\(\int \frac {1}{\sqrt {x} (a+b \csc (c+d \sqrt {x}))^2} \, dx\) [69]

Optimal result

Integrand size = 22, antiderivative size = 125 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {2 \sqrt {x}}{a^2}+\frac {4 b \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2} d}-\frac {2 b^2 \cot \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \] Output:

2*x^(1/2)/a^2+4*b*(2*a^2-b^2)*arctanh((a+b*tan(1/2*c+1/2*d*x^(1/2)))/(a^2- 
b^2)^(1/2))/a^2/(a^2-b^2)^(3/2)/d-2*b^2*cot(c+d*x^(1/2))/a/(a^2-b^2)/d/(a+ 
b*csc(c+d*x^(1/2)))
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {2 \csc \left (c+d \sqrt {x}\right ) \left (\frac {a b^2 \cot \left (c+d \sqrt {x}\right )}{(-a+b) (a+b)}+\left (c+d \sqrt {x}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )-\frac {2 b \left (-2 a^2+b^2\right ) \arctan \left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {-a^2+b^2}}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}{\left (-a^2+b^2\right )^{3/2}}\right ) \left (b+a \sin \left (c+d \sqrt {x}\right )\right )}{a^2 d \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.28, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {4693, 3042, 4272, 25, 3042, 4407, 3042, 4318, 3042, 3139, 1083, 219}

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.

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

\(\Big \downarrow \) 4693

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4272

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle 2 \left (\frac {\frac {4 \left (2 a^2-b^2\right ) \int \frac {1}{-4 \left (1-\frac {a^2}{b^2}\right )-x}d\left (\frac {2 a}{b}+2 \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{a d}+\frac {\sqrt {x} \left (a^2-b^2\right )}{a}}{a \left (a^2-b^2\right )}-\frac {b^2 \cot \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}\right )\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

2*((((a^2 - b^2)*Sqrt[x])/a + (2*b*(2*a^2 - b^2)*ArcTanh[(b*((2*a)/b + 2*T 
an[(c + d*Sqrt[x])/2]))/(2*Sqrt[a^2 - b^2])])/(a*Sqrt[a^2 - b^2]*d))/(a*(a 
^2 - b^2)) - (b^2*Cot[c + d*Sqrt[x]])/(a*(a^2 - b^2)*d*(a + b*Csc[c + d*Sq 
rt[x]])))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 4272

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[b^2*Cot[ 
c + d*x]*((a + b*Csc[c + d*x])^(n + 1)/(a*d*(n + 1)*(a^2 - b^2))), x] + Sim 
p[1/(a*(n + 1)*(a^2 - b^2))   Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^2 - 
b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x 
], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ 
erQ[2*n]
 

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4407

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
 (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a   Int[Csc[e + f* 
x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0]
 

rule 4693

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Csc[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.43

method	result	size
derivativedivides	\(\frac {-\frac {4 b \left (\frac {\frac {a^{2} \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{2 a^{2}-2 b^{2}}+\frac {b a}{2 a^{2}-2 b^{2}}}{\frac {\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )^{2} b}{2}+a \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+\frac {b}{2}}+\frac {2 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 b \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (2 a^{2}-2 b^{2}\right ) \sqrt {-a^{2}+b^{2}}}\right )}{a^{2}}+\frac {4 \arctan \left (\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right )}{a^{2}}}{d}\)	\(179\)
default	\(\frac {-\frac {4 b \left (\frac {\frac {a^{2} \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{2 a^{2}-2 b^{2}}+\frac {b a}{2 a^{2}-2 b^{2}}}{\frac {\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )^{2} b}{2}+a \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+\frac {b}{2}}+\frac {2 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 b \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (2 a^{2}-2 b^{2}\right ) \sqrt {-a^{2}+b^{2}}}\right )}{a^{2}}+\frac {4 \arctan \left (\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right )}{a^{2}}}{d}\)	\(179\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (112) = 224\).

