3.355 \(\int \frac {\cot ^5(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=140 \[ \frac {b^4}{2 a^2 f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac {b^3 (4 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^4}+\frac {\left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))}{f (a+b)^4}-\frac {\csc ^4(e+f x)}{4 f (a+b)^2}+\frac {(a+2 b) \csc ^2(e+f x)}{f (a+b)^3} \]

[Out]

1/2*b^4/a^2/(a+b)^3/f/(b+a*cos(f*x+e)^2)+(a+2*b)*csc(f*x+e)^2/(a+b)^3/f-1/4*csc(f*x+e)^4/(a+b)^2/f+1/2*b^3*(4*
a+b)*ln(b+a*cos(f*x+e)^2)/a^2/(a+b)^4/f+(a^2+4*a*b+6*b^2)*ln(sin(f*x+e))/(a+b)^4/f

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Rubi [A]  time = 0.20, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4138, 446, 88} \[ \frac {b^4}{2 a^2 f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac {\left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))}{f (a+b)^4}+\frac {b^3 (4 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^4}-\frac {\csc ^4(e+f x)}{4 f (a+b)^2}+\frac {(a+2 b) \csc ^2(e+f x)}{f (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

b^4/(2*a^2*(a + b)^3*f*(b + a*Cos[e + f*x]^2)) + ((a + 2*b)*Csc[e + f*x]^2)/((a + b)^3*f) - Csc[e + f*x]^4/(4*
(a + b)^2*f) + (b^3*(4*a + b)*Log[b + a*Cos[e + f*x]^2])/(2*a^2*(a + b)^4*f) + ((a^2 + 4*a*b + 6*b^2)*Log[Sin[
e + f*x]])/((a + b)^4*f)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4138

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

Rubi steps

\begin {align*} \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^9}{\left (1-x^2\right )^3 \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^4}{(1-x)^3 (b+a x)^2} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{(a+b)^2 (-1+x)^3}-\frac {2 (a+2 b)}{(a+b)^3 (-1+x)^2}+\frac {-a^2-4 a b-6 b^2}{(a+b)^4 (-1+x)}+\frac {b^4}{a (a+b)^3 (b+a x)^2}-\frac {b^3 (4 a+b)}{a (a+b)^4 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac {b^4}{2 a^2 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}+\frac {(a+2 b) \csc ^2(e+f x)}{(a+b)^3 f}-\frac {\csc ^4(e+f x)}{4 (a+b)^2 f}+\frac {b^3 (4 a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 (a+b)^4 f}+\frac {\left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))}{(a+b)^4 f}\\ \end {align*}

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Mathematica [A]  time = 1.84, size = 162, normalized size = 1.16 \[ \frac {\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \left (\frac {2 b^4 (a+b)}{a^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {2 b^3 (4 a+b) \log \left (-a \sin ^2(e+f x)+a+b\right )}{a^2}+4 \left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))-(a+b)^2 \csc ^4(e+f x)+4 (a+b) (a+2 b) \csc ^2(e+f x)\right )}{16 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[e + f*x]^4*(4*(a + b)*(a + 2*b)*Csc[e + f*x]^2 - (a + b)^2*Csc[e + f*x]^
4 + 4*(a^2 + 4*a*b + 6*b^2)*Log[Sin[e + f*x]] + (2*b^3*(4*a + b)*Log[a + b - a*Sin[e + f*x]^2])/a^2 + (2*b^4*(
a + b))/(a^2*(a + b - a*Sin[e + f*x]^2))))/(16*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^2)

