\(\int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2} \, dx\) [241]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 125 \[ \int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2} \, dx=\frac {4 c d^3 \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}{b}-\frac {4 c^2 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{b \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}} \] Output:

4*c*d^3*(c*sec(b*x+a))^(1/2)/b/(d*csc(b*x+a))^(3/2)-2*c*d*(d*csc(b*x+a))^( 
1/2)*(c*sec(b*x+a))^(1/2)/b+4*c^2*d^2*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2)) 
/b/(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 = 0.70 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.79 \[ \int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2} \, dx=-\frac {2 c d \sqrt {d \csc (a+b x)} \left (\cos (2 (a+b x)) \cot ^2(a+b x)+2 \cos ^2(a+b x) \sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\csc ^2(a+b x)\right )\right ) \sqrt {c \sec (a+b x)} \tan ^2(a+b x)}{b} \] Input:

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

Output:

(-2*c*d*Sqrt[d*Csc[a + b*x]]*(Cos[2*(a + b*x)]*Cot[a + b*x]^2 + 2*Cos[a + 
b*x]^2*(-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]]*Tan[a + b*x]^2)/b
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3105, 3042, 3106, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3105

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3106

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3110

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

(-2*c*d*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]])/b + 2*d^2*((2*c*d*Sqrt[ 
c*Sec[a + b*x]])/(b*(d*Csc[a + b*x])^(3/2)) - (2*c^2*EllipticE[a - Pi/4 + 
b*x, 2])/(b*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 3105
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*(m - 1))), x] + Simp[a^2*((m + n - 2)/(m - 1))   Int[(a*Csc[e + f* 
x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[ 
m, 1] && IntegersQ[2*m, 2*n] &&  !GtQ[n, m]
 

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 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 [A] (verified)

Time = 1.60 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.74

method result size
default \(\frac {\left (2-4 \cos \left (b x +a \right )+\left (4 \cos \left (b x +a \right )+4\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+\left (-2 \cos \left (b x +a \right )-2\right ) \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\right ) \sqrt {c \sec \left (b x +a \right )}\, c d \sqrt {d \csc \left (b x +a \right )}}{b}\) \(217\)

Input:

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

Output:

1/b*(2-4*cos(b*x+a)+(4*cos(b*x+a)+4)*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(2*c 
ot(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))+(-2*cos(b*x+a)-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))*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2))*(c*s 
ec(b*x+a))^(1/2)*c*d*(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 = 158, normalized size of antiderivative = 1.26 \[ \int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2} \, dx=\frac {\sqrt {-4 i \, c d} c d E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {4 i \, c d} c d E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {-4 i \, c d} c d F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {4 i \, c d} c d F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - 2 \, {\left (2 \, c d \cos \left (b x + a\right )^{2} - c d\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}}}{b} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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