\(\int \cot ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx\) [319]

Optimal result

Integrand size = 23, antiderivative size = 153 \[ \int \cot ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {(4 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{4 \sqrt {a-b} d}+\frac {(4 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{4 \sqrt {a+b} d}-\frac {\cot ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 2.10 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.39 \[ \int \cot ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\frac {-\frac {b \arctan \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {-a+b}}\right )}{\sqrt {-a+b}}-8 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )+\frac {4 \sqrt {-(a-b)^2} \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {-a+b}}+\frac {4 a \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-2 \cot ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{4 d} \] Input:

Integrate[Cot[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]],x]
 

Output:

(-((b*ArcTan[Sqrt[a + b*Sec[c + d*x]]/Sqrt[-a + b]])/Sqrt[-a + b]) - 8*Sqr 
t[a]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]] + (4*Sqrt[-(a - b)^2]*ArcTa 
nh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]])/Sqrt[-a + b] + (4*a*ArcTanh[Sqrt 
[a + b*Sec[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] + (3*b*ArcTanh[Sqrt[a + b*S 
ec[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] - 2*Cot[c + d*x]^2*Sqrt[a + b*Sec[c 
 + d*x]])/(4*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.55 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.75, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 25, 4373, 561, 25, 1652, 25, 1484, 1492, 27, 1406, 220, 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.

\(\displaystyle \int \cot ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4373

\(\displaystyle \frac {b^4 \int \frac {\cos (c+d x) \sqrt {a+b \sec (c+d x)}}{b \left (b^2-b^2 \sec ^2(c+d x)\right )^2}d(b \sec (c+d x))}{d}\)

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1652

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1484

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

\(\Big \downarrow \) 1492

\(\displaystyle \frac {2 b^4 \left (\frac {a \int \left (\frac {1}{2 b^2 \left (-b^2 \sec ^2(c+d x)+a+b\right )}-\frac {1}{2 b^2 \left (b^2 \sec ^2(c+d x)-a+b\right )}-\frac {1}{b^2 \left (a-b^2 \sec ^2(c+d x)\right )}\right )d\sqrt {a+b \sec (c+d x)}}{b^2}-\frac {\frac {\sqrt {a+b \sec (c+d x)}}{4 \left (a^2-2 a b^2 \sec ^2(c+d x)+b^4 \sec ^4(c+d x)-b^2\right )}-\frac {\int -\frac {6 b^2 \left (a^2-b^2\right )}{b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sec (c+d x)}}{8 b^2 \left (a^2-b^2\right )}}{b^2}\right )}{d}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1406

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

\(\Big \downarrow \) 220

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

\(\Big \downarrow \) 2009

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

Input:

Int[Cot[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]],x]
 

Output:

(2*b^4*((a*(-(ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]/(Sqrt[a]*b^2)) + A 
rcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(2*Sqrt[a - b]*b^2) + ArcTanh 
[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/(2*b^2*Sqrt[a + b])))/b^2 - ((3*(Ar 
cTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(2*Sqrt[a - b]*b) - ArcTanh[Sq 
rt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/(2*b*Sqrt[a + b])))/4 + Sqrt[a + b*Sec 
[c + d*x]]/(4*(a^2 - b^2 - 2*a*b^2*Sec[c + d*x]^2 + b^4*Sec[c + d*x]^4)))/ 
b^2))/d
 

Defintions of rubi rules used

rule 25

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

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 220

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

rule 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1406

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^ 
2 - 4*a*c, 2]}, Simp[c/q   Int[1/(b/2 - q/2 + c*x^2), x], x] - Simp[c/q   I 
nt[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c 
, 0] && PosQ[b^2 - 4*a*c]
 

rule 1484

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb 
ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 
 0] && IntegerQ[q]
 

rule 1492

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb 
ol] :> Simp[x*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)*((a + b*x^2 + 
 c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 
 - 4*a*c))   Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 
7)*(d*b - 2*a*e)*c*x^2, x]*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, 
 b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && 
 LtQ[p, -1] && IntegerQ[2*p]
 

rule 1652

Int[(((f_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_))/((d_) + 
 (e_.)*(x_)^2), x_Symbol] :> Simp[f^2/(c*d^2 - b*d*e + a*e^2)   Int[(f*x)^( 
m - 2)*(a*e + c*d*x^2)*(a + b*x^2 + c*x^4)^p, x], x] - Simp[d*e*(f^2/(c*d^2 
 - b*d*e + a*e^2))   Int[(f*x)^(m - 2)*((a + b*x^2 + c*x^4)^(p + 1)/(d + e* 
x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ 
[p, -1] && GtQ[m, 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 4373

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1))   Subst[Int[(b^2 - x^ 
2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, 
 d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1427\) vs. \(2(127)=254\).

