3.242 \(\int \frac {\cot ^5(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=210 \[ -\frac {b^3 (4 a-3 b)}{2 a^4 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac {(a+3 b) \cot ^2(e+f x)}{2 a^4 f}-\frac {b^3}{4 a^3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\cot ^4(e+f x)}{4 a^3 f}+\frac {\left (a^2+3 a b+6 b^2\right ) \log (\tan (e+f x))}{a^5 f}+\frac {b^3 \left (10 a^2-15 a b+6 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^5 f (a-b)^3}+\frac {\log (\cos (e+f x))}{f (a-b)^3} \]

[Out]

1/2*(a+3*b)*cot(f*x+e)^2/a^4/f-1/4*cot(f*x+e)^4/a^3/f+ln(cos(f*x+e))/(a-b)^3/f+(a^2+3*a*b+6*b^2)*ln(tan(f*x+e)
)/a^5/f+1/2*b^3*(10*a^2-15*a*b+6*b^2)*ln(a+b*tan(f*x+e)^2)/a^5/(a-b)^3/f-1/4*b^3/a^3/(a-b)/f/(a+b*tan(f*x+e)^2
)^2-1/2*(4*a-3*b)*b^3/a^4/(a-b)^2/f/(a+b*tan(f*x+e)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 210, 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 = {3670, 446, 88} \[ -\frac {b^3 (4 a-3 b)}{2 a^4 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {b^3}{4 a^3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}+\frac {b^3 \left (10 a^2-15 a b+6 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^5 f (a-b)^3}+\frac {\left (a^2+3 a b+6 b^2\right ) \log (\tan (e+f x))}{a^5 f}+\frac {(a+3 b) \cot ^2(e+f x)}{2 a^4 f}-\frac {\cot ^4(e+f x)}{4 a^3 f}+\frac {\log (\cos (e+f x))}{f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

((a + 3*b)*Cot[e + f*x]^2)/(2*a^4*f) - Cot[e + f*x]^4/(4*a^3*f) + Log[Cos[e + f*x]]/((a - b)^3*f) + ((a^2 + 3*
a*b + 6*b^2)*Log[Tan[e + f*x]])/(a^5*f) + (b^3*(10*a^2 - 15*a*b + 6*b^2)*Log[a + b*Tan[e + f*x]^2])/(2*a^5*(a
- b)^3*f) - b^3/(4*a^3*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) - ((4*a - 3*b)*b^3)/(2*a^4*(a - b)^2*f*(a + b*Tan[e
+ f*x]^2))

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 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 ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^5 \left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^3 (1+x) (a+b x)^3} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a^3 x^3}+\frac {-a-3 b}{a^4 x^2}+\frac {a^2+3 a b+6 b^2}{a^5 x}-\frac {1}{(a-b)^3 (1+x)}+\frac {b^4}{a^3 (a-b) (a+b x)^3}+\frac {(4 a-3 b) b^4}{a^4 (a-b)^2 (a+b x)^2}+\frac {b^4 \left (10 a^2-15 a b+6 b^2\right )}{a^5 (a-b)^3 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {(a+3 b) \cot ^2(e+f x)}{2 a^4 f}-\frac {\cot ^4(e+f x)}{4 a^3 f}+\frac {\log (\cos (e+f x))}{(a-b)^3 f}+\frac {\left (a^2+3 a b+6 b^2\right ) \log (\tan (e+f x))}{a^5 f}+\frac {b^3 \left (10 a^2-15 a b+6 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^5 (a-b)^3 f}-\frac {b^3}{4 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(4 a-3 b) b^3}{2 a^4 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.65, size = 178, normalized size = 0.85 \[ \frac {\frac {(a+3 b) \cot ^2(e+f x)}{a^4}-\frac {\cot ^4(e+f x)}{2 a^3}+\frac {4 \left (a^2+3 a b+6 b^2\right ) \log (\tan (e+f x))+\frac {b^3 \left (2 \left (10 a^2-15 a b+6 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )-\frac {a (a-b) \left (2 b (4 a-3 b) \tan ^2(e+f x)+a (9 a-7 b)\right )}{\left (a+b \tan ^2(e+f x)\right )^2}\right )}{(a-b)^3}}{2 a^5}+\frac {2 \log (\cos (e+f x))}{(a-b)^3}}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(((a + 3*b)*Cot[e + f*x]^2)/a^4 - Cot[e + f*x]^4/(2*a^3) + (2*Log[Cos[e + f*x]])/(a - b)^3 + (4*(a^2 + 3*a*b +
6*b^2)*Log[Tan[e + f*x]] + (b^3*(2*(10*a^2 - 15*a*b + 6*b^2)*Log[a + b*Tan[e + f*x]^2] - (a*(a - b)*(a*(9*a -
7*b) + 2*(4*a - 3*b)*b*Tan[e + f*x]^2))/(a + b*Tan[e + f*x]^2)^2))/(a - b)^3)/(2*a^5))/(2*f)

