3.4.38 \(\int \frac {\cot (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [338]

3.4.38.1 Optimal result

Integrand size = 21, antiderivative size = 142 \[ \int \frac {\cot (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2} d}+\frac {2 b^2}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

output

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

3.4.38.2 Mathematica [C] (verified)

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

Time = 0.67 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.49 \[ \int \frac {\cot (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sec (c+d x)}{a-b}\right )}{(a-b) \sqrt {a+b \sec (c+d x)}}+\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sec (c+d x)}{a+b}\right )}{(a+b) \sqrt {a+b \sec (c+d x)}}+\frac {2 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sec (c+d x)}{a}\right )}{a \sqrt {a+b \sec (c+d x)}}}{b d} \]

input

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

output

-((-(ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/Sqrt[a - b]) + ArcTanh[ 
Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/Sqrt[a + b] - (a*Hypergeometric2F1[- 
1/2, 1, 1/2, (a + b*Sec[c + d*x])/(a - b)])/((a - b)*Sqrt[a + b*Sec[c + d* 
x]]) + (a*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sec[c + d*x])/(a + b)])/( 
(a + b)*Sqrt[a + b*Sec[c + d*x]]) + (2*b*Hypergeometric2F1[-1/2, 1, 1/2, 1 
 + (b*Sec[c + d*x])/a])/(a*Sqrt[a + b*Sec[c + d*x]]))/(b*d))
 

3.4.38.3 Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4373, 561, 1610, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 1610

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

\(\Big \downarrow \) 2009

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

input

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

output

(-2*b^2*(-(ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]/(a^(3/2)*b^2)) + ArcT 
anh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(2*(a - b)^(3/2)*b^2) + ArcTanh[ 
Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/(2*b^2*(a + b)^(3/2)) - Cos[c + d*x] 
/(a*b*(a^2 - b^2))))/d
 

3.4.38.3.1 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 1610

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

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]
 

3.4.38.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2691\) vs. \(2(122)=244\).

Time = 2.22 (sec) , antiderivative size = 2692, normalized size of antiderivative = 18.96

method	result	size
default	\(\text {Expression too large to display}\)	\(2692\)

input

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

output

-1/2/d/(a-b)^(5/2)/a^2/(a+b)^2*(-2*(a-b)^(3/2)*ln(4*cos(d*x+c)*((b+a*cos(d 
*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2) 
*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*cos(d*x+c)*a^(9 
/2)-2*(a-b)^(3/2)*a^(7/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(co 
s(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*co 
s(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b*cos(d*x+c)-2*(a-b)^(3/2)*a^(7/2)*l 
n(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2 
)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^ 
(1/2)+2*b)*b+2*(a-b)^(3/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(c 
os(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*c 
os(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*cos(d*x+c)*a^(5/2)*b^2-2*(a-b)^(3/2 
)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^( 
1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^ 
2)^(1/2)+2*b)*a^(5/2)*b^2+2*(a-b)^(3/2)*a^(3/2)*ln(4*cos(d*x+c)*((b+a*cos( 
d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2 
)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b^3*cos(d*x+c) 
+(a+b)^(1/2)*(a-b)^(3/2)*ln(-2*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/ 
(cos(d*x+c)+1)^2)^(1/2)*(a+b)^(1/2)+2*a*cos(d*x+c)+cos(d*x+c)*b+2*(a+b)^(1 
/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+b)/(cos(d*x+c)-1) 
)*cos(d*x+c)*a^4-(a+b)^(1/2)*(a-b)^(3/2)*ln(-2*(2*cos(d*x+c)*((b+a*cos(...
 

3.4.38.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (122) = 244\).

Time = 24.54 (sec) , antiderivative size = 3924, normalized size of antiderivative = 27.63 \[ \int \frac {\cot (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

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

output

[1/4*(8*(a^3*b^2 - a*b^4)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x 
+ c) + 2*(a^4*b - 2*a^2*b^3 + b^5 + (a^5 - 2*a^3*b^2 + a*b^4)*cos(d*x + c) 
)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 - 4*(2*a*co 
s(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + 
 c))) - (a^4*b + 2*a^3*b^2 + a^2*b^3 + (a^5 + 2*a^4*b + a^3*b^2)*cos(d*x + 
 c))*sqrt(a - b)*log(-((8*a^2 - 8*a*b + b^2)*cos(d*x + c)^2 + b^2 + 4*((2* 
a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*sqrt((a*cos(d*x + c) + 
 b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 + 2*co 
s(d*x + c) + 1)) + (a^4*b - 2*a^3*b^2 + a^2*b^3 + (a^5 - 2*a^4*b + a^3*b^2 
)*cos(d*x + c))*sqrt(a + b)*log(-((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^2 + b 
^2 - 4*((2*a + b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a + b)*sqrt((a*cos 
(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b + 3*b^2)*cos(d*x + c))/(cos(d*x + 
c)^2 - 2*cos(d*x + c) + 1)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + 
 (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), -1/4*(4*(a^4*b - 2*a^2*b^3 + b^5 + (a^5 
 - 2*a^3*b^2 + a*b^4)*cos(d*x + c))*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos 
(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) - 8*(a^3 
*b^2 - a*b^4)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c) + (a^4* 
b + 2*a^3*b^2 + a^2*b^3 + (a^5 + 2*a^4*b + a^3*b^2)*cos(d*x + c))*sqrt(a - 
 b)*log(-((8*a^2 - 8*a*b + b^2)*cos(d*x + c)^2 + b^2 + 4*((2*a - b)*cos(d* 
x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*sqrt((a*cos(d*x + c) + b)/cos(d*...
 

3.4.38.6 Sympy [F]

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

input

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

output

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

3.4.38.7 Maxima [F]

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

input

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

output

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

3.4.38.8 Giac [F(-2)]

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

input

integrate(cot(d*x+c)/(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:Error: Bad Argument Type
 

3.4.38.9 Mupad [F(-1)]

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

input

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

output

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