3.356 \(\int \frac{\cot ^2(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=186 \[ -\frac{(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a^3 f (a-b)^2}-\frac{b (7 a-4 b) \cot (e+f x)}{3 a^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac{b \cot (e+f x)}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
[Out]
-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) - (b*Cot[e + f*x])/(3*a*(a
- b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - ((7*a - 4*b)*b*Cot[e + f*x])/(3*a^2*(a - b)^2*f*Sqrt[a + b*Tan[e + f*x]
^2]) - ((a - 4*b)*(3*a - 2*b)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(3*a^3*(a - b)^2*f)
________________________________________________________________________________________
Rubi [A] time = 0.278731, antiderivative size = 186, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used =
{3670, 472, 579, 583, 12, 377, 203} \[ -\frac{(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a^3 f (a-b)^2}-\frac{b (7 a-4 b) \cot (e+f x)}{3 a^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac{b \cot (e+f x)}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]
[Out]
-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) - (b*Cot[e + f*x])/(3*a*(a
- b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - ((7*a - 4*b)*b*Cot[e + f*x])/(3*a^2*(a - b)^2*f*Sqrt[a + b*Tan[e + f*x]
^2]) - ((a - 4*b)*(3*a - 2*b)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(3*a^3*(a - b)^2*f)
Rule 3670
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))
Rule 472
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 579
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]
Rule 583
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{\cot ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{3 a-4 b-4 b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a (a-b) f}\\ &=-\frac{b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{(a-4 b) (3 a-2 b)-2 (7 a-4 b) b x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 (a-b)^2 f}\\ &=-\frac{b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{3 a^3}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^3 (a-b)^2 f}\\ &=-\frac{b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}\\ &=-\frac{b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-(-a+b) x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{(a-b)^2 f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac{b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f}\\ \end{align*}
Mathematica [C] time = 15.9142, size = 1890, normalized size = 10.16 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]
[Out]
-((Cos[e + f*x]^2*Cot[e + f*x]*((20*a*Csc[e + f*x]^2)/(3*(a - b)) - (5*a^2*Csc[e + f*x]^4)/(a - b)^2 + (40*b*S
ec[e + f*x]^2)/(a - b) - (30*a*b*Csc[e + f*x]^2*Sec[e + f*x]^2)/(a - b)^2 - (40*b^2*Sec[e + f*x]^4)/(a - b)^2
+ (92*(a - b)*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(105*a) + (24*(a - b)*H
ypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(35*a) + (16*(a - b)*Hyperge
ometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(105*a) + (160*b^2*Sec[e + f
*x]^2*Tan[e + f*x]^2)/(3*a*(a - b)) + (124*(a - b)*b*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*
Sin[e + f*x]^2*Tan[e + f*x]^2)/(35*a^2) + (16*(a - b)*b*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e
+ f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(7*a^2) + (16*(a - b)*b*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2
}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(35*a^2) + (64*b^3*Sec[e + f*x]^2*Tan[e + f*x]^4
)/(3*a^2*(a - b)) + (152*(a - b)*b^2*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*T
an[e + f*x]^4)/(35*a^3) + (88*(a - b)*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*S
in[e + f*x]^2*Tan[e + f*x]^4)/(35*a^3) + (16*(a - b)*b^2*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)
*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^4)/(35*a^3) + (176*(a - b)*b^3*Hypergeometric2F1[2, 2, 9/2, ((
a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^6)/(105*a^4) + (32*(a - b)*b^3*HypergeometricPFQ[{2, 2,
2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^6)/(35*a^4) + (16*(a - b)*b^3*Hypergeome
tricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^6)/(105*a^4) + (5*A
rcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]])/((((a - b)*Sin[e + f*x]^2)/a)^(5/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e
+ f*x]^2))/a]) + (30*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^2)/(a*(((a - b)*Sin[e + f*x]^2)/
a)^(5/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (40*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*T
an[e + f*x]^4)/(a^2*(((a - b)*Sin[e + f*x]^2)/a)^(5/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (16*
b^3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^6)/(a^3*(((a - b)*Sin[e + f*x]^2)/a)^(5/2)*Sqrt[(Cos
[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (5*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]])/Sqrt[((a - b)*Cos[e + f
*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2] - (10*a*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Csc[e + f*x
]^2)/((a - b)*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]) - (60*b*ArcSin[Sqrt[((
a - b)*Sin[e + f*x]^2)/a]]*Sec[e + f*x]^2)/((a - b)*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e +
f*x]^2))/a^2]) + (30*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^2)/(a*Sqrt[((a - b)*Cos[e + f*x]
^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]) - (80*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sec[e + f*x
]^2*Tan[e + f*x]^2)/(a*(a - b)*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]) + (40
*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^4)/(a^2*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2
*(a + b*Tan[e + f*x]^2))/a^2]) - (32*b^3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sec[e + f*x]^2*Tan[e + f*x]^
4)/(a^2*(a - b)*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]) + (16*b^3*ArcSin[Sqr
t[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^6)/(a^3*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e +
f*x]^2))/a^2]) - (16*b^3*(Tan[e + f*x] + Tan[e + f*x]^3)^2)/(a*(a - b)^2)))/(a^2*f*Sqrt[a + b*Tan[e + f*x]^2]
*(1 + (b*Tan[e + f*x]^2)/a)))
________________________________________________________________________________________
Maple [F] time = 0.219, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cot \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x)
[Out]
int(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x)
________________________________________________________________________________________
Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [B] time = 3.00555, size = 1693, normalized size = 9.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")
[Out]
[-1/12*(3*(a^3*b^2*tan(f*x + e)^5 + 2*a^4*b*tan(f*x + e)^3 + a^5*tan(f*x + e))*sqrt(-a + b)*log(-((a^2 - 8*a*b
+ 8*b^2)*tan(f*x + e)^4 - 2*(3*a^2 - 4*a*b)*tan(f*x + e)^2 + a^2 + 4*((a - 2*b)*tan(f*x + e)^3 - a*tan(f*x +
e))*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b))/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) + 4*(3*a^5 - 9*a^4*b + 9
*a^3*b^2 - 3*a^2*b^3 + (3*a^3*b^2 - 17*a^2*b^3 + 22*a*b^4 - 8*b^5)*tan(f*x + e)^4 + 3*(2*a^4*b - 9*a^3*b^2 + 1
1*a^2*b^3 - 4*a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*
f*tan(f*x + e)^5 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^3 + (a^8 - 3*a^7*b + 3*a^6*b^2 -
a^5*b^3)*f*tan(f*x + e)), -1/6*(3*(a^3*b^2*tan(f*x + e)^5 + 2*a^4*b*tan(f*x + e)^3 + a^5*tan(f*x + e))*sqrt(a
- b)*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b)*tan(f*x + e)/((a - 2*b)*tan(f*x + e)^2 - a)) + 2*(3*a^5
- 9*a^4*b + 9*a^3*b^2 - 3*a^2*b^3 + (3*a^3*b^2 - 17*a^2*b^3 + 22*a*b^4 - 8*b^5)*tan(f*x + e)^4 + 3*(2*a^4*b -
9*a^3*b^2 + 11*a^2*b^3 - 4*a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b
^4 - a^3*b^5)*f*tan(f*x + e)^5 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^3 + (a^8 - 3*a^7*b
+ 3*a^6*b^2 - a^5*b^3)*f*tan(f*x + e))]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{2}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)**2/(a+b*tan(f*x+e)**2)**(5/2),x)
[Out]
Integral(cot(e + f*x)**2/(a + b*tan(e + f*x)**2)**(5/2), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")
[Out]
integrate(cot(f*x + e)^2/(b*tan(f*x + e)^2 + a)^(5/2), x)