\(\int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx\) [250]

Optimal result

Integrand size = 25, antiderivative size = 93 \[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}+\frac {2 c^2 \sqrt {d \csc (a+b x)} \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}{3 b} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.70 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.73 \[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=-\frac {2 c d \left (-1+\left (-\cot ^2(a+b x)\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\csc ^2(a+b x)\right )\right ) (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 93, 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, 3106, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3106

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3110

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3053

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

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 3106

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[a*b*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 
1)/(f*(n - 1))), x] + Simp[b^2*((m + n - 2)/(n - 1))   Int[(a*Csc[e + f*x]) 
^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[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 = 1.72 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {\left (\frac {2 \left (\cos \left (b x +a \right )+1\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 \tan \left (b x +a \right )}{3}\right ) \sqrt {c \sec \left (b x +a \right )}\, c^{2} \sqrt {d \csc \left (b x +a \right )}}{b}\)	\(126\)

Input:

int((d*csc(b*x+a))^(1/2)*(c*sec(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.28 \[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=\frac {-i \, \sqrt {-4 i \, c d} c^{2} \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + i \, \sqrt {4 i \, c d} c^{2} \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, c^{2} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}} \sin \left (b x + a\right )}{3 \, b \cos \left (b x + a\right )} \] Input:

integrate((d*csc(b*x+a))^(1/2)*(c*sec(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

1/3*(-I*sqrt(-4*I*c*d)*c^2*cos(b*x + a)*elliptic_f(arcsin(cos(b*x + a) + I 
*sin(b*x + a)), -1) + I*sqrt(4*I*c*d)*c^2*cos(b*x + a)*elliptic_f(arcsin(c 
os(b*x + a) - I*sin(b*x + a)), -1) + 2*c^2*sqrt(c/cos(b*x + a))*sqrt(d/sin 
(b*x + a))*sin(b*x + a))/(b*cos(b*x + a))
                                                                                    
                                                                                    
 

Sympy [F(-1)]

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

integrate((d*csc(b*x+a))**(1/2)*(c*sec(b*x+a))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((d*csc(b*x+a))^(1/2)*(c*sec(b*x+a))^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*csc(b*x+a))^(1/2)*(c*sec(b*x+a))^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((d*csc(b*x+a))^(1/2)*(c*sec(b*x+a))^(5/2),x)
 

Output:

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