________________________________________________________________________________________

fricas [B]  time = 0.62, size = 611, normalized size = 2.91 \[ \frac {3 \, {\left (a^{5} b^{2} - a^{4} b^{3} - 3 \, a^{3} b^{4} + 8 \, a^{2} b^{5} - 4 \, a b^{6}\right )} \tan \left (f x + e\right )^{8} - a^{7} + 3 \, a^{6} b - 3 \, a^{5} b^{2} + a^{4} b^{3} + 2 \, {\left (3 \, a^{6} b - 2 \, a^{5} b^{2} - 9 \, a^{4} b^{3} + 14 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - 6 \, a b^{6}\right )} \tan \left (f x + e\right )^{6} + {\left (3 \, a^{7} + a^{6} b - 10 \, a^{5} b^{2} - 6 \, a^{4} b^{3} + 33 \, a^{3} b^{4} - 18 \, a^{2} b^{5}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{7} - a^{6} b - 3 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left ({\left (a^{5} b^{2} - 10 \, a^{2} b^{5} + 15 \, a b^{6} - 6 \, b^{7}\right )} \tan \left (f x + e\right )^{8} + 2 \, {\left (a^{6} b - 10 \, a^{3} b^{4} + 15 \, a^{2} b^{5} - 6 \, a b^{6}\right )} \tan \left (f x + e\right )^{6} + {\left (a^{7} - 10 \, a^{4} b^{3} + 15 \, a^{3} b^{4} - 6 \, a^{2} b^{5}\right )} \tan \left (f x + e\right )^{4}\right )} \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left ({\left (10 \, a^{2} b^{5} - 15 \, a b^{6} + 6 \, b^{7}\right )} \tan \left (f x + e\right )^{8} + 2 \, {\left (10 \, a^{3} b^{4} - 15 \, a^{2} b^{5} + 6 \, a b^{6}\right )} \tan \left (f x + e\right )^{6} + {\left (10 \, a^{4} b^{3} - 15 \, a^{3} b^{4} + 6 \, a^{2} b^{5}\right )} \tan \left (f x + e\right )^{4}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{8} b^{2} - 3 \, a^{7} b^{3} + 3 \, a^{6} b^{4} - a^{5} b^{5}\right )} f \tan \left (f x + e\right )^{8} + 2 \, {\left (a^{9} b - 3 \, a^{8} b^{2} + 3 \, a^{7} b^{3} - a^{6} b^{4}\right )} f \tan \left (f x + e\right )^{6} + {\left (a^{10} - 3 \, a^{9} b + 3 \, a^{8} b^{2} - a^{7} b^{3}\right )} f \tan \left (f x + e\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

