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

Optimal. Leaf size=189 \[ -\frac {b (9 a-5 b) \cot (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac {b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{7/2} f (a-b)^3}-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 f (a-b)^2}-\frac {b \cot (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {x}{(a-b)^3} \]

[Out]

-x/(a-b)^3+1/8*b^(3/2)*(35*a^2-42*a*b+15*b^2)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))/a^(7/2)/(a-b)^3/f-1/8*(8*a^2-
27*a*b+15*b^2)*cot(f*x+e)/a^3/(a-b)^2/f-1/4*b*cot(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^2-1/8*(9*a-5*b)*b*cot(f*
x+e)/a^2/(a-b)^2/f/(a+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.29, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3670, 472, 579, 583, 522, 203, 205} \[ \frac {b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{7/2} f (a-b)^3}-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 f (a-b)^2}-\frac {b (9 a-5 b) \cot (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {b \cot (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {x}{(a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(a - b)^3) + (b^(3/2)*(35*a^2 - 42*a*b + 15*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(7/2)*(a - b
)^3*f) - ((8*a^2 - 27*a*b + 15*b^2)*Cot[e + f*x])/(8*a^3*(a - b)^2*f) - (b*Cot[e + f*x])/(4*a*(a - b)*f*(a + b
*Tan[e + f*x]^2)^2) - ((9*a - 5*b)*b*Cot[e + f*x])/(8*a^2*(a - b)^2*f*(a + b*Tan[e + f*x]^2))

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]

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 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 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 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]))

Rubi steps

\begin {align*} \int \frac {\cot ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right ) \left (a+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 \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {4 a-5 b-5 b x^2}{x^2 \left (1+x^2\right ) \left (a+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 \tan ^2(e+f x)\right )^2}-\frac {(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {8 a^2-27 a b+15 b^2-3 (9 a-5 b) b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a-b)^2 f}\\ &=-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac {b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {8 a^3+8 a^2 b-27 a b^2+15 b^3+b \left (8 a^2-27 a b+15 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a-b)^2 f}\\ &=-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac {b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+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-b)^3 f}+\frac {\left (b^2 \left (35 a^2-42 a b+15 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a-b)^3 f}\\ &=-\frac {x}{(a-b)^3}+\frac {b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{7/2} (a-b)^3 f}-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac {b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 2.40, size = 174, normalized size = 0.92 \[ \frac {\frac {b^2 (13 a-7 b) \sin (2 (e+f x))}{a^3 (a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)}-\frac {8 \cot (e+f x)}{a^3}-\frac {4 b^3 \sin (2 (e+f x))}{a^2 (a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)^2}+\frac {b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} (a-b)^3}+\frac {8 (e+f x)}{(b-a)^3}}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*(e + f*x))/(-a + b)^3 + (b^(3/2)*(35*a^2 - 42*a*b + 15*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(7/
2)*(a - b)^3) - (8*Cot[e + f*x])/a^3 - (4*b^3*Sin[2*(e + f*x)])/(a^2*(a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x
)])^2) + ((13*a - 7*b)*b^2*Sin[2*(e + f*x)])/(a^3*(a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x)])))/(8*f)

