Integrand size = 25, antiderivative size = 153 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \csc (c+d x)}} \, dx=-\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a^2 \tan (c+d x)}{d \sqrt {e \csc (c+d x)}} \]
-2*a^2*arctan(sin(d*x+c)^(1/2))/d/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+2* a^2*arctanh(sin(d*x+c)^(1/2))/d/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)-a^2* (sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(co s(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/d/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+a ^2*tan(d*x+c)/d/(e*csc(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 18.33 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.88 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \csc (c+d x)}} \, dx=-\frac {\left (1+\cos \left (2 \left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )^2 \cos (c+d x) \left (-1+\csc ^2(c+d x)\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (-\arctan \left (\sqrt {\csc (c+d x)}\right )-\text {arctanh}\left (\sqrt {\csc (c+d x)}\right )-\frac {2 \sqrt {\csc (c+d x)} \sqrt {1-\csc ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\csc ^2(c+d x)\right )}{3 \sqrt {1-\sin ^2(c+d x)}}+\frac {\sqrt {\csc (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},\csc ^2(c+d x)\right ) \sqrt {1-\sin ^2(c+d x)}}{\sqrt {1-\csc ^2(c+d x)}}\right )}{2 d \left (1+\cos \left (2 \left (\frac {c}{2}+\frac {1}{2} \left (-c+\csc ^{-1}(\csc (c+d x))\right )\right )\right )\right )^2 \csc ^{\frac {3}{2}}(c+d x) \sqrt {e \csc (c+d x)} \sqrt {1-\sin ^2(c+d x)}} \]
-1/2*((1 + Cos[2*(c/2 + (d*x)/2)])^2*Cos[c + d*x]*(-1 + Csc[c + d*x]^2)*Se c[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(-ArcTan[Sqrt[Csc[c + d*x]]] - A rcTanh[Sqrt[Csc[c + d*x]]] - (2*Sqrt[Csc[c + d*x]]*Sqrt[1 - Csc[c + d*x]^2 ]*Hypergeometric2F1[3/4, 3/2, 7/4, Csc[c + d*x]^2])/(3*Sqrt[1 - Sin[c + d* x]^2]) + (Sqrt[Csc[c + d*x]]*Hypergeometric2F1[-1/4, 3/2, 3/4, Csc[c + d*x ]^2]*Sqrt[1 - Sin[c + d*x]^2])/Sqrt[1 - Csc[c + d*x]^2]))/(d*(1 + Cos[2*(c /2 + (-c + ArcCsc[Csc[c + d*x]])/2)])^2*Csc[c + d*x]^(3/2)*Sqrt[e*Csc[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])
Time = 0.56 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.71, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4366, 3042, 4360, 3042, 3352, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{\sqrt {e \csc (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\sqrt {e \sec \left (c+d x-\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4366 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^2 \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx}{\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \frac {\int (-\cos (c+d x) a-a)^2 \sec ^2(c+d x) \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {-\cos \left (c+d x+\frac {\pi }{2}\right )} \left (-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (\sec ^2(c+d x) \sqrt {\sin (c+d x)} a^2+2 \sec (c+d x) \sqrt {\sin (c+d x)} a^2+\sqrt {\sin (c+d x)} a^2\right )dx}{\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d}+\frac {a^2 \sin ^{\frac {3}{2}}(c+d x) \sec (c+d x)}{d}}{\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
((-2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/d + (2*a^2*ArcTanh[Sqrt[Sin[c + d*x]] ])/d + (a^2*EllipticE[(c - Pi/2 + d*x)/2, 2])/d + (a^2*Sec[c + d*x]*Sin[c + d*x]^(3/2))/d)/(Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])
3.3.90.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*( x_)])^(p_), x_Symbol] :> Simp[g^IntPart[p]*(g*Sec[e + f*x])^FracPart[p]*Cos [e + f*x]^FracPart[p] Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x], x] / ; FreeQ[{a, b, e, f, g, m, p}, x] && !IntegerQ[p]
Result contains complex when optimal does not.
