\(\int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx\) [276]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx=\frac {d}{3 b c (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b c^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.54 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx=\frac {d \left (1+\cos (2 (a+b x))+3 \sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\csc ^2(a+b x)\right )\right ) \sqrt {c \sec (a+b x)}}{6 b c^3 (d \csc (a+b x))^{3/2}} \] Input:

Integrate[1/(Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x])^(5/2)),x]
 

Output:

(d*(1 + Cos[2*(a + b*x)] + 3*(-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[-1/ 
2, 1/4, 1/2, Csc[a + b*x]^2])*Sqrt[c*Sec[a + b*x]])/(6*b*c^3*(d*Csc[a + b* 
x])^(3/2))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 95, 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, 3108, 3042, 3110, 3042, 3052, 3042, 3119}

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.

\(\displaystyle \int \frac {1}{(c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}dx\)

\(\Big \downarrow \) 3108

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3110

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

Int[1/(Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x])^(5/2)),x]
 

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3052

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

rule 3108

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)/(b*f*(m + n))), x] + Simp[(n + 1)/(b^2*(m + n))   Int[(a*Csc[e + f*x])^ 
m*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, - 
1] && NeQ[m + n, 0] && 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 3119

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(248\) vs. \(2(82)=164\).

Time = 1.78 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.62

method	result	size
default	\(-\frac {\left (\left (6 \cos \left (b x +a \right )+6\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}+\left (-3 \cos \left (b x +a \right )-3\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+\cos \left (b x +a \right ) \left (4 \cos \left (b x +a \right )^{3}+2 \cos \left (b x +a \right )-6\right )\right ) \sec \left (b x +a \right ) \csc \left (b x +a \right )}{12 b \sqrt {c \sec \left (b x +a \right )}\, \sqrt {d \csc \left (b x +a \right )}\, c^{2}}\)	\(249\)

Input:

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

Output:

-1/12/b*((6*cos(b*x+a)+6)*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*EllipticE((-cot 
(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(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)+(-3*cos(b*x+a)-3)*(-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))+cos(b*x+a)*(4*co 
s(b*x+a)^3+2*cos(b*x+a)-6))/(c*sec(b*x+a))^(1/2)/(d*csc(b*x+a))^(1/2)/c^2* 
sec(b*x+a)*csc(b*x+a)
 

Fricas [F]

\[ \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {d \csc \left (b x + a\right )} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

integral(sqrt(d*csc(b*x + a))*sqrt(c*sec(b*x + a))/(c^3*d*csc(b*x + a)*sec 
(b*x + a)^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {d \csc \left (b x + a\right )} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {d \csc \left (b x + a\right )} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sec \left (b x +a \right )}\, \sqrt {\csc \left (b x +a \right )}}{\csc \left (b x +a \right ) \sec \left (b x +a \right )^{3}}d x \right )}{c^{3} d} \] Input:

int(1/(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)))/(csc(a + b*x) 
*sec(a + b*x)**3),x))/(c**3*d)