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fricas [B]  time = 1.56, size = 557, normalized size = 3.98 \[ \frac {3 \, a^{4} b + 10 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + 2 \, a b^{4} + 2 \, b^{5} - 2 \, {\left (2 \, a^{5} + 6 \, a^{4} b + 4 \, a^{3} b^{2} - a b^{4} - b^{5}\right )} \cos \left (f x + e\right )^{4} + {\left (3 \, a^{5} + 6 \, a^{4} b - 5 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 4 \, a b^{4} - 4 \, b^{5}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left ({\left (4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (f x + e\right )^{6} + 4 \, a b^{4} + b^{5} - {\left (8 \, a^{2} b^{3} - 2 \, a b^{4} - b^{5}\right )} \cos \left (f x + e\right )^{4} + {\left (4 \, a^{2} b^{3} - 7 \, a b^{4} - 2 \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 4 \, {\left ({\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2}\right )} \cos \left (f x + e\right )^{6} + a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - {\left (2 \, a^{5} + 7 \, a^{4} b + 8 \, a^{3} b^{2} - 6 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{4} + {\left (a^{5} + 2 \, a^{4} b - 2 \, a^{3} b^{2} - 12 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{4 \, {\left ({\left (a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{7} + 7 \, a^{6} b + 8 \, a^{5} b^{2} + 2 \, a^{4} b^{3} - 2 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{7} + 2 \, a^{6} b - 2 \, a^{5} b^{2} - 8 \, a^{4} b^{3} - 7 \, a^{3} b^{4} - 2 \, a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/4*(3*a^4*b + 10*a^3*b^2 + 7*a^2*b^3 + 2*a*b^4 + 2*b^5 - 2*(2*a^5 + 6*a^4*b + 4*a^3*b^2 - a*b^4 - b^5)*cos(f*
x + e)^4 + (3*a^5 + 6*a^4*b - 5*a^3*b^2 - 8*a^2*b^3 - 4*a*b^4 - 4*b^5)*cos(f*x + e)^2 + 2*((4*a^2*b^3 + a*b^4)
*cos(f*x + e)^6 + 4*a*b^4 + b^5 - (8*a^2*b^3 - 2*a*b^4 - b^5)*cos(f*x + e)^4 + (4*a^2*b^3 - 7*a*b^4 - 2*b^5)*c
os(f*x + e)^2)*log(a*cos(f*x + e)^2 + b) + 4*((a^5 + 4*a^4*b + 6*a^3*b^2)*cos(f*x + e)^6 + a^4*b + 4*a^3*b^2 +
6*a^2*b^3 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 - 6*a^2*b^3)*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 12*a^2*b^
3)*cos(f*x + e)^2)*log(1/2*sin(f*x + e)))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6
- (2*a^7 + 7*a^6*b + 8*a^5*b^2 + 2*a^4*b^3 - 2*a^3*b^4 - a^2*b^5)*f*cos(f*x + e)^4 + (a^7 + 2*a^6*b - 2*a^5*b^
2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^2 + (a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5
)*f)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*((-((1-cos(f*x
+exp(1)))/(1+cos(f*x+exp(1))))^2*b^5-5*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^4*a-4*((1-cos(f*x+exp(1))
)/(1+cos(f*x+exp(1))))^2*b^3*a^2-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^5-2*(1-cos(f*x+exp(1)))/(1+cos(f*
x+exp(1)))*b^4*a+8*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^3*a^2-b^5-5*b^4*a-4*b^3*a^2)/(4*b^4*a^2+16*b^3*a^
3+24*b^2*a^4+16*b*a^5+4*a^6)/(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b+((1-cos(f*x+exp(1)))/(1+cos(f*x+ex
p(1))))^2*a+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+b+a)+(-288
*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2-192*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b*a-48*((1-co
s(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2+28*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^2+40*(1-cos(f*x+exp(1))
)/(1+cos(f*x+exp(1)))*b*a+12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2-b^2-2*b*a-a^2)/(128*b^4+512*b^3*a+768
*b^2*a^2+512*b*a^3+128*a^4)/((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2+(-32*((1-cos(f*x+exp(1)))/(1+cos(f*x+e
xp(1))))^2*b^2-64*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b*a-32*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))
^2*a^2+896*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^2+1280*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b*a+384*(1
-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2)/(4096*b^4+16384*b^3*a+24576*b^2*a^2+16384*b*a^3+4096*a^4)-1/2/a^2*l
n(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1))+(b^4+4*b^3*a)/(4*b^4*a^2+16*b^3*a^3+24*b^2*a^4+16*b*a^5+4*a^
6)*ln(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+2*(1-cos(f*x
+exp(1)))/(1+cos(f*x+exp(1)))*b-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+b+a)+(6*b^2+4*b*a+a^2)/(4*b^4+16*b
^3*a+24*b^2*a^2+16*b*a^3+4*a^4)*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1)))))

