Integrand size = 25, antiderivative size = 188 \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\frac {4 e^3}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos (c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos ^3(c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {4 e}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {16 e \cos (c+d x)}{5 a^2 d \sqrt {e \sin (c+d x)}}+\frac {28 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}} \]
4/5*e^3/a^2/d/(e*sin(d*x+c))^(5/2)-2/5*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c)) ^(5/2)-2/5*e^3*cos(d*x+c)^3/a^2/d/(e*sin(d*x+c))^(5/2)-4*e/a^2/d/(e*sin(d* x+c))^(1/2)+16/5*e*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(1/2)-28/5*(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(cos(1/2*c+1/4* Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/a^2/d/sin(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.27 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {e \sin (c+d x)} \left (\sec (c) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (49 \sin \left (\frac {1}{2} (c-d x)\right )+35 \sin \left (\frac {1}{2} (3 c+d x)\right )-23 \sin \left (\frac {1}{2} (c+3 d x)\right )+5 \sin \left (\frac {1}{2} (5 c+3 d x)\right )\right )-56 \sqrt {\csc ^2(c)} \csc (c+d x) \csc (d x-\arctan (\cot (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x-\arctan (\cot (c)))\right ) \sin (c) \sqrt {\sin ^2(d x-\arctan (\cot (c)))}-28 \sqrt {\csc ^2(c)} \csc (c+d x) (\sin (c+d x-\arctan (\cot (c)))+3 \sin (c-d x+\arctan (\cot (c)))) \tan (c)\right )}{5 a^2 d (1+\sec (c+d x))^2} \]
(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*Sqrt[e*Sin[c + d*x]]*(Sec[c]*Sec[(c + d *x)/2]^3*(49*Sin[(c - d*x)/2] + 35*Sin[(3*c + d*x)/2] - 23*Sin[(c + 3*d*x) /2] + 5*Sin[(5*c + 3*d*x)/2]) - 56*Sqrt[Csc[c]^2]*Csc[c + d*x]*Csc[d*x - A rcTan[Cot[c]]]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x - ArcTan[Cot [c]]]^2]*Sin[c]*Sqrt[Sin[d*x - ArcTan[Cot[c]]]^2] - 28*Sqrt[Csc[c]^2]*Csc[ c + d*x]*(Sin[c + d*x - ArcTan[Cot[c]]] + 3*Sin[c - d*x + ArcTan[Cot[c]]]) *Tan[c]))/(5*a^2*d*(1 + Sec[c + d*x])^2)
Time = 0.88 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4360, 3042, 3354, 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.
| \(\displaystyle \int \frac {\sqrt {e \sin (c+d x)}}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) \sqrt {e \sin (c+d x)}}{(a (-\cos (c+d x))-a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )}}{\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {e^4 \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}}dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^4 \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}dx}{a^4}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {e^4 \int \left (\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{7/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{7/2}}+\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{7/2}}\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^4 \left (\frac {28 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^4 \sqrt {\sin (c+d x)}}-\frac {4 a^2}{d e^3 \sqrt {e \sin (c+d x)}}+\frac {16 a^2 \cos (c+d x)}{5 d e^3 \sqrt {e \sin (c+d x)}}+\frac {4 a^2}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a^2 \cos ^3(c+d x)}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a^2 \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}\right )}{a^4}\) |
(e^4*((4*a^2)/(5*d*e*(e*Sin[c + d*x])^(5/2)) - (2*a^2*Cos[c + d*x])/(5*d*e *(e*Sin[c + d*x])^(5/2)) - (2*a^2*Cos[c + d*x]^3)/(5*d*e*(e*Sin[c + d*x])^ (5/2)) - (4*a^2)/(d*e^3*Sqrt[e*Sin[c + d*x]]) + (16*a^2*Cos[c + d*x])/(5*d *e^3*Sqrt[e*Sin[c + d*x]]) + (28*a^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt [e*Sin[c + d*x]])/(5*d*e^4*Sqrt[Sin[c + d*x]])))/a^4
3.2.30.3.1 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_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[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]
Time = 5.92 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.09
| method | result | size |
| default | \(\frac {-\frac {2 e \left (-\frac {2 e^{2}}{5 \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {2}{\sqrt {e \sin \left (d x +c \right )}}\right )}{a^{2}}-\frac {2 e \left (14 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-7 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+9 \sin \left (d x +c \right )^{5}-11 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{5 a^{2} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(205\) |
(-2*e/a^2*(-2/5*e^2/(e*sin(d*x+c))^(5/2)+2/(e*sin(d*x+c))^(1/2))-2/5*e*(14 *(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE(( -sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-7*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2) ^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+9*sin (d*x+c)^5-11*sin(d*x+c)^3+2*sin(d*x+c))/a^2/sin(d*x+c)^3/cos(d*x+c)/(e*sin (d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {e \sin \left (d x + c\right )} {\left (9 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right ) + 7 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} \sqrt {-i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 7 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} \sqrt {i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )\right )}}{5 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
-2/5*(sqrt(e*sin(d*x + c))*(9*cos(d*x + c) + 8)*sin(d*x + c) + 7*(-I*sqrt( 2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*sqrt(-I*e)*weier strassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 7*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*sqr t(I*e)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*si n(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\sqrt {e \sin {\left (c + d x \right )}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\sqrt {e\,\sin \left (c+d\,x\right )}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]