Time = 0.11 (sec) , antiderivative size = 576, normalized size of antiderivative = 4.61 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx=\left [\frac {2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sqrt {x} \sin \left (d \sqrt {x} + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sqrt {x} - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d \sqrt {x} + c\right ) + {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt {a^{2} - b^{2}} \sin \left (d \sqrt {x} + c\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}\right )} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d \sqrt {x} + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} a \cos \left (d \sqrt {x} + c\right ) + a^{2} + b^{2} + 2 \, {\left (\sqrt {a^{2} - b^{2}} b \cos \left (d \sqrt {x} + c\right ) + a b\right )} \sin \left (d \sqrt {x} + c\right )}{a^{2} \cos \left (d \sqrt {x} + c\right )^{2} - 2 \, a b \sin \left (d \sqrt {x} + c\right ) - a^{2} - b^{2}}\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d \sqrt {x} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}, \frac {2 \, {\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sqrt {x} \sin \left (d \sqrt {x} + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sqrt {x} + {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt {-a^{2} + b^{2}} \sin \left (d \sqrt {x} + c\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}\right )} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} b \sin \left (d \sqrt {x} + c\right ) + \sqrt {-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \cos \left (d \sqrt {x} + c\right )}\right ) - {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d \sqrt {x} + c\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d \sqrt {x} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \] Input:

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

Output:

[(2*(a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*sin(d*sqrt(x) + c) + 2*(a^4*b - 2* 
a^2*b^3 + b^5)*d*sqrt(x) - 2*(a^3*b^2 - a*b^4)*cos(d*sqrt(x) + c) + ((2*a^ 
3*b - a*b^3)*sqrt(a^2 - b^2)*sin(d*sqrt(x) + c) + (2*a^2*b^2 - b^4)*sqrt(a 
^2 - b^2))*log(((a^2 - 2*b^2)*cos(d*sqrt(x) + c)^2 + 2*sqrt(a^2 - b^2)*a*c 
os(d*sqrt(x) + c) + a^2 + b^2 + 2*(sqrt(a^2 - b^2)*b*cos(d*sqrt(x) + c) + 
a*b)*sin(d*sqrt(x) + c))/(a^2*cos(d*sqrt(x) + c)^2 - 2*a*b*sin(d*sqrt(x) + 
 c) - a^2 - b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*sin(d*sqrt(x) + c) + (a^ 
6*b - 2*a^4*b^3 + a^2*b^5)*d), 2*((a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*sin( 
d*sqrt(x) + c) + (a^4*b - 2*a^2*b^3 + b^5)*d*sqrt(x) + ((2*a^3*b - a*b^3)* 
sqrt(-a^2 + b^2)*sin(d*sqrt(x) + c) + (2*a^2*b^2 - b^4)*sqrt(-a^2 + b^2))* 
arctan(-(sqrt(-a^2 + b^2)*b*sin(d*sqrt(x) + c) + sqrt(-a^2 + b^2)*a)/((a^2 
 - b^2)*cos(d*sqrt(x) + c))) - (a^3*b^2 - a*b^4)*cos(d*sqrt(x) + c))/((a^7 
 - 2*a^5*b^2 + a^3*b^4)*d*sin(d*sqrt(x) + c) + (a^6*b - 2*a^4*b^3 + a^2*b^ 
5)*d)]
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt(x)*(a + b*csc(c + d*sqrt(x)))**2), x)
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx=-\frac {4 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d \sqrt {x} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt {-a^{2} + b^{2}}} - \frac {4 \, {\left (a b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + b^{2}\right )}}{{\left (a^{3} d - a b^{2} d\right )} {\left (b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + b\right )}} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a^{2} d} \] Input:

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

Output:

-4*(2*a^2*b - b^3)*(pi*floor(1/2*(d*sqrt(x) + c)/pi + 1/2)*sgn(b) + arctan 
((b*tan(1/2*d*sqrt(x) + 1/2*c) + a)/sqrt(-a^2 + b^2)))/((a^4*d - a^2*b^2*d 
)*sqrt(-a^2 + b^2)) - 4*(a*b*tan(1/2*d*sqrt(x) + 1/2*c) + b^2)/((a^3*d - a 
*b^2*d)*(b*tan(1/2*d*sqrt(x) + 1/2*c)^2 + 2*a*tan(1/2*d*sqrt(x) + 1/2*c) + 
 b)) + 2*(d*sqrt(x) + c)/(a^2*d)
 

Mupad [B] (verification not implemented)

