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

Optimal. Leaf size=181 \[ \frac{b^{3/2} \left (35 a^2+28 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 f (a+b)^{7/2}}-\frac{\left (8 a^2-11 a b-4 b^2\right ) \cot (e+f x)}{8 a^2 f (a+b)^3}-\frac{b (9 a+4 b) \cot (e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{x}{a^3}-\frac{b \cot (e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

-(x/a^3) + (b^(3/2)*(35*a^2 + 28*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^3*(a + b)^(7/2)
*f) - ((8*a^2 - 11*a*b - 4*b^2)*Cot[e + f*x])/(8*a^2*(a + b)^3*f) - (b*Cot[e + f*x])/(4*a*(a + b)*f*(a + b + b
*Tan[e + f*x]^2)^2) - (b*(9*a + 4*b)*Cot[e + f*x])/(8*a^2*(a + b)^2*f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.380387, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4141, 1975, 472, 579, 583, 522, 203, 205} \[ \frac{b^{3/2} \left (35 a^2+28 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 f (a+b)^{7/2}}-\frac{\left (8 a^2-11 a b-4 b^2\right ) \cot (e+f x)}{8 a^2 f (a+b)^3}-\frac{b (9 a+4 b) \cot (e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{x}{a^3}-\frac{b \cot (e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^3) + (b^(3/2)*(35*a^2 + 28*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^3*(a + b)^(7/2)
*f) - ((8*a^2 - 11*a*b - 4*b^2)*Cot[e + f*x])/(8*a^2*(a + b)^3*f) - (b*Cot[e + f*x])/(4*a*(a + b)*f*(a + b + b
*Tan[e + f*x]^2)^2) - (b*(9*a + 4*b)*Cot[e + f*x])/(8*a^2*(a + b)^2*f*(a + b + b*Tan[e + f*x]^2))

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

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 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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])

