3.3.29 \(\int \frac {\cot ^5(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\) [229]

Optimal. Leaf size=161 \[ \frac {(a+2 b) \cot ^2(e+f x)}{2 a^3 f}-\frac {\cot ^4(e+f x)}{4 a^2 f}+\frac {\log (\cos (e+f x))}{(a-b)^2 f}+\frac {\left (a^2+2 a b+3 b^2\right ) \log (\tan (e+f x))}{a^4 f}+\frac {(4 a-3 b) b^3 \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 (a-b)^2 f}-\frac {b^3}{2 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 161, 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 = {3751, 457, 90} \begin {gather*} \frac {b^3 (4 a-3 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 f (a-b)^2}-\frac {b^3}{2 a^3 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {(a+2 b) \cot ^2(e+f x)}{2 a^3 f}-\frac {\cot ^4(e+f x)}{4 a^2 f}+\frac {\left (a^2+2 a b+3 b^2\right ) \log (\tan (e+f x))}{a^4 f}+\frac {\log (\cos (e+f x))}{f (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b)*Cot[e + f*x]^2)/(2*a^3*f) - Cot[e + f*x]^4/(4*a^2*f) + Log[Cos[e + f*x]]/((a - b)^2*f) + ((a^2 + 2*
a*b + 3*b^2)*Log[Tan[e + f*x]])/(a^4*f) + ((4*a - 3*b)*b^3*Log[a + b*Tan[e + f*x]^2])/(2*a^4*(a - b)^2*f) - b^
3/(2*a^3*(a - b)*f*(a + b*Tan[e + f*x]^2))

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 3751

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 ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x^5 \left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x^3 (1+x) (a+b x)^2} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{a^2 x^3}+\frac {-a-2 b}{a^3 x^2}+\frac {a^2+2 a b+3 b^2}{a^4 x}-\frac {1}{(a-b)^2 (1+x)}+\frac {b^4}{a^3 (a-b) (a+b x)^2}+\frac {(4 a-3 b) b^4}{a^4 (a-b)^2 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {(a+2 b) \cot ^2(e+f x)}{2 a^3 f}-\frac {\cot ^4(e+f x)}{4 a^2 f}+\frac {\log (\cos (e+f x))}{(a-b)^2 f}+\frac {\left (a^2+2 a b+3 b^2\right ) \log (\tan (e+f x))}{a^4 f}+\frac {(4 a-3 b) b^3 \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 (a-b)^2 f}-\frac {b^3}{2 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.71, size = 121, normalized size = 0.75 \begin {gather*} -\frac {-\frac {(a+2 b) \cot ^2(e+f x)}{a^3}+\frac {\cot ^4(e+f x)}{2 a^2}-\frac {b^4}{a^4 (a-b) \left (b+a \cot ^2(e+f x)\right )}-\frac {(4 a-3 b) b^3 \log \left (b+a \cot ^2(e+f x)\right )}{a^4 (a-b)^2}-\frac {2 \log (\sin (e+f x))}{(a-b)^2}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.37, size = 217, normalized size = 1.35

