\(\int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\) [240]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=-\frac {\csc (a+b x)}{b d (d \cos (a+b x))^{3/2}}+\frac {5 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}+\frac {5 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}} \] Output:

-csc(b*x+a)/b/d/(d*cos(b*x+a))^(3/2)+5/3*cos(b*x+a)^(1/2)*InverseJacobiAM( 
1/2*a+1/2*b*x,2^(1/2))/b/d^2/(d*cos(b*x+a))^(1/2)+5/3*sin(b*x+a)/b/d/(d*co 
s(b*x+a))^(3/2)
 

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.63 \[ \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {-3 \cot (a+b x)+5 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )+2 \tan (a+b x)}{3 b d^2 \sqrt {d \cos (a+b x)}} \] Input:

Integrate[Csc[a + b*x]^2/(d*Cos[a + b*x])^(5/2),x]
 

Output:

(-3*Cot[a + b*x] + 5*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2] + 2*Tan[ 
a + b*x])/(3*b*d^2*Sqrt[d*Cos[a + b*x]])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3050, 3042, 3116, 3042, 3121, 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.

\(\displaystyle \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (a+b x)^2 (d \cos (a+b x))^{5/2}}dx\)

\(\Big \downarrow \) 3050

\(\displaystyle \frac {5}{2} \int \frac {1}{(d \cos (a+b x))^{5/2}}dx-\frac {\csc (a+b x)}{b d (d \cos (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{2} \int \frac {1}{\left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{5/2}}dx-\frac {\csc (a+b x)}{b d (d \cos (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {5}{2} \left (\frac {\int \frac {1}{\sqrt {d \cos (a+b x)}}dx}{3 d^2}+\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}\right )-\frac {\csc (a+b x)}{b d (d \cos (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{2} \left (\frac {\int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 d^2}+\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}\right )-\frac {\csc (a+b x)}{b d (d \cos (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {5}{2} \left (\frac {\sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 d^2 \sqrt {d \cos (a+b x)}}+\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}\right )-\frac {\csc (a+b x)}{b d (d \cos (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{2} \left (\frac {\sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 d^2 \sqrt {d \cos (a+b x)}}+\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}\right )-\frac {\csc (a+b x)}{b d (d \cos (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {5}{2} \left (\frac {2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}+\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}\right )-\frac {\csc (a+b x)}{b d (d \cos (a+b x))^{3/2}}\)

Input:

Int[Csc[a + b*x]^2/(d*Cos[a + b*x])^(5/2),x]
 

Output:

-(Csc[a + b*x]/(b*d*(d*Cos[a + b*x])^(3/2))) + (5*((2*Sqrt[Cos[a + b*x]]*E 
llipticF[(a + b*x)/2, 2])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]) + (2*Sin[a + b*x] 
)/(3*b*d*(d*Cos[a + b*x])^(3/2))))/2
 

Defintions of rubi rules used

rule 3042

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

rule 3050

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

rule 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[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 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(189\) vs. \(2(87)=174\).

Time = 5.30 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.94

method	result	size
default	\(\frac {\sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \left (10 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{\frac {3}{2}} \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-20 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+20 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-3\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )}{6 d \left (-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4} d +\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d \right )^{\frac {3}{2}} \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\)	\(190\)

Input:

int(csc(b*x+a)^2/(d*cos(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/d/(-2*sin(1/ 
2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2*a)^2*d)^(3/2)/cos(1/2*b*x+1/2*a)*(10*cos( 
1/2*b*x+1/2*a)*(2*sin(1/2*b*x+1/2*a)^2-1)^(3/2)*(sin(1/2*b*x+1/2*a)^2)^(1/ 
2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))-20*sin(1/2*b*x+1/2*a)^4+20*sin(1/ 
2*b*x+1/2*a)^2-3)*sin(1/2*b*x+1/2*a)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/ 
b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.32 \[ \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {-5 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 5 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} - 2\right )}}{3 \, b d^{3} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )} \] Input:

integrate(csc(b*x+a)^2/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

1/3*(-5*I*sqrt(1/2)*sqrt(d)*cos(b*x + a)^2*sin(b*x + a)*weierstrassPInvers 
e(-4, 0, cos(b*x + a) + I*sin(b*x + a)) + 5*I*sqrt(1/2)*sqrt(d)*cos(b*x + 
a)^2*sin(b*x + a)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a) 
) - sqrt(d*cos(b*x + a))*(5*cos(b*x + a)^2 - 2))/(b*d^3*cos(b*x + a)^2*sin 
(b*x + a))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(csc(b*x+a)**2/(d*cos(b*x+a))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(csc(b*x+a)^2/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")
 

Output:

integrate(csc(b*x + a)^2/(d*cos(b*x + a))^(5/2), x)
 

Giac [F]

\[ \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(csc(b*x+a)^2/(d*cos(b*x+a))^(5/2),x, algorithm="giac")
 

Output:

integrate(csc(b*x + a)^2/(d*cos(b*x + a))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^2\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \] Input:

int(1/(sin(a + b*x)^2*(d*cos(a + b*x))^(5/2)),x)
 

Output:

int(1/(sin(a + b*x)^2*(d*cos(a + b*x))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\cos \left (b x +a \right )}\, \csc \left (b x +a \right )^{2}}{\cos \left (b x +a \right )^{3}}d x \right )}{d^{3}} \] Input:

int(csc(b*x+a)^2/(d*cos(b*x+a))^(5/2),x)
 

Output:

(sqrt(d)*int((sqrt(cos(a + b*x))*csc(a + b*x)**2)/cos(a + b*x)**3,x))/d**3