Integrand size = 25, antiderivative size = 375 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \cot (c+d x))^{3/2}} \, dx=\frac {2 a^2 \cot (c+d x)}{d (e \cot (c+d x))^{3/2}}+\frac {4 a^2 \csc (c+d x)}{3 d (e \cot (c+d x))^{3/2}}-\frac {2 a^2 \cot (c+d x) \csc (c+d x) \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sqrt {\sin (2 c+2 d x)}}{3 d (e \cot (c+d x))^{3/2}}+\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a^2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}+\frac {a^2 \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a^2 \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 \tan (c+d x)}{5 d (e \cot (c+d x))^{3/2}} \]
2*a^2*cot(d*x+c)/d/(e*cot(d*x+c))^(3/2)+4/3*a^2*csc(d*x+c)/d/(e*cot(d*x+c) )^(3/2)+2/3*a^2*cot(d*x+c)*csc(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/ 4*Pi+d*x)*EllipticF(cos(c+1/4*Pi+d*x),2^(1/2))*sin(2*d*x+2*c)^(1/2)/d/(e*c ot(d*x+c))^(3/2)-1/2*a^2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d/(e*cot(d*x+ c))^(3/2)*2^(1/2)/tan(d*x+c)^(3/2)-1/2*a^2*arctan(1+2^(1/2)*tan(d*x+c)^(1/ 2))/d/(e*cot(d*x+c))^(3/2)*2^(1/2)/tan(d*x+c)^(3/2)+1/4*a^2*ln(1-2^(1/2)*t an(d*x+c)^(1/2)+tan(d*x+c))/d/(e*cot(d*x+c))^(3/2)*2^(1/2)/tan(d*x+c)^(3/2 )-1/4*a^2*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d/(e*cot(d*x+c))^(3/2) *2^(1/2)/tan(d*x+c)^(3/2)+2/5*a^2*tan(d*x+c)/d/(e*cot(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 13.32 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.34 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \cot (c+d x))^{3/2}} \, dx=\frac {a^2 \left (\operatorname {Hypergeometric2F1}\left (-\frac {5}{4},1,-\frac {1}{4},-\cot ^2(c+d x)\right )+2 \left (5 \cot ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\cot ^2(c+d x)\right )+\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},-\tan ^2(c+d x)\right )\right )\right ) (1+\sec (c+d x))^2 \sec ^4\left (\frac {1}{2} \cot ^{-1}(\cot (c+d x))\right ) \sin ^2(c+d x)}{10 d e \sqrt {e \cot (c+d x)}} \]
(a^2*(Hypergeometric2F1[-5/4, 1, -1/4, -Cot[c + d*x]^2] + 2*(5*Cot[c + d*x ]^2*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2] + Hypergeometric2F1[1 /2, 5/4, 9/4, -Tan[c + d*x]^2]))*(1 + Sec[c + d*x])^2*Sec[ArcCot[Cot[c + d *x]]/2]^4*Sin[c + d*x]^2)/(10*d*e*Sqrt[e*Cot[c + d*x]])
Time = 0.65 (sec) , antiderivative size = 284, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{(e \cot (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{(e \cot (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4388 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^2 \tan ^{\frac {3}{2}}(c+d x)dx}{\tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (-\cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2dx}{\tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle \frac {\int \left (\sec ^2(c+d x) \tan ^{\frac {3}{2}}(c+d x) a^2+2 \sec (c+d x) \tan ^{\frac {3}{2}}(c+d x) a^2+\tan ^{\frac {3}{2}}(c+d x) a^2\right )dx}{\tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {a^2 \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a^2 \sqrt {\tan (c+d x)}}{d}+\frac {a^2 \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {a^2 \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {4 a^2 \sqrt {\tan (c+d x)} \sec (c+d x)}{3 d}-\frac {2 a^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d \sqrt {\tan (c+d x)}}}{\tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2}}\) |
((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*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))/((e*Cot[c + d*x])^(3/2)*Tan[c + d*x]^(3/ 2))
3.3.42.3.1 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 = 11.11 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.08
| method | result | size |
| parts | \(-\frac {2 a^{2} e \left (-\frac {1}{e^{2} \sqrt {e \cot \left (d x +c \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e^{2} \left (e^{2}\right )^{\frac {1}{4}}}\right )}{d}+\frac {2 a^{2} e}{5 d \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {2 a^{2} \sqrt {2}\, \left (-\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )^{2}-\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )+\sqrt {2}\, \sin \left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d e \sqrt {e \cot \left (d x +c \right )}\, \left (\cos \left (d x +c \right )^{2}-1\right )}\) | \(405\) |
| default | \(\text {Expression too large to display}\) | \(933\) |
-2*a^2/d*e*(-1/e^2/(e*cot(d*x+c))^(1/2)-1/8/e^2/(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*c ot(d*x+c))^(1/2)+1)))+2/5*a^2*e/d/(e*cot(d*x+c))^(5/2)-2/3*a^2/d*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) )*cos(d*x+c)^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))*cos(d*x+c)+2^(1/2)*sin(d*x+c))/e/(e*cot(d*x+c))^(1/2)/(cos (d*x+c)^2-1)*tan(d*x+c)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \cot (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \cot (c+d x))^{3/2}} \, dx=a^{2} \left (\int \frac {1}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \]
a**2*(Integral((e*cot(c + d*x))**(-3/2), x) + Integral(2*sec(c + d*x)/(e*c ot(c + d*x))**(3/2), x) + Integral(sec(c + d*x)**2/(e*cot(c + d*x))**(3/2) , x))
Exception generated. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \cot (c+d x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \cot (c+d x))^{3/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \cot \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \cot (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]