\(\int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [406]

Optimal result

Integrand size = 25, antiderivative size = 116 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f}+\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{3/2} f}-\frac {\cot ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) f} \] Output:

-arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/a^(1/2)/f+1/2*(2*a+3*b)*arctanh 
((a+b*sec(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)/f-1/2*cot(f*x+e)^2*(a+b 
*sec(f*x+e)^2)^(1/2)/(a+b)/f
 

Mathematica [F]

\[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx \] Input:

Integrate[Cot[e + f*x]^3/Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

Integrate[Cot[e + f*x]^3/Sqrt[a + b*Sec[e + f*x]^2], x]
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4627, 354, 114, 27, 174, 73, 221}

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.

\(\displaystyle \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (e+f x)^3 \sqrt {a+b \sec (e+f x)^2}}dx\)

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Cot[e + f*x]^3/Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(((-4*(a + b)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/Sqrt[a] + (2*(2 
*a + 3*b)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]])/Sqrt[a + b])/(2 
*(a + b)) + Sqrt[a + b*Sec[e + f*x]^2]/((a + b)*(1 - Sec[e + f*x]^2)))/(2* 
f)
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 221

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

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 3042

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

rule 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1047\) vs. \(2(98)=196\).

Time = 0.43 (sec) , antiderivative size = 1048, normalized size of antiderivative = 9.03

method	result	size
default	\(\text {Expression too large to display}\)	\(1048\)

Input:

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

Output:

-1/2/f/(a+b)^(5/2)/a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)/(a+ 
b*sec(f*x+e)^2)^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)/(1-cos(f*x+e))^2*( 
2*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e 
)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+cos(f*x+e)*a+b)/ 
(cos(f*x+e)-1))*a^(5/2)*(1-cos(f*x+e))^2-2*ln(2/(a+b)^(1/2)*((a+b)^(1/2)*( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+((b+a*cos(f*x+e)^2)/ 
(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/(1+cos(f*x+e)))*a^(5/2 
)*(1-cos(f*x+e))^2+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(1-cos(f*x+ 
e))^2*(a+b)^(3/2)*a^(1/2)-4*ln(4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 
2)*a^(1/2)*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 
)+4*cos(f*x+e)*a)*(1-cos(f*x+e))^2*(a+b)^(3/2)*a-4*ln(4*((b+a*cos(f*x+e)^2 
)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2) 
/(1+cos(f*x+e))^2)^(1/2)+4*cos(f*x+e)*a)*(1-cos(f*x+e))^2*(a+b)^(3/2)*b+5* 
ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+ 
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+cos(f*x+e)*a+b)/(c 
os(f*x+e)-1))*(1-cos(f*x+e))^2*a^(3/2)*b-5*ln(2/(a+b)^(1/2)*((a+b)^(1/2)*( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+((b+a*cos(f*x+e)^2)/ 
(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/(1+cos(f*x+e)))*(1-cos 
(f*x+e))^2*a^(3/2)*b+3*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e 
))^2)^(1/2)*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (98) = 196\).

Time = 0.42 (sec) , antiderivative size = 1550, normalized size of antiderivative = 13.36 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/8*(4*(a^2 + a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + 
e)^2 + ((a^2 + 2*a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sqrt(a)*lo 
g(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x 
+ e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 - 8*(16*a^3*cos(f*x + e)^8 + 24*a^2 
*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)* 
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + ((2*a^2 + 3*a*b)*cos(f*x + 
e)^2 - 2*a^2 - 3*a*b)*sqrt(a + b)*log(2*((8*a^2 + 8*a*b + b^2)*cos(f*x + e 
)^4 + 2*(4*a*b + 3*b^2)*cos(f*x + e)^2 + b^2 + 4*((2*a + b)*cos(f*x + e)^4 
 + b*cos(f*x + e)^2)*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^ 
2))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)))/((a^3 + 2*a^2*b + a*b^2)*f*c 
os(f*x + e)^2 - (a^3 + 2*a^2*b + a*b^2)*f), 1/8*(4*(a^2 + a*b)*sqrt((a*cos 
(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)^2 - 2*((2*a^2 + 3*a*b)*cos(f 
*x + e)^2 - 2*a^2 - 3*a*b)*sqrt(-a - b)*arctan(1/2*((2*a + b)*cos(f*x + e) 
^2 + b)*sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^2 + a 
*b)*cos(f*x + e)^2 + a*b + b^2)) + ((a^2 + 2*a*b + b^2)*cos(f*x + e)^2 - a 
^2 - 2*a*b - b^2)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + 
 e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 - 8*(16 
*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + 
b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)))/ 
((a^3 + 2*a^2*b + a*b^2)*f*cos(f*x + e)^2 - (a^3 + 2*a^2*b + a*b^2)*f),...
 

Sympy [F]

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

integrate(cot(f*x+e)**3/(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Integral(cot(e + f*x)**3/sqrt(a + b*sec(e + f*x)**2), x)
 

Maxima [F]

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

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

Output:

integrate(cot(f*x + e)^3/sqrt(b*sec(f*x + e)^2 + a), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 595 vs. \(2 (98) = 196\).

Time = 0.71 (sec) , antiderivative size = 595, normalized size of antiderivative = 5.13 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/8*(16*arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f 
*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2* 
b*tan(1/2*f*x + 1/2*e)^2 + a + b) + sqrt(a + b))/sqrt(-a))/sqrt(-a) - 4*(2 
*a + 3*b)*arctan(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x 
 + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b* 
tan(1/2*f*x + 1/2*e)^2 + a + b))/sqrt(-a - b))/((a + b)*sqrt(-a - b)) + 2* 
(2*a + 3*b)*log(abs(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2* 
f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2 
*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*(a + b) + sqrt(a + b)*(a - b)))/(a + b 
)^(3/2) + sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*t 
an(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)/(a + b) - 2*(( 
sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan 
(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e 
)^2 + a + b))*(a - b) - (a + b)^(3/2))/(((sqrt(a + b)*tan(1/2*f*x + 1/2*e) 
^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/ 
2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2 - a - b)*(a + b) 
))/(f*sgn(cos(f*x + e)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(e + f*x)**2*b + a)*cot(e + f*x)**3)/(sec(e + f*x)**2*b + a), 
x)