method	result	size
norman	\(\frac {-\frac {1}{4 a f}+\frac {\left (2 a +3 b \right ) \left (\tan ^{2}\left (f x +e \right )\right )}{4 a^{2} f}+\frac {\left (-a^{2} b -a \,b^{2}+3 b^{3}\right ) b \left (\tan ^{6}\left (f x +e \right )\right )}{2 a^{4} f \left (a -b \right )}}{\tan \left (f x +e \right )^{4} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {\left (a^{2}+2 a b +3 b^{2}\right ) \ln \left (\tan \left (f x +e \right )\right )}{a^{4} f}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}+\frac {b^{3} \left (4 a -3 b \right ) \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 a^{4} f \left (a^{2}-2 a b +b^{2}\right )}\)	\(199\)
derivativedivides	\(\frac {-\frac {1}{16 a^{2} \left (\cos \left (f x +e \right )+1\right )^{2}}-\frac {-7 a -8 b}{16 a^{3} \left (\cos \left (f x +e \right )+1\right )}+\frac {\left (a^{2}+2 a b +3 b^{2}\right ) \ln \left (\cos \left (f x +e \right )+1\right )}{2 a^{4}}-\frac {1}{16 a^{2} \left (\cos \left (f x +e \right )-1\right )^{2}}-\frac {7 a +8 b}{16 a^{3} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (a^{2}+2 a b +3 b^{2}\right ) \ln \left (\cos \left (f x +e \right )-1\right )}{2 a^{4}}+\frac {b^{3} \left (\frac {a b}{\left (a -b \right )^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {\left (4 a -3 b \right ) \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{\left (a -b \right )^{2}}\right )}{2 a^{4}}}{f}\)	\(217\)
default	\(\frac {-\frac {1}{16 a^{2} \left (\cos \left (f x +e \right )+1\right )^{2}}-\frac {-7 a -8 b}{16 a^{3} \left (\cos \left (f x +e \right )+1\right )}+\frac {\left (a^{2}+2 a b +3 b^{2}\right ) \ln \left (\cos \left (f x +e \right )+1\right )}{2 a^{4}}-\frac {1}{16 a^{2} \left (\cos \left (f x +e \right )-1\right )^{2}}-\frac {7 a +8 b}{16 a^{3} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (a^{2}+2 a b +3 b^{2}\right ) \ln \left (\cos \left (f x +e \right )-1\right )}{2 a^{4}}+\frac {b^{3} \left (\frac {a b}{\left (a -b \right )^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {\left (4 a -3 b \right ) \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{\left (a -b \right )^{2}}\right )}{2 a^{4}}}{f}\)	\(217\)
risch	\(-\frac {6 i b^{2} x}{a^{4}}-\frac {6 i b^{2} e}{a^{4} f}-\frac {2 i e}{a^{2} f}-\frac {4 i b e}{a^{3} f}-\frac {2 i x}{a^{2}}-\frac {8 i b^{3} e}{a^{3} f \left (a^{2}-2 a b +b^{2}\right )}+\frac {6 i b^{4} x}{a^{4} \left (a^{2}-2 a b +b^{2}\right )}+\frac {i x}{a^{2}-2 a b +b^{2}}-\frac {8 i b^{3} x}{a^{3} \left (a^{2}-2 a b +b^{2}\right )}+\frac {6 i b^{4} e}{a^{4} f \left (a^{2}-2 a b +b^{2}\right )}-\frac {4 i b x}{a^{3}}-\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 )}+4 a \,b^{3} {\mathrm e}^{10 i \left (f x +e \right )}-3 b^{4} {\mathrm e}^{10 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{8 i \left (f x +e \right )}+2 a^{3} b \,{\mathrm e}^{8 i \left (f x +e \right )}-2 a^{2} b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-10 a \,b^{3} {\mathrm e}^{8 i \left (f x +e \right )}+12 b^{4} {\mathrm e}^{8 i \left (f x +e \right )}-12 a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}+12 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+12 a \,b^{3} {\mathrm e}^{6 i \left (f x +e \right )}-18 b^{4} {\mathrm e}^{6 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}+2 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-10 a \,b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+12 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{2 i \left (f x +e \right )}-4 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}+4 a \,b^{3} {\mathrm e}^{2 i \left (f x +e \right )}-3 b^{4} {\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \,a^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4} \left (a -b \right )^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{a^{2} f}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{a^{3} f}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b^{2}}{a^{4} f}+\frac {2 b^{3} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{a^{3} f \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 b^{4} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 a^{4} f \left (a^{2}-2 a b +b^{2}\right )}\)	\(763\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(-1/16/a^2/(cos(f*x+e)+1)^2-1/16*(-7*a-8*b)/a^3/(cos(f*x+e)+1)+1/2*(a^2+2*a*b+3*b^2)/a^4*ln(cos(f*x+e)+1)-
1/16/a^2/(cos(f*x+e)-1)^2-1/16*(7*a+8*b)/a^3/(cos(f*x+e)-1)+1/2*(a^2+2*a*b+3*b^2)/a^4*ln(cos(f*x+e)-1)+1/2*b^3
/a^4*(a*b/(a-b)^2/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)+(4*a-3*b)/(a-b)^2*ln(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)))

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 242, normalized size = 1.50 \begin {gather*} \frac {\frac {2 \, {\left (4 \, a b^{3} - 3 \, b^{4}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{6} - 2 \, a^{5} b + a^{4} b^{2}} + \frac {2 \, {\left (2 \, a^{4} - 4 \, a^{3} b + 4 \, a b^{3} - 3 \, b^{4}\right )} \sin \left (f x + e\right )^{4} + a^{4} - 2 \, a^{3} b + a^{2} b^{2} - {\left (5 \, a^{4} - 7 \, a^{3} b - a^{2} b^{2} + 3 \, a b^{3}\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} - 2 \, a^{5} b + a^{4} b^{2}\right )} \sin \left (f x + e\right )^{4}} + \frac {2 \, {\left (a^{2} + 2 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{4}}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(4*a*b^3 - 3*b^4)*log(-(a - b)*sin(f*x + e)^2 + a)/(a^6 - 2*a^5*b + a^4*b^2) + (2*(2*a^4 - 4*a^3*b + 4*
a*b^3 - 3*b^4)*sin(f*x + e)^4 + a^4 - 2*a^3*b + a^2*b^2 - (5*a^4 - 7*a^3*b - a^2*b^2 + 3*a*b^3)*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 - 2*a^5*b + a^4*b^2)*sin(f*x + e)^4) + 2*(a^2 +
2*a*b + 3*b^2)*log(sin(f*x + e)^2)/a^4)/f