1/4*(3*(a^5*b^2 - a^4*b^3 - 3*a^3*b^4 + 8*a^2*b^5 - 4*a*b^6)*tan(f*x + e)^8 - a^7 + 3*a^6*b - 3*a^5*b^2 + a^4*
b^3 + 2*(3*a^6*b - 2*a^5*b^2 - 9*a^4*b^3 + 14*a^3*b^4 + 3*a^2*b^5 - 6*a*b^6)*tan(f*x + e)^6 + (3*a^7 + a^6*b -
10*a^5*b^2 - 6*a^4*b^3 + 33*a^3*b^4 - 18*a^2*b^5)*tan(f*x + e)^4 + 2*(a^7 - a^6*b - 3*a^5*b^2 + 5*a^4*b^3 - 2
*a^3*b^4)*tan(f*x + e)^2 + 2*((a^5*b^2 - 10*a^2*b^5 + 15*a*b^6 - 6*b^7)*tan(f*x + e)^8 + 2*(a^6*b - 10*a^3*b^4
+ 15*a^2*b^5 - 6*a*b^6)*tan(f*x + e)^6 + (a^7 - 10*a^4*b^3 + 15*a^3*b^4 - 6*a^2*b^5)*tan(f*x + e)^4)*log(tan(
f*x + e)^2/(tan(f*x + e)^2 + 1)) + 2*((10*a^2*b^5 - 15*a*b^6 + 6*b^7)*tan(f*x + e)^8 + 2*(10*a^3*b^4 - 15*a^2*
b^5 + 6*a*b^6)*tan(f*x + e)^6 + (10*a^4*b^3 - 15*a^3*b^4 + 6*a^2*b^5)*tan(f*x + e)^4)*log((b*tan(f*x + e)^2 +
a)/(tan(f*x + e)^2 + 1)))/((a^8*b^2 - 3*a^7*b^3 + 3*a^6*b^4 - a^5*b^5)*f*tan(f*x + e)^8 + 2*(a^9*b - 3*a^8*b^2
+ 3*a^7*b^3 - a^6*b^4)*f*tan(f*x + e)^6 + (a^10 - 3*a^9*b + 3*a^8*b^2 - a^7*b^3)*f*tan(f*x + e)^4)

________________________________________________________________________________________

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*tan(f*x+e)^2)^3,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*((-32*((1-cos(
f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^3+384*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^3+768*(1-cos(f*x+exp(1))
)/(1+cos(f*x+exp(1)))*a^2*b)*1/4096/a^6+(-16*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^6*a^7-160*((1-cos(f*x+e
xp(1)))/(1+cos(f*x+exp(1))))^6*a^4*b^3+240*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^6*a^3*b^4-96*((1-cos(f*x+
exp(1)))/(1+cos(f*x+exp(1))))^6*a^2*b^5+76*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^5*a^7-140*((1-cos(f*x+exp
(1)))/(1+cos(f*x+exp(1))))^5*a^6*b-36*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^5*a^5*b^2+700*((1-cos(f*x+exp(
1)))/(1+cos(f*x+exp(1))))^5*a^4*b^3-1624*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^5*a^3*b^4+1152*((1-cos(f*x+
exp(1)))/(1+cos(f*x+exp(1))))^5*a^2*b^5-256*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^5*a*b^6-145*((1-cos(f*x+
exp(1)))/(1+cos(f*x+exp(1))))^4*a^7+403*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a^6*b-211*((1-cos(f*x+exp(
1)))/(1+cos(f*x+exp(1))))^4*a^5*b^2-1487*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a^4*b^3+3296*((1-cos(f*x+
exp(1)))/(1+cos(f*x+exp(1))))^4*a^3*b^4-2560*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a^2*b^5+256*((1-cos(f
*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a*b^6+256*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*b^7+140*((1-cos(f*x+e
xp(1)))/(1+cos(f*x+exp(1))))^3*a^7-412*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^6*b+204*((1-cos(f*x+exp(1
)))/(1+cos(f*x+exp(1))))^3*a^5*b^2+1356*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^4*b^3-3272*((1-cos(f*x+e
xp(1)))/(1+cos(f*x+exp(1))))^3*a^3*b^4+2496*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^2*b^5-640*((1-cos(f*
x+exp(1)))/(1+cos(f*x+exp(1))))^3*a*b^6-70*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^7+178*((1-cos(f*x+exp
(1)))/(1+cos(f*x+exp(1))))^2*a^6*b-34*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^5*b^2-586*((1-cos(f*x+exp(
1)))/(1+cos(f*x+exp(1))))^2*a^4*b^3+752*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^3*b^4-272*((1-cos(f*x+ex
p(1)))/(1+cos(f*x+exp(1))))^2*a^2*b^5+16*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^7-32*(1-cos(f*x+exp(1)))/(1
+cos(f*x+exp(1)))*a^6*b+32*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^4*b^3-16*(1-cos(f*x+exp(1)))/(1+cos(f*x+e
xp(1)))*a^3*b^4-a^7+3*a^6*b-3*a^5*b^2+a^4*b^3)/(128*a^8-384*a^7*b+384*a^6*b^2-128*a^5*b^3)/(((1-cos(f*x+exp(1)
))/(1+cos(f*x+exp(1))))^3*a-2*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+4*((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)))*a)^2-1/(2*a^3-6*a^2*b+6*a*b^2-2*b^3)*ln(abs((1-cos(f*x+e
xp(1)))/(1+cos(f*x+exp(1)))+1))+(a^2+3*a*b+6*b^2)*1/4/a^5*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1))))+(1
0*a^2*b^3-15*a*b^4+6*b^5)/(4*a^8-12*a^7*b+12*a^6*b^2-4*a^5*b^3)*ln(((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)))*a+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+a))

