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

Optimal. Leaf size=128 \[ \frac {b^{3/2} (5 a-3 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{5/2} f (a-b)^2}-\frac {(2 a-3 b) \cot (e+f x)}{2 a^2 f (a-b)}-\frac {b \cot (e+f x)}{2 a f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac {x}{(a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3670, 472, 583, 522, 203, 205} \[ \frac {b^{3/2} (5 a-3 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{5/2} f (a-b)^2}-\frac {(2 a-3 b) \cot (e+f x)}{2 a^2 f (a-b)}-\frac {b \cot (e+f x)}{2 a f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac {x}{(a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(a - b)^2) + ((5*a - 3*b)*b^(3/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*a^(5/2)*(a - b)^2*f) - ((2*a
- 3*b)*Cot[e + f*x])/(2*a^2*(a - b)*f) - (b*Cot[e + f*x])/(2*a*(a - b)*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 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 )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {b \cot (e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {2 a-3 b-3 b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a (a-b) f}\\ &=-\frac {(2 a-3 b) \cot (e+f x)}{2 a^2 (a-b) f}-\frac {b \cot (e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {2 a^2+2 a b-3 b^2+(2 a-3 b) b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a^2 (a-b) f}\\ &=-\frac {(2 a-3 b) \cot (e+f x)}{2 a^2 (a-b) f}-\frac {b \cot (e+f x)}{2 a (a-b) 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)^2 f}+\frac {\left ((5 a-3 b) b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 (a-b)^2 f}\\ &=-\frac {x}{(a-b)^2}+\frac {(5 a-3 b) b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{5/2} (a-b)^2 f}-\frac {(2 a-3 b) \cot (e+f x)}{2 a^2 (a-b) f}-\frac {b \cot (e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 3.22, size = 117, normalized size = 0.91 \[ \frac {\frac {b^{3/2} (5 a-3 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{5/2} (a-b)^2}+\frac {\frac {b^2 (a-b) \sin (2 (e+f x))}{a^2 ((a-b) \cos (2 (e+f x))+a+b)}-2 (e+f x)}{(a-b)^2}-\frac {2 \cot (e+f x)}{a^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((5*a - 3*b)*b^(3/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(5/2)*(a - b)^2) - (2*Cot[e + f*x])/a^2 + (-2
*(e + f*x) + ((a - b)*b^2*Sin[2*(e + f*x)])/(a^2*(a + b + (a - b)*Cos[2*(e + f*x)])))/(a - b)^2)/(2*f)

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fricas [A]  time = 0.48, size = 503, normalized size = 3.93 \[ \left [-\frac {8 \, a^{2} b f x \tan \left (f x + e\right )^{3} + 8 \, a^{3} f x \tan \left (f x + e\right ) + 8 \, a^{3} - 16 \, a^{2} b + 8 \, a b^{2} + 4 \, {\left (2 \, a^{2} b - 5 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (f x + e\right )^{2} + {\left ({\left (5 \, a b^{2} - 3 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a^{2} b - 3 \, a b^{2}\right )} \tan \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left (a b \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right )}{8 \, {\left ({\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \tan \left (f x + e\right )\right )}}, -\frac {4 \, a^{2} b f x \tan \left (f x + e\right )^{3} + 4 \, a^{3} f x \tan \left (f x + e\right ) + 4 \, a^{3} - 8 \, a^{2} b + 4 \, a b^{2} + 2 \, {\left (2 \, a^{2} b - 5 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (f x + e\right )^{2} - {\left ({\left (5 \, a b^{2} - 3 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a^{2} b - 3 \, a b^{2}\right )} \tan \left (f x + e\right )\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {b}{a}}}{2 \, b \tan \left (f x + e\right )}\right )}{4 \, {\left ({\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \tan \left (f x + e\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(8*a^2*b*f*x*tan(f*x + e)^3 + 8*a^3*f*x*tan(f*x + e) + 8*a^3 - 16*a^2*b + 8*a*b^2 + 4*(2*a^2*b - 5*a*b^2
+ 3*b^3)*tan(f*x + e)^2 + ((5*a*b^2 - 3*b^3)*tan(f*x + e)^3 + (5*a^2*b - 3*a*b^2)*tan(f*x + e))*sqrt(-b/a)*lo
g((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^4*b - 2*a^3*b^2 + a^2*b^3)*f*tan(f*x + e)^3 + (a^5 - 2*a^
4*b + a^3*b^2)*f*tan(f*x + e)), -1/4*(4*a^2*b*f*x*tan(f*x + e)^3 + 4*a^3*f*x*tan(f*x + e) + 4*a^3 - 8*a^2*b +
4*a*b^2 + 2*(2*a^2*b - 5*a*b^2 + 3*b^3)*tan(f*x + e)^2 - ((5*a*b^2 - 3*b^3)*tan(f*x + e)^3 + (5*a^2*b - 3*a*b^
2)*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^4*b - 2*a^3*b^2
+ a^2*b^3)*f*tan(f*x + e)^3 + (a^5 - 2*a^4*b + a^3*b^2)*f*tan(f*x + e))]