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (159) = 318\).
time = 1.30, size = 360, normalized size = 2.24 \begin {gather*} \frac {{\left (3 \, a^{4} b - 2 \, a^{3} b^{2} - 5 \, a^{2} b^{3} + 6 \, a b^{4}\right )} \tan \left (f x + e\right )^{6} - a^{5} + 2 \, a^{4} b - a^{3} b^{2} + {\left (3 \, a^{5} - 5 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + 6 \, a b^{4}\right )} \tan \left (f x + e\right )^{4} + {\left (2 \, a^{5} - a^{4} b - 4 \, a^{3} b^{2} + 3 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left ({\left (a^{4} b - 4 \, a b^{4} + 3 \, b^{5}\right )} \tan \left (f x + e\right )^{6} + {\left (a^{5} - 4 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \tan \left (f x + e\right )^{4}\right )} \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left ({\left (4 \, a b^{4} - 3 \, b^{5}\right )} \tan \left (f x + e\right )^{6} + {\left (4 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \tan \left (f x + e\right )^{4}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{6} b - 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \tan \left (f x + e\right )^{6} + {\left (a^{7} - 2 \, a^{6} b + a^{5} b^{2}\right )} f \tan \left (f x + e\right )^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((3*a^4*b - 2*a^3*b^2 - 5*a^2*b^3 + 6*a*b^4)*tan(f*x + e)^6 - a^5 + 2*a^4*b - a^3*b^2 + (3*a^5 - 5*a^3*b^2
- 2*a^2*b^3 + 6*a*b^4)*tan(f*x + e)^4 + (2*a^5 - a^4*b - 4*a^3*b^2 + 3*a^2*b^3)*tan(f*x + e)^2 + 2*((a^4*b -
4*a*b^4 + 3*b^5)*tan(f*x + e)^6 + (a^5 - 4*a^2*b^3 + 3*a*b^4)*tan(f*x + e)^4)*log(tan(f*x + e)^2/(tan(f*x + e)
^2 + 1)) + 2*((4*a*b^4 - 3*b^5)*tan(f*x + e)^6 + (4*a^2*b^3 - 3*a*b^4)*tan(f*x + e)^4)*log((b*tan(f*x + e)^2 +
a)/(tan(f*x + e)^2 + 1)))/((a^6*b - 2*a^5*b^2 + a^4*b^3)*f*tan(f*x + e)^6 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(f
*x + e)^4)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**5/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 685 vs. \(2 (159) = 318\).
time = 1.11, size = 685, normalized size = 4.25 \begin {gather*} \frac {\frac {32 \, {\left (4 \, a b^{3} - 3 \, b^{4}\right )} \log \left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, 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}}\right )}{a^{6} - 2 \, a^{5} b + a^{4} b^{2}} - \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} - 2 \, a b + b^{2}} - \frac {32 \, {\left (4 \, a^{2} b^{3} - 3 \, a b^{4} + \frac {8 \, a^{2} b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {18 \, a b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {8 \, 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 {3 \, a b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} {\left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, 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}}\right )}} + \frac {32 \, {\left (a^{2} + 2 \, a b + 3 \, 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}} - \frac {\frac {12 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {16 \, a b {\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}}}{a^{4}} - \frac {{\left (a^{2} + \frac {12 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {16 \, a b {\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 {96 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {144 \, 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}}{a^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}}{64 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(32*(4*a*b^3 - 3*b^4)*log(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x
+ e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^6 - 2*a^5*b + a^4*b^2) - 64*log(abs(-(cos(f*x + e
) - 1)/(cos(f*x + e) + 1) + 1))/(a^2 - 2*a*b + b^2) - 32*(4*a^2*b^3 - 3*a*b^4 + 8*a^2*b^3*(cos(f*x + e) - 1)/(
cos(f*x + e) + 1) - 18*a*b^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 8*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 - 3*a*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/
((a^6 - 2*a^5*b + a^4*b^2)*(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)) + 32*(a^2 + 2*a*b + 3*b^2)*log(abs(-cos(f*x + e) + 1)/
abs(cos(f*x + e) + 1))/a^4 - (12*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 16*a*b*(cos(f*x + e) - 1)/(cos(f*
x + e) + 1) + a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/a^4 - (a^2 + 12*a^2*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) + 16*a*b*(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 + 9
6*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 144*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x
+ e) + 1)^2/(a^4*(cos(f*x + e) - 1)^2))/f

________________________________________________________________________________________

Mupad [B]
time = 12.37, size = 191, normalized size = 1.19 \begin {gather*} \frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a+3\,b\right )}{4\,a^2}-\frac {1}{4\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2\,b+a\,b^2-3\,b^3\right )}{2\,a^3\,\left (a-b\right )}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^6+a\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,{\left (a-b\right )}^2}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+2\,a\,b+3\,b^2\right )}{a^4\,f}+\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (4\,a\,b^3-3\,b^4\right )}{f\,\left (2\,a^6-4\,a^5\,b+2\,a^4\,b^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________