Integrand size = 25, antiderivative size = 343 \[ \int (e \cot (c+d x))^{3/2} (a+a \sec (c+d x))^2 \, dx=-\frac {4 a^2 (e \cot (c+d x))^{3/2} \sin (c+d x)}{d}-\frac {4 a^2 (e \cot (c+d x))^{3/2} \tan (c+d x)}{d}-\frac {4 a^2 (e \cot (c+d x))^{3/2} E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sin (c+d x) \tan (c+d x)}{d \sqrt {\sin (2 c+2 d x)}}+\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right ) (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}{\sqrt {2} d}-\frac {a^2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right ) (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}{\sqrt {2} d}-\frac {a^2 (e \cot (c+d x))^{3/2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac {3}{2}}(c+d x)}{2 \sqrt {2} d}+\frac {a^2 (e \cot (c+d x))^{3/2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac {3}{2}}(c+d x)}{2 \sqrt {2} d} \]
-4*a^2*(e*cot(d*x+c))^(3/2)*sin(d*x+c)/d-4*a^2*(e*cot(d*x+c))^(3/2)*tan(d* x+c)/d+4*a^2*(e*cot(d*x+c))^(3/2)*(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))*sin(d*x+c)*tan(d*x+c)/d/sin(2*d *x+2*c)^(1/2)-1/2*a^2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*(e*cot(d*x+c))^( 3/2)*tan(d*x+c)^(3/2)/d*2^(1/2)-1/2*a^2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2)) *(e*cot(d*x+c))^(3/2)*tan(d*x+c)^(3/2)/d*2^(1/2)-1/4*a^2*(e*cot(d*x+c))^(3 /2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))*tan(d*x+c)^(3/2)/d*2^(1/2)+1 /4*a^2*(e*cot(d*x+c))^(3/2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))*tan( d*x+c)^(3/2)/d*2^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 14.44 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.64 \[ \int (e \cot (c+d x))^{3/2} (a+a \sec (c+d x))^2 \, dx=-\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) (e \cot (c+d x))^{3/2} \left (2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )+16 \sqrt {\cot (c+d x)}+16 \sqrt {\cot (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\tan ^2(c+d x)\right )+\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )\right ) \sec ^4\left (\frac {1}{2} \cot ^{-1}(\cot (c+d x))\right )}{4 d \cot ^{\frac {3}{2}}(c+d x)} \]
-1/4*(a^2*Cos[(c + d*x)/2]^4*(e*Cot[c + d*x])^(3/2)*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d* x]]] + 16*Sqrt[Cot[c + d*x]] + 16*Sqrt[Cot[c + d*x]]*Hypergeometric2F1[-1/ 4, 1/2, 3/4, -Tan[c + d*x]^2] + Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d* x]])*Sec[ArcCot[Cot[c + d*x]]/2]^4)/(d*Cot[c + d*x]^(3/2))
Time = 0.62 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.76, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4388, 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.
\(\displaystyle \int (a \sec (c+d x)+a)^2 (e \cot (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sec (c+d x)+a)^2 (e \cot (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 4388 |
\(\displaystyle \tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2} \int \frac {(\sec (c+d x) a+a)^2}{\tan ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}{\left (-\cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4374 |
\(\displaystyle \tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2} \int \left (\frac {\sec ^2(c+d x) a^2}{\tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \sec (c+d x) a^2}{\tan ^{\frac {3}{2}}(c+d x)}+\frac {a^2}{\tan ^{\frac {3}{2}}(c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2} \left (\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {a^2 \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {4 a^2}{d \sqrt {\tan (c+d x)}}-\frac {a^2 \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {a^2 \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {4 a^2 \cos (c+d x)}{d \sqrt {\tan (c+d x)}}-\frac {4 a^2 \cos (c+d x) \sqrt {\tan (c+d x)} E\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{d \sqrt {\sin (2 c+2 d x)}}\right )\) |
(e*Cot[c + d*x])^(3/2)*((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*L og[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (a^2*Lo g[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]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[Tan[c + d*x]])/(d*Sqrt[ Sin[2*c + 2*d*x]]))*Tan[c + d*x]^(3/2)
3.3.39.3.1 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_)*((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]
Time = 10.33 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.53
method | result | size |
parts | \(-\frac {2 a^{2} e \left (\sqrt {e \cot \left (d x +c \right )}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{d}-\frac {2 a^{2} e \sqrt {e \cot \left (d x +c \right )}}{d}-\frac {2 a^{2} \sqrt {2}\, e \sqrt {e \cot \left (d x +c \right )}\, \left (-2 \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 )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\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 )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \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 )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sec \left (d x +c \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 )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sec \left (d x +c \right )+\sqrt {2}\right )}{d}\) | \(524\) |
default | \(\text {Expression too large to display}\) | \(1037\) |
-2*a^2/d*e*((e*cot(d*x+c))^(1/2)-1/8*(e^2)^(1/4)*2^(1/2)*(ln((e*cot(d*x+c) +(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)-(e^2) ^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^(1/2)/(e^2)^( 1/4)*(e*cot(d*x+c))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^ (1/2)+1)))-2*a^2/d*e*(e*cot(d*x+c))^(1/2)-2*a^2/d*2^(1/2)*e*(e*cot(d*x+c)) ^(1/2)*(-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)*EllipticE((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)*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2), 1/2*2^(1/2))-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)*EllipticE((csc(d*x+c)-cot(d*x+c)+1)^(1/ 2),1/2*2^(1/2))*sec(d*x+c)+(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)*EllipticF((csc(d*x+c)-cot(d *x+c)+1)^(1/2),1/2*2^(1/2))*sec(d*x+c)+2^(1/2))
Timed out. \[ \int (e \cot (c+d x))^{3/2} (a+a \sec (c+d x))^2 \, dx=\text {Timed out} \]
Timed out. \[ \int (e \cot (c+d x))^{3/2} (a+a \sec (c+d x))^2 \, dx=\text {Timed out} \]
Exception generated. \[ \int (e \cot (c+d x))^{3/2} (a+a \sec (c+d x))^2 \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int (e \cot (c+d x))^{3/2} (a+a \sec (c+d x))^2 \, dx=\int { \left (e \cot \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int (e \cot (c+d x))^{3/2} (a+a \sec (c+d x))^2 \, dx=\int {\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \]