\(\int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\) [861]

Optimal result

Integrand size = 25, antiderivative size = 220 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}-\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a+b} d}+\frac {4 b \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{3 a^2 d}-\frac {2 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 a d} \] Output:

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

Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.87 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {3 (-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {3 (-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}-\frac {2 (a-2 b \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}{a^2 \tan ^{\frac {3}{2}}(c+d x)}\right )}{3 d} \] Input:

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

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.94, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4729, 3042, 4052, 27, 3042, 4132, 27, 3042, 4058, 615, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {-\frac {2 \int -\frac {3 a^2}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {3 a \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {3 a \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4058

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\frac {3 a \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 615

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {3 a \int \left (\frac {i}{2 (i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {i}{2 \sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}\right )d\tan (c+d x)}{d}}{3 a}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {3 a \left (\frac {\arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}+\frac {\text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}\right )}{d}}{3 a}\right )\)

Input:

Int[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]]*((-2*Sqrt[a + b*Tan[c + d*x]])/(3*a* 
d*Tan[c + d*x]^(3/2)) - ((3*a*(ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/S 
qrt[a + b*Tan[c + d*x]]]/Sqrt[I*a - b] + ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c 
 + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[I*a + b]))/d - (4*b*Sqrt[a + b*Ta 
n[c + d*x]])/(a*d*Sqrt[Tan[c + d*x]]))/(3*a))
 

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 615

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
 

rule 2009

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

rule 3042

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

rule 4052

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 + 1)*((c 
+ d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 
/((m + 1)*(a^2 + b^2)*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + 
d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - 
 a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*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] && LtQ[m, -1] && (LtQ[n, 0] || Integ 
erQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4058

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), 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/(1 + ff^2*x^2)), x], x, 
 Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a 
*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
 

rule 4132

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] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ 
[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 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. \(1192\) vs. \(2(180)=360\).

Time = 3.92 (sec) , antiderivative size = 1193, normalized size of antiderivative = 5.42

method	result	size
default	\(\text {Expression too large to display}\)	\(1193\)

Input:

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

Output:

1/12/d*2^(1/2)/a^2/(a^2+b^2)^(1/2)/(-b+(a^2+b^2)^(1/2))^(1/2)*((3*cos(d*x+ 
c)-3)*sin(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)))*a^2+(-3*cos(d*x+c)+3)*sin(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)))*a^2+(-6*cos(d*x+c)+6)*sin(d*x+c)*(a^2+b^2)^(1/2)*arctan( 
(-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)* 
sin(d*x+c)+(b+(a^2+b^2)^(1/2))^(1/2)*cos(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2)) 
/(-b+(a^2+b^2)^(1/2))^(1/2)/(-1+cos(d*x+c)))*a^2+(6*cos(d*x+c)-6)*sin(d*x+ 
c)*arctan((-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)*sin(d*x+c)+(b+(a^2+b^2)^(1/2))^(1/2)*cos(d*x+c)-(b+(a^2+b^2)^(1/ 
2))^(1/2))/(-b+(a^2+b^2)^(1/2))^(1/2)/(-1+cos(d*x+c)))*a^2*b+(6*cos(d*x+c) 
-6)*sin(d*x+c)*(a^2+b^2)^(1/2)*arctan((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)*sin(d*x+c)+(b+(a^2+b^2)^(1/2))^(1/2)* 
cos(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2))/(-b+(a^2+b^2)^(1/2))^(1/2)/(-1+co...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4220 vs. \(2 (176) = 352\).

Time = 0.52 (sec) , antiderivative size = 4220, normalized size of antiderivative = 19.18 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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