\(\int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx\) [302]

Optimal result

Integrand size = 25, antiderivative size = 201 \[ \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\frac {16 \cot (c+d x) \sqrt {e \csc (c+d x)}}{21 a^2 d}-\frac {2 \cot ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {4 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 a^2 d}-\frac {2 \cot (c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}+\frac {4 \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}+\frac {20 \sqrt {e \csc (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 a^2 d} \] Output:

16/21*cot(d*x+c)*(e*csc(d*x+c))^(1/2)/a^2/d-2/7*cot(d*x+c)^3*(e*csc(d*x+c) 
)^(1/2)/a^2/d-4/3*csc(d*x+c)*(e*csc(d*x+c))^(1/2)/a^2/d-2/7*cot(d*x+c)*csc 
(d*x+c)^2*(e*csc(d*x+c))^(1/2)/a^2/d+4/7*csc(d*x+c)^3*(e*csc(d*x+c))^(1/2) 
/a^2/d+20/21*(e*csc(d*x+c))^(1/2)*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^( 
1/2))*sin(d*x+c)^(1/2)/a^2/d
 

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=-\frac {4 \csc ^3(c+d x) \sqrt {e \csc (c+d x)} \left (2 (8+11 \cos (c+d x)) \sin ^4\left (\frac {1}{2} (c+d x)\right )+5 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {7}{2}}(c+d x)\right )}{21 a^2 d} \] Input:

Integrate[Sqrt[e*Csc[c + d*x]]/(a + a*Sec[c + d*x])^2,x]
 

Output:

(-4*Csc[c + d*x]^3*Sqrt[e*Csc[c + d*x]]*(2*(8 + 11*Cos[c + d*x])*Sin[(c + 
d*x)/2]^4 + 5*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(7/2)))/(21 
*a^2*d)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.86, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4366, 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.

\(\displaystyle \int \frac {\sqrt {e \csc (c+d x)}}{(a \sec (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {e \sec \left (c+d x-\frac {\pi }{2}\right )}}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 4366

\(\displaystyle \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {1}{(\sec (c+d x) a+a)^2 \sqrt {\sin (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {1}{\sqrt {\cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\cos ^2(c+d x)}{(-\cos (c+d x) a-a)^2 \sqrt {\sin (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\sqrt {\cos \left (c+d x-\frac {\pi }{2}\right )} \left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}dx\)

\(\Big \downarrow \) 3354

\(\displaystyle \frac {\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{\sin ^{\frac {9}{2}}(c+d x)}dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^{9/2}}dx}{a^4}\)

\(\Big \downarrow \) 3352

\(\displaystyle \frac {\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \left (\frac {a^2 \cos ^4(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}-\frac {2 a^2 \cos ^3(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}+\frac {a^2 \cos ^2(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (-\frac {4 a^2}{3 d \sin ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2}{7 d \sin ^{\frac {7}{2}}(c+d x)}+\frac {20 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{21 d}-\frac {2 a^2 \cos ^3(c+d x)}{7 d \sin ^{\frac {7}{2}}(c+d x)}+\frac {16 a^2 \cos (c+d x)}{21 d \sin ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2 \cos (c+d x)}{7 d \sin ^{\frac {7}{2}}(c+d x)}\right )}{a^4}\)

Input:

Int[Sqrt[e*Csc[c + d*x]]/(a + a*Sec[c + d*x])^2,x]
 

Output:

(Sqrt[e*Csc[c + d*x]]*((20*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2])/(21*d) + 
(4*a^2)/(7*d*Sin[c + d*x]^(7/2)) - (2*a^2*Cos[c + d*x])/(7*d*Sin[c + d*x]^ 
(7/2)) - (2*a^2*Cos[c + d*x]^3)/(7*d*Sin[c + d*x]^(7/2)) - (4*a^2)/(3*d*Si 
n[c + d*x]^(3/2)) + (16*a^2*Cos[c + d*x])/(21*d*Sin[c + d*x]^(3/2)))*Sqrt[ 
Sin[c + d*x]])/a^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3352

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]
 

rule 3354

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]
 

rule 4360

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]
 

rule 4366

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.00 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\sqrt {2}\, \left (\sin \left (d x +c \right ) \left (11 \cos \left (d x +c \right )+8\right ) \sqrt {2}+i \left (-10 \cos \left (d x +c \right )^{3}-30 \cos \left (d x +c \right )^{2}-30 \cos \left (d x +c \right )-10\right ) \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}\, \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {-i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {e \csc \left (d x +c \right )}}{21 a^{2} d \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\)	\(188\)

Input:

int((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

-1/21/a^2/d*2^(1/2)*(sin(d*x+c)*(11*cos(d*x+c)+8)*2^(1/2)+I*(-10*cos(d*x+c 
)^3-30*cos(d*x+c)^2-30*cos(d*x+c)-10)*(1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2)* 
(1-I*cot(d*x+c)+I*csc(d*x+c))^(1/2)*(-I*(-csc(d*x+c)+cot(d*x+c)))^(1/2)*El 
lipticF((1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2),1/2*2^(1/2)))*(e*csc(d*x+c))^( 
1/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {\frac {e}{\sin \left (d x + c\right )}} {\left (11 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right ) + 5 \, {\left (i \, \cos \left (d x + c\right )^{2} + 2 i \, \cos \left (d x + c\right ) + i\right )} \sqrt {2 i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-i \, \cos \left (d x + c\right )^{2} - 2 i \, \cos \left (d x + c\right ) - i\right )} \sqrt {-2 i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}}{21 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

-2/21*(sqrt(e/sin(d*x + c))*(11*cos(d*x + c) + 8)*sin(d*x + c) + 5*(I*cos( 
d*x + c)^2 + 2*I*cos(d*x + c) + I)*sqrt(2*I*e)*weierstrassPInverse(4, 0, c 
os(d*x + c) + I*sin(d*x + c)) + 5*(-I*cos(d*x + c)^2 - 2*I*cos(d*x + c) - 
I)*sqrt(-2*I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)))/ 
(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F]

\[ \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\sqrt {e \csc {\left (c + d x \right )}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:

integrate((e*csc(d*x+c))**(1/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Integral(sqrt(e*csc(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a 
**2
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\sqrt {e \csc \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate(sqrt(e*csc(d*x + c))/(a*sec(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\sqrt {\frac {e}{\sin \left (c+d\,x\right )}}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \] Input:

int((e/sin(c + d*x))^(1/2)/(a + a/cos(c + d*x))^2,x)
 

Output:

int((cos(c + d*x)^2*(e/sin(c + d*x))^(1/2))/(a^2*(cos(c + d*x) + 1)^2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\csc \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:

int((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x)
 

Output:

(sqrt(e)*int(sqrt(csc(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)) 
/a**2