\(\int \frac {1}{(a+b \csc ^2(c+d x))^{3/2}} \, dx\) [13]

Optimal result

Integrand size = 16, antiderivative size = 77 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{a^{3/2} d}+\frac {b \cot (c+d x)}{a (a+b) d \sqrt {a+b+b \cot ^2(c+d x)}} \] Output:

-arctan(a^(1/2)*cot(d*x+c)/(a+b+b*cot(d*x+c)^2)^(1/2))/a^(3/2)/d+b*cot(d*x 
+c)/a/(a+b)/d/(a+b+b*cot(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.90 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{3/2}} \, dx=\frac {\csc ^2(c+d x) \left (-\frac {2 \sqrt {a} b (-a-2 b+a \cos (2 (c+d x))) \cot (c+d x)}{a+b}+\sqrt {2} (-a-2 b+a \cos (2 (c+d x)))^{3/2} \csc (c+d x) \log \left (\sqrt {2} \sqrt {a} \cos (c+d x)+\sqrt {-a-2 b+a \cos (2 (c+d x))}\right )\right )}{4 a^{3/2} d \left (a+b \csc ^2(c+d x)\right )^{3/2}} \] Input:

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

Output:

(Csc[c + d*x]^2*((-2*Sqrt[a]*b*(-a - 2*b + a*Cos[2*(c + d*x)])*Cot[c + d*x 
])/(a + b) + Sqrt[2]*(-a - 2*b + a*Cos[2*(c + d*x)])^(3/2)*Csc[c + d*x]*Lo 
g[Sqrt[2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]]))/(4 
*a^(3/2)*d*(a + b*Csc[c + d*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3042, 4616, 296, 291, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

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

\(\Big \downarrow \) 296

\(\displaystyle -\frac {\frac {\int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)}{a}-\frac {b \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{d}\)

\(\Big \downarrow \) 291

\(\displaystyle -\frac {\frac {\int \frac {1}{\frac {a \cot ^2(c+d x)}{b \cot ^2(c+d x)+a+b}+1}d\frac {\cot (c+d x)}{\sqrt {b \cot ^2(c+d x)+a+b}}}{a}-\frac {b \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\frac {\arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{a^{3/2}}-\frac {b \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{d}\)

Input:

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

Output:

-((ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]]/a^(3/2) - 
 (b*Cot[c + d*x])/(a*(a + b)*Sqrt[a + b + b*Cot[c + d*x]^2]))/d)
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 296

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[p, -1]
 

rule 3042

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

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Subst[Int[(a + b + b*ff^2*x^2)^p 
/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
&& NeQ[a + b, 0] && NeQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(488\) vs. \(2(69)=138\).

Time = 1.55 (sec) , antiderivative size = 489, normalized size of antiderivative = 6.35

method	result	size
default	\(-\frac {\sqrt {4}\, b \left (-\sqrt {-a}\, \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) a b -\sqrt {-a}\, \cos \left (d x +c \right ) b^{2}+\left (1+\cos \left (d x +c \right )\right ) \sin \left (d x +c \right )^{2} \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \ln \left (4 \sqrt {-a}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \cos \left (d x +c \right )+4 \sqrt {-a}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}-4 \cos \left (d x +c \right ) a \right ) a^{2}+\left (2-\cos \left (d x +c \right )^{3}-\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )\right ) \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \ln \left (4 \sqrt {-a}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \cos \left (d x +c \right )+4 \sqrt {-a}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}-4 \cos \left (d x +c \right ) a \right ) a b +\left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \ln \left (4 \sqrt {-a}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \cos \left (d x +c \right )+4 \sqrt {-a}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}-4 \cos \left (d x +c \right ) a \right ) b^{2}\right ) \csc \left (d x +c \right )^{3}}{2 d \left (a +b \right ) \left (\sqrt {a \left (a +b \right )}-a \right ) \left (\sqrt {a \left (a +b \right )}+a \right ) \sqrt {-a}\, \left (a +b \csc \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}\)	\(489\)

Input:

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

Output:

-1/2/d*4^(1/2)*b/(a+b)/((a*(a+b))^(1/2)-a)/((a*(a+b))^(1/2)+a)/(-a)^(1/2)* 
(-(-a)^(1/2)*sin(d*x+c)^2*cos(d*x+c)*a*b-(-a)^(1/2)*cos(d*x+c)*b^2+(1+cos( 
d*x+c))*sin(d*x+c)^2*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*ln(4*(-a) 
^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)+4*(-a)^(1/2) 
*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)-4*cos(d*x+c)*a)*a^2+(2-cos(d* 
x+c)^3-cos(d*x+c)^2+2*cos(d*x+c))*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1 
/2)*ln(4*(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c) 
+4*(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)-4*cos(d*x+c)*a)* 
a*b+(1+cos(d*x+c))*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*ln(4*(-a)^( 
1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)+4*(-a)^(1/2)*( 
(a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)-4*cos(d*x+c)*a)*b^2)/(a+b*csc(d 
*x+c)^2)^(3/2)*csc(d*x+c)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (69) = 138\).