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fricas [B]  time = 0.51, size = 881, normalized size = 4.66 \[ \left [-\frac {32 \, a^{3} b^{2} f x \tan \left (f x + e\right )^{5} + 64 \, a^{4} b f x \tan \left (f x + e\right )^{3} + 32 \, a^{5} f x \tan \left (f x + e\right ) + 32 \, a^{5} - 96 \, a^{4} b + 96 \, a^{3} b^{2} - 32 \, a^{2} b^{3} + 4 \, {\left (8 \, a^{3} b^{2} - 35 \, a^{2} b^{3} + 42 \, a b^{4} - 15 \, b^{5}\right )} \tan \left (f x + e\right )^{4} + 4 \, {\left (16 \, a^{4} b - 61 \, a^{3} b^{2} + 70 \, a^{2} b^{3} - 25 \, a b^{4}\right )} \tan \left (f x + e\right )^{2} + {\left ({\left (35 \, a^{2} b^{3} - 42 \, a b^{4} + 15 \, b^{5}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (35 \, a^{3} b^{2} - 42 \, a^{2} b^{3} + 15 \, a b^{4}\right )} \tan \left (f x + e\right )^{3} + {\left (35 \, a^{4} b - 42 \, a^{3} b^{2} + 15 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left (a b \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right )}{32 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} - a^{3} b^{5}\right )} f \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{7} b - 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} - a^{4} b^{4}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} f \tan \left (f x + e\right )\right )}}, -\frac {16 \, a^{3} b^{2} f x \tan \left (f x + e\right )^{5} + 32 \, a^{4} b f x \tan \left (f x + e\right )^{3} + 16 \, a^{5} f x \tan \left (f x + e\right ) + 16 \, a^{5} - 48 \, a^{4} b + 48 \, a^{3} b^{2} - 16 \, a^{2} b^{3} + 2 \, {\left (8 \, a^{3} b^{2} - 35 \, a^{2} b^{3} + 42 \, a b^{4} - 15 \, b^{5}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (16 \, a^{4} b - 61 \, a^{3} b^{2} + 70 \, a^{2} b^{3} - 25 \, a b^{4}\right )} \tan \left (f x + e\right )^{2} - {\left ({\left (35 \, a^{2} b^{3} - 42 \, a b^{4} + 15 \, b^{5}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (35 \, a^{3} b^{2} - 42 \, a^{2} b^{3} + 15 \, a b^{4}\right )} \tan \left (f x + e\right )^{3} + {\left (35 \, a^{4} b - 42 \, a^{3} b^{2} + 15 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {b}{a}}}{2 \, b \tan \left (f x + e\right )}\right )}{16 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} - a^{3} b^{5}\right )} f \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{7} b - 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} - a^{4} b^{4}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} f \tan \left (f x + e\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*(32*a^3*b^2*f*x*tan(f*x + e)^5 + 64*a^4*b*f*x*tan(f*x + e)^3 + 32*a^5*f*x*tan(f*x + e) + 32*a^5 - 96*a^
4*b + 96*a^3*b^2 - 32*a^2*b^3 + 4*(8*a^3*b^2 - 35*a^2*b^3 + 42*a*b^4 - 15*b^5)*tan(f*x + e)^4 + 4*(16*a^4*b -
61*a^3*b^2 + 70*a^2*b^3 - 25*a*b^4)*tan(f*x + e)^2 + ((35*a^2*b^3 - 42*a*b^4 + 15*b^5)*tan(f*x + e)^5 + 2*(35*
a^3*b^2 - 42*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^3 + (35*a^4*b - 42*a^3*b^2 + 15*a^2*b^3)*tan(f*x + e))*sqrt(-b/a
)*log((b^2*tan(f*x + e)^4 - 6*a*b*tan(f*x + e)^2 + a^2 - 4*(a*b*tan(f*x + e)^3 - a^2*tan(f*x + e))*sqrt(-b/a))
/(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)))/((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)), -1/16*(16*a^3*b^2*f*x*tan(f*x + e)^5 + 32*a^4*b*f*x*tan(f*x + e)^3 + 16*a^5*f*x*tan(f*x + e) +
16*a^5 - 48*a^4*b + 48*a^3*b^2 - 16*a^2*b^3 + 2*(8*a^3*b^2 - 35*a^2*b^3 + 42*a*b^4 - 15*b^5)*tan(f*x + e)^4 +
2*(16*a^4*b - 61*a^3*b^2 + 70*a^2*b^3 - 25*a*b^4)*tan(f*x + e)^2 - ((35*a^2*b^3 - 42*a*b^4 + 15*b^5)*tan(f*x +
e)^5 + 2*(35*a^3*b^2 - 42*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^3 + (35*a^4*b - 42*a^3*b^2 + 15*a^2*b^3)*tan(f*x +
e))*sqrt(b/a)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(b/a)/(b*tan(f*x + e))))/((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))]

