Integrand size = 25, antiderivative size = 389 \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x) \csc (c+d x) \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)} \]
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)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*Elli pticF(cos(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))/a^2/d/(e*cot(d*x+c))^(11/2 )*2^(1/2)/tan(d*x+c)^(11/2)-1/2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d/( e*cot(d*x+c))^(11/2)*2^(1/2)/tan(d*x+c)^(11/2)+1/4*ln(1-2^(1/2)*tan(d*x+c) ^(1/2)+tan(d*x+c))/a^2/d/(e*cot(d*x+c))^(11/2)*2^(1/2)/tan(d*x+c)^(11/2)-1 /4*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^2/d/(e*cot(d*x+c))^(11/2)*2 ^(1/2)/tan(d*x+c)^(11/2)
\[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx \]
Time = 0.75 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.74, 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 |
| \(\displaystyle \frac {\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {a^2 \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a^2 \sqrt {\tan (c+d x)}}{d}+\frac {a^2 \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {a^2 \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {4 a^2 \sqrt {\tan (c+d x)} \sec (c+d x)}{3 d}+\frac {2 a^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d \sqrt {\tan (c+d x)}}}{a^4 \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}\) |
((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))
3.3.55.3.1 Defintions of rubi rules used
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]
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]
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]
Result contains complex when optimal does not.
Time = 9.69 (sec) , antiderivative size = 1007, normalized size of antiderivative = 2.59
1/30/a^2/d*2^(1/2)*(15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d *x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d* x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^3-15*I*(cot(d*x+c)-csc(d*x +c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2) *EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x +c)^3+15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*( csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2), 1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^2-15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*( cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((c sc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^2+15*(cot( d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d *x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^ (1/2))*cos(d*x+c)^3+15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x +c))^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2 ))*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*cos(d*x+c)^3-50*(csc(d*x+c)-cot(d*x+c)+ 1)^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*Ell ipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*cos(d*x+c)^3+15*(cot(d *x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d* x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^( 1/2))*cos(d*x+c)^2+15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d...
Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cot \left (d x + c\right )\right )^{\frac {11}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{11/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]