Time = 0.20 (sec) , antiderivative size = 652, normalized size of antiderivative = 8.47 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/8*(8*a*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x 
 + c)*sin(d*x + c) + ((a^2 + a*b)*cos(d*x + c)^2 - a^2 - 2*a*b - b^2)*sqrt 
(-a)*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)^6 + 160*( 
a^4 + 2*a^3*b + a^2*b^2)*cos(d*x + c)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a* 
b^3 + b^4 - 32*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c)^2 + 8*(16* 
a^3*cos(d*x + c)^7 - 24*(a^3 + a^2*b)*cos(d*x + c)^5 + 10*(a^3 + 2*a^2*b + 
 a*b^2)*cos(d*x + c)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c))*sqr 
t(-a)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c))) 
/((a^4 + a^3*b)*d*cos(d*x + c)^2 - (a^4 + 2*a^3*b + a^2*b^2)*d), -1/4*(4*a 
*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c)*sin( 
d*x + c) - ((a^2 + a*b)*cos(d*x + c)^2 - a^2 - 2*a*b - b^2)*sqrt(a)*arctan 
(1/4*(8*a^2*cos(d*x + c)^4 - 8*(a^2 + a*b)*cos(d*x + c)^2 + a^2 + 2*a*b + 
b^2)*sqrt(a)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x 
 + c)/(2*a^3*cos(d*x + c)^5 - 3*(a^3 + a^2*b)*cos(d*x + c)^3 + (a^3 + 2*a^ 
2*b + a*b^2)*cos(d*x + c))))/((a^4 + a^3*b)*d*cos(d*x + c)^2 - (a^4 + 2*a^ 
3*b + a^2*b^2)*d)]
 

Sympy [F]

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

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

Output:

Integral((a + b*csc(c + d*x)**2)**(-3/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2050 vs. \(2 (69) = 138\).

Time = 0.39 (sec) , antiderivative size = 2050, normalized size of antiderivative = 26.62 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(2*a*b*cos(1/2*arctan2(a*sin(4*d*x + 4*c) - 2*(a + 2*b)*sin(2*d*x + 2 
*c), a*cos(4*d*x + 4*c) - 2*(a + 2*b)*cos(2*d*x + 2*c) + a))*sin(2*d*x + 2 
*c) + 2*(a^2 + a*b)*sin(1/2*arctan2(a*sin(4*d*x + 4*c) - 2*(a + 2*b)*sin(2 
*d*x + 2*c), a*cos(4*d*x + 4*c) - 2*(a + 2*b)*cos(2*d*x + 2*c) + a))^3 - 2 
*(a*b*cos(2*d*x + 2*c) - (a^2 + a*b)*cos(1/2*arctan2(a*sin(4*d*x + 4*c) - 
2*(a + 2*b)*sin(2*d*x + 2*c), a*cos(4*d*x + 4*c) - 2*(a + 2*b)*cos(2*d*x + 
 2*c) + a))^2 + a^2 + 2*a*b)*sin(1/2*arctan2(a*sin(4*d*x + 4*c) - 2*(a + 2 
*b)*sin(2*d*x + 2*c), a*cos(4*d*x + 4*c) - 2*(a + 2*b)*cos(2*d*x + 2*c) + 
a)) - (a^2*cos(4*d*x + 4*c)^2 + a^2*sin(4*d*x + 4*c)^2 + 4*(a^2 + 4*a*b + 
4*b^2)*cos(2*d*x + 2*c)^2 - 4*(a^2 + 2*a*b)*sin(4*d*x + 4*c)*sin(2*d*x + 2 
*c) + 4*(a^2 + 4*a*b + 4*b^2)*sin(2*d*x + 2*c)^2 + a^2 + 2*(a^2 - 2*(a^2 + 
 2*a*b)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - 4*(a^2 + 2*a*b)*cos(2*d*x + 2 
*c))^(1/4)*(((a + b)*cos(1/2*arctan2(a*sin(4*d*x + 4*c) - 2*(a + 2*b)*sin( 
2*d*x + 2*c), a*cos(4*d*x + 4*c) - 2*(a + 2*b)*cos(2*d*x + 2*c) + a))^2 + 
(a + b)*sin(1/2*arctan2(a*sin(4*d*x + 4*c) - 2*(a + 2*b)*sin(2*d*x + 2*c), 
 a*cos(4*d*x + 4*c) - 2*(a + 2*b)*cos(2*d*x + 2*c) + a))^2)*arctan2(2*a*si 
n(2*d*x + 2*c) + 2*(a^2*cos(4*d*x + 4*c)^2 + a^2*sin(4*d*x + 4*c)^2 + 4*(a 
^2 + 4*a*b + 4*b^2)*cos(2*d*x + 2*c)^2 - 4*(a^2 + 2*a*b)*sin(4*d*x + 4*c)* 
sin(2*d*x + 2*c) + 4*(a^2 + 4*a*b + 4*b^2)*sin(2*d*x + 2*c)^2 + a^2 + 2*(a 
^2 - 2*(a^2 + 2*a*b)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - 4*(a^2 + 2*a*...
                                                                                    
                                                                                    
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\csc \left (d x +c \right )^{2} b +a}}{\csc \left (d x +c \right )^{4} b^{2}+2 \csc \left (d x +c \right )^{2} a b +a^{2}}d x \] Input:

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

Output:

int(sqrt(csc(c + d*x)**2*b + a)/(csc(c + d*x)**4*b**2 + 2*csc(c + d*x)**2* 
a*b + a**2),x)