Integrand size = 25, antiderivative size = 355 \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx=-\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right )}{a^2 d (e \cot (c+d x))^{5/2} \sqrt {\sin (2 c+2 d x)}}+\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}-\frac {\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}+\frac {\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)} \]
-4*cot(d*x+c)^3/a^2/d/(e*cot(d*x+c))^(5/2)+4*cos(d*x+c)*cot(d*x+c)^3/a^2/d /(e*cot(d*x+c))^(5/2)-4*cos(d*x+c)*cot(d*x+c)^2*(sin(c+1/4*Pi+d*x)^2)^(1/2 )/sin(c+1/4*Pi+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))/a^2/d/(e*cot(d*x+ c))^(5/2)/sin(2*d*x+2*c)^(1/2)-1/2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^2 /d/(e*cot(d*x+c))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)-1/2*arctan(1+2^(1/2)*tan( d*x+c)^(1/2))/a^2/d/(e*cot(d*x+c))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)-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))^(5/2)*2^(1/2)/t an(d*x+c)^(5/2)+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))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)
\[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx=\int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx \]
Time = 0.74 (sec) , antiderivative size = 263, 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.
| \(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^2 (e \cot (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^2 (e \cot (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4388 |
| \(\displaystyle \frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(\sec (c+d x) a+a)^2}dx}{\tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (-\cot \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{\tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4376 |
| \(\displaystyle \frac {\int \frac {(a-a \sec (c+d x))^2}{\tan ^{\frac {3}{2}}(c+d x)}dx}{a^4 \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\left (-\cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{a^4 \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle \frac {\int \left (\frac {\sec ^2(c+d x) a^2}{\tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \sec (c+d x) a^2}{\tan ^{\frac {3}{2}}(c+d x)}+\frac {a^2}{\tan ^{\frac {3}{2}}(c+d x)}\right )dx}{a^4 \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {a^2 \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {4 a^2}{d \sqrt {\tan (c+d x)}}-\frac {a^2 \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {a^2 \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {4 a^2 \cos (c+d x)}{d \sqrt {\tan (c+d x)}}+\frac {4 a^2 \cos (c+d x) \sqrt {\tan (c+d x)} E\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{d \sqrt {\sin (2 c+2 d x)}}}{a^4 \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}\) |
((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) - (4*a^2)/(d*Sqrt[Tan[c + d*x]]) + (4*a^2*Cos[c + d*x])/(d*Sqrt[Tan[c + d*x]]) + (4*a^2*Cos[c + d*x]*Ellipt icE[c - Pi/4 + d*x, 2]*Sqrt[Tan[c + d*x]])/(d*Sqrt[Sin[2*c + 2*d*x]]))/(a^ 4*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))
3.3.52.3.1 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 = 8.10 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.74
| method | result | size |
| default | \(\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \left (i \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-4 i \operatorname {EllipticE}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 i \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+4 \operatorname {EllipticE}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sin \left (d x +c \right )}{a^{2} d \left (\cos \left (d x +c \right )-1\right ) e^{2} \sqrt {e \cot \left (d x +c \right )}}\) | \(262\) |
(1/2+1/2*I)/a^2/d*2^(1/2)*(I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/ 2-1/2*I,1/2*2^(1/2))-4*I*EllipticE((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^( 1/2))+2*I*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))-EllipticP i((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))+4*EllipticE((csc( d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*EllipticF((csc(d*x+c)-cot(d*x+c) +1)^(1/2),1/2*2^(1/2)))*(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)/(cos(d*x+c)-1)/e^2/(e*cot(d*x+ c))^(1/2)*sin(d*x+c)
Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]