\(\int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}} \, dx\) [269]

Optimal result

Integrand size = 25, antiderivative size = 135 \[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}} \, dx=-\frac {c}{3 b d \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2}}+\frac {1}{6 b c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}+\frac {\sqrt {d \csc (a+b x)} \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}{12 b c^2 d^2} \] Output:

-1/3*c/b/d/(d*csc(b*x+a))^(1/2)/(c*sec(b*x+a))^(5/2)+1/6/b/c/d/(d*csc(b*x+ 
a))^(1/2)/(c*sec(b*x+a))^(1/2)+1/12*(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/d^2
 

Mathematica [C] (verified)

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

Time = 0.71 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}} \, dx=\frac {-2 \cos (2 (a+b x))+\frac {\csc ^2(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\csc ^2(a+b x)\right )}{\sqrt [4]{-\cot ^2(a+b x)}}}{12 b c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}} \] Input:

Integrate[1/((d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(3/2)),x]
 

Output:

(-2*Cos[2*(a + b*x)] + (Csc[a + b*x]^2*Hypergeometric2F1[1/2, 3/4, 3/2, Cs 
c[a + b*x]^2])/(-Cot[a + b*x]^2)^(1/4))/(12*b*c*d*Sqrt[d*Csc[a + b*x]]*Sqr 
t[c*Sec[a + b*x]])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3107, 3042, 3108, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3107

\(\displaystyle \frac {\int \frac {\sqrt {d \csc (a+b x)}}{(c \sec (a+b x))^{3/2}}dx}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {d \csc (a+b x)}}{(c \sec (a+b x))^{3/2}}dx}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3108

\(\displaystyle \frac {\frac {\int \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}dx}{2 c^2}+\frac {d}{b c \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}dx}{2 c^2}+\frac {d}{b c \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3110

\(\displaystyle \frac {\frac {\sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}}dx}{2 c^2}+\frac {d}{b c \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}}dx}{2 c^2}+\frac {d}{b c \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {\frac {\sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 c^2}+\frac {d}{b c \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 c^2}+\frac {d}{b c \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{2 b c^2}+\frac {d}{b c \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}}{6 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}}\)

Input:

Int[1/((d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(3/2)),x]
 

Output:

-1/3*c/(b*d*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x])^(5/2)) + (d/(b*c*Sqrt[d* 
Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]]) + (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]])/(2*b*c^2))/(6* 
d^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 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 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 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 [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.86 (sec) , antiderivative size = 998, normalized size of antiderivative = 7.39

method	result	size
default	\(\text {Expression too large to display}\)	\(998\)

Input:

int(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/384/b/c/d*(sin(b*x+a)*cos(b*x+a)*(16*cos(b*x+a)^2-8)+I*(-6*cos(b*x+a)-6 
)*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(co 
t(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2 
-1/2*I,1/2*2^(1/2))+I*(6*cos(b*x+a)+6)*(-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)*EllipticPi 
((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))+(-6*cos(b*x+a)-6) 
*(-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)*EllipticPi((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2- 
1/2*I,1/2*2^(1/2))+(-6*cos(b*x+a)-6)*(-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)*EllipticPi(( 
-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))+(8*cos(b*x+a)+8)*(- 
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))+(3*cos(b*x+a)+3)*ln(-(cos(b*x+a)*cot(b*x+a)-2*cot(b*x+a)+2*(-2*sin(b* 
x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*sin(b*x+a)-2*cos(b*x+a)-sin(b*x+a) 
+csc(b*x+a)+2)/(cos(b*x+a)-1))*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2) 
^(1/2)+(6*cos(b*x+a)+6)*arctan(((-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2 
)^(1/2)*sin(b*x+a)-cos(b*x+a)+1)/(cos(b*x+a)-1))*(-2*sin(b*x+a)*cos(b*x+a) 
/(cos(b*x+a)+1)^2)^(1/2)+(6*cos(b*x+a)+6)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b 
*x+a)+1)^2)^(1/2)*arctan(((-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(...
 

Fricas [F]

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

integrate(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(3/2),x, algorithm="fricas 
")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

integrate(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(3/2),x, algorithm="maxima 
")
 

Output:

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

Giac [F]

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

integrate(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(1/((c/cos(a + b*x))^(3/2)*(d/sin(a + b*x))^(3/2)),x)
 

Output:

int(1/((c/cos(a + b*x))^(3/2)*(d/sin(a + b*x))^(3/2)), x)
 

Reduce [F]

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

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