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

Optimal. Leaf size=132 \[ \frac{b^2}{2 a^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{b^2 (3 a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^3 f (a-b)^2}-\frac{(a+2 b) \log (\tan (e+f x))}{a^3 f}-\frac{\cot ^2(e+f x)}{2 a^2 f}-\frac{\log (\cos (e+f x))}{f (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.155696, antiderivative size = 132, 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} \[ \frac{b^2}{2 a^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{b^2 (3 a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^3 f (a-b)^2}-\frac{(a+2 b) \log (\tan (e+f x))}{a^3 f}-\frac{\cot ^2(e+f x)}{2 a^2 f}-\frac{\log (\cos (e+f x))}{f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cot[e + f*x]^2/(2*a^2*f) - Log[Cos[e + f*x]]/((a - b)^2*f) - ((a + 2*b)*Log[Tan[e + f*x]])/(a^3*f) - ((3*a -
2*b)*b^2*Log[a + b*Tan[e + f*x]^2])/(2*a^3*(a - b)^2*f) + b^2/(2*a^2*(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 ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^3 \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^2 (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^2}+\frac{-a-2 b}{a^3 x}+\frac{1}{(a-b)^2 (1+x)}-\frac{b^3}{a^2 (a-b) (a+b x)^2}-\frac{(3 a-2 b) b^3}{a^3 (a-b)^2 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{\cot ^2(e+f x)}{2 a^2 f}-\frac{\log (\cos (e+f x))}{(a-b)^2 f}-\frac{(a+2 b) \log (\tan (e+f x))}{a^3 f}-\frac{(3 a-2 b) b^2 \log \left (a+b \tan ^2(e+f x)\right )}{2 a^3 (a-b)^2 f}+\frac{b^2}{2 a^2 (a-b) f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.825586, size = 98, normalized size = 0.74 \[ -\frac{\frac{b^3}{a^3 (a-b) \left (a \cot ^2(e+f x)+b\right )}+\frac{b^2 (3 a-2 b) \log \left (a \cot ^2(e+f x)+b\right )}{a^3 (a-b)^2}+\frac{\cot ^2(e+f x)}{a^2}+\frac{2 \log (\sin (e+f x))}{(a-b)^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.101, size = 234, normalized size = 1.8 \begin{align*} -{\frac{1}{4\,f{a}^{2} \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{{b}^{3}}{2\,f{a}^{2} \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 ) }}-{\frac{3\,{b}^{2}\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}^{2} \left ( a-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{1}{4\,f{a}^{2} \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16446, size = 252, normalized size = 1.91 \begin{align*} -\frac{\frac{{\left (3 \, a b^{2} - 2 \, b^{3}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}} - \frac{a^{3} - 2 \, a^{2} b + a b^{2} -{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - 2 \, b^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \sin \left (f x + e\right )^{4} -{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} \sin \left (f x + e\right )^{2}} + \frac{{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.38986, size = 648, normalized size = 4.91 \begin{align*} -\frac{{\left (a^{3} b - 2 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{4} + a^{4} - 2 \, a^{3} b + a^{2} b^{2} +{\left (a^{4} - a^{3} b - a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2} +{\left ({\left (a^{3} b - 3 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (f x + e\right )^{4} +{\left (a^{4} - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac{\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) +{\left ({\left (3 \, a b^{3} - 2 \, b^{4}\right )} \tan \left (f x + e\right )^{4} +{\left (3 \, a^{2} b^{2} - 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac{b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \,{\left ({\left (a^{5} b - 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \tan \left (f x + e\right )^{4} +{\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} f \tan \left (f x + e\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**3/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.46413, size = 918, normalized size = 6.95 \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)^3/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/24*(12*(3*a*b^2 - 2*b^3)*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^5 - 2*a^4*b + a^3*b^2) - 24*log(-(cos(f*x + e) -
1)/(cos(f*x + e) + 1) + 1)/(a^2 - 2*a*b + b^2) - (3*a^4 - 6*a^3*b + 3*a^2*b^2 + 10*a^4*(cos(f*x + e) - 1)/(co
s(f*x + e) + 1) - 24*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 42*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e)
+ 1) - 20*a*b^3*(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 + 12*a^2*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 8*a*b^3*(cos(f*
x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 - 2*a^4*b + a^3*b^2)*(a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2*a*
(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 4*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)) + 12*(a + 2*b)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/a^3 - 3*(cos(f*x + e) -
1)/(a^2*(cos(f*x + e) + 1)))/f