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giac [A]  time = 2.60, size = 171, normalized size = 1.34 \[ \frac {\frac {{\left (5 \, a b^{2} - 3 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {a b}} - \frac {2 \, {\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}} - \frac {2 \, a b \tan \left (f x + e\right )^{2} - 3 \, b^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{2} - 2 \, a b}{{\left (b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right )\right )} {\left (a^{3} - a^{2} b\right )}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((5*a*b^2 - 3*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))/((a^4 - 2*a^3*
b + a^2*b^2)*sqrt(a*b)) - 2*(f*x + e)/(a^2 - 2*a*b + b^2) - (2*a*b*tan(f*x + e)^2 - 3*b^2*tan(f*x + e)^2 + 2*a
^2 - 2*a*b)/((b*tan(f*x + e)^3 + a*tan(f*x + e))*(a^3 - a^2*b)))/f

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maple [A]  time = 0.83, size = 187, normalized size = 1.46 \[ \frac {b^{2} \tan \left (f x +e \right )}{2 f \left (a -b \right )^{2} a \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{3} \tan \left (f x +e \right )}{2 f \,a^{2} \left (a -b \right )^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \left (a -b \right )^{2} a \sqrt {a b}}-\frac {3 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \,a^{2} \left (a -b \right )^{2} \sqrt {a b}}-\frac {1}{f \,a^{2} \tan \left (f x +e \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \left (a -b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*b^2/(a-b)^2/a*tan(f*x+e)/(a+b*tan(f*x+e)^2)-1/2/f*b^3/a^2/(a-b)^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)+5/2/f*b^
2/(a-b)^2/a/(a*b)^(1/2)*arctan(tan(f*x+e)*b/(a*b)^(1/2))-3/2/f*b^3/a^2/(a-b)^2/(a*b)^(1/2)*arctan(tan(f*x+e)*b
/(a*b)^(1/2))-1/f/a^2/tan(f*x+e)-1/f/(a-b)^2*arctan(tan(f*x+e))

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maxima [A]  time = 0.71, size = 151, normalized size = 1.18 \[ \frac {\frac {{\left (5 \, a b^{2} - 3 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {a b}} - \frac {{\left (2 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2} - 2 \, a b}{{\left (a^{3} b - a^{2} b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{4} - a^{3} b\right )} \tan \left (f x + e\right )} - \frac {2 \, {\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((5*a*b^2 - 3*b^3)*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^4 - 2*a^3*b + a^2*b^2)*sqrt(a*b)) - ((2*a*b - 3*b^
2)*tan(f*x + e)^2 + 2*a^2 - 2*a*b)/((a^3*b - a^2*b^2)*tan(f*x + e)^3 + (a^4 - a^3*b)*tan(f*x + e)) - 2*(f*x +
e)/(a^2 - 2*a*b + b^2))/f