________________________________________________________________________________________

maple [B]  time = 0.92, size = 477, normalized size = 2.27 \[ \frac {5 b^{4}}{2 f \,a^{3} \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}-\frac {3 b^{5}}{2 f \,a^{4} \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {5 b^{3} \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{f \,a^{3} \left (a -b \right )^{3}}-\frac {15 b^{4} \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{2 f \,a^{4} \left (a -b \right )^{3}}+\frac {3 b^{5} \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{f \,a^{5} \left (a -b \right )^{3}}-\frac {b^{5}}{4 f \,a^{3} \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}-\frac {1}{16 f \,a^{3} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7}{16 f \,a^{3} \left (-1+\cos \left (f x +e \right )\right )}-\frac {3 b}{4 f \,a^{4} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 f \,a^{3}}+\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{4}}+\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b^{2}}{f \,a^{5}}-\frac {1}{16 f \,a^{3} \left (1+\cos \left (f x +e \right )\right )^{2}}+\frac {7}{16 f \,a^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {3 b}{4 f \,a^{4} \left (1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 f \,a^{3}}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{4}}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b^{2}}{f \,a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.38, size = 416, normalized size = 1.98 \[ \frac {\frac {2 \, {\left (10 \, a^{2} b^{3} - 15 \, a b^{4} + 6 \, b^{5}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}} + \frac {2 \, {\left (2 \, a^{6} - 7 \, a^{5} b + 5 \, a^{4} b^{2} + 10 \, a^{3} b^{3} - 25 \, a^{2} b^{4} + 21 \, a b^{5} - 6 \, b^{6}\right )} \sin \left (f x + e\right )^{6} - a^{6} + 3 \, a^{5} b - 3 \, a^{4} b^{2} + a^{3} b^{3} - {\left (9 \, a^{6} - 25 \, a^{5} b + 10 \, a^{4} b^{2} + 30 \, a^{3} b^{3} - 45 \, a^{2} b^{4} + 18 \, a b^{5}\right )} \sin \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{6} - 7 \, a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - 2 \, a^{2} b^{4}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{9} - 5 \, a^{8} b + 10 \, a^{7} b^{2} - 10 \, a^{6} b^{3} + 5 \, a^{5} b^{4} - a^{4} b^{5}\right )} \sin \left (f x + e\right )^{8} - 2 \, {\left (a^{9} - 4 \, a^{8} b + 6 \, a^{7} b^{2} - 4 \, a^{6} b^{3} + a^{5} b^{4}\right )} \sin \left (f x + e\right )^{6} + {\left (a^{9} - 3 \, a^{8} b + 3 \, a^{7} b^{2} - a^{6} b^{3}\right )} \sin \left (f x + e\right )^{4}} + \frac {2 \, {\left (a^{2} + 3 \, a b + 6 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{5}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

