\(\int \frac {\cot ^3(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [337]

Optimal result

Integrand size = 25, antiderivative size = 157 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{5/2} f}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2} f}-\frac {(a-3 b) b}{2 a^2 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 a f \sqrt {a+b \tan ^2(e+f x)}} \] Output:

1/2*(2*a+3*b)*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)^(3/2)/f-1/2*(a-3*b)*b/a^2/(a-b 
)/f/(a+b*tan(f*x+e)^2)^(1/2)-1/2*cot(f*x+e)^2/a/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 = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.73 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {-2 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan ^2(e+f x)}{a-b}\right )+(a-b) \left (a \cot ^2(e+f x)+(2 a+3 b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \tan ^2(e+f x)}{a}\right )\right )}{2 a^2 (-a+b) f \sqrt {a+b \tan ^2(e+f x)}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.64 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4153, 354, 114, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-(Cot[e + f*x]/(a*Sqrt[a + b*Tan[e + f*x]^2])) - (((-2*(a - b)*(2*a + 3*b 
)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/Sqrt[a] + (4*a^2*ArcTanh[Sq 
rt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/Sqrt[a - b])/(a*(a - b)) + (2*(a - 
3*b)*b)/(a*(a - b)*Sqrt[a + b*Tan[e + f*x]^2]))/(2*a))/(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 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(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] + Simp[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*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

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. \(9633\) vs. \(2(135)=270\).

Time = 7.73 (sec) , antiderivative size = 9634, normalized size of antiderivative = 61.36

method	result	size
default	\(\text {Expression too large to display}\)	\(9634\)

Input:

int(cot(f*x+e)^3/(a+b*tan(f*x+e)^2)^(3/2),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. 281 vs. \(2 (135) = 270\).

Time = 0.15 (sec) , antiderivative size = 1229, normalized size of antiderivative = 7.83 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(2*(a^3*b*tan(f*x + e)^4 + a^4*tan(f*x + e)^2)*sqrt(a - b)*log((b*ta 
n(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)) - ((2*a^3*b - a^2*b^2 - 4*a*b^3 + 3*b^4)*tan(f*x + e)^4 + ( 
2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*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 
*(a^4 - 2*a^3*b + a^2*b^2 + (a^3*b - 4*a^2*b^2 + 3*a*b^3)*tan(f*x + e)^2)* 
sqrt(b*tan(f*x + e)^2 + a))/((a^5*b - 2*a^4*b^2 + a^3*b^3)*f*tan(f*x + e)^ 
4 + (a^6 - 2*a^5*b + a^4*b^2)*f*tan(f*x + e)^2), 1/4*(4*(a^3*b*tan(f*x + e 
)^4 + a^4*tan(f*x + e)^2)*sqrt(-a + b)*arctan(sqrt(-a + b)/sqrt(b*tan(f*x 
+ e)^2 + a)) + ((2*a^3*b - a^2*b^2 - 4*a*b^3 + 3*b^4)*tan(f*x + e)^4 + (2* 
a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*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*( 
a^4 - 2*a^3*b + a^2*b^2 + (a^3*b - 4*a^2*b^2 + 3*a*b^3)*tan(f*x + e)^2)*sq 
rt(b*tan(f*x + e)^2 + a))/((a^5*b - 2*a^4*b^2 + a^3*b^3)*f*tan(f*x + e)^4 
+ (a^6 - 2*a^5*b + a^4*b^2)*f*tan(f*x + e)^2), -1/2*(((2*a^3*b - a^2*b^2 - 
 4*a*b^3 + 3*b^4)*tan(f*x + e)^4 + (2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*t 
an(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*tan(f*x + e)^2 + a)) + (a^3 
*b*tan(f*x + e)^4 + a^4*tan(f*x + e)^2)*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) 
) + (a^4 - 2*a^3*b + a^2*b^2 + (a^3*b - 4*a^2*b^2 + 3*a*b^3)*tan(f*x + ...
 

Sympy [F]

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

integrate(cot(f*x+e)**3/(a+b*tan(f*x+e)**2)**(3/2),x)
 

Output:

Integral(cot(e + f*x)**3/(a + b*tan(e + f*x)**2)**(3/2), x)
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 8.48 (sec) , antiderivative size = 2483, normalized size of antiderivative = 15.82 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

(b^2/(a*b - a^2) + (b*(a + b*tan(e + f*x)^2)*(a - 3*b))/(2*a*(a*b - a^2))) 
/(f*(a + b*tan(e + f*x)^2)^(3/2) - a*f*(a + b*tan(e + f*x)^2)^(1/2)) - (at 
an((((((a + b*tan(e + f*x)^2)^(1/2)*(144*a^6*b^9*f^3 - 528*a^7*b^8*f^3 + 5 
44*a^8*b^7*f^3 + 160*a^9*b^6*f^3 - 496*a^10*b^5*f^3 - 16*a^11*b^4*f^3 + 32 
0*a^12*b^3*f^3 - 128*a^13*b^2*f^3))/2 + (((a - b)^3)^(1/2)*(512*a^9*b^8*f^ 
4 - 96*a^8*b^9*f^4 - 1056*a^10*b^7*f^4 + 1024*a^11*b^6*f^4 - 416*a^12*b^5* 
f^4 + 32*a^14*b^3*f^4 + ((a + b*tan(e + f*x)^2)^(1/2)*((a - b)^3)^(1/2)*(2 
56*a^10*b^8*f^5 - 1792*a^11*b^7*f^5 + 5120*a^12*b^6*f^5 - 7680*a^13*b^5*f^ 
5 + 6400*a^14*b^4*f^5 - 2816*a^15*b^3*f^5 + 512*a^16*b^2*f^5))/(4*f*(a - b 
)^3)))/(2*f*(a - b)^3))*((a - b)^3)^(1/2)*1i)/(f*(a - b)^3) + ((((a + b*ta 
n(e + f*x)^2)^(1/2)*(144*a^6*b^9*f^3 - 528*a^7*b^8*f^3 + 544*a^8*b^7*f^3 + 
 160*a^9*b^6*f^3 - 496*a^10*b^5*f^3 - 16*a^11*b^4*f^3 + 320*a^12*b^3*f^3 - 
 128*a^13*b^2*f^3))/2 + (((a - b)^3)^(1/2)*(96*a^8*b^9*f^4 - 512*a^9*b^8*f 
^4 + 1056*a^10*b^7*f^4 - 1024*a^11*b^6*f^4 + 416*a^12*b^5*f^4 - 32*a^14*b^ 
3*f^4 + ((a + b*tan(e + f*x)^2)^(1/2)*((a - b)^3)^(1/2)*(256*a^10*b^8*f^5 
- 1792*a^11*b^7*f^5 + 5120*a^12*b^6*f^5 - 7680*a^13*b^5*f^5 + 6400*a^14*b^ 
4*f^5 - 2816*a^15*b^3*f^5 + 512*a^16*b^2*f^5))/(4*f*(a - b)^3)))/(2*f*(a - 
 b)^3))*((a - b)^3)^(1/2)*1i)/(f*(a - b)^3))/(144*a^6*b^8*f^2 - 384*a^7*b^ 
7*f^2 + 256*a^8*b^6*f^2 + 96*a^9*b^5*f^2 - 144*a^10*b^4*f^2 + 32*a^11*b^3* 
f^2 - ((((a + b*tan(e + f*x)^2)^(1/2)*(144*a^6*b^9*f^3 - 528*a^7*b^8*f^...
 

Reduce [F]

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

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

Output:

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