\(\int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) [866]

Optimal result

Integrand size = 25, antiderivative size = 248 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=-\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{b^{3/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a+b} d}+\frac {\sqrt {a+b \tan (c+d x)}}{b d \sqrt {\cot (c+d x)}} \] Output:

-I*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c 
)^(1/2)*tan(d*x+c)^(1/2)/(I*a-b)^(1/2)/d-a*arctanh(b^(1/2)*tan(d*x+c)^(1/2 
)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/b^(3/2)/d+I*ar 
ctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1 
/2)*tan(d*x+c)^(1/2)/(I*a+b)^(1/2)/d+(a+b*tan(d*x+c))^(1/2)/b/d/cot(d*x+c) 
^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.61 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b}-\frac {a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{b^{3/2} \sqrt {a+b \tan (c+d x)}}\right )}{d} \] Input:

Integrate[1/(Cot[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]]),x]
 

Output:

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqr 
t[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] 
- ((-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a 
+ b*Tan[c + d*x]]])/Sqrt[a + I*b] + (Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + 
 d*x]])/b - (a^(3/2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 
+ (b*Tan[c + d*x])/a])/(b^(3/2)*Sqrt[a + b*Tan[c + d*x]])))/d
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.85, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4729, 3042, 4049, 27, 3042, 4138, 2035, 2257, 2009}

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.

\(\displaystyle \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cot (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4729

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan (c+d x)^{5/2}}{\sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {i b \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}+\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-\frac {i b \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}}{b d}\right )\)

Input:

Int[1/(Cot[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]]),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-(((I*b*ArcTan[(Sqrt[I*a - b]*Sqrt[ 
Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a - b] + (a*ArcTanh[(Sqrt 
[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] - (I*b*ArcTanh[ 
(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a + b 
])/(b*d)) + (Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(b*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4138

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

rule 4729

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1309\) vs. \(2(202)=404\).

Time = 4.18 (sec) , antiderivative size = 1310, normalized size of antiderivative = 5.28

method	result	size
default	\(\text {Expression too large to display}\)	\(1310\)

Input:

int(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*2^(1/2)/(-b+(a^2+b^2)^(1/2))^(1/2)/(a^2+b^2)^(1/2)/a/b^(3/2)*(b^(5/2 
)*cos(d*x+c)*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*ln((a*co 
t(d*x+c)*cos(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos 
(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*a*cot(d*x+c)+2* 
(a^2+b^2)^(1/2)*cos(d*x+c)+2*cos(d*x+c)*b-sin(d*x+c)*a+a*csc(d*x+c)-2*(a^2 
+b^2)^(1/2)-2*b)/(-1+cos(d*x+c)))-b^(5/2)*cos(d*x+c)*(b+(a^2+b^2)^(1/2))^( 
1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*ln((a*cot(d*x+c)*cos(d*x+c)-2*2^(1/2)*((a* 
cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^( 
1/2))^(1/2)*sin(d*x+c)-2*a*cot(d*x+c)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*cos(d 
*x+c)*b-sin(d*x+c)*a+a*csc(d*x+c)-2*(a^2+b^2)^(1/2)-2*b)/(-1+cos(d*x+c)))- 
b^(3/2)*cos(d*x+c)*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*(a 
^2+b^2)^(1/2)*ln((a*cot(d*x+c)*cos(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d 
*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d* 
x+c)-2*a*cot(d*x+c)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*cos(d*x+c)*b-sin(d*x+c) 
*a+a*csc(d*x+c)-2*(a^2+b^2)^(1/2)-2*b)/(-1+cos(d*x+c)))+b^(3/2)*cos(d*x+c) 
*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*(a^2+b^2)^(1/2)*ln(( 
a*cot(d*x+c)*cos(d*x+c)-2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/ 
(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*a*cot(d*x+c 
)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*cos(d*x+c)*b-sin(d*x+c)*a+a*csc(d*x+c)-2* 
(a^2+b^2)^(1/2)-2*b)/(-1+cos(d*x+c)))+2*b^(3/2)*cos(d*x+c)*arctan((-2^(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4173 vs. \(2 (196) = 392\).

Time = 0.85 (sec) , antiderivative size = 8379, normalized size of antiderivative = 33.79 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \tan {\left (c + d x \right )}} \cot ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate(1/cot(d*x+c)**(5/2)/(a+b*tan(d*x+c))**(1/2),x)
 

Output:

Integral(1/(sqrt(a + b*tan(c + d*x))*cot(c + d*x)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(b*tan(d*x + c) + a)*cot(d*x + c)^(5/2)), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \] Input:

int(1/(cot(c + d*x)^(5/2)*(a + b*tan(c + d*x))^(1/2)),x)
 

Output:

int(1/(cot(c + d*x)^(5/2)*(a + b*tan(c + d*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )^{3} \tan \left (d x +c \right ) b +\cot \left (d x +c \right )^{3} a}d x \] Input:

int(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(1/2),x)
 

Output:

int((sqrt(tan(c + d*x)*b + a)*sqrt(cot(c + d*x)))/(cot(c + d*x)**3*tan(c + 
 d*x)*b + cot(c + d*x)**3*a),x)