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{4 a-b-5 b x^2}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a (a+b) f}\\ &=-\frac{b \cot (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (9 a+4 b) \cot (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2-11 a b-4 b^2-3 b (9 a+4 b) x^2}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a+b)^2 f}\\ &=-\frac{\left (8 a^2-11 a b-4 b^2\right ) \cot (e+f x)}{8 a^2 (a+b)^3 f}-\frac{b \cot (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (9 a+4 b) \cot (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{8 a^3+32 a^2 b+13 a b^2+4 b^3+b \left (8 a^2-11 a b-4 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a+b)^3 f}\\ &=-\frac{\left (8 a^2-11 a b-4 b^2\right ) \cot (e+f x)}{8 a^2 (a+b)^3 f}-\frac{b \cot (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (9 a+4 b) \cot (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}+\frac{\left (b^2 \left (35 a^2+28 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a+b)^3 f}\\ &=-\frac{x}{a^3}+\frac{b^{3/2} \left (35 a^2+28 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 (a+b)^{7/2} f}-\frac{\left (8 a^2-11 a b-4 b^2\right ) \cot (e+f x)}{8 a^2 (a+b)^3 f}-\frac{b \cot (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (9 a+4 b) \cot (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.99894, size = 2089, normalized size = 11.54 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((35*a^2 + 28*a*b + 8*b^2)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(-(b^2*ArcTan[Sec[f*x]*(Cos[2*e]/(2
*Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]])
)*(-(a*Sin[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Cos[2*e])/(64*a^3*Sqrt[a + b]*f*Sqrt[b*Cos[4*e] - I*b*Sin
[4*e]]) + ((I/64)*b^2*ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2
*e])/(Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Sin[2*e
])/(a^3*Sqrt[a + b]*f*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]])))/((a + b)^3*(a + b*Sec[e + f*x]^2)^3) + ((a + 2*b + a*
Cos[2*e + 2*f*x])*Csc[e]*Csc[e + f*x]*Sec[2*e]*Sec[e + f*x]^6*(8*a^5*f*x*Cos[f*x] + 56*a^4*b*f*x*Cos[f*x] + 18
4*a^3*b^2*f*x*Cos[f*x] + 296*a^2*b^3*f*x*Cos[f*x] + 224*a*b^4*f*x*Cos[f*x] + 64*b^5*f*x*Cos[f*x] - 12*a^5*f*x*
Cos[3*f*x] - 68*a^4*b*f*x*Cos[3*f*x] - 132*a^3*b^2*f*x*Cos[3*f*x] - 108*a^2*b^3*f*x*Cos[3*f*x] - 32*a*b^4*f*x*
Cos[3*f*x] - 8*a^5*f*x*Cos[2*e - f*x] - 56*a^4*b*f*x*Cos[2*e - f*x] - 184*a^3*b^2*f*x*Cos[2*e - f*x] - 296*a^2
*b^3*f*x*Cos[2*e - f*x] - 224*a*b^4*f*x*Cos[2*e - f*x] - 64*b^5*f*x*Cos[2*e - f*x] - 8*a^5*f*x*Cos[2*e + f*x]
