tan(f*x + e)^2 + 1)) + 3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^3* 
b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3 
*a^2*b^3 - a*b^4)*tan(f*x + e)^2)*sqrt(a)*log((b*tan(f*x + e)^2 - 2*sqrt(b 
*tan(f*x + e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2) - 2*(7*a^4*b - 11*a^3* 
b^2 + 4*a^2*b^3 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(f*x + e)^2)*sqrt(b 
*tan(f*x + e)^2 + a))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f 
*x + e)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^2 + 
 (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*f), -1/6*(6*(a^3*b^2*tan(f*x + e)^4 
 + 2*a^4*b*tan(f*x + e)^2 + a^5)*sqrt(-a + b)*arctan(sqrt(-a + b)/sqrt(b*t 
an(f*x + e)^2 + a)) - 3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^3*b^2 - 
3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b 
^3 - a*b^4)*tan(f*x + e)^2)*sqrt(a)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f 
*x + e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2) + 2*(7*a^4*b - 11*a^3*b^2 + 
4*a^2*b^3 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f 
*x + e)^2 + a))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f*x + e 
)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^2 + (a^8 
- 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*f), 1/6*(6*(a^5 - 3*a^4*b + 3*a^3*b^2 - a 
^2*b^3 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*...
 

Sympy [F]

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

integrate(cot(f*x+e)/(a+b*tan(f*x+e)**2)**(5/2),x)
 

Output:

Integral(cot(e + f*x)/(a + b*tan(e + f*x)**2)**(5/2), x)
                                                                                    
                                                                                    
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Giac [F(-1)]

Timed out. \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 8.02 (sec) , antiderivative size = 2788, normalized size of antiderivative = 18.97 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

int(cot(e + f*x)/(a + b*tan(e + f*x)^2)^(5/2),x)
 

Output:

(b/(3*(a*b - a^2)) - (b*(a + b*tan(e + f*x)^2)*(2*a - b))/(a*b - a^2)^2)/( 
f*(a + b*tan(e + f*x)^2)^(3/2)) - atanh((2*a^5*b^13*f^2*(a + b*tan(e + f*x 
)^2)^(1/2))/((a^5)^(1/2)*(2*a^3*b^13*f^2 - 22*a^4*b^12*f^2 + 110*a^5*b^11* 
f^2 - 330*a^6*b^10*f^2 + 660*a^7*b^9*f^2 - 922*a^8*b^8*f^2 + 912*a^9*b^7*f 
^2 - 630*a^10*b^6*f^2 + 290*a^11*b^5*f^2 - 80*a^12*b^4*f^2 + 10*a^13*b^3*f 
^2)) - (22*a^6*b^12*f^2*(a + b*tan(e + f*x)^2)^(1/2))/((a^5)^(1/2)*(2*a^3* 
b^13*f^2 - 22*a^4*b^12*f^2 + 110*a^5*b^11*f^2 - 330*a^6*b^10*f^2 + 660*a^7 
*b^9*f^2 - 922*a^8*b^8*f^2 + 912*a^9*b^7*f^2 - 630*a^10*b^6*f^2 + 290*a^11 
*b^5*f^2 - 80*a^12*b^4*f^2 + 10*a^13*b^3*f^2)) + (110*a^7*b^11*f^2*(a + b* 
tan(e + f*x)^2)^(1/2))/((a^5)^(1/2)*(2*a^3*b^13*f^2 - 22*a^4*b^12*f^2 + 11 
0*a^5*b^11*f^2 - 330*a^6*b^10*f^2 + 660*a^7*b^9*f^2 - 922*a^8*b^8*f^2 + 91 
2*a^9*b^7*f^2 - 630*a^10*b^6*f^2 + 290*a^11*b^5*f^2 - 80*a^12*b^4*f^2 + 10 
*a^13*b^3*f^2)) - (330*a^8*b^10*f^2*(a + b*tan(e + f*x)^2)^(1/2))/((a^5)^( 
1/2)*(2*a^3*b^13*f^2 - 22*a^4*b^12*f^2 + 110*a^5*b^11*f^2 - 330*a^6*b^10*f 
^2 + 660*a^7*b^9*f^2 - 922*a^8*b^8*f^2 + 912*a^9*b^7*f^2 - 630*a^10*b^6*f^ 
2 + 290*a^11*b^5*f^2 - 80*a^12*b^4*f^2 + 10*a^13*b^3*f^2)) + (660*a^9*b^9* 
f^2*(a + b*tan(e + f*x)^2)^(1/2))/((a^5)^(1/2)*(2*a^3*b^13*f^2 - 22*a^4*b^ 
12*f^2 + 110*a^5*b^11*f^2 - 330*a^6*b^10*f^2 + 660*a^7*b^9*f^2 - 922*a^8*b 
^8*f^2 + 912*a^9*b^7*f^2 - 630*a^10*b^6*f^2 + 290*a^11*b^5*f^2 - 80*a^12*b 
^4*f^2 + 10*a^13*b^3*f^2)) - (922*a^10*b^8*f^2*(a + b*tan(e + f*x)^2)^(...
 

Reduce [F]

\[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )}{\tan \left (f x +e \right )^{6} b^{3}+3 \tan \left (f x +e \right )^{4} a \,b^{2}+3 \tan \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(cot(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(tan(e + f*x)**2*b + a)*cot(e + f*x))/(tan(e + f*x)**6*b**3 + 3*t 
an(e + f*x)**4*a*b**2 + 3*tan(e + f*x)**2*a**2*b + a**3),x)