\(\int \frac {\cot ^4(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [344]

Optimal result

Integrand size = 25, antiderivative size = 184 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {b \cot ^3(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b) f}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 (a-b) f} \]

[Out]

arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f-b*cot(f*x+e)^3/a/(a-b)/f/(a+b*tan(f*x+e)
^2)^(1/2)+1/3*(3*a-4*b)*(a+2*b)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a^3/(a-b)/f-1/3*(a-4*b)*cot(f*x+e)^3*(a+b*
tan(f*x+e)^2)^(1/2)/a^2/(a-b)/f

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3751, 483, 597, 12, 385, 209} \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 f (a-b)}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f (a-b)}+\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac {b \cot ^3(e+f x)}{a f (a-b) \sqrt {a+b \tan ^2(e+f x)}} \]

[In]

Int[Cot[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(3/2)*f) - (b*Cot[e + f*x]^3)/(a*(a - b
)*f*Sqrt[a + b*Tan[e + f*x]^2]) + ((3*a - 4*b)*(a + 2*b)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(3*a^3*(a -
b)*f) - ((a - 4*b)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(3*a^2*(a - b)*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

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 483

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 597

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 3751

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]))

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 17.45 (sec) , antiderivative size = 802, normalized size of antiderivative = 4.36 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {-\frac {b \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac {4 b \sqrt {1+\cos (2 (e+f x))} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{(a-b) f}+\frac {\sqrt {\frac {a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {(4 a \cos (e+f x)+5 b \cos (e+f x)) \csc (e+f x)}{3 a^3}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2}-\frac {b^3 \sin (2 (e+f x))}{a^3 (a-b) (a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x)))}\right )}{f} \]

[In]

Integrate[Cot[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(-((b*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*
(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(
e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*
x]^4)/(a*(a + b + (a - b)*Cos[2*(e + f*x)]))) - (4*b*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[(a + b + (a - b)*Cos[2*(e
+ f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x
]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((
a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(4*a*Sqrt[1 + Cos[2*(e + f*x
)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]) - (Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*
Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticPi[-(
b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(2
*(a - b)*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])))/Sqrt[a + b + (a - b)*Cos[2*(e +
f*x)]])/((a - b)*f) + (Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(((4*a*C
os[e + f*x] + 5*b*Cos[e + f*x])*Csc[e + f*x])/(3*a^3) - (Cot[e + f*x]*Csc[e + f*x]^2)/(3*a^2) - (b^3*Sin[2*(e
+ f*x)])/(a^3*(a - b)*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)]))))/f

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(972\) vs. \(2(168)=336\).

Time = 6.85 (sec) , antiderivative size = 973, normalized size of antiderivative = 5.29

method	result	size
default	\(\text {Expression too large to display}\)	\(973\)

[In]

int(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/24/f/a^3/(a-b)^(5/2)*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1
)^2*csc(f*x+e)^2+a)*((a-b)^(3/2)*a^3*(-cos(f*x+e)+1)^8*csc(f*x+e)^8-(a-b)^(3/2)*a^2*b*(-cos(f*x+e)+1)^8*csc(f*
x+e)^8-16*(a-b)^(3/2)*a^3*(-cos(f*x+e)+1)^6*csc(f*x+e)^6+16*(a-b)^(3/2)*a*b^2*(-cos(f*x+e)+1)^6*csc(f*x+e)^6+3
0*(a-b)^(3/2)*a^3*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-46*(a-b)^(3/2)*a^2*b*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-64*(a-b)^
(3/2)*a*b^2*(-cos(f*x+e)+1)^4*csc(f*x+e)^4+128*b^3*(-cos(f*x+e)+1)^4*(a-b)^(3/2)*csc(f*x+e)^4+24*arctan(1/2*(a
*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)
/(-cos(f*x+e)+1)*sin(f*x+e)/(a-b)^(1/2))*a^4*(-cos(f*x+e)+1)^3*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x
+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*csc(f*x+e)^3-24*arctan(1/2*(a*(-cos(f*x+e)+1
)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)/(-cos(f*x+e)+1
)*sin(f*x+e)/(a-b)^(1/2))*a^3*(-cos(f*x+e)+1)^3*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*
x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*b*csc(f*x+e)^3-16*(a-b)^(3/2)*a^3*(-cos(f*x+e)+1)^2*csc(f*x
+e)^2+16*(a-b)^(3/2)*a*b^2*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+(a-b)^(3/2)*a^3-(a-b)^(3/2)*a^2*b)/((a*(-cos(f*x+e)+
1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)/((-cos(f*x+e)+1)^2*
csc(f*x+e)^2-1)^2)^(3/2)/((-cos(f*x+e)+1)^2*csc(f*x+e)^2-1)^3/(-cos(f*x+e)+1)^3*sin(f*x+e)^3

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 579, normalized size of antiderivative = 3.15 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a^{3} b \tan \left (f x + e\right )^{5} + a^{4} \tan \left (f x + e\right )^{3}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left ({\left (3 \, a^{3} b - a^{2} b^{2} - 10 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{4} - a^{4} + 2 \, a^{3} b - a^{2} b^{2} + {\left (3 \, a^{4} - 2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left ({\left (a^{5} b - 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \tan \left (f x + e\right )^{5} + {\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} f \tan \left (f x + e\right )^{3}\right )}}, \frac {3 \, {\left (a^{3} b \tan \left (f x + e\right )^{5} + a^{4} \tan \left (f x + e\right )^{3}\right )} \sqrt {a - b} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) + 2 \, {\left ({\left (3 \, a^{3} b - a^{2} b^{2} - 10 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{4} - a^{4} + 2 \, a^{3} b - a^{2} b^{2} + {\left (3 \, a^{4} - 2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{5} b - 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \tan \left (f x + e\right )^{5} + {\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} f \tan \left (f x + e\right )^{3}\right )}}\right ] \]

[In]

integrate(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/12*(3*(a^3*b*tan(f*x + e)^5 + a^4*tan(f*x + e)^3)*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^3*b - a^2*b^2 - 10*a*b^3 + 8*b^4)*tan(f*
x + e)^4 - a^4 + 2*a^3*b - a^2*b^2 + (3*a^4 - 2*a^3*b - 5*a^2*b^2 + 4*a*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x +
e)^2 + a))/((a^5*b - 2*a^4*b^2 + a^3*b^3)*f*tan(f*x + e)^5 + (a^6 - 2*a^5*b + a^4*b^2)*f*tan(f*x + e)^3), 1/6*
(3*(a^3*b*tan(f*x + e)^5 + a^4*tan(f*x + e)^3)*sqrt(a - b)*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b)*ta
n(f*x + e)/((a - 2*b)*tan(f*x + e)^2 - a)) + 2*((3*a^3*b - a^2*b^2 - 10*a*b^3 + 8*b^4)*tan(f*x + e)^4 - a^4 +
2*a^3*b - a^2*b^2 + (3*a^4 - 2*a^3*b - 5*a^2*b^2 + 4*a*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^5*
b - 2*a^4*b^2 + a^3*b^3)*f*tan(f*x + e)^5 + (a^6 - 2*a^5*b + a^4*b^2)*f*tan(f*x + e)^3)]

Sympy [F]

\[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(cot(f*x+e)**4/(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral(cot(e + f*x)**4/(a + b*tan(e + f*x)**2)**(3/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-1)]

Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

[In]

int(cot(e + f*x)^4/(a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

int(cot(e + f*x)^4/(a + b*tan(e + f*x)^2)^(3/2), x)