3.4.4 \(\int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))^2} \, dx\) [304]

3.4.4.1 Optimal result
3.4.4.2 Mathematica [A] (verified)
3.4.4.3 Rubi [A] (verified)
3.4.4.4 Maple [C] (verified)
3.4.4.5 Fricas [C] (verification not implemented)
3.4.4.6 Sympy [F]
3.4.4.7 Maxima [F]
3.4.4.8 Giac [F]
3.4.4.9 Mupad [F(-1)]

3.4.4.1 Optimal result

Integrand size = 25, antiderivative size = 213 \[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))^2} \, dx=\frac {4}{a^2 d e \sqrt {e \csc (c+d x)}}-\frac {4 \cos (c+d x)}{3 a^2 d e \sqrt {e \csc (c+d x)}}-\frac {2 \cos (c+d x) \cot ^2(c+d x)}{3 a^2 d e \sqrt {e \csc (c+d x)}}-\frac {2 \cot (c+d x) \csc (c+d x)}{3 a^2 d e \sqrt {e \csc (c+d x)}}+\frac {4 \csc ^2(c+d x)}{3 a^2 d e \sqrt {e \csc (c+d x)}}-\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right )}{a^2 d e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \]

output
4/a^2/d/e/(e*csc(d*x+c))^(1/2)-4/3*cos(d*x+c)/a^2/d/e/(e*csc(d*x+c))^(1/2) 
-2/3*cos(d*x+c)*cot(d*x+c)^2/a^2/d/e/(e*csc(d*x+c))^(1/2)-2/3*cot(d*x+c)*c 
sc(d*x+c)/a^2/d/e/(e*csc(d*x+c))^(1/2)+4/3*csc(d*x+c)^2/a^2/d/e/(e*csc(d*x 
+c))^(1/2)+4*(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) 
*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/a^2/d/e/(e*csc(d*x+c))^(1/2) 
/sin(d*x+c)^(1/2)
 
3.4.4.2 Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.47 \[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))^2} \, dx=\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (12 (1+\cos (c+d x)) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )+(15+10 \cos (c+d x)-\cos (2 (c+d x))) \sqrt {\sin (c+d x)}\right )}{6 a^2 d (e \csc (c+d x))^{3/2} \sin ^{\frac {3}{2}}(c+d x)} \]

input
Integrate[1/((e*Csc[c + d*x])^(3/2)*(a + a*Sec[c + d*x])^2),x]
 
output
(Sec[(c + d*x)/2]^2*(12*(1 + Cos[c + d*x])*EllipticF[(-2*c + Pi - 2*d*x)/4 
, 2] + (15 + 10*Cos[c + d*x] - Cos[2*(c + d*x)])*Sqrt[Sin[c + d*x]]))/(6*a 
^2*d*(e*Csc[c + d*x])^(3/2)*Sin[c + d*x]^(3/2))
 
3.4.4.3 Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.81, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4366

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3354

\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{\sin ^{\frac {5}{2}}(c+d x)}dx}{a^4 e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^{5/2}}dx}{a^4 e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

\(\Big \downarrow \) 3352

\(\displaystyle \frac {\int \left (\frac {a^2 \cos ^4(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)}-\frac {2 a^2 \cos ^3(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)}+\frac {a^2 \cos ^2(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)}\right )dx}{a^4 e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

\(\Big \downarrow \) 2009

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

input
Int[1/((e*Csc[c + d*x])^(3/2)*(a + a*Sec[c + d*x])^2),x]
 
output
((-4*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2])/d + (4*a^2)/(3*d*Sin[c + d*x]^( 
3/2)) - (2*a^2*Cos[c + d*x])/(3*d*Sin[c + d*x]^(3/2)) - (2*a^2*Cos[c + d*x 
]^3)/(3*d*Sin[c + d*x]^(3/2)) + (4*a^2*Sqrt[Sin[c + d*x]])/d - (4*a^2*Cos[ 
c + d*x]*Sqrt[Sin[c + d*x]])/(3*d))/(a^4*e*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c 
 + d*x]])
 

3.4.4.3.1 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]
 
3.4.4.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 9.99 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.86

method result size
default \(\frac {\sqrt {2}\, \left (6 i \sqrt {i \left (-i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \sqrt {i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {-i \left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \cos \left (d x +c \right )^{2}+12 i \sqrt {i \left (-i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \sqrt {i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \sqrt {-i \left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )+6 i \sqrt {i \left (-i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \sqrt {i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \sqrt {-i \left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}, \frac {\sqrt {2}}{2}\right )+\cos \left (d x +c \right )^{2} \sqrt {2}\, \sin \left (d x +c \right )-5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}-8 \sqrt {2}\, \sin \left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 a^{2} d \left (\cos \left (d x +c \right )-1\right ) \left (\cos \left (d x +c \right )+1\right )^{2} e \sqrt {e \csc \left (d x +c \right )}}\) \(397\)

input
int(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 
output
1/3/a^2/d*2^(1/2)*(6*I*(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))*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*cos(d*x+c)^2+12*I*(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))*(-I*(I-cot(d*x+c)+csc(d*x+c) 
))^(1/2)*cos(d*x+c)+6*I*(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))*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)+cos(d*x+c)^2*2^(1/2)*sin(d*x+ 
c)-5*cos(d*x+c)*sin(d*x+c)*2^(1/2)-8*2^(1/2)*sin(d*x+c))/(cos(d*x+c)-1)/(c 
os(d*x+c)+1)^2/e/(e*csc(d*x+c))^(1/2)*sin(d*x+c)
 
3.4.4.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left ({\left (\cos \left (d x + c\right )^{2} - 5 \, \cos \left (d x + c\right ) - 8\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}} \sin \left (d x + c\right ) + 3 \, \sqrt {2 i \, e} {\left (-i \, \cos \left (d x + c\right ) - i\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {-2 i \, e} {\left (i \, \cos \left (d x + c\right ) + i\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}}{3 \, {\left (a^{2} d e^{2} \cos \left (d x + c\right ) + a^{2} d e^{2}\right )}} \]

input
integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 
output
-2/3*((cos(d*x + c)^2 - 5*cos(d*x + c) - 8)*sqrt(e/sin(d*x + c))*sin(d*x + 
 c) + 3*sqrt(2*I*e)*(-I*cos(d*x + c) - I)*weierstrassPInverse(4, 0, cos(d* 
x + c) + I*sin(d*x + c)) + 3*sqrt(-2*I*e)*(I*cos(d*x + c) + I)*weierstrass 
PInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)))/(a^2*d*e^2*cos(d*x + c) + a 
^2*d*e^2)
 
3.4.4.6 Sympy [F]

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

input
integrate(1/(e*csc(d*x+c))**(3/2)/(a+a*sec(d*x+c))**2,x)
 
output
Integral(1/((e*csc(c + d*x))**(3/2)*sec(c + d*x)**2 + 2*(e*csc(c + d*x))** 
(3/2)*sec(c + d*x) + (e*csc(c + d*x))**(3/2)), x)/a**2
 
3.4.4.7 Maxima [F]

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

input
integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 
output
integrate(1/((e*csc(d*x + c))^(3/2)*(a*sec(d*x + c) + a)^2), x)
 
3.4.4.8 Giac [F]

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

input
integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 
output
integrate(1/((e*csc(d*x + c))^(3/2)*(a*sec(d*x + c) + a)^2), x)
 
3.4.4.9 Mupad [F(-1)]

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

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