\(\int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx\) [261]

Optimal result

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

-1/3*c/b/d/(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/d^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 = 0.57 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx=-\frac {\left (1+\cos (2 (a+b x))-3 \sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\csc ^2(a+b x)\right )\right ) \tan (a+b x)}{6 b d^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}} \] Input:

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

Output:

-1/6*((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])*Tan[a + b*x])/(b*d^2*Sqrt[d*Csc[a + b*x]] 
*Sqrt[c*Sec[a + b*x]])
 

Rubi [A] (verified)

Time = 0.53 (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, 3107, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3107

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3110

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

-1/3*c/(b*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(3/2)) + EllipticE[a - 
 Pi/4 + b*x, 2]/(2*b*d^2*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Si 
n[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 3107

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_.), x_Symbol] :> Simp[b*(a*Csc[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1) 
/(a*f*(m + n))), x] + Simp[(m + 1)/(a^2*(m + n))   Int[(a*Csc[e + f*x])^(m 
+ 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -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 = 2.64 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.62

method	result	size
default	\(\frac {\left (\left (-6 \cos \left (b x +a \right )-6\right ) \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}+\left (3 \cos \left (b x +a \right )+3\right ) \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}+\cos \left (b x +a \right ) \left (4 \cos \left (b x +a \right )^{3}-10 \cos \left (b x +a \right )+6\right )\right ) \sec \left (b x +a \right ) \csc \left (b x +a \right )}{12 b \sqrt {c \sec \left (b x +a \right )}\, \sqrt {d \csc \left (b x +a \right )}\, d^{2}}\)	\(249\)

Input:

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

Output:

1/12/b*((-6*cos(b*x+a)-6)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)- 
csc(b*x+a))^(1/2)*EllipticE((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))* 
(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)+(3*cos(b*x+a)+3)*(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))*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)+cos(b*x+a)*(4*cos 
(b*x+a)^3-10*cos(b*x+a)+6))/(c*sec(b*x+a))^(1/2)/(d*csc(b*x+a))^(1/2)/d^2* 
sec(b*x+a)*csc(b*x+a)
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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