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mupad [B]  time = 14.23, size = 2674, normalized size = 20.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (1/a + (tan(e + f*x)^2*(2*a*b - 3*b^2))/(2*a^2*(a - b)))/(f*(a*tan(e + f*x) + b*tan(e + f*x)^3)) - (2*atan((
(((1280*a^9*b^9 - 192*a^8*b^10 - 3520*a^10*b^8 + 4992*a^11*b^7 - 3520*a^12*b^6 + 512*a^13*b^5 + 960*a^14*b^4 -
640*a^15*b^3 + 128*a^16*b^2 + (tan(e + f*x)*(256*a^10*b^10 - 1536*a^11*b^9 + 3584*a^12*b^8 - 3584*a^13*b^7 +
3584*a^15*b^5 - 3584*a^16*b^4 + 1536*a^17*b^3 - 256*a^18*b^2)*1i)/(2*a^2 - 4*a*b + 2*b^2))*1i)/(2*a^2 - 4*a*b
+ 2*b^2) + tan(e + f*x)*(144*a^6*b^10 - 912*a^7*b^9 + 2272*a^8*b^8 - 2784*a^9*b^7 + 1744*a^10*b^6 - 592*a^11*b
^5 + 192*a^12*b^4 - 64*a^13*b^3))/(2*a^2 - 4*a*b + 2*b^2) + (((192*a^8*b^10 - 1280*a^9*b^9 + 3520*a^10*b^8 - 4
992*a^11*b^7 + 3520*a^12*b^6 - 512*a^13*b^5 - 960*a^14*b^4 + 640*a^15*b^3 - 128*a^16*b^2 + (tan(e + f*x)*(256*
a^10*b^10 - 1536*a^11*b^9 + 3584*a^12*b^8 - 3584*a^13*b^7 + 3584*a^15*b^5 - 3584*a^16*b^4 + 1536*a^17*b^3 - 25
6*a^18*b^2)*1i)/(2*a^2 - 4*a*b + 2*b^2))*1i)/(2*a^2 - 4*a*b + 2*b^2) + tan(e + f*x)*(144*a^6*b^10 - 912*a^7*b^
9 + 2272*a^8*b^8 - 2784*a^9*b^7 + 1744*a^10*b^6 - 592*a^11*b^5 + 192*a^12*b^4 - 64*a^13*b^3))/(2*a^2 - 4*a*b +
2*b^2))/(((((192*a^8*b^10 - 1280*a^9*b^9 + 3520*a^10*b^8 - 4992*a^11*b^7 + 3520*a^12*b^6 - 512*a^13*b^5 - 960
*a^14*b^4 + 640*a^15*b^3 - 128*a^16*b^2 + (tan(e + f*x)*(256*a^10*b^10 - 1536*a^11*b^9 + 3584*a^12*b^8 - 3584*
a^13*b^7 + 3584*a^15*b^5 - 3584*a^16*b^4 + 1536*a^17*b^3 - 256*a^18*b^2)*1i)/(2*a^2 - 4*a*b + 2*b^2))*1i)/(2*a
^2 - 4*a*b + 2*b^2) + tan(e + f*x)*(144*a^6*b^10 - 912*a^7*b^9 + 2272*a^8*b^8 - 2784*a^9*b^7 + 1744*a^10*b^6 -
592*a^11*b^5 + 192*a^12*b^4 - 64*a^13*b^3))*1i)/(2*a^2 - 4*a*b + 2*b^2) - ((((1280*a^9*b^9 - 192*a^8*b^10 - 3
520*a^10*b^8 + 4992*a^11*b^7 - 3520*a^12*b^6 + 512*a^13*b^5 + 960*a^14*b^4 - 640*a^15*b^3 + 128*a^16*b^2 + (ta
n(e + f*x)*(256*a^10*b^10 - 1536*a^11*b^9 + 3584*a^12*b^8 - 3584*a^13*b^7 + 3584*a^15*b^5 - 3584*a^16*b^4 + 15
36*a^17*b^3 - 256*a^18*b^2)*1i)/(2*a^2 - 4*a*b + 2*b^2))*1i)/(2*a^2 - 4*a*b + 2*b^2) + tan(e + f*x)*(144*a^6*b
^10 - 912*a^7*b^9 + 2272*a^8*b^8 - 2784*a^9*b^7 + 1744*a^10*b^6 - 592*a^11*b^5 + 192*a^12*b^4 - 64*a^13*b^3))*
1i)/(2*a^2 - 4*a*b + 2*b^2) + 144*a^6*b^8 - 624*a^7*b^7 + 976*a^8*b^6 - 656*a^9*b^5 + 160*a^10*b^4)))/(f*(2*a^
2 - 4*a*b + 2*b^2)) - (atan((((tan(e + f*x)*(144*a^6*b^10 - 912*a^7*b^9 + 2272*a^8*b^8 - 2784*a^9*b^7 + 1744*a
^10*b^6 - 592*a^11*b^5 + 192*a^12*b^4 - 64*a^13*b^3) + ((5*a - 3*b)*(-a^5*b^3)^(1/2)*(1280*a^9*b^9 - 192*a^8*b
