Integrand size = 25, antiderivative size = 311 \[ \int \sqrt {e \cot (c+d x)} (a+a \sec (c+d x))^2 \, dx=\frac {2 a^2 \sqrt {e \cot (c+d x)} \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{d}-\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {e \cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {2} d}+\frac {a^2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {e \cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {2} d}-\frac {a^2 \sqrt {e \cot (c+d x)} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \sqrt {\tan (c+d x)}}{2 \sqrt {2} d}+\frac {a^2 \sqrt {e \cot (c+d x)} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \sqrt {\tan (c+d x)}}{2 \sqrt {2} d}+\frac {2 a^2 \sqrt {e \cot (c+d x)} \tan (c+d x)}{d} \]
-2*a^2*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*EllipticF(cos(c+1/4*P i+d*x),2^(1/2))*sec(d*x+c)*(e*cot(d*x+c))^(1/2)*sin(2*d*x+2*c)^(1/2)/d+1/2 *a^2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*(e*cot(d*x+c))^(1/2)*tan(d*x+c)^( 1/2)/d*2^(1/2)+1/2*a^2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*(e*cot(d*x+c))^( 1/2)*tan(d*x+c)^(1/2)/d*2^(1/2)-1/4*a^2*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan( d*x+c))*(e*cot(d*x+c))^(1/2)*tan(d*x+c)^(1/2)/d*2^(1/2)+1/4*a^2*ln(1+2^(1/ 2)*tan(d*x+c)^(1/2)+tan(d*x+c))*(e*cot(d*x+c))^(1/2)*tan(d*x+c)^(1/2)/d*2^ (1/2)+2*a^2*(e*cot(d*x+c))^(1/2)*tan(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 12.01 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.38 \[ \int \sqrt {e \cot (c+d x)} (a+a \sec (c+d x))^2 \, dx=\frac {a^2 e (1+\cos (c+d x))^2 \left (3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\cot ^2(c+d x)\right )+6 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(c+d x)\right )-2 \cot ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\cot ^2(c+d x)\right )\right ) \sec ^4\left (\frac {1}{2} \cot ^{-1}(\cot (c+d x))\right )}{6 d \sqrt {e \cot (c+d x)}} \]
(a^2*e*(1 + Cos[c + d*x])^2*(3*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d* x]^2] + 6*Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[c + d*x]^2] - 2*Cot[c + d* x]^2*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2])*Sec[ArcCot[Cot[c + d *x]]/2]^4)/(6*d*Sqrt[e*Cot[c + d*x]])
Time = 0.58 (sec) , antiderivative size = 236, 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.
\(\displaystyle \int (a \sec (c+d x)+a)^2 \sqrt {e \cot (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sec (c+d x)+a)^2 \sqrt {e \cot (c+d x)}dx\) |
\(\Big \downarrow \) 4388 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)} \int \frac {(\sec (c+d x) a+a)^2}{\sqrt {\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}{\sqrt {-\cot \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4374 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)} \int \left (\frac {\sec ^2(c+d x) a^2}{\sqrt {\tan (c+d x)}}+\frac {2 \sec (c+d x) a^2}{\sqrt {\tan (c+d x)}}+\frac {a^2}{\sqrt {\tan (c+d x)}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)} \left (-\frac {a^2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {a^2 \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 a^2 \sqrt {\tan (c+d x)}}{d}-\frac {a^2 \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {a^2 \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 a^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {\tan (c+d x)}}\right )\) |
Sqrt[e*Cot[c + d*x]]*(-((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*L og[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]])/(d*Sqrt[ Tan[c + d*x]]) + (2*a^2*Sqrt[Tan[c + d*x]])/d)*Sqrt[Tan[c + d*x]]
3.3.40.3.1 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 = 11.14 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.85
method | result | size |
parts | \(-\frac {a^{2} e \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d \left (e^{2}\right )^{\frac {1}{4}}}+\frac {2 a^{2} e}{d \sqrt {e \cot \left (d x +c \right )}}+\frac {2 a^{2} \sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}\, \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (1+\sec \left (d x +c \right )\right )}{d}\) | \(265\) |
default | \(\text {Expression too large to display}\) | \(931\) |
-1/4*a^2/d*e/(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*c ot(d*x+c))^(1/2)+2*a^2/d*2^(1/2)*(e*cot(d*x+c))^(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))*(1+sec(d*x+c))
Timed out. \[ \int \sqrt {e \cot (c+d x)} (a+a \sec (c+d x))^2 \, dx=\text {Timed out} \]
\[ \int \sqrt {e \cot (c+d x)} (a+a \sec (c+d x))^2 \, dx=a^{2} \left (\int \sqrt {e \cot {\left (c + d x \right )}}\, dx + \int 2 \sqrt {e \cot {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx + \int \sqrt {e \cot {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(sqrt(e*cot(c + d*x)), x) + Integral(2*sqrt(e*cot(c + d*x))* sec(c + d*x), x) + Integral(sqrt(e*cot(c + d*x))*sec(c + d*x)**2, x))
Exception generated. \[ \int \sqrt {e \cot (c+d x)} (a+a \sec (c+d x))^2 \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \sqrt {e \cot (c+d x)} (a+a \sec (c+d x))^2 \, dx=\int { \sqrt {e \cot \left (d x + c\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int \sqrt {e \cot (c+d x)} (a+a \sec (c+d x))^2 \, dx=\int \sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \]