\(\int \frac {(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{3/2}} \, dx\) [265]

Optimal result

Integrand size = 25, antiderivative size = 98 \[ \int \frac {(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{3/2}} \, dx=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt {c \sec (a+b x)}}-\frac {d^2 \sqrt {d \csc (a+b x)} \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}{3 b c^2} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.95 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07 \[ \int \frac {(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{3/2}} \, dx=-\frac {d \cos (2 (a+b x)) (d \csc (a+b x))^{3/2} \left (2 \cot ^2(a+b x)-\left (-\cot ^2(a+b x)\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\csc ^2(a+b x)\right )\right ) \sec ^3(a+b x)}{3 b \left (-2+\csc ^2(a+b x)\right ) (c \sec (a+b x))^{3/2}} \] Input:

Integrate[(d*Csc[a + b*x])^(5/2)/(c*Sec[a + b*x])^(3/2),x]
 

Output:

-1/3*(d*Cos[2*(a + b*x)]*(d*Csc[a + b*x])^(3/2)*(2*Cot[a + b*x]^2 - (-Cot[ 
a + b*x]^2)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, Csc[a + b*x]^2])*Sec[a 
+ b*x]^3)/(b*(-2 + Csc[a + b*x]^2)*(c*Sec[a + b*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3103, 3042, 3110, 3042, 3053, 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.

\(\displaystyle \int \frac {(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3103

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3110

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3053

\(\displaystyle -\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 c^2}-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt {c \sec (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 c^2}-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt {c \sec (a+b x)}}\)

\(\Big \downarrow \) 3120

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

Input:

Int[(d*Csc[a + b*x])^(5/2)/(c*Sec[a + b*x])^(3/2),x]
 

Output:

(-2*d*(d*Csc[a + b*x])^(3/2))/(3*b*c*Sqrt[c*Sec[a + b*x]]) - (d^2*Sqrt[d*C 
sc[a + b*x]]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin[2* 
a + 2*b*x]])/(3*b*c^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b 
*Cos[e + f*x]])   Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f 
}, x]
 

rule 3103

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(-a)*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n + 
1)/(f*b*(m - 1))), x] + Simp[a^2*((n + 1)/(b^2*(m - 1)))   Int[(a*Csc[e + f 
*x])^(m - 2)*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && 
GtQ[m, 1] && LtQ[n, -1] && IntegersQ[2*m, 2*n]
 

rule 3110

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(a*Csc[e + f*x])^m*(b*Sec[e + f*x])^n*(a*Sin[e + f*x] 
)^m*(b*Cos[e + f*x])^n   Int[1/((a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n), x], 
 x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[m - 1/2] && IntegerQ[n - 1/ 
2]
 

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]
 

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.32

method	result	size
default	\(-\frac {\sqrt {d \csc \left (b x +a \right )}\, d^{2} \left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (1+\sec \left (b x +a \right )\right )+2 \csc \left (b x +a \right )\right )}{3 b \sqrt {c \sec \left (b x +a \right )}\, c}\)	\(129\)

Input:

int((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3/b*(d*csc(b*x+a))^(1/2)*d^2/(c*sec(b*x+a))^(1/2)/c*((-cot(b*x+a)+csc(b 
*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a)) 
^(1/2)*EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*(1+sec(b*x+ 
a))+2*csc(b*x+a))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.24 \[ \int \frac {(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{3/2}} \, dx=\frac {i \, \sqrt {-4 i \, c d} d^{2} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) - i \, \sqrt {4 i \, c d} d^{2} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) - 4 \, d^{2} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}} \cos \left (b x + a\right )}{6 \, b c^{2} \sin \left (b x + a\right )} \] Input:

integrate((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(3/2),x, algorithm="fricas")
 

Output:

1/6*(I*sqrt(-4*I*c*d)*d^2*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)) 
, -1)*sin(b*x + a) - I*sqrt(4*I*c*d)*d^2*elliptic_f(arcsin(cos(b*x + a) - 
I*sin(b*x + a)), -1)*sin(b*x + a) - 4*d^2*sqrt(c/cos(b*x + a))*sqrt(d/sin( 
b*x + a))*cos(b*x + a))/(b*c^2*sin(b*x + a))
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((d*csc(b*x+a))**(5/2)/(c*sec(b*x+a))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*csc(b*x + a))^(5/2)/(c*sec(b*x + a))^(3/2), x)
 

Giac [F]

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

integrate((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

integrate((d*csc(b*x + a))^(5/2)/(c*sec(b*x + a))^(3/2), x)
 

Mupad [F(-1)]

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

int((d/sin(a + b*x))^(5/2)/(c/cos(a + b*x))^(3/2),x)
 

Output:

int((d/sin(a + b*x))^(5/2)/(c/cos(a + b*x))^(3/2), x)
 

Reduce [F]

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

int((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(3/2),x)
 

Output:

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