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maple [B]  time = 1.22, size = 374, normalized size = 2.67 \[ \frac {2 b^{3} \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{f \left (a +b \right )^{4} a}+\frac {b^{4} \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{2 f \left (a +b \right )^{4} a^{2}}+\frac {b^{4}}{2 f \left (a +b \right )^{4} a \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {b^{5}}{2 f \left (a +b \right )^{4} a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}-\frac {1}{16 f \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a}{16 f \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}-\frac {15 b}{16 f \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right ) a^{2}}{2 f \left (a +b \right )^{4}}+\frac {2 \ln \left (-1+\cos \left (f x +e \right )\right ) a b}{f \left (a +b \right )^{4}}+\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b^{2}}{f \left (a +b \right )^{4}}-\frac {1}{16 f \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}+\frac {7 a}{16 f \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {15 b}{16 f \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (1+\cos \left (f x +e \right )\right ) a^{2}}{2 f \left (a +b \right )^{4}}+\frac {2 \ln \left (1+\cos \left (f x +e \right )\right ) a b}{f \left (a +b \right )^{4}}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b^{2}}{f \left (a +b \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x)

[Out]

2/f*b^3/(a+b)^4/a*ln(b+a*cos(f*x+e)^2)+1/2/f*b^4/(a+b)^4/a^2*ln(b+a*cos(f*x+e)^2)+1/2/f*b^4/(a+b)^4/a/(b+a*cos
(f*x+e)^2)+1/2/f*b^5/(a+b)^4/a^2/(b+a*cos(f*x+e)^2)-1/16/f/(a+b)^2/(-1+cos(f*x+e))^2-7/16/f/(a+b)^3/(-1+cos(f*
x+e))*a-15/16/f/(a+b)^3/(-1+cos(f*x+e))*b+1/2/f/(a+b)^4*ln(-1+cos(f*x+e))*a^2+2/f/(a+b)^4*ln(-1+cos(f*x+e))*a*
b+3/f/(a+b)^4*ln(-1+cos(f*x+e))*b^2-1/16/f/(a+b)^2/(1+cos(f*x+e))^2+7/16/f/(a+b)^3/(1+cos(f*x+e))*a+15/16/f/(a
+b)^3/(1+cos(f*x+e))*b+1/2/f/(a+b)^4*ln(1+cos(f*x+e))*a^2+2/f/(a+b)^4*ln(1+cos(f*x+e))*a*b+3/f/(a+b)^4*ln(1+co
s(f*x+e))*b^2

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maxima [B]  time = 0.34, size = 279, normalized size = 1.99 \[ \frac {\frac {2 \, {\left (4 \, a b^{3} + b^{4}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}} + \frac {2 \, {\left (a^{2} + 4 \, a b + 6 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {2 \, {\left (2 \, a^{4} + 4 \, a^{3} b - b^{4}\right )} \sin \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - {\left (5 \, a^{4} + 13 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \sin \left (f x + e\right )^{6} - {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \sin \left (f x + e\right )^{4}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/4*(2*(4*a*b^3 + b^4)*log(a*sin(f*x + e)^2 - a - b)/(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4) + 2*(a^
2 + 4*a*b + 6*b^2)*log(sin(f*x + e)^2)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + (2*(2*a^4 + 4*a^3*b - b^4
)*sin(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - (5*a^4 + 13*a^3*b + 8*a^2*b^2)*sin(f*x + e)^2)/((a^6 + 3*a^5*b +
3*a^4*b^2 + a^3*b^3)*sin(f*x + e)^6 - (a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*sin(f*x + e)^4))/f

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mupad [B]  time = 6.09, size = 206, normalized size = 1.47 \[ \frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a+5\,b\right )}{4\,{\left (a+b\right )}^2}-\frac {1}{4\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2\,b+3\,a\,b^2-b^3\right )}{2\,a\,{\left (a+b\right )}^3}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^6+\left (a+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^2\,f}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+4\,a\,b+6\,b^2\right )}{f\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )}+\frac {b^3\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (4\,a+b\right )}{2\,a^2\,f\,{\left (a+b\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^5/(a + b/cos(e + f*x)^2)^2,x)

[Out]

((tan(e + f*x)^2*(2*a + 5*b))/(4*(a + b)^2) - 1/(4*(a + b)) + (tan(e + f*x)^4*(3*a*b^2 + a^2*b - b^3))/(2*a*(a
+ b)^3))/(f*(tan(e + f*x)^4*(a + b) + b*tan(e + f*x)^6)) - log(tan(e + f*x)^2 + 1)/(2*a^2*f) + (log(tan(e + f
*x))*(4*a*b + a^2 + 6*b^2))/(f*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)) + (b^3*log(a + b + b*tan(e + f*x)^
2)*(4*a + b))/(2*a^2*f*(a + b)^4)

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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)**5/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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