\(\int \frac {1}{(a+b \cot ^2(c+d x))^{3/2}} \, dx\) [35]

Optimal result

Integrand size = 16, antiderivative size = 85 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{(a-b)^{3/2} d}+\frac {b \cot (c+d x)}{a (a-b) d \sqrt {a+b \cot ^2(c+d x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.73 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.72 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\cos ^2(c+d x) \cot (c+d x) \left (4 (a-b)^2 \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},\frac {(a-b) \cos ^2(c+d x)}{a}\right ) \left (b+a \tan ^2(c+d x)\right )-\frac {15 a \left (2 b+3 a \tan ^2(c+d x)\right ) \left (\arcsin \left (\sqrt {\frac {(a-b) \cos ^2(c+d x)}{a}}\right ) \left (b+a \tan ^2(c+d x)\right )-a \sec ^2(c+d x) \sqrt {\frac {(a-b) \cos ^4(c+d x) \left (b+a \tan ^2(c+d x)\right )}{a^2}}\right )}{\sqrt {\frac {(a-b) \cos ^4(c+d x) \left (b+a \tan ^2(c+d x)\right )}{a^2}}}\right )}{15 a^3 (a-b) d \sqrt {a+b \cot ^2(c+d x)}} \] Input:

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

Output:

-1/15*(Cos[c + d*x]^2*Cot[c + d*x]*(4*(a - b)^2*Cos[c + d*x]^2*Hypergeomet 
ric2F1[2, 2, 7/2, ((a - b)*Cos[c + d*x]^2)/a]*(b + a*Tan[c + d*x]^2) - (15 
*a*(2*b + 3*a*Tan[c + d*x]^2)*(ArcSin[Sqrt[((a - b)*Cos[c + d*x]^2)/a]]*(b 
 + a*Tan[c + d*x]^2) - a*Sec[c + d*x]^2*Sqrt[((a - b)*Cos[c + d*x]^4*(b + 
a*Tan[c + d*x]^2))/a^2]))/Sqrt[((a - b)*Cos[c + d*x]^4*(b + a*Tan[c + d*x] 
^2))/a^2]))/(a^3*(a - b)*d*Sqrt[a + b*Cot[c + d*x]^2])
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 84, 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, 4144, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4144

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-((ArcTan[(Sqrt[a - b]*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]]/(a - b)^( 
3/2) - (b*Cot[c + d*x])/(a*(a - b)*Sqrt[a + 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 4144

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> 
With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f)   Subst[Int[(a + b* 
(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, 
 b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || 
EqQ[n^2, 16])
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \cot \left (d x +c \right )^{2}}}\right )}{\left (a -b \right )^{2} b^{2}}+\frac {b \cot \left (d x +c \right )}{\left (a -b \right ) a \sqrt {a +b \cot \left (d x +c \right )^{2}}}}{d}\)	\(102\)
default	\(\frac {-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \cot \left (d x +c \right )^{2}}}\right )}{\left (a -b \right )^{2} b^{2}}+\frac {b \cot \left (d x +c \right )}{\left (a -b \right ) a \sqrt {a +b \cot \left (d x +c \right )^{2}}}}{d}\)	\(102\)

Input:

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

Output:

1/d*(-1/(a-b)^2*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/( 
a+b*cot(d*x+c)^2)^(1/2)*cot(d*x+c))+b/(a-b)*cot(d*x+c)/a/(a+b*cot(d*x+c)^2 
)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (77) = 154\).

Time = 0.14 (sec) , antiderivative size = 578, normalized size of antiderivative = 6.80 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^{3/2}} \, dx=\left [-\frac {{\left (a^{2} + a b - {\left (a^{2} - a b\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt {-a + b} \log \left (-2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, {\left ({\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - b\right )} \sqrt {-a + b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right ) + a^{2} - 2 \, b^{2} + 4 \, {\left (a b - b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )\right ) + 4 \, {\left (a b - b^{2}\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{4 \, {\left ({\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} d \cos \left (2 \, d x + 2 \, c\right ) - {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3}\right )} d\right )}}, \frac {{\left (a^{2} + a b - {\left (a^{2} - a b\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt {a - b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - b\right )} \sqrt {a - b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} - a^{2} + b^{2} - 2 \, {\left (a b - b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}\right ) - 2 \, {\left (a b - b^{2}\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{2 \, {\left ({\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} d \cos \left (2 \, d x + 2 \, c\right ) - {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3}\right )} d\right )}}\right ] \] Input:

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

Output:

[-1/4*((a^2 + a*b - (a^2 - a*b)*cos(2*d*x + 2*c))*sqrt(-a + b)*log(-2*(a^2 
 - 2*a*b + b^2)*cos(2*d*x + 2*c)^2 + 2*((a - b)*cos(2*d*x + 2*c) - b)*sqrt 
(-a + b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*s 
in(2*d*x + 2*c) + a^2 - 2*b^2 + 4*(a*b - b^2)*cos(2*d*x + 2*c)) + 4*(a*b - 
 b^2)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin( 
2*d*x + 2*c))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d*cos(2*d*x + 2*c) - (a 
^4 - a^3*b - a^2*b^2 + a*b^3)*d), 1/2*((a^2 + a*b - (a^2 - a*b)*cos(2*d*x 
+ 2*c))*sqrt(a - b)*arctan(-((a - b)*cos(2*d*x + 2*c) - b)*sqrt(a - b)*sqr 
t(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2 
*c)/((a^2 - 2*a*b + b^2)*cos(2*d*x + 2*c)^2 - a^2 + b^2 - 2*(a*b - b^2)*co 
s(2*d*x + 2*c))) - 2*(a*b - b^2)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/( 
cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b 
^3)*d*cos(2*d*x + 2*c) - (a^4 - a^3*b - a^2*b^2 + a*b^3)*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(a+b*cot(d*x+c)^2)^(3/2),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(b-a>0)', see `assume?` for more 
details)Is
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (77) = 154\).

Time = 0.61 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.53 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\frac {\frac {{\left (a^{2} b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 2 \, a b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}} - \frac {a^{2} b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 2 \, a b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )}{a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}}}{\sqrt {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b}} - \frac {2 \, \arctan \left (-\frac {\sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b} + \sqrt {b}}{2 \, \sqrt {a - b}}\right )}{{\left (a \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - b \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )} \sqrt {a - b}}}{d} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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