^10 - 3520*a^10*b^8 + 4992*a^11*b^7 - 3520*a^12*b^6 + 512*a^13*b^5 + 960*a^14*b^4 - 640*a^15*b^3 + 128*a^16*b^
2 + (tan(e + f*x)*(5*a - 3*b)*(-a^5*b^3)^(1/2)*(256*a^10*b^10 - 1536*a^11*b^9 + 3584*a^12*b^8 - 3584*a^13*b^7
+ 3584*a^15*b^5 - 3584*a^16*b^4 + 1536*a^17*b^3 - 256*a^18*b^2))/(4*(a^7 - 2*a^6*b + a^5*b^2))))/(4*(a^7 - 2*a
^6*b + a^5*b^2)))*(5*a - 3*b)*(-a^5*b^3)^(1/2)*1i)/(4*(a^7 - 2*a^6*b + a^5*b^2)) + ((tan(e + f*x)*(144*a^6*b^1
0 - 912*a^7*b^9 + 2272*a^8*b^8 - 2784*a^9*b^7 + 1744*a^10*b^6 - 592*a^11*b^5 + 192*a^12*b^4 - 64*a^13*b^3) + (
(5*a - 3*b)*(-a^5*b^3)^(1/2)*(192*a^8*b^10 - 1280*a^9*b^9 + 3520*a^10*b^8 - 4992*a^11*b^7 + 3520*a^12*b^6 - 51
2*a^13*b^5 - 960*a^14*b^4 + 640*a^15*b^3 - 128*a^16*b^2 + (tan(e + f*x)*(5*a - 3*b)*(-a^5*b^3)^(1/2)*(256*a^10
*b^10 - 1536*a^11*b^9 + 3584*a^12*b^8 - 3584*a^13*b^7 + 3584*a^15*b^5 - 3584*a^16*b^4 + 1536*a^17*b^3 - 256*a^
18*b^2))/(4*(a^7 - 2*a^6*b + a^5*b^2))))/(4*(a^7 - 2*a^6*b + a^5*b^2)))*(5*a - 3*b)*(-a^5*b^3)^(1/2)*1i)/(4*(a
^7 - 2*a^6*b + a^5*b^2)))/(144*a^6*b^8 - 624*a^7*b^7 + 976*a^8*b^6 - 656*a^9*b^5 + 160*a^10*b^4 - ((tan(e + f*
x)*(144*a^6*b^10 - 912*a^7*b^9 + 2272*a^8*b^8 - 2784*a^9*b^7 + 1744*a^10*b^6 - 592*a^11*b^5 + 192*a^12*b^4 - 6
4*a^13*b^3) + ((5*a - 3*b)*(-a^5*b^3)^(1/2)*(1280*a^9*b^9 - 192*a^8*b^10 - 3520*a^10*b^8 + 4992*a^11*b^7 - 352
0*a^12*b^6 + 512*a^13*b^5 + 960*a^14*b^4 - 640*a^15*b^3 + 128*a^16*b^2 + (tan(e + f*x)*(5*a - 3*b)*(-a^5*b^3)^
(1/2)*(256*a^10*b^10 - 1536*a^11*b^9 + 3584*a^12*b^8 - 3584*a^13*b^7 + 3584*a^15*b^5 - 3584*a^16*b^4 + 1536*a^
17*b^3 - 256*a^18*b^2))/(4*(a^7 - 2*a^6*b + a^5*b^2))))/(4*(a^7 - 2*a^6*b + a^5*b^2)))*(5*a - 3*b)*(-a^5*b^3)^
(1/2))/(4*(a^7 - 2*a^6*b + a^5*b^2)) + ((tan(e + f*x)*(144*a^6*b^10 - 912*a^7*b^9 + 2272*a^8*b^8 - 2784*a^9*b^
7 + 1744*a^10*b^6 - 592*a^11*b^5 + 192*a^12*b^4 - 64*a^13*b^3) + ((5*a - 3*b)*(-a^5*b^3)^(1/2)*(192*a^8*b^10 -
1280*a^9*b^9 + 3520*a^10*b^8 - 4992*a^11*b^7 + 3520*a^12*b^6 - 512*a^13*b^5 - 960*a^14*b^4 + 640*a^15*b^3 - 1
28*a^16*b^2 + (tan(e + f*x)*(5*a - 3*b)*(-a^5*b^3)^(1/2)*(256*a^10*b^10 - 1536*a^11*b^9 + 3584*a^12*b^8 - 3584
*a^13*b^7 + 3584*a^15*b^5 - 3584*a^16*b^4 + 1536*a^17*b^3 - 256*a^18*b^2))/(4*(a^7 - 2*a^6*b + a^5*b^2))))/(4*
(a^7 - 2*a^6*b + a^5*b^2)))*(5*a - 3*b)*(-a^5*b^3)^(1/2))/(4*(a^7 - 2*a^6*b + a^5*b^2))))*(5*a - 3*b)*(-a^5*b^
3)^(1/2)*1i)/(2*f*(a^7 - 2*a^6*b + a^5*b^2))

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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)**2,x)

[Out]

Timed out

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