\(\int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [336]

Optimal result

Integrand size = 23, antiderivative size = 217 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {(4 a-7 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}+\frac {(4 a+7 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}+\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {\cot ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (a^2+b^2-2 a b \sec (c+d x)\right )}{2 \left (a^2-b^2\right )^2 d} \] Output:

-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+1/4*(4*a-7*b)*arctanh 
((a+b*sec(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/d+1/4*(4*a+7*b)*arctanh(( 
a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d+2*b^4/a/(a^2-b^2)^2/d/(a+ 
b*sec(d*x+c))^(1/2)-1/2*cot(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*(a^2+b^2-2*a*b 
*sec(d*x+c))/(a^2-b^2)^2/d
 

Mathematica [C] (verified)

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

Time = 1.00 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.45 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sec (c+d x)}{a-b}\right )}{(a-b) \sqrt {a+b \sec (c+d x)}}+\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sec (c+d x)}{a+b}\right )}{(a+b) \sqrt {a+b \sec (c+d x)}}+\frac {4 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sec (c+d x)}{a}\right )}{a \sqrt {a+b \sec (c+d x)}}+\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {a+b \sec (c+d x)}{a-b}\right )}{(a-b)^2 \sqrt {a+b \sec (c+d x)}}-\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {a+b \sec (c+d x)}{a+b}\right )}{(a+b)^2 \sqrt {a+b \sec (c+d x)}}}{2 b d} \] Input:

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

Output:

((-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] + (2*ArcTa 
nh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] - (2*a*Hypergeometri 
c2F1[-1/2, 1, 1/2, (a + b*Sec[c + d*x])/(a - b)])/((a - b)*Sqrt[a + b*Sec[ 
c + d*x]]) + (2*a*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sec[c + d*x])/(a 
+ b)])/((a + b)*Sqrt[a + b*Sec[c + d*x]]) + (4*b*Hypergeometric2F1[-1/2, 1 
, 1/2, 1 + (b*Sec[c + d*x])/a])/(a*Sqrt[a + b*Sec[c + d*x]]) + (b^2*Hyperg 
eometric2F1[-1/2, 2, 1/2, (a + b*Sec[c + d*x])/(a - b)])/((a - b)^2*Sqrt[a 
 + b*Sec[c + d*x]]) - (b^2*Hypergeometric2F1[-1/2, 2, 1/2, (a + b*Sec[c + 
d*x])/(a + b)])/((a + b)^2*Sqrt[a + b*Sec[c + d*x]]))/(2*b*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.64 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.43, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4373, 561, 25, 1674, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1674

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 b^4 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} b^4}+\frac {\cos (c+d x)}{a b \left (a^2-b^2\right )^2}+\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{4 b^4 (a-b)^{5/2}}+\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{4 b^4 (a+b)^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{8 b^3 (a-b)^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{8 b^3 (a+b)^{5/2}}-\frac {\sec (c+d x)}{8 b^2 (a-b)^2 \left (a-b^2 \sec ^2(c+d x)-b\right )}+\frac {\sec (c+d x)}{8 b^2 (a+b)^2 \left (a-b^2 \sec ^2(c+d x)+b\right )}\right )}{d}\)

Input:

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

Output:

(2*b^4*(-(ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]/(a^(3/2)*b^4)) + ((2*a 
 - 3*b)*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(5/2)*b^ 
4) - ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(8*(a - b)^(5/2)*b^3) + 
 ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/(8*b^3*(a + b)^(5/2)) + ((2 
*a + 3*b)*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]])/(4*b^4*(a + b)^(5 
/2)) + Cos[c + d*x]/(a*b*(a^2 - b^2)^2) - Sec[c + d*x]/(8*(a - b)^2*b^2*(a 
 - b - b^2*Sec[c + d*x]^2)) + Sec[c + d*x]/(8*b^2*(a + b)^2*(a + b - b^2*S 
ec[c + d*x]^2))))/d
 

Defintions of rubi rules used

rule 25

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

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 1674

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && N 
eQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])
 

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. \(2982\) vs. \(2(189)=378\).

Time = 0.93 (sec) , antiderivative size = 2983, normalized size of antiderivative = 13.75

method	result	size
default	\(\text {Expression too large to display}\)	\(2983\)

Input:

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

Output:

-1/8/d/a^2/(a-b)^(7/2)/(a+b)^3*(cos(d*x+c)*(4*cos(d*x+c)-4)*(a+b)^(1/2)*(a 
-b)^(3/2)*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)))*a^6+cos(d*x+ 
c)*(-4*cos(d*x+c)+4)*ln(1/(a-b)^(1/2)*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+c 
os(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)))*a^8+4*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)^(3/2)*a^6+(-cos(d*x+c)^2-3*cos(d*x+c)+4)*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*c 
os(d*x+c)+b*cos(d*x+c)-b)/(1+cos(d*x+c)))*a^7*b+(14*cos(d*x+c)^2-15*cos(d* 
x+c)+1)*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)))*a^6* 
b^2+(8*cos(d*x+c)^2+6*cos(d*x+c)-14)*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)))*a^5*b^3+(-10*cos(d*x+c)^2+18*cos(d*x+c)-8)*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...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 936 vs. \(2 (193) = 386\).

Time = 44.54 (sec) , antiderivative size = 8098, normalized size of antiderivative = 37.32 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{32,[2,6]%%%}+%%%{-32,[1,7]%%%},[6,1]%%%}+%%%{%%{[%%%{6 
4,[2,6]%%
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(c + d*x)*b + a)*cot(c + d*x)**3)/(sec(c + d*x)**2*b**2 + 2*s 
ec(c + d*x)*a*b + a**2),x)