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giac [A]  time = 5.48, size = 231, normalized size = 1.22 \[ \frac {\frac {{\left (35 \, a^{2} b^{2} - 42 \, a b^{3} + 15 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} \sqrt {a b}} - \frac {8 \, {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {11 \, a b^{3} \tan \left (f x + e\right )^{3} - 7 \, b^{4} \tan \left (f x + e\right )^{3} + 13 \, a^{2} b^{2} \tan \left (f x + e\right ) - 9 \, a b^{3} \tan \left (f x + e\right )}{{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2}} - \frac {8}{a^{3} \tan \left (f x + e\right )}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((35*a^2*b^2 - 42*a*b^3 + 15*b^4)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))
/((a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*sqrt(a*b)) - 8*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (11*a*b^3*t
an(f*x + e)^3 - 7*b^4*tan(f*x + e)^3 + 13*a^2*b^2*tan(f*x + e) - 9*a*b^3*tan(f*x + e))/((a^5 - 2*a^4*b + a^3*b
^2)*(b*tan(f*x + e)^2 + a)^2) - 8/(a^3*tan(f*x + e)))/f

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maple [B]  time = 0.98, size = 379, normalized size = 2.01 \[ \frac {11 b^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} a}-\frac {9 b^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{4 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} a^{2}}+\frac {7 b^{5} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{3} \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {13 \tan \left (f x +e \right ) b^{2}}{8 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {11 b^{3} \tan \left (f x +e \right )}{4 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} a}+\frac {9 b^{4} \tan \left (f x +e \right )}{8 f \,a^{2} \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {35 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right ) b^{2}}{8 f \left (a -b \right )^{3} a \sqrt {a b}}-\frac {21 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{4 f \left (a -b \right )^{3} a^{2} \sqrt {a b}}+\frac {15 b^{4} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{8 f \,a^{3} \left (a -b \right )^{3} \sqrt {a b}}-\frac {1}{f \,a^{3} \tan \left (f x +e \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/8/f/(a-b)^3/(a+b*tan(f*x+e)^2)^2*b^3/a*tan(f*x+e)^3-9/4/f*b^4/(a-b)^3/(a+b*tan(f*x+e)^2)^2/a^2*tan(f*x+e)^3
+7/8/f*b^5/a^3/(a-b)^3/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3+13/8/f/(a-b)^3/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)*b^2-11
/4/f*b^3/(a-b)^3/(a+b*tan(f*x+e)^2)^2/a*tan(f*x+e)+9/8/f*b^4/a^2/(a-b)^3/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)+35/8/
f/(a-b)^3/a/(a*b)^(1/2)*arctan(tan(f*x+e)*b/(a*b)^(1/2))*b^2-21/4/f*b^3/(a-b)^3/a^2/(a*b)^(1/2)*arctan(tan(f*x
+e)*b/(a*b)^(1/2))+15/8/f*b^4/a^3/(a-b)^3/(a*b)^(1/2)*arctan(tan(f*x+e)*b/(a*b)^(1/2))-1/f/a^3/tan(f*x+e)-1/f/
(a-b)^3*arctan(tan(f*x+e))

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maxima [A]  time = 0.82, size = 275, normalized size = 1.46 \[ \frac {\frac {{\left (35 \, a^{2} b^{2} - 42 \, a b^{3} + 15 \, b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} \sqrt {a b}} - \frac {{\left (8 \, a^{2} b^{2} - 27 \, a b^{3} + 15 \, b^{4}\right )} \tan \left (f x + e\right )^{4} + 8 \, a^{4} - 16 \, a^{3} b + 8 \, a^{2} b^{2} + {\left (16 \, a^{3} b - 45 \, a^{2} b^{2} + 25 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{5} b^{2} - 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{6} b - 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{7} - 2 \, a^{6} b + a^{5} b^{2}\right )} \tan \left (f x + e\right )} - \frac {8 \, {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((35*a^2*b^2 - 42*a*b^3 + 15*b^4)*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*
sqrt(a*b)) - ((8*a^2*b^2 - 27*a*b^3 + 15*b^4)*tan(f*x + e)^4 + 8*a^4 - 16*a^3*b + 8*a^2*b^2 + (16*a^3*b - 45*a
^2*b^2 + 25*a*b^3)*tan(f*x + e)^2)/((a^5*b^2 - 2*a^4*b^3 + a^3*b^4)*tan(f*x + e)^5 + 2*(a^6*b - 2*a^5*b^2 + a^
4*b^3)*tan(f*x + e)^3 + (a^7 - 2*a^6*b + a^5*b^2)*tan(f*x + e)) - 8*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3))
/f

