\(\int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx\) [255]

Optimal result

Integrand size = 25, antiderivative size = 326 \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x) \csc (c+d x) \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)} \] Output:

2/5*cot(d*x+c)^3/a^2/d/(e*cot(d*x+c))^(11/2)+2*cot(d*x+c)^5/a^2/d/(e*cot(d 
*x+c))^(11/2)-4/3*cot(d*x+c)^4*csc(d*x+c)/a^2/d/(e*cot(d*x+c))^(11/2)+2/3* 
cot(d*x+c)^5*csc(d*x+c)*InverseJacobiAM(c-1/4*Pi+d*x,2^(1/2))*sin(2*d*x+2* 
c)^(1/2)/a^2/d/(e*cot(d*x+c))^(11/2)-1/2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2 
))*2^(1/2)/a^2/d/(e*cot(d*x+c))^(11/2)/tan(d*x+c)^(11/2)-1/2*arctan(1+2^(1 
/2)*tan(d*x+c)^(1/2))*2^(1/2)/a^2/d/(e*cot(d*x+c))^(11/2)/tan(d*x+c)^(11/2 
)-1/2*arctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/a^2/d/(e*co 
t(d*x+c))^(11/2)/tan(d*x+c)^(11/2)
 

Mathematica [F]

\[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx \] Input:

Integrate[1/((e*Cot[c + d*x])^(11/2)*(a + a*Sec[c + d*x])^2),x]
 

Output:

Integrate[1/((e*Cot[c + d*x])^(11/2)*(a + a*Sec[c + d*x])^2), x]
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4388, 3042, 4376, 3042, 4374, 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.

\(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^2 (e \cot (c+d x))^{11/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^2 (e \cot (c+d x))^{11/2}}dx\)

\(\Big \downarrow \) 4388

\(\displaystyle \frac {\int \frac {\tan ^{\frac {11}{2}}(c+d x)}{(\sec (c+d x) a+a)^2}dx}{\tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (-\cot \left (c+d x+\frac {\pi }{2}\right )\right )^{11/2}}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{\tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}\)

\(\Big \downarrow \) 4376

\(\displaystyle \frac {\int (a-a \sec (c+d x))^2 \tan ^{\frac {3}{2}}(c+d x)dx}{a^4 \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (-\cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx}{a^4 \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}\)

\(\Big \downarrow \) 4374

\(\displaystyle \frac {\int \left (\sec ^2(c+d x) \tan ^{\frac {3}{2}}(c+d x) a^2-2 \sec (c+d x) \tan ^{\frac {3}{2}}(c+d x) a^2+\tan ^{\frac {3}{2}}(c+d x) a^2\right )dx}{a^4 \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {a^2 \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a^2 \sqrt {\tan (c+d x)}}{d}+\frac {a^2 \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {a^2 \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {4 a^2 \sqrt {\tan (c+d x)} \sec (c+d x)}{3 d}+\frac {2 a^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d \sqrt {\tan (c+d x)}}}{a^4 \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}\)

Input:

Int[1/((e*Cot[c + d*x])^(11/2)*(a + a*Sec[c + d*x])^2),x]
 

Output:

((a^2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - (a^2*ArcTan[1 
+ Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) + (a^2*Log[1 - Sqrt[2]*Sqrt[Tan 
[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (a^2*Log[1 + Sqrt[2]*Sqrt[Tan[ 
c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*a^2*EllipticF[c - Pi/4 + d*x 
, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(3*d*Sqrt[Tan[c + d*x]]) + (2*a^ 
2*Sqrt[Tan[c + d*x]])/d - (4*a^2*Sec[c + d*x]*Sqrt[Tan[c + d*x]])/(3*d) + 
(2*a^2*Tan[c + d*x]^(5/2))/(5*d))/(a^4*(e*Cot[c + d*x])^(11/2)*Tan[c + d*x 
]^(11/2))
 

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 4374

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ 
c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
 

rule 4376

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

rule 4388

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*((a_) + (b_.)*sec[(c_.) + (d_.)*(x 
_)])^(n_.), x_Symbol] :> Simp[(e*Cot[c + d*x])^m*Tan[c + d*x]^m   Int[(a + 
b*Sec[c + d*x])^n/Tan[c + d*x]^m, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] 
 &&  !IntegerQ[m]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.11 (sec) , antiderivative size = 589, normalized size of antiderivative = 1.81

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {-\frac {2 \sin \left (d x +c \right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \left (-48+40 \sec \left (d x +c \right )-12 \sec \left (d x +c \right )^{2}+15 \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )+15 \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )+50 \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+15 i \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )+15 i \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{60 a^{2} d \,e^{5} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}\)	\(589\)

Input:

int(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

-1/60/a^2/d*2^(1/2)/e^5*(-2*sin(d*x+c)*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)/ 
(e*cot(d*x+c))^(1/2)/(-sin(d*x+c)*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(-48+ 
40*sec(d*x+c)-12*sec(d*x+c)^2+15*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d 
*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*EllipticPi((-co 
t(d*x+c)+csc(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(csc(d*x+c)+cot(d*x+c) 
)+15*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)* 
(-csc(d*x+c)+cot(d*x+c))^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2) 
,1/2+1/2*I,1/2*2^(1/2))*(csc(d*x+c)+cot(d*x+c))+50*(-cot(d*x+c)+csc(d*x+c) 
+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/ 
2)*EllipticF((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))*(-cot(d*x+c)-cs 
c(d*x+c))+15*I*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c) 
+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c 
)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(csc(d*x+c)+cot(d*x+c))+15*I*(-cot(d*x+c 
)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot 
(d*x+c))^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2 
^(1/2))*(-cot(d*x+c)-csc(d*x+c)))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas" 
)
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(1/(e*cot(d*x+c))**(11/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima" 
)
 

Output:

Timed out
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{11/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \] Input:

int(1/((e*cot(c + d*x))^(11/2)*(a + a/cos(c + d*x))^2),x)
 

Output:

int(cos(c + d*x)^2/(a^2*(e*cot(c + d*x))^(11/2)*(cos(c + d*x) + 1)^2), x)
 

Reduce [F]

\[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )^{6} \sec \left (d x +c \right )^{2}+2 \cot \left (d x +c \right )^{6} \sec \left (d x +c \right )+\cot \left (d x +c \right )^{6}}d x \right )}{a^{2} e^{6}} \] Input:

int(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x)
 

Output:

(sqrt(e)*int(sqrt(cot(c + d*x))/(cot(c + d*x)**6*sec(c + d*x)**2 + 2*cot(c 
 + d*x)**6*sec(c + d*x) + cot(c + d*x)**6),x))/(a**2*e**6)