\(\int \frac {\cot ^3(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [419]

Optimal result

Integrand size = 25, antiderivative size = 153 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} f}-\frac {(a-2 b) b}{2 a (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}} \]

[Out]

-arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/a^(3/2)/f+1/2*(2*a+5*b)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/(a+b)^(1/2
))/(a+b)^(5/2)/f-1/2*(a-2*b)*b/a/(a+b)^2/f/(a+b*sec(f*x+e)^2)^(1/2)-1/2*cot(f*x+e)^2/(a+b)/f/(a+b*sec(f*x+e)^2
)^(1/2)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4224, 457, 105, 157, 162, 65, 214} \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f (a+b)^{5/2}}-\frac {b (a-2 b)}{2 a f (a+b)^2 \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 f (a+b) \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

-(ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(a^(3/2)*f)) + ((2*a + 5*b)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/S
qrt[a + b]])/(2*(a + b)^(5/2)*f) - ((a - 2*b)*b)/(2*a*(a + b)^2*f*Sqrt[a + b*Sec[e + f*x]^2]) - Cot[e + f*x]^2
/(2*(a + b)*f*Sqrt[a + b*Sec[e + f*x]^2])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] + Dist[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 157

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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] + Dist[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*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(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 214

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 457

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

Rule 4224

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]}, Dist[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] || IntegersQ[2*n, p])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x \left (-1+x^2\right )^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{(-1+x)^2 x (a+b x)^{3/2}} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = -\frac {\cot ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\text {Subst}\left (\int \frac {a+b+\frac {3 b x}{2}}{(-1+x) x (a+b x)^{3/2}} \, dx,x,\sec ^2(e+f x)\right )}{2 (a+b) f} \\ & = -\frac {(a-2 b) b}{2 a (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{2} (a+b)^2-\frac {1}{4} (a-2 b) b x}{(-1+x) x \sqrt {a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{a (a+b)^2 f} \\ & = -\frac {(a-2 b) b}{2 a (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 a f}-\frac {(2 a+5 b) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{4 (a+b)^2 f} \\ & = -\frac {(a-2 b) b}{2 a (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sec ^2(e+f x)}\right )}{a b f}-\frac {(2 a+5 b) \text {Subst}\left (\int \frac {1}{-1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sec ^2(e+f x)}\right )}{2 b (a+b)^2 f} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} f}-\frac {(a-2 b) b}{2 a (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(13609\) vs. \(2(131)=262\).

Time = 1.16 (sec) , antiderivative size = 13610, normalized size of antiderivative = 88.95

method	result	size
default	\(\text {Expression too large to display}\)	\(13610\)

[In]

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

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 540 vs. \(2 (131) = 262\).

Time = 1.49 (sec) , antiderivative size = 2347, normalized size of antiderivative = 15.34 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/8*(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3*a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b
- 2*a*b^3 - b^4)*cos(f*x + e)^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)) + ((2*a^4 + 5*a^3*b
)*cos(f*x + e)^4 - 2*a^3*b - 5*a^2*b^2 - (2*a^4 + 3*a^3*b - 5*a^2*b^2)*cos(f*x + e)^2)*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*co
s(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)
) + 4*((a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^4 + (a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)^2)*sqrt((
a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 - (a^6 + 2*a^5*
b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f), -1/8*(2*((2*a^4 + 5*
a^3*b)*cos(f*x + e)^4 - 2*a^3*b - 5*a^2*b^2 - (2*a^4 + 3*a^3*b - 5*a^2*b^2)*cos(f*x + e)^2)*sqrt(-a - b)*arcta
n(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^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3*a^2*b^2 - 3*a*b^3
- b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*cos(f*x + e)^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
)) - 4*((a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^4 + (a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)^2)*sqrt(
(a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 - (a^6 + 2*a^5
*b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f), 1/8*(2*((a^4 + 3*a^
3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3*a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*
cos(f*x + e)^2)*sqrt(-a)*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f
*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) + ((2*a^4 + 5*a^3*b)*c
os(f*x + e)^4 - 2*a^3*b - 5*a^2*b^2 - (2*a^4 + 3*a^3*b - 5*a^2*b^2)*cos(f*x + e)^2)*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)) +
4*((a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^4 + (a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*c
os(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 - (a^6 + 2*a^5*b -
2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f), 1/4*(((a^4 + 3*a^3*b +
3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3*a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*cos(f*
x + e)^2)*sqrt(-a)*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e
)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) - ((2*a^4 + 5*a^3*b)*cos(f*x
+ e)^4 - 2*a^3*b - 5*a^2*b^2 - (2*a^4 + 3*a^3*b - 5*a^2*b^2)*cos(f*x + e)^2)*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)) + 2*((a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^4 + (a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)
^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 - (a^
6 + 2*a^5*b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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