\(\int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx\) [297]

Optimal result

Integrand size = 25, antiderivative size = 106 \[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx=\frac {2}{a d e \sqrt {e \csc (c+d x)}}-\frac {2 \cos (c+d x)}{3 a d e \sqrt {e \csc (c+d x)}}-\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right )}{3 a d e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \] Output:

2/a/d/e/(e*csc(d*x+c))^(1/2)-2/3*cos(d*x+c)/a/d/e/(e*csc(d*x+c))^(1/2)-4/3 
*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2))/a/d/e/(e*csc(d*x+c))^(1/2)/ 
sin(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx=\frac {4 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )-2 (-3+\cos (c+d x)) \sqrt {\sin (c+d x)}}{3 a d (e \csc (c+d x))^{3/2} \sin ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[1/((e*Csc[c + d*x])^(3/2)*(a + a*Sec[c + d*x])),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4366, 3042, 4360, 25, 25, 3042, 3318, 3042, 3044, 15, 3049, 3042, 3120}

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.

\(\displaystyle \int \frac {1}{(a \sec (c+d x)+a) (e \csc (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right ) \left (e \sec \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4366

\(\displaystyle \frac {\int \frac {\sin ^{\frac {3}{2}}(c+d x)}{\sec (c+d x) a+a}dx}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 3049

\(\displaystyle \frac {\frac {2 \sqrt {\sin (c+d x)}}{a d}-\frac {\frac {2}{3} \int \frac {1}{\sqrt {\sin (c+d x)}}dx+\frac {2 \sqrt {\sin (c+d x)} \cos (c+d x)}{3 d}}{a}}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \sqrt {\sin (c+d x)}}{a d}-\frac {\frac {2}{3} \int \frac {1}{\sqrt {\sin (c+d x)}}dx+\frac {2 \sqrt {\sin (c+d x)} \cos (c+d x)}{3 d}}{a}}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

\(\Big \downarrow \) 3120

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

Input:

Int[1/((e*Csc[c + d*x])^(3/2)*(a + a*Sec[c + d*x])),x]
 

Output:

(-(((4*EllipticF[(c - Pi/2 + d*x)/2, 2])/(3*d) + (2*Cos[c + d*x]*Sqrt[Sin[ 
c + d*x]])/(3*d))/a) + (2*Sqrt[Sin[c + d*x]])/(a*d))/(e*Sqrt[e*Csc[c + d*x 
]]*Sqrt[Sin[c + d*x]])
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3049

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a*(b*Sin[e + f*x])^(n + 1)*((a*Cos[e + f*x])^(m - 1)/ 
(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Sin[e + f*x])^n*(a 
*Cos[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && 
 NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3318

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d)   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, 
d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 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 = 150, normalized size of antiderivative = 1.42

method	result	size
default	\(-\frac {\sqrt {2}\, \left (\left (\cos \left (d x +c \right )-3\right ) \sqrt {2}+i \sqrt {-i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}\, \left (2 \csc \left (d x +c \right )+2 \cot \left (d x +c \right )\right )\right )}{3 a d e \sqrt {e \csc \left (d x +c \right )}}\)	\(150\)

Input:

int(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

-1/3/a/d*2^(1/2)/e/(e*csc(d*x+c))^(1/2)*((cos(d*x+c)-3)*2^(1/2)+I*(-I*(-cs 
c(d*x+c)+cot(d*x+c)))^(1/2)*EllipticF((1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2), 
1/2*2^(1/2))*(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)*(2*csc(d*x+c)+2*cot(d*x+c)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx=-\frac {2 \, {\left (\sqrt {\frac {e}{\sin \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 3\right )} \sin \left (d x + c\right ) - i \, \sqrt {2 i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {-2 i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}}{3 \, a d e^{2}} \] Input:

integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
 

Output:

-2/3*(sqrt(e/sin(d*x + c))*(cos(d*x + c) - 3)*sin(d*x + c) - I*sqrt(2*I*e) 
*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(-2*I*e) 
*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)))/(a*d*e^2)
 

Sympy [F]

\[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx=\frac {\int \frac {1}{\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )} + \left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \] Input:

integrate(1/(e*csc(d*x+c))**(3/2)/(a+a*sec(d*x+c)),x)
 

Output:

Integral(1/((e*csc(c + d*x))**(3/2)*sec(c + d*x) + (e*csc(c + d*x))**(3/2) 
), x)/a
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx=\int { \frac {1}{\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}} \,d x } \] Input:

integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
 

Output:

integrate(1/((e*csc(d*x + c))^(3/2)*(a*sec(d*x + c) + a)), x)
 

Giac [F]

\[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx=\int { \frac {1}{\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}} \,d x } \] Input:

integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="giac")
 

Output:

integrate(1/((e*csc(d*x + c))^(3/2)*(a*sec(d*x + c) + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx=\int \frac {\cos \left (c+d\,x\right )}{a\,{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{3/2}\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \] Input:

int(1/((a + a/cos(c + d*x))*(e/sin(c + d*x))^(3/2)),x)
 

Output:

int(cos(c + d*x)/(a*(e/sin(c + d*x))^(3/2)*(cos(c + d*x) + 1)), x)
 

Reduce [F]

\[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\csc \left (d x +c \right )}}{\csc \left (d x +c \right )^{2} \sec \left (d x +c \right )+\csc \left (d x +c \right )^{2}}d x \right )}{a \,e^{2}} \] Input:

int(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x)
 

Output:

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