- 56*a^4*b*f*x*Cos[2*e + f*x] - 184*a^3*b^2*f*x*Cos[2*e + f*x] - 296*a^2*b^3*f*x*Cos[2*e + f*x] - 224*a*b^4*f*
x*Cos[2*e + f*x] - 64*b^5*f*x*Cos[2*e + f*x] + 8*a^5*f*x*Cos[4*e + f*x] + 56*a^4*b*f*x*Cos[4*e + f*x] + 184*a^
3*b^2*f*x*Cos[4*e + f*x] + 296*a^2*b^3*f*x*Cos[4*e + f*x] + 224*a*b^4*f*x*Cos[4*e + f*x] + 64*b^5*f*x*Cos[4*e
+ f*x] + 12*a^5*f*x*Cos[2*e + 3*f*x] + 68*a^4*b*f*x*Cos[2*e + 3*f*x] + 132*a^3*b^2*f*x*Cos[2*e + 3*f*x] + 108*
a^2*b^3*f*x*Cos[2*e + 3*f*x] + 32*a*b^4*f*x*Cos[2*e + 3*f*x] - 12*a^5*f*x*Cos[4*e + 3*f*x] - 68*a^4*b*f*x*Cos[
4*e + 3*f*x] - 132*a^3*b^2*f*x*Cos[4*e + 3*f*x] - 108*a^2*b^3*f*x*Cos[4*e + 3*f*x] - 32*a*b^4*f*x*Cos[4*e + 3*
f*x] + 12*a^5*f*x*Cos[6*e + 3*f*x] + 68*a^4*b*f*x*Cos[6*e + 3*f*x] + 132*a^3*b^2*f*x*Cos[6*e + 3*f*x] + 108*a^
2*b^3*f*x*Cos[6*e + 3*f*x] + 32*a*b^4*f*x*Cos[6*e + 3*f*x] - 4*a^5*f*x*Cos[2*e + 5*f*x] - 12*a^4*b*f*x*Cos[2*e
+ 5*f*x] - 12*a^3*b^2*f*x*Cos[2*e + 5*f*x] - 4*a^2*b^3*f*x*Cos[2*e + 5*f*x] + 4*a^5*f*x*Cos[4*e + 5*f*x] + 12
*a^4*b*f*x*Cos[4*e + 5*f*x] + 12*a^3*b^2*f*x*Cos[4*e + 5*f*x] + 4*a^2*b^3*f*x*Cos[4*e + 5*f*x] - 4*a^5*f*x*Cos
[6*e + 5*f*x] - 12*a^4*b*f*x*Cos[6*e + 5*f*x] - 12*a^3*b^2*f*x*Cos[6*e + 5*f*x] - 4*a^2*b^3*f*x*Cos[6*e + 5*f*
x] + 4*a^5*f*x*Cos[8*e + 5*f*x] + 12*a^4*b*f*x*Cos[8*e + 5*f*x] + 12*a^3*b^2*f*x*Cos[8*e + 5*f*x] + 4*a^2*b^3*
f*x*Cos[8*e + 5*f*x] - 32*a^5*Sin[f*x] - 64*a^4*b*Sin[f*x] - 30*a^2*b^3*Sin[f*x] - 120*a*b^4*Sin[f*x] - 48*b^5
*Sin[f*x] + 32*a^5*Sin[3*f*x] + 64*a^4*b*Sin[3*f*x] + 26*a^3*b^2*Sin[3*f*x] + 86*a^2*b^3*Sin[3*f*x] + 32*a*b^4
*Sin[3*f*x] - 48*a^5*Sin[2*e - f*x] - 128*a^4*b*Sin[2*e - f*x] - 128*a^3*b^2*Sin[2*e - f*x] - 30*a^2*b^3*Sin[2
*e - f*x] - 120*a*b^4*Sin[2*e - f*x] - 48*b^5*Sin[2*e - f*x] + 48*a^5*Sin[2*e + f*x] + 128*a^4*b*Sin[2*e + f*x
] + 102*a^3*b^2*Sin[2*e + f*x] - 86*a^2*b^3*Sin[2*e + f*x] - 136*a*b^4*Sin[2*e + f*x] - 48*b^5*Sin[2*e + f*x]
- 32*a^5*Sin[4*e + f*x] - 64*a^4*b*Sin[4*e + f*x] + 26*a^3*b^2*Sin[4*e + f*x] + 86*a^2*b^3*Sin[4*e + f*x] + 13
6*a*b^4*Sin[4*e + f*x] + 48*b^5*Sin[4*e + f*x] - 8*a^5*Sin[2*e + 3*f*x] - 26*a^3*b^2*Sin[2*e + 3*f*x] - 86*a^2
*b^3*Sin[2*e + 3*f*x] - 32*a*b^4*Sin[2*e + 3*f*x] + 32*a^5*Sin[4*e + 3*f*x] + 64*a^4*b*Sin[4*e + 3*f*x] - 13*a
^3*b^2*Sin[4*e + 3*f*x] - 36*a^2*b^3*Sin[4*e + 3*f*x] - 16*a*b^4*Sin[4*e + 3*f*x] - 8*a^5*Sin[6*e + 3*f*x] + 1
3*a^3*b^2*Sin[6*e + 3*f*x] + 36*a^2*b^3*Sin[6*e + 3*f*x] + 16*a*b^4*Sin[6*e + 3*f*x] + 8*a^5*Sin[2*e + 5*f*x]
+ 13*a^3*b^2*Sin[2*e + 5*f*x] + 6*a^2*b^3*Sin[2*e + 5*f*x] - 13*a^3*b^2*Sin[4*e + 5*f*x] - 6*a^2*b^3*Sin[4*e +
5*f*x] + 8*a^5*Sin[6*e + 5*f*x]))/(512*a^3*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^3)