Time = 0.89 (sec) , antiderivative size = 1428, normalized size of antiderivative = 9.33

method	result	size
default	\(\text {Expression too large to display}\)	\(1428\)

Input:

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

Output:

1/16/d/(a-b)^(3/2)/(a+b)*(8*a^(3/2)*ln(4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+c 
os(d*x+c))^2)^(1/2)*a^(1/2)*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x 
+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*cos(d*x+c)+2*b)*(1-cos(d*x+c))^2*(a-b)^(3/ 
2)*csc(d*x+c)^2+8*a^(1/2)*ln(4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c)) 
^2)^(1/2)*a^(1/2)*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos 
(d*x+c))^2)^(1/2)+4*a*cos(d*x+c)+2*b)*(1-cos(d*x+c))^2*(a-b)^(3/2)*b*csc(d 
*x+c)^2-4*(a+b)^(1/2)*ln(-2*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c)) 
^2)^(1/2)*(a+b)^(1/2)*cos(d*x+c)+2*(a+b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c 
)/(1+cos(d*x+c))^2)^(1/2)+2*a*cos(d*x+c)+b*cos(d*x+c)+b)/(-1+cos(d*x+c)))* 
(1-cos(d*x+c))^2*(a-b)^(3/2)*a*csc(d*x+c)^2-3*(a+b)^(1/2)*ln(-2*(2*((b+a*c 
os(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a+b)^(1/2)*cos(d*x+c)+2*(a+ 
b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+2*a*cos(d*x+ 
c)+b*cos(d*x+c)+b)/(-1+cos(d*x+c)))*(1-cos(d*x+c))^2*(a-b)^(3/2)*b*csc(d*x 
+c)^2+4*a^3*ln(1/(a-b)^(1/2)*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c) 
)^2)^(1/2)*(a-b)^(1/2)*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d* 
x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/(1+cos(d*x+c)))* 
(1-cos(d*x+c))^2*csc(d*x+c)^2-3*a^2*ln(1/(a-b)^(1/2)*(2*((b+a*cos(d*x+c))* 
cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*cos(d*x+c)+2*((b+a*cos(d*x+ 
c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d* 
x+c)-b)/(1+cos(d*x+c)))*(1-cos(d*x+c))^2*b*csc(d*x+c)^2-4*ln(1/(a-b)^(1...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (127) = 254\).

Time = 1.67 (sec) , antiderivative size = 3523, normalized size of antiderivative = 23.03 \[ \int \cot ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

Integral(sqrt(a + b*sec(c + d*x))*cot(c + d*x)**3, x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(b*sec(d*x + c) + a)*cot(d*x + c)^3, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 514 vs. \(2 (127) = 254\).

Time = 0.62 (sec) , antiderivative size = 514, normalized size of antiderivative = 3.36 \[ \int \cot ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/8*(16*a*arctan(-1/2*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2 
*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + 
a + b) + sqrt(a - b))/sqrt(-a))/sqrt(-a) - 2*(4*a + 3*b)*arctan(-(sqrt(a - 
 b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x 
 + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))/sqrt(-a - b))/sqrt(-a - 
 b) + (4*a - 3*b)*log(abs((sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan 
(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^ 
2 + a + b))*(a - b) - sqrt(a - b)*a))/sqrt(a - b) + sqrt(a*tan(1/2*d*x + 1 
/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) - 
 2*((sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - 
b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))*a - (a + b 
)*sqrt(a - b))/((sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 
 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) 
)^2 - a - b))*sgn(cos(d*x + c))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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