3.4.54 \(\int \frac {\cot ^3(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [354]

Optimal. Leaf size=111 \[ -\frac {b^3}{2 a^2 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\csc ^2(e+f x)}{2 (a+b)^2 f}-\frac {b^2 (3 a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 (a+b)^3 f}-\frac {(a+3 b) \log (\sin (e+f x))}{(a+b)^3 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 111, 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^3}{2 a^2 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}-\frac {b^2 (3 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^3}-\frac {\csc ^2(e+f x)}{2 f (a+b)^2}-\frac {(a+3 b) \log (\sin (e+f x))}{f (a+b)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*b^3/(a^2*(a + b)^2*f*(b + a*Cos[e + f*x]^2)) - Csc[e + f*x]^2/(2*(a + b)^2*f) - (b^2*(3*a + b)*Log[b + a*
Cos[e + f*x]^2])/(2*a^2*(a + b)^3*f) - ((a + 3*b)*Log[Sin[e + f*x]])/((a + b)^3*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 ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=-\frac {\text {Subst}\left (\int \frac {x^7}{\left (1-x^2\right )^2 \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\text {Subst}\left (\int \frac {x^3}{(1-x)^2 (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)^2}+\frac {a+3 b}{(a+b)^3 (-1+x)}-\frac {b^3}{a (a+b)^2 (b+a x)^2}+\frac {b^2 (3 a+b)}{a (a+b)^3 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {b^3}{2 a^2 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\csc ^2(e+f x)}{2 (a+b)^2 f}-\frac {b^2 (3 a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 (a+b)^3 f}-\frac {(a+3 b) \log (\sin (e+f x))}{(a+b)^3 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.47, size = 130, normalized size = 1.17 \begin {gather*} -\frac {(a+2 b+a \cos (2 (e+f x)))^2 \left ((a+b) \csc ^2(e+f x)+2 (a+3 b) \log (\sin (e+f x))+\frac {b^2 \left (\frac {2 b (a+b)}{a+2 b+a \cos (2 (e+f x))}+(3 a+b) \log \left (a+b-a \sin ^2(e+f x)\right )\right )}{a^2}\right ) \sec ^4(e+f x)}{8 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((a + 2*b + a*Cos[2*(e + f*x)])^2*((a + b)*Csc[e + f*x]^2 + 2*(a + 3*b)*Log[Sin[e + f*x]] + (b^2*((2*b*(a
 + b))/(a + 2*b + a*Cos[2*(e + f*x)]) + (3*a + b)*Log[a + b - a*Sin[e + f*x]^2]))/a^2)*Sec[e + f*x]^4)/((a + b
)^3*f*(a + b*Sec[e + f*x]^2)^2)

________________________________________________________________________________________

Maple [A]
time = 0.13, size = 141, normalized size = 1.27

method result size
derivativedivides \(\frac {-\frac {b^{2} \left (\frac {\left (a +b \right ) b}{a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {\left (3 a +b \right ) \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{a^{2}}\right )}{2 \left (a +b \right )^{3}}+\frac {1}{4 \left (a +b \right )^{2} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (-a -3 b \right ) \ln \left (\cos \left (f x +e \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {1}{4 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -3 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}}{f}\) \(141\)
default \(\frac {-\frac {b^{2} \left (\frac {\left (a +b \right ) b}{a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {\left (3 a +b \right ) \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{a^{2}}\right )}{2 \left (a +b \right )^{3}}+\frac {1}{4 \left (a +b \right )^{2} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (-a -3 b \right ) \ln \left (\cos \left (f x +e \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {1}{4 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -3 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}}{f}\) \(141\)
risch \(-\frac {i x}{a^{2}}+\frac {2 i a x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {2 i a e}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 i b x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {6 i b e}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 i b^{2} x}{a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 i b^{2} e}{f a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 i b^{3} x}{a^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 i b^{3} e}{f \,a^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}-2 b^{3} {\mathrm e}^{6 i \left (f x +e \right )}+4 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+8 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+4 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+2 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}-2 b^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{a^{2} f \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 b^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 f a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {b^{3} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 f \,a^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\) \(616\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 197, normalized size = 1.77 \begin {gather*} -\frac {\frac {{\left (3 \, a b^{2} + b^{3}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}} + \frac {{\left (a + 3 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {a^{3} + a^{2} b - {\left (a^{3} - b^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sin \left (f x + e\right )^{4} - {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \sin \left (f x + e\right )^{2}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (109) = 218\).
time = 4.39, size = 321, normalized size = 2.89 \begin {gather*} \frac {a^{3} b + a^{2} b^{2} + a b^{3} + b^{4} + {\left (a^{4} + a^{3} b - a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2} - {\left ({\left (3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{4} - 3 \, a b^{3} - b^{4} - {\left (3 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, {\left ({\left (a^{4} + 3 \, a^{3} b\right )} \cos \left (f x + e\right )^{4} - a^{3} b - 3 \, a^{2} b^{2} - {\left (a^{4} + 2 \, a^{3} b - 3 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{2 \, {\left ({\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{6} + 2 \, a^{5} b - 2 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{3}{\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)**3/(a+b*sec(f*x+e)**2)**2,x)

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 811 vs. \(2 (105) = 210\).
time = 0.59, size = 811, normalized size = 7.31 \begin {gather*} -\frac {\frac {12 \, {\left (3 \, a b^{2} + b^{3}\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^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}} + \frac {12 \, {\left (a + 3 \, b\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {3 \, a^{4} + 6 \, a^{3} b + 3 \, a^{2} b^{2} + \frac {10 \, a^{4} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {16 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {30 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {32 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {8 \, b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {11 \, a^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {22 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {27 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {16 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {16 \, b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {4 \, a^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {16 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {36 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {32 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {8 \, b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{{\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} {\left (\frac {a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}} - \frac {3 \, {\left (\cos \left (f x + e\right ) - 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}} - \frac {24 \, \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right )}{a^{2}}}{24 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(12*(3*a*b^2 + b^3)*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^
5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3) + 12*(a + 3*b)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1))/(a^3 + 3*a
^2*b + 3*a*b^2 + b^3) - (3*a^4 + 6*a^3*b + 3*a^2*b^2 + 10*a^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 16*a^3*b
*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 30*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 32*a*b^3*(cos(f*x
+ e) - 1)/(cos(f*x + e) + 1) + 8*b^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 11*a^4*(cos(f*x + e) - 1)^2/(cos(
f*x + e) + 1)^2 + 22*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 27*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*
x + e) + 1)^2 + 16*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 16*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e)
 + 1)^2 + 4*a^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 16*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3
 + 36*a^2*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 32*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 +
 8*b^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*(a*(cos(f*x + e) - 1)
/(cos(f*x + e) + 1) + b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2*a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2
- 2*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + a*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + b*(cos(f*x + e
) - 1)^3/(cos(f*x + e) + 1)^3)) - 3*(cos(f*x + e) - 1)/((a^2 + 2*a*b + b^2)*(cos(f*x + e) + 1)) - 24*log(abs(-
(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1))/a^2)/f

________________________________________________________________________________________

Mupad [B]
time = 5.05, size = 160, normalized size = 1.44 \begin {gather*} \frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^2\,f}-\frac {\frac {1}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a\,b-b^2\right )}{2\,a\,{\left (a+b\right )}^2}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4+\left (a+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a+3\,b\right )}{f\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {b^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (3\,a+b\right )}{2\,a^2\,f\,{\left (a+b\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________