3.4.34 \(\int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [334]

3.4.34.1 Optimal result

Integrand size = 23, antiderivative size = 361 \[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {\cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{\sqrt {a+b} d}-\frac {\cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{\sqrt {a+b} d}+\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d}-\frac {\cot (c+d x)}{d \sqrt {a+b \sec (c+d x)}}+\frac {b^2 \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

output

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)/d/(a+b)^ 
(1/2)-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)/d/ 
(a+b)^(1/2)+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))*(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/d-cot(d*x+c)/d/(a+b*sec(d*x+c))^(1/2)+b^2*tan 
(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
 

3.4.34.2 Mathematica [A] (verified)

Time = 13.15 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.01 \[ \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {\left (-8 b (a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \cot \left (\frac {1}{2} (c+d x)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-8 \left (2 a^2-a b-3 b^2\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \cot \left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+4 (a-b) \csc ^2(c+d x) \left (2 \cos (c+d x) (b+a \cos (c+d x))+32 (a+b) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \tan (c+d x)}{8 \left (-a^2+b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

input

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

output

((-8*b*(a + b)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d 
*x])/((a + b)*(1 + Cos[c + d*x]))]*Cot[(c + d*x)/2]*EllipticE[ArcSin[Tan[( 
c + d*x)/2]], (a - b)/(a + b)] - 8*(2*a^2 - a*b - 3*b^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]) 
)]*Cot[(c + d*x)/2]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 
 4*(a - b)*Csc[c + d*x]^2*(2*Cos[c + d*x]*(b + a*Cos[c + d*x]) + 32*(a + b 
)*Cos[(c + d*x)/2]^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]))]*EllipticPi[-1, ArcSin[Tan[(c + d* 
x)/2]], (a - b)/(a + b)]*Sin[(c + d*x)/2]))*Tan[c + d*x])/(8*(-a^2 + b^2)* 
d*Sqrt[a + b*Sec[c + d*x]])
 

3.4.34.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4384

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

\(\Big \downarrow \) 2009

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

input

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

output

(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))])/(Sqrt[a + b]*d) - (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))])/(Sqrt[a + b]*d) + (2*S 
qrt[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)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sq 
rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d) - Cot[c + d*x]/(d*Sqrt[a + b* 
Sec[c + d*x]]) + (b^2*Tan[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x] 
])
 

3.4.34.3.1 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]
 

3.4.34.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1373\) vs. \(2(330)=660\).

Time = 7.10 (sec) , antiderivative size = 1374, normalized size of antiderivative = 3.81

method	result	size
default	\(\text {Expression too large to display}\)	\(1374\)

input

int(cot(d*x+c)^2/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

output

-1/2/d/(a-b)/(a+b)*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*cs 
c(d*x+c)^2-a-b)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^2 
*csc(d*x+c)^2-1)*(-4*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos( 
d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*E 
llipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*(-cot(d*x+c)+csc(d 
*x+c))+2*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*cs 
c(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticF(cot 
(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b*(-cot(d*x+c)+csc(d*x+c))+6*(-( 
1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b 
*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc( 
d*x+c),((a-b)/(a+b))^(1/2))*b^2*(-cot(d*x+c)+csc(d*x+c))-2*(-(1-cos(d*x+c) 
)^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+ 
c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b 
)/(a+b))^(1/2))*a*b*(-cot(d*x+c)+csc(d*x+c))-2*(-(1-cos(d*x+c))^2*csc(d*x+ 
c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d* 
x+c)^2-a-b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/ 
2))*b^2*(-cot(d*x+c)+csc(d*x+c))+8*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2 
)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/ 
(a+b))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^2* 
(-cot(d*x+c)+csc(d*x+c))-8*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(...
 

3.4.34.5 Fricas [F]

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

input

integrate(cot(d*x+c)^2/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
 

output

integral(cot(d*x + c)^2/sqrt(b*sec(d*x + c) + a), x)
 

3.4.34.6 Sympy [F]

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

input

integrate(cot(d*x+c)**2/(a+b*sec(d*x+c))**(1/2),x)
 

output

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

3.4.34.7 Maxima [F]

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

input

integrate(cot(d*x+c)^2/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
 

output

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

3.4.34.8 Giac [F]

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

input

integrate(cot(d*x+c)^2/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
 

output

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

3.4.34.9 Mupad [F(-1)]

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

input

int(cot(c + d*x)^2/(a + b/cos(c + d*x))^(1/2),x)
 

output

int(cot(c + d*x)^2/(a + b/cos(c + d*x))^(1/2), x)