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

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

[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))

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Rubi [A]  time = 0.179065, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3670, 446, 88} \[ -\frac{b^3}{2 a^3 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac{b^3 (4 a-3 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 f (a-b)^2}+\frac{\left (a^2+2 a b+3 b^2\right ) \log (\tan (e+f x))}{a^4 f}+\frac{(a+2 b) \cot ^2(e+f x)}{2 a^3 f}-\frac{\cot ^4(e+f x)}{4 a^2 f}+\frac{\log (\cos (e+f x))}{f (a-b)^2} \]

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 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]))

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 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]))

Rubi steps

\begin{align*} \int \frac{\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{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{\operatorname{Subst}\left (\int \frac{1}{x^3 (1+x) (a+b x)^2} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{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 = 1.01116, size = 121, normalized size = 0.75 \[ -\frac{-\frac{b^4}{a^4 (a-b) \left (a \cot ^2(e+f x)+b\right )}-\frac{b^3 (4 a-3 b) \log \left (a \cot ^2(e+f x)+b\right )}{a^4 (a-b)^2}-\frac{(a+2 b) \cot ^2(e+f x)}{a^3}+\frac{\cot ^4(e+f x)}{2 a^2}-\frac{2 \log (\sin (e+f x))}{(a-b)^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-(((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)/(2*f)

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Maple [B]  time = 0.109, size = 347, normalized size = 2.2 \begin{align*} -{\frac{1}{16\,f{a}^{2} \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}+{\frac{7}{16\,f{a}^{2} \left ( \cos \left ( fx+e \right ) +1 \right ) }}+{\frac{b}{2\,f{a}^{3} \left ( \cos \left ( fx+e \right ) +1 \right ) }}+{\frac{\ln \left ( \cos \left ( fx+e \right ) +1 \right ) }{2\,f{a}^{2}}}+{\frac{\ln \left ( \cos \left ( fx+e \right ) +1 \right ) b}{f{a}^{3}}}+{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) +1 \right ){b}^{2}}{2\,f{a}^{4}}}+{\frac{{b}^{4}}{2\,f{a}^{3} \left ( a-b \right ) ^{2} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }}+2\,{\frac{{b}^{3}\ln \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }{f{a}^{3} \left ( a-b \right ) ^{2}}}-{\frac{3\,{b}^{4}\ln \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }{2\,f{a}^{4} \left ( a-b \right ) ^{2}}}-{\frac{1}{16\,f{a}^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2}}}-{\frac{7}{16\,f{a}^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) }}-{\frac{b}{2\,f{a}^{3} \left ( \cos \left ( fx+e \right ) -1 \right ) }}+{\frac{\ln \left ( \cos \left ( fx+e \right ) -1 \right ) }{2\,f{a}^{2}}}+{\frac{\ln \left ( \cos \left ( fx+e \right ) -1 \right ) b}{f{a}^{3}}}+{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) -1 \right ){b}^{2}}{2\,f{a}^{4}}} \end{align*}

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)

[Out]

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

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Maxima [A]  time = 1.15325, size = 319, normalized size = 1.98 \begin{align*} \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{align*}

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

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Fricas [B]  time = 1.47444, size = 759, normalized size = 4.71 \begin{align*} \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{align*}

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)

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

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

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Giac [B]  time = 1.49274, size = 919, normalized size = 5.71 \begin{align*} \text{result too large to display} \end{align*}

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(-(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(-(cos(f*x + e) - 1)/(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 + 96*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