Time = 20.04 (sec) , antiderivative size = 2737, normalized size of antiderivative = 21.90 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- (4*atan((512*a^3*b^3*tan(c/2 + (d*x^(1/2))/2))/((512*a^3*b^9)/(a^6 + a^2 
*b^4 - 2*a^4*b^2) - (1536*a^5*b^7)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (1024*a^7 
*b^5)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (512*a^9*b^3)/(a^6 + a^2*b^4 - 2*a^4*b 
^2) - (512*a^11*b)/(a^6 + a^2*b^4 - 2*a^4*b^2)) - (512*a*b^5*tan(c/2 + (d* 
x^(1/2))/2))/((512*a^3*b^9)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (1536*a^5*b^7)/( 
a^6 + a^2*b^4 - 2*a^4*b^2) + (1024*a^7*b^5)/(a^6 + a^2*b^4 - 2*a^4*b^2) + 
(512*a^9*b^3)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (512*a^11*b)/(a^6 + a^2*b^4 - 
2*a^4*b^2)) + (512*a^5*b*tan(c/2 + (d*x^(1/2))/2))/((512*a^3*b^9)/(a^6 + a 
^2*b^4 - 2*a^4*b^2) - (1536*a^5*b^7)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (1024*a 
^7*b^5)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (512*a^9*b^3)/(a^6 + a^2*b^4 - 2*a^4 
*b^2) - (512*a^11*b)/(a^6 + a^2*b^4 - 2*a^4*b^2))))/(a^2*d) - ((4*b^2)/(a* 
(a^2 - b^2)) + (4*b*tan(c/2 + (d*x^(1/2))/2))/(a^2 - b^2))/(d*(b + b*tan(c 
/2 + (d*x^(1/2))/2)^2 + 2*a*tan(c/2 + (d*x^(1/2))/2))) - (b*atan(((b*(2*a^ 
2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x^(1/2))/2)*(8*a*b^ 
7 - 8*a^7*b - 32*a^3*b^5 + 36*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (32* 
(4*a*b^6 - 8*a^3*b^4 + 4*a^5*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (2*b*(2*a 
^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(2*a^8*b - 2*a^6*b^3))/(a^6 + a 
^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(4*a^4*b^6 - 12*a^6*b^4 
 + 8*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (2*b*((32*(a^5*b^6 - 2*a^7*b^ 
4 + a^9*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.86 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {8 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right ) \sin \left (\sqrt {x}\, d +c \right ) a^{3} b -4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right ) \sin \left (\sqrt {x}\, d +c \right ) a \,b^{3}+8 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2} b^{2}-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b^{4}-2 \cos \left (\sqrt {x}\, d +c \right ) a^{3} b^{2}+2 \cos \left (\sqrt {x}\, d +c \right ) a \,b^{4}+2 \sqrt {x}\, \sin \left (\sqrt {x}\, d +c \right ) a^{5} d -4 \sqrt {x}\, \sin \left (\sqrt {x}\, d +c \right ) a^{3} b^{2} d +2 \sqrt {x}\, \sin \left (\sqrt {x}\, d +c \right ) a \,b^{4} d +2 \sqrt {x}\, a^{4} b d -4 \sqrt {x}\, a^{2} b^{3} d +2 \sqrt {x}\, b^{5} d}{a^{2} d \left (\sin \left (\sqrt {x}\, d +c \right ) a^{5}-2 \sin \left (\sqrt {x}\, d +c \right ) a^{3} b^{2}+\sin \left (\sqrt {x}\, d +c \right ) a \,b^{4}+a^{4} b -2 a^{2} b^{3}+b^{5}\right )} \] Input:

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

Output:

(2*(4*sqrt( - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*b + a)/sqrt( - a** 
2 + b**2))*sin(sqrt(x)*d + c)*a**3*b - 2*sqrt( - a**2 + b**2)*atan((tan((s 
qrt(x)*d + c)/2)*b + a)/sqrt( - a**2 + b**2))*sin(sqrt(x)*d + c)*a*b**3 + 
4*sqrt( - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*b + a)/sqrt( - a**2 + 
b**2))*a**2*b**2 - 2*sqrt( - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*b + 
 a)/sqrt( - a**2 + b**2))*b**4 - cos(sqrt(x)*d + c)*a**3*b**2 + cos(sqrt(x 
)*d + c)*a*b**4 + sqrt(x)*sin(sqrt(x)*d + c)*a**5*d - 2*sqrt(x)*sin(sqrt(x 
)*d + c)*a**3*b**2*d + sqrt(x)*sin(sqrt(x)*d + c)*a*b**4*d + sqrt(x)*a**4* 
b*d - 2*sqrt(x)*a**2*b**3*d + sqrt(x)*b**5*d))/(a**2*d*(sin(sqrt(x)*d + c) 
*a**5 - 2*sin(sqrt(x)*d + c)*a**3*b**2 + sin(sqrt(x)*d + c)*a*b**4 + a**4* 
b - 2*a**2*b**3 + b**5))