Time = 13.83 (sec) , antiderivative size = 996, normalized size of antiderivative = 6.51
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(996\) |
| default | \(\text {Expression too large to display}\) | \(1290\) |
-a^2/d*2^(1/2)*(2*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-c sc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticE((-I*(I-cot( d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cos(d*x+c)-(-I*(I-cot(d*x+c)+csc(d* x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+ c)))^(1/2)*EllipticF((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cos (d*x+c)+2*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c )))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticE((-I*(I-cot(d*x+c)+c sc(d*x+c)))^(1/2),1/2*2^(1/2))-(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I +cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*Elliptic F((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))+2^(1/2)*cos(d*x+c)-2^( 1/2))/(e*csc(d*x+c))^(1/2)*csc(d*x+c)-1/2*a^2/d*2^(1/2)/(e*csc(d*x+c))^(1/ 2)*(-2*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c))) ^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticE((-I*(I-cot(d*x+c)+csc( d*x+c)))^(1/2),1/2*2^(1/2))*cot(d*x+c)+(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2 )*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)* EllipticF((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cot(d*x+c)-2*( -I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*( -I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticE((-I*(I-cot(d*x+c)+csc(d*x+c))) ^(1/2),1/2*2^(1/2))*csc(d*x+c)+(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I +cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*Ellip...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.42 (sec) , antiderivative size = 698, normalized size of antiderivative = 4.56 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \csc (c+d x)}} \, dx=\left [-\frac {2 \, a^{2} \sqrt {-e} \arctan \left (-\frac {{\left (\cos \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {-e} \sqrt {\frac {e}{\sin \left (d x + c\right )}}}{4 \, {\left (e \sin \left (d x + c\right ) + e\right )}}\right ) \cos \left (d x + c\right ) + a^{2} \sqrt {-e} \cos \left (d x + c\right ) \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} + 8 \, {\left (\cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} + {\left (7 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {-e} \sqrt {\frac {e}{\sin \left (d x + c\right )}} + 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) - 2 \, a^{2} \sqrt {2 i \, e} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 2 \, a^{2} \sqrt {-2 i \, e} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 4 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}}}{4 \, d e \cos \left (d x + c\right )}, \frac {2 \, a^{2} \sqrt {e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e} \sqrt {\frac {e}{\sin \left (d x + c\right )}}}{4 \, {\left (e \sin \left (d x + c\right ) - e\right )}}\right ) \cos \left (d x + c\right ) + a^{2} \sqrt {e} \cos \left (d x + c\right ) \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} + 8 \, {\left (\cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} - {\left (7 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {e} \sqrt {\frac {e}{\sin \left (d x + c\right )}} - 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) + 2 \, a^{2} \sqrt {2 i \, e} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 2 \, a^{2} \sqrt {-2 i \, e} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 4 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}}}{4 \, d e \cos \left (d x + c\right )}\right ] \]
[-1/4*(2*a^2*sqrt(-e)*arctan(-1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2)*sq rt(-e)*sqrt(e/sin(d*x + c))/(e*sin(d*x + c) + e))*cos(d*x + c) + a^2*sqrt( -e)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 + 8*(cos(d*x + c)^4 - 9*cos(d*x + c)^2 + (7*cos(d*x + c)^2 - 8)*sin(d*x + c) + 8)*sqrt( -e)*sqrt(e/sin(d*x + c)) + 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e )/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 2*a^2*sqrt(2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassP Inverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) - 2*a^2*sqrt(-2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin (d*x + c))) + 4*(a^2*cos(d*x + c)^2 - a^2)*sqrt(e/sin(d*x + c)))/(d*e*cos( d*x + c)), 1/4*(2*a^2*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2)*sqrt(e)*sqrt(e/sin(d*x + c))/(e*sin(d*x + c) - e))*cos(d*x + c) + a^2 *sqrt(e)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 + 8*(cos (d*x + c)^4 - 9*cos(d*x + c)^2 - (7*cos(d*x + c)^2 - 8)*sin(d*x + c) + 8)* sqrt(e)*sqrt(e/sin(d*x + c)) - 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) + 2*a^2*sqrt(2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weierstr assPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 2*a^2*sqrt(-2*I*e)*cos (d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I *sin(d*x + c))) - 4*(a^2*cos(d*x + c)^2 - a^2)*sqrt(e/sin(d*x + c)))/(d...
\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \csc (c+d x)}} \, dx=a^{2} \left (\int \frac {1}{\sqrt {e \csc {\left (c + d x \right )}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\sqrt {e \csc {\left (c + d x \right )}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {e \csc {\left (c + d x \right )}}}\, dx\right ) \]
a**2*(Integral(1/sqrt(e*csc(c + d*x)), x) + Integral(2*sec(c + d*x)/sqrt(e *csc(c + d*x)), x) + Integral(sec(c + d*x)**2/sqrt(e*csc(c + d*x)), x))
\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \csc (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \csc \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \csc (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \csc \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \csc (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{\sqrt {\frac {e}{\sin \left (c+d\,x\right )}}} \,d x \]