3.4.55 \(\int \frac {\cot ^5(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [355]

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

[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 = 0.14, 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 = {4223, 457, 90} \begin {gather*} \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} \end {gather*}

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 90

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 457

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 4223

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 {\text {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 {\text {Subst}\left (\int \frac {x^4}{(1-x)^3 (b+a x)^2} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {\text {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 = 2.03, size = 162, normalized size = 1.16 \begin {gather*} \frac {(a+2 b+a \cos (2 (e+f x)))^2 \sec ^4(e+f x) \left (4 (a+b) (a+2 b) \csc ^2(e+f x)-(a+b)^2 \csc ^4(e+f x)+4 \left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))+\frac {2 b^3 (4 a+b) \log \left (a+b-a \sin ^2(e+f x)\right )}{a^2}+\frac {2 b^4 (a+b)}{a^2 \left (a+b-a \sin ^2(e+f x)\right )}\right )}{16 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^2} \end {gather*}

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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Maple [A]
time = 0.15, size = 201, normalized size = 1.44

method	result	size
derivativedivides	\(\frac {\frac {b^{3} \left (\frac {\left (a +b \right ) b}{a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {\left (4 a +b \right ) \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{a^{2}}\right )}{2 \left (a +b \right )^{4}}-\frac {1}{16 \left (a +b \right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2}}-\frac {7 a +15 b}{16 \left (a +b \right )^{3} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (a^{2}+4 a b +6 b^{2}\right ) \ln \left (\cos \left (f x +e \right )-1\right )}{2 \left (a +b \right )^{4}}-\frac {1}{16 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -15 b}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+4 a b +6 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}}{f}\)	\(201\)
default	\(\frac {\frac {b^{3} \left (\frac {\left (a +b \right ) b}{a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {\left (4 a +b \right ) \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{a^{2}}\right )}{2 \left (a +b \right )^{4}}-\frac {1}{16 \left (a +b \right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2}}-\frac {7 a +15 b}{16 \left (a +b \right )^{3} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (a^{2}+4 a b +6 b^{2}\right ) \ln \left (\cos \left (f x +e \right )-1\right )}{2 \left (a +b \right )^{4}}-\frac {1}{16 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -15 b}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+4 a b +6 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}}{f}\)	\(201\)
risch	\(-\frac {12 i b^{2} x}{a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {8 i b^{3} x}{a \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {12 i b^{2} e}{f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {2 i b^{4} x}{a^{2} \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {8 i a b x}{a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {8 i b^{3} e}{a f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {8 i a b e}{f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {2 i a^{2} e}{f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {2 i b^{4} e}{a^{2} f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {2 i a^{2} x}{a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {i x}{a^{2}}-\frac {2 \left (2 a^{4} {\mathrm e}^{10 i \left (f x +e \right )}+4 a^{3} b \,{\mathrm e}^{10 i \left (f x +e \right )}-b^{4} {\mathrm e}^{10 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{8 i \left (f x +e \right )}+10 a^{3} b \,{\mathrm e}^{8 i \left (f x +e \right )}+16 a^{2} b^{2} {\mathrm e}^{8 i \left (f x +e \right )}+4 b^{4} {\mathrm e}^{8 i \left (f x +e \right )}-12 a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}-24 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-6 b^{4} {\mathrm e}^{6 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}+10 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}+16 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+4 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{4}+4 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}-b^{4} {\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4} a^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a^{2}}{f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {4 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a b}{f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b^{2}}{f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {2 b^{3} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{a f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {b^{4} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a^{2} f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}\)	\(999\)

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,method=_RETURNVERBOSE)

[Out]

1/f*(1/2*b^3/(a+b)^4*((a+b)*b/a^2/(b+a*cos(f*x+e)^2)+(4*a+b)/a^2*ln(b+a*cos(f*x+e)^2))-1/16/(a+b)^2/(cos(f*x+e
)-1)^2-1/16*(7*a+15*b)/(a+b)^3/(cos(f*x+e)-1)+1/2*(a^2+4*a*b+6*b^2)/(a+b)^4*ln(cos(f*x+e)-1)-1/16/(a+b)^2/(1+c
os(f*x+e))^2-1/16*(-7*a-15*b)/(a+b)^3/(1+cos(f*x+e))+1/2*(a^2+4*a*b+6*b^2)/(a+b)^4*ln(1+cos(f*x+e)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (139) = 278\).
time = 0.27, size = 285, normalized size = 2.04 \begin {gather*} \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} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 570 vs. \(2 (139) = 278\).
time = 4.84, size = 570, normalized size = 4.07 \begin {gather*} \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 )}} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{5}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \end {gather*}

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]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 913 vs. \(2 (134) = 268\).
time = 0.62, size = 913, normalized size = 6.52 \begin {gather*} \frac {\frac {32 \, {\left (4 \, a b^{3} + b^{4}\right )} \log \left (a + b + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {2 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}} + \frac {32 \, {\left (a^{2} + 4 \, a b + 6 \, b^{2}\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {\frac {12 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {40 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {28 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {2 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {{\left (a^{2} + 2 \, a b + b^{2} + \frac {12 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {40 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {28 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {48 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {192 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {288 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}} - \frac {32 \, {\left (4 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5} + \frac {8 \, a^{2} b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {2 \, b^{5} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {4 \, a^{2} b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, a b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {b^{5} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} {\left (a + b + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {2 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}} - \frac {64 \, \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right )}{a^{2}}}{64 \, f} \end {gather*}

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]

1/64*(32*(4*a*b^3 + b^4)*log(a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f
*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^6
+ 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4) + 32*(a^2 + 4*a*b + 6*b^2)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*
x + e) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - (12*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 40*
a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 28*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a^2*(cos(f*x + e) -
1)^2/(cos(f*x + e) + 1)^2 + 2*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b^2*(cos(f*x + e) - 1)^2/(cos(f
*x + e) + 1)^2)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - (a^2 + 2*a*b + b^2 + 12*a^2*(cos(f*x + e) - 1)/(
cos(f*x + e) + 1) + 40*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 28*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1
) + 48*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 192*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 288
*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b
^4)*(cos(f*x + e) - 1)^2) - 32*(4*a^2*b^3 + 5*a*b^4 + b^5 + 8*a^2*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) -
2*a*b^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b^5*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 4*a^2*b^3*(cos(f
*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 5*a*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b^5*(cos(f*x + e) -
1)^2/(cos(f*x + e) + 1)^2)/((a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*(a + b + 2*a*(cos(f*x + e) - 1)/
(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 +
b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)) - 64*log(abs(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1))/a^2)/
f

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Mupad [B]
time = 6.09, size = 206, normalized size = 1.47 \begin {gather*} \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} \end {gather*}

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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