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mupad [B]  time = 15.05, size = 915, normalized size = 4.84 \[ -\frac {2\,\mathrm {atan}\left (\frac {2\,\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {262144\,a^{28}\,b^2-2883584\,a^{27}\,b^3+14155776\,a^{26}\,b^4-40370176\,a^{25}\,b^5+72089600\,a^{24}\,b^6-77856768\,a^{23}\,b^7+34603008\,a^{22}\,b^8+34603008\,a^{21}\,b^9-77856768\,a^{20}\,b^{10}[phone]\,a^{19}\,b^{11}-40370176\,a^{18}\,b^{12}[phone]\,a^{17}\,b^{13}-2883584\,a^{16}\,b^{14}+262144\,a^{15}\,b^{15}}{{\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}^2}-230400\,a^9\,b^{15}+2672640\,a^{10}\,b^{14}-14078976\,a^{11}\,b^{13}[phone]\,a^{12}\,b^{12}-91801600\,a^{13}\,b^{11}[phone]\,a^{14}\,b^{10}-130287616\,a^{15}\,b^9+89219072\,a^{16}\,b^8-40743936\,a^{17}\,b^7+11847680\,a^{18}\,b^6-2237440\,a^{19}\,b^5+393216\,a^{20}\,b^4-65536\,a^{21}\,b^3\right )}{\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )\,\left (\frac {2\,\left (131072\,a^{25}\,b^2-1179648\,a^{24}\,b^3+4145152\,a^{23}\,b^4-5160960\,a^{22}\,b^5-10567680\,a^{21}\,b^6+58638336\,a^{20}\,b^7-127893504\,a^{19}\,b^8+174882816\,a^{18}\,b^9-164659200\,a^{17}\,b^{10}[phone]\,a^{16}\,b^{11}-50577408\,a^{15}\,b^{12}[phone]\,a^{14}\,b^{13}-2899968\,a^{13}\,b^{14}+245760\,a^{12}\,b^{15}\right )}{{\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}^2}+230400\,a^9\,b^{12}-1981440\,a^{10}\,b^{11}+7443456\,a^{11}\,b^{10}-15879168\,a^{12}\,b^9+20933632\,a^{13}\,b^8-17363968\,a^{14}\,b^7+8788992\,a^{15}\,b^6-2458624\,a^{16}\,b^5+286720\,a^{17}\,b^4\right )}\right )}{f\,\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}-\frac {\frac {1}{a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (8\,a^2\,b^2-27\,a\,b^3+15\,b^4\right )}{8\,a^3\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (16\,a^2\,b-45\,a\,b^2+25\,b^3\right )}{8\,a^2\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left (a^2\,\mathrm {tan}\left (e+f\,x\right )+2\,a\,b\,{\mathrm {tan}\left (e+f\,x\right )}^3+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )}+\frac {\mathrm {atan}\left (\frac {b^5\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,225{}\mathrm {i}-a\,b^4\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,1260{}\mathrm {i}+a^4\,b\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,1225{}\mathrm {i}+a^{14}\,b\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^7\,b^3}\,64{}\mathrm {i}+a^2\,b^3\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,2814{}\mathrm {i}-a^3\,b^2\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,2940{}\mathrm {i}}{-64\,a^{18}\,b^2+1225\,a^{15}\,b^5-2940\,a^{14}\,b^6+2814\,a^{13}\,b^7-1260\,a^{12}\,b^8+225\,a^{11}\,b^9}\right )\,\sqrt {-a^7\,b^3}\,\left (35\,a^2-42\,a\,b+15\,b^2\right )\,1{}\mathrm {i}}{8\,f\,\left (-a^{10}+3\,a^9\,b-3\,a^8\,b^2+a^7\,b^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((b^5*tan(e + f*x)*(-a^7*b^3)^(3/2)*225i - a*b^4*tan(e + f*x)*(-a^7*b^3)^(3/2)*1260i + a^4*b*tan(e + f*x)
*(-a^7*b^3)^(3/2)*1225i + a^14*b*tan(e + f*x)*(-a^7*b^3)^(1/2)*64i + a^2*b^3*tan(e + f*x)*(-a^7*b^3)^(3/2)*281
4i - a^3*b^2*tan(e + f*x)*(-a^7*b^3)^(3/2)*2940i)/(225*a^11*b^9 - 1260*a^12*b^8 + 2814*a^13*b^7 - 2940*a^14*b^
6 + 1225*a^15*b^5 - 64*a^18*b^2))*(-a^7*b^3)^(1/2)*(35*a^2 - 42*a*b + 15*b^2)*1i)/(8*f*(3*a^9*b - a^10 + a^7*b
^3 - 3*a^8*b^2)) - (1/a + (tan(e + f*x)^4*(15*b^4 - 27*a*b^3 + 8*a^2*b^2))/(8*a^3*(a^2 - 2*a*b + b^2)) + (tan(
e + f*x)^2*(16*a^2*b - 45*a*b^2 + 25*b^3))/(8*a^2*(a^2 - 2*a*b + b^2)))/(f*(a^2*tan(e + f*x) + b^2*tan(e + f*x
)^5 + 2*a*b*tan(e + f*x)^3)) - (2*atan((2*tan(e + f*x)*((262144*a^15*b^15 - 2883584*a^16*b^14 + 14155776*a^17*
b^13 - 40370176*a^18*b^12 + 72089600*a^19*b^11 - 77856768*a^20*b^10 + 34603008*a^21*b^9 + 34603008*a^22*b^8 -
77856768*a^23*b^7 + 72089600*a^24*b^6 - 40370176*a^25*b^5 + 14155776*a^26*b^4 - 2883584*a^27*b^3 + 262144*a^28
*b^2)/(6*a*b^2 - 6*a^2*b + 2*a^3 - 2*b^3)^2 - 230400*a^9*b^15 + 2672640*a^10*b^14 - 14078976*a^11*b^13 + 44261
376*a^12*b^12 - 91801600*a^13*b^11 + 131051520*a^14*b^10 - 130287616*a^15*b^9 + 89219072*a^16*b^8 - 40743936*a
^17*b^7 + 11847680*a^18*b^6 - 2237440*a^19*b^5 + 393216*a^20*b^4 - 65536*a^21*b^3))/((6*a*b^2 - 6*a^2*b + 2*a^
3 - 2*b^3)*((2*(245760*a^12*b^15 - 2899968*a^13*b^14 + 15613952*a^14*b^13 - 50577408*a^15*b^12 + 109281280*a^1
6*b^11 - 164659200*a^17*b^10 + 174882816*a^18*b^9 - 127893504*a^19*b^8 + 58638336*a^20*b^7 - 10567680*a^21*b^6
- 5160960*a^22*b^5 + 4145152*a^23*b^4 - 1179648*a^24*b^3 + 131072*a^25*b^2))/(6*a*b^2 - 6*a^2*b + 2*a^3 - 2*b
^3)^2 + 230400*a^9*b^12 - 1981440*a^10*b^11 + 7443456*a^11*b^10 - 15879168*a^12*b^9 + 20933632*a^13*b^8 - 1736
3968*a^14*b^7 + 8788992*a^15*b^6 - 2458624*a^16*b^5 + 286720*a^17*b^4))))/(f*(6*a*b^2 - 6*a^2*b + 2*a^3 - 2*b^
3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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