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Maple [B]  time = 0.113, size = 337, normalized size = 1.9 \begin{align*} -{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{3}}}-{\frac{1}{f \left ( a+b \right ) ^{3}\tan \left ( fx+e \right ) }}+{\frac{11\,{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a+b \right ) ^{3}a \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{2\,f \left ( a+b \right ) ^{3}{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{13\,{b}^{2}\tan \left ( fx+e \right ) }{8\,f \left ( a+b \right ) ^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{17\,{b}^{3}\tan \left ( fx+e \right ) }{8\,f \left ( a+b \right ) ^{3}a \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{4}\tan \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{3}{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{35\,{b}^{2}}{8\,f \left ( a+b \right ) ^{3}a}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+{\frac{7\,{b}^{3}}{2\,f \left ( a+b \right ) ^{3}{a}^{2}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+{\frac{{b}^{4}}{f \left ( a+b \right ) ^{3}{a}^{3}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f/a^3*arctan(tan(f*x+e))-1/f/(a+b)^3/tan(f*x+e)+11/8/f*b^3/(a+b)^3/a/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3+1/
2/f*b^4/(a+b)^3/a^2/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3+13/8/f*b^2/(a+b)^3/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)+1
7/8/f*b^3/(a+b)^3/a/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)+1/2/f*b^4/(a+b)^3/a^2/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)+
35/8/f*b^2/(a+b)^3/a/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+7/2/f*b^3/(a+b)^3/a^2/((a+b)*b)^(1/2
)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+1/f*b^4/(a+b)^3/a^3/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2)
)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.811265, size = 2372, normalized size = 13.1 \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*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/32*(4*(8*a^5 + 13*a^3*b^2 + 6*a^2*b^3)*cos(f*x + e)^5 + 4*(16*a^4*b - 13*a^3*b^2 + 5*a^2*b^3 + 4*a*b^4)*co
s(f*x + e)^3 - (35*a^2*b^3 + 28*a*b^4 + 8*b^5 + (35*a^4*b + 28*a^3*b^2 + 8*a^2*b^3)*cos(f*x + e)^4 + 2*(35*a^3
*b^2 + 28*a^2*b^3 + 8*a*b^4)*cos(f*x + e)^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3
*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a
+ b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) + 4*(8*a^3*b^2 - 11*
a^2*b^3 - 4*a*b^4)*cos(f*x + e) + 32*((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*f*x*cos(f*x + e)^4 + 2*(a^4*b + 3*
a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*x*cos(f*x + e)^2 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f*x)*sin(f*x + e))/(((
a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x
+ e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)*sin(f*x + e)), -1/16*(2*(8*a^5 + 13*a^3*b^2 + 6*a^2*b^
3)*cos(f*x + e)^5 + 2*(16*a^4*b - 13*a^3*b^2 + 5*a^2*b^3 + 4*a*b^4)*cos(f*x + e)^3 + (35*a^2*b^3 + 28*a*b^4 +
8*b^5 + (35*a^4*b + 28*a^3*b^2 + 8*a^2*b^3)*cos(f*x + e)^4 + 2*(35*a^3*b^2 + 28*a^2*b^3 + 8*a*b^4)*cos(f*x + e
)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*
sin(f*x + e) + 2*(8*a^3*b^2 - 11*a^2*b^3 - 4*a*b^4)*cos(f*x + e) + 16*((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*f
*x*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*x*cos(f*x + e)^2 + (a^3*b^2 + 3*a^2*b^3 + 3*a*
b^4 + b^5)*f*x)*sin(f*x + e))/(((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2
+ 3*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)*sin(f*x + e))]

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Sympy [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*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.44737, size = 347, normalized size = 1.92 \begin{align*} \frac{\frac{{\left (35 \, a^{2} b^{2} + 28 \, a b^{3} + 8 \, b^{4}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \sqrt{a b + b^{2}}} + \frac{11 \, a b^{3} \tan \left (f x + e\right )^{3} + 4 \, b^{4} \tan \left (f x + e\right )^{3} + 13 \, a^{2} b^{2} \tan \left (f x + e\right ) + 17 \, a b^{3} \tan \left (f x + e\right ) + 4 \, b^{4} \tan \left (f x + e\right )}{{\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac{8 \,{\left (f x + e\right )}}{a^{3}} - \frac{8}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((35*a^2*b^2 + 28*a*b^3 + 8*b^4)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b
^2)))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*sqrt(a*b + b^2)) + (11*a*b^3*tan(f*x + e)^3 + 4*b^4*tan(f*x + e)^
3 + 13*a^2*b^2*tan(f*x + e) + 17*a*b^3*tan(f*x + e) + 4*b^4*tan(f*x + e))/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^
3)*(b*tan(f*x + e)^2 + a + b)^2) - 8*(f*x + e)/a^3 - 8/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f*x + e)))/f