1/4*(2*(10*a^2*b^3 - 15*a*b^4 + 6*b^5)*log(-(a - b)*sin(f*x + e)^2 + a)/(a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)
+ (2*(2*a^6 - 7*a^5*b + 5*a^4*b^2 + 10*a^3*b^3 - 25*a^2*b^4 + 21*a*b^5 - 6*b^6)*sin(f*x + e)^6 - a^6 + 3*a^5*b
- 3*a^4*b^2 + a^3*b^3 - (9*a^6 - 25*a^5*b + 10*a^4*b^2 + 30*a^3*b^3 - 45*a^2*b^4 + 18*a*b^5)*sin(f*x + e)^4 +
2*(3*a^6 - 7*a^5*b + 3*a^4*b^2 + 3*a^3*b^3 - 2*a^2*b^4)*sin(f*x + e)^2)/((a^9 - 5*a^8*b + 10*a^7*b^2 - 10*a^6
*b^3 + 5*a^5*b^4 - a^4*b^5)*sin(f*x + e)^8 - 2*(a^9 - 4*a^8*b + 6*a^7*b^2 - 4*a^6*b^3 + a^5*b^4)*sin(f*x + e)^
6 + (a^9 - 3*a^8*b + 3*a^7*b^2 - a^6*b^3)*sin(f*x + e)^4) + 2*(a^2 + 3*a*b + 6*b^2)*log(sin(f*x + e)^2)/a^5)/f

________________________________________________________________________________________

mupad [B]  time = 13.51, size = 269, normalized size = 1.28 \[ \frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a+2\,b\right )}{2\,a^2}-\frac {1}{4\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (a^3\,b^2+a^2\,b^3-9\,a\,b^4+6\,b^5\right )}{2\,a^4\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (4\,a^3\,b+3\,a^2\,b^2-27\,a\,b^3+18\,b^4\right )}{4\,a^3\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left (a^2\,{\mathrm {tan}\left (e+f\,x\right )}^4+2\,a\,b\,{\mathrm {tan}\left (e+f\,x\right )}^6+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^8\right )}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (\frac {3\,b}{2\,a^4}+\frac {1}{2\,a^3}-\frac {1}{2\,{\left (a-b\right )}^3}+\frac {3\,b^2}{a^5}\right )}{f}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,{\left (a-b\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+3\,a\,b+6\,b^2\right )}{a^5\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^5/(a + b*tan(e + f*x)^2)^3,x)

[Out]

((tan(e + f*x)^2*(a + 2*b))/(2*a^2) - 1/(4*a) + (tan(e + f*x)^6*(6*b^5 - 9*a*b^4 + a^2*b^3 + a^3*b^2))/(2*a^4*
(a^2 - 2*a*b + b^2)) + (tan(e + f*x)^4*(4*a^3*b - 27*a*b^3 + 18*b^4 + 3*a^2*b^2))/(4*a^3*(a^2 - 2*a*b + b^2)))
/(f*(a^2*tan(e + f*x)^4 + b^2*tan(e + f*x)^8 + 2*a*b*tan(e + f*x)^6)) - (log(a + b*tan(e + f*x)^2)*((3*b)/(2*a
^4) + 1/(2*a^3) - 1/(2*(a - b)^3) + (3*b^2)/a^5))/f - log(tan(e + f*x)^2 + 1)/(2*f*(a - b)^3) + (log(tan(e + f
*x))*(3*a*b + a^2 + 6*b^2))/(a^5*f)

________________________________________________________________________________________

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*tan(f*x+e)**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________