\(\int \frac {\cot ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [340]

Optimal result

Integrand size = 23, antiderivative size = 449 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {2 \left (a^2+b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {-\frac {b (-1+\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a (a-b) (a+b)^{3/2} d}-\frac {\left (a^2-a b+2 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {-\frac {b (-1+\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a (a-b) (a+b)^{3/2} d}-\frac {\cot (c+d x)}{d (a+b \sec (c+d x))^{3/2}}+\frac {b^2 \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (a^2+b^2\right ) \tan (c+d x)}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

2*(a+b)^(1/2)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+ 
b)/a,((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c) 
)/(a-b))^(1/2)/a^2/d+2*(a^2+b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/ 
2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(-b*(-1+sec(d*x+c))/(a+b))^(1/2)*(-b*( 
1+sec(d*x+c))/(a-b))^(1/2)/a/(a-b)/(a+b)^(3/2)/d-(a^2-a*b+2*b^2)*cot(d*x+c 
)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(-b*(- 
1+sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a/(a-b)/(a+b)^( 
3/2)/d-cot(d*x+c)/d/(a+b*sec(d*x+c))^(3/2)+b^2*tan(d*x+c)/(a^2-b^2)/d/(a+b 
*sec(d*x+c))^(3/2)+2*b^2*(a^2+b^2)*tan(d*x+c)/a/(a^2-b^2)^2/d/(a+b*sec(d*x 
+c))^(1/2)
 

Mathematica [A] (verified)

Time = 9.66 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {(b+a \cos (c+d x)) \sec ^2(c+d x) \left (-a (b+a \cos (c+d x)) \left (-2 a b+\left (a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)+2 b^4 \sin (c+d x)-2 b \left (a^2+b^2\right ) (b+a \cos (c+d x)) \sin (c+d x)+2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (2 b \left (a^3+a^2 b+a b^2+b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+a \left (2 a^3-a^2 b-6 a b^2-3 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-4 \left (a^2-b^2\right )^2 \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+b \left (a^2+b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}} \] Input:

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

Output:

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(-(a*(b + a*Cos[c + d*x])*(-2*a*b + ( 
a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]) + 2*b^4*Sin[c + d*x] - 2*b*(a^2 + b 
^2)*(b + a*Cos[c + d*x])*Sin[c + d*x] + 2*Cos[(c + d*x)/2]^2*(2*b*(a^3 + a 
^2*b + a*b^2 + b^3)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[ 
c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]] 
, (a - b)/(a + b)] + a*(2*a^3 - a^2*b - 6*a*b^2 - 3*b^3)*Sqrt[Cos[c + d*x] 
/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]) 
)]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 4*(a^2 - b^2)^2* 
Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*( 
1 + Cos[c + d*x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + 
b)] + b*(a^2 + b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*T 
an[(c + d*x)/2])))/(a*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 664, normalized size of antiderivative = 1.48, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4384, 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.

\(\displaystyle \int \frac {\cot ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4384

\(\displaystyle \int \left (\frac {\csc ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}}-\frac {1}{(a+b \sec (c+d x))^{3/2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 b^2 \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {4 a b^2 \tan (c+d x)}{d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}}-\frac {(3 a-b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d (a-b) (a+b)^{3/2}}-\frac {2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}}+\frac {4 a \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d (a-b) (a+b)^{3/2}}-\frac {\cot (c+d x)}{d (a+b \sec (c+d x))^{3/2}}\)

Input:

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

Output:

(4*a*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], 
(a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c 
 + d*x]))/(a - b))])/((a - b)*(a + b)^(3/2)*d) - (2*Cot[c + d*x]*EllipticE 
[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 
 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*Sqr 
t[a + b]*d) - ((3*a - b)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + 
d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]* 
Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/((a - b)*(a + b)^(3/2)*d) + (2*Co 
t[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b) 
/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x] 
))/(a - b))])/(a*Sqrt[a + b]*d) + (2*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[( 
a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S 
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b)) 
])/(a^2*d) - Cot[c + d*x]/(d*(a + b*Sec[c + d*x])^(3/2)) + (b^2*Tan[c + d* 
x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (4*a*b^2*Tan[c + d*x])/(( 
a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]]) - (2*b^2*Tan[c + d*x])/(a*(a^2 - 
b^2)*d*Sqrt[a + b*Sec[c + d*x]])
 

Defintions of rubi rules used

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 4384

Int[cot[(c_.) + (d_.)*(x_)]^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_ 
), x_Symbol] :> Int[ExpandIntegrand[(a + b*Csc[c + d*x])^n, (-1 + Sec[c + d 
*x]^2)^(-m/2), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0] && 
ILtQ[m/2, 0] && IntegerQ[n - 1/2] && EqQ[m, -2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1150\) vs. \(2(416)=832\).

Time = 7.32 (sec) , antiderivative size = 1151, normalized size of antiderivative = 2.56

method	result	size
default	\(\text {Expression too large to display}\)	\(1151\)

Input:

int(cot(d*x+c)^2/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/d/a/(a-b)^2/(a+b)^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))*((-4*cos(d* 
x+c)-4)*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))*(cos(d*x 
+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)* 
a^4+(8*cos(d*x+c)+8)*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1 
/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d* 
x+c)))^(1/2)*a^2*b^2+(-4*cos(d*x+c)-4)*EllipticPi(-csc(d*x+c)+cot(d*x+c),- 
1,((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos 
(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^4+(2*cos(d*x+c)+2)*EllipticE(-csc(d*x+c)+ 
cot(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b) 
*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^3*b+(2*cos(d*x+c)+2)*EllipticE(- 
csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 
2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b^2+(2*cos(d*x+c)+2 
)*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos 
(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b^3+(2*c 
os(d*x+c)+2)*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d* 
x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2) 
*b^4+(2*cos(d*x+c)+2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos( 
d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b 
))^(1/2))*a^4+(-1-cos(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*( 
b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+cot(d*x+c),...
 

Fricas [F(-1)]

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

integrate(cot(d*x+c)^2/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate(cot(d*x+c)**2/(a+b*sec(d*x+c))**(3/2),x)
 

Output:

Integral(cot(c + d*x)**2/(a + b*sec(c + d*x))**(3/2), x)
 

Maxima [F]

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

integrate(cot(d*x+c)^2/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(cot(d*x + c)^2/(b*sec(d*x + c) + a)^(3/2), x)
 

Giac [F]

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

integrate(cot(d*x+c)^2/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate(cot(d*x + c)^2/(b*sec(d*x + c) + a)^(3/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(c + d*x)*b + a)*cot(c + d*x)**2)/(sec(c + d*x)**2*b**2 + 2*s 
ec(c + d*x)*a*b + a**2),x)