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

Optimal. Leaf size=181 \[ \frac{b^2 (3 a-2 b)}{2 a^3 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{b^2}{4 a^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b^2 \left (6 a^2-8 a b+3 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 f (a-b)^3}-\frac{(a+3 b) \log (\tan (e+f x))}{a^4 f}-\frac{\cot ^2(e+f x)}{2 a^3 f}-\frac{\log (\cos (e+f x))}{f (a-b)^3} \]

[Out]

-Cot[e + f*x]^2/(2*a^3*f) - Log[Cos[e + f*x]]/((a - b)^3*f) - ((a + 3*b)*Log[Tan[e + f*x]])/(a^4*f) - (b^2*(6*
a^2 - 8*a*b + 3*b^2)*Log[a + b*Tan[e + f*x]^2])/(2*a^4*(a - b)^3*f) + b^2/(4*a^2*(a - b)*f*(a + b*Tan[e + f*x]
^2)^2) + ((3*a - 2*b)*b^2)/(2*a^3*(a - b)^2*f*(a + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.213556, antiderivative size = 181, 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} \[ \frac{b^2 (3 a-2 b)}{2 a^3 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{b^2}{4 a^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b^2 \left (6 a^2-8 a b+3 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 f (a-b)^3}-\frac{(a+3 b) \log (\tan (e+f x))}{a^4 f}-\frac{\cot ^2(e+f x)}{2 a^3 f}-\frac{\log (\cos (e+f x))}{f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cot[e + f*x]^2/(2*a^3*f) - Log[Cos[e + f*x]]/((a - b)^3*f) - ((a + 3*b)*Log[Tan[e + f*x]])/(a^4*f) - (b^2*(6*
a^2 - 8*a*b + 3*b^2)*Log[a + b*Tan[e + f*x]^2])/(2*a^4*(a - b)^3*f) + b^2/(4*a^2*(a - b)*f*(a + b*Tan[e + f*x]
^2)^2) + ((3*a - 2*b)*b^2)/(2*a^3*(a - b)^2*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 )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^3 \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^2 (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^2}+\frac{-a-3 b}{a^4 x}+\frac{1}{(a-b)^3 (1+x)}-\frac{b^3}{a^2 (a-b) (a+b x)^3}-\frac{(3 a-2 b) b^3}{a^3 (a-b)^2 (a+b x)^2}-\frac{b^3 \left (6 a^2-8 a b+3 b^2\right )}{a^4 (a-b)^3 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{\cot ^2(e+f x)}{2 a^3 f}-\frac{\log (\cos (e+f x))}{(a-b)^3 f}-\frac{(a+3 b) \log (\tan (e+f x))}{a^4 f}-\frac{b^2 \left (6 a^2-8 a b+3 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 (a-b)^3 f}+\frac{b^2}{4 a^2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{(3 a-2 b) b^2}{2 a^3 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.92266, size = 144, normalized size = 0.8 \[ -\frac{-\frac{b^4}{2 a^4 (a-b) \left (a \cot ^2(e+f x)+b\right )^2}+\frac{b^3 (4 a-3 b)}{a^4 (a-b)^2 \left (a \cot ^2(e+f x)+b\right )}+\frac{b^2 \left (6 a^2-8 a b+3 b^2\right ) \log \left (a \cot ^2(e+f x)+b\right )}{a^4 (a-b)^3}+\frac{\cot ^2(e+f x)}{a^3}+\frac{2 \log (\sin (e+f x))}{(a-b)^3}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cot[e + f*x]^2/a^3 - b^4/(2*a^4*(a - b)*(b + a*Cot[e + f*x]^2)^2) + ((4*a - 3*b)*b^3)/(a^4*(a - b)^2*(b + a*
Cot[e + f*x]^2)) + (b^2*(6*a^2 - 8*a*b + 3*b^2)*Log[b + a*Cot[e + f*x]^2])/(a^4*(a - b)^3) + (2*Log[Sin[e + f*
x]])/(a - b)^3)/(2*f)

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Maple [B]  time = 0.105, size = 362, normalized size = 2. \begin{align*} -{\frac{1}{4\,f{a}^{3} \left ( \cos \left ( fx+e \right ) +1 \right ) }}-{\frac{\ln \left ( \cos \left ( fx+e \right ) +1 \right ) }{2\,f{a}^{3}}}-{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) +1 \right ) b}{2\,f{a}^{4}}}-3\,{\frac{{b}^{2}\ln \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }{f{a}^{2} \left ( a-b \right ) ^{3}}}+4\,{\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 ) ^{3}}}-{\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 ) ^{3}}}-2\,{\frac{{b}^{3}}{f{a}^{2} \left ( a-b \right ) ^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }}+{\frac{{b}^{4}}{f{a}^{3} \left ( a-b \right ) ^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }}+{\frac{{b}^{4}}{4\,f{a}^{2} \left ( a-b \right ) ^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}+{\frac{1}{4\,f{a}^{3} \left ( \cos \left ( fx+e \right ) -1 \right ) }}-{\frac{\ln \left ( \cos \left ( fx+e \right ) -1 \right ) }{2\,f{a}^{3}}}-{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) -1 \right ) b}{2\,f{a}^{4}}} \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)^3,x)

[Out]

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

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Maxima [A]  time = 1.13208, size = 466, normalized size = 2.57 \begin{align*} -\frac{\frac{2 \,{\left (6 \, a^{2} b^{2} - 8 \, a b^{3} + 3 \, b^{4}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}} + \frac{2 \, a^{5} - 6 \, a^{4} b + 6 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + 2 \,{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 14 \, a^{2} b^{3} + 11 \, a b^{4} - 3 \, b^{5}\right )} \sin \left (f x + e\right )^{4} -{\left (4 \, a^{5} - 16 \, a^{4} b + 24 \, a^{3} b^{2} - 24 \, a^{2} b^{3} + 9 \, a b^{4}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{8} - 5 \, a^{7} b + 10 \, a^{6} b^{2} - 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} - a^{3} b^{5}\right )} \sin \left (f x + e\right )^{6} - 2 \,{\left (a^{8} - 4 \, a^{7} b + 6 \, a^{6} b^{2} - 4 \, a^{5} b^{3} + a^{4} b^{4}\right )} \sin \left (f x + e\right )^{4} +{\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac{2 \,{\left (a + 3 \, b\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)^3/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

-1/4*(2*(6*a^2*b^2 - 8*a*b^3 + 3*b^4)*log(-(a - b)*sin(f*x + e)^2 + a)/(a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3) +
(2*a^5 - 6*a^4*b + 6*a^3*b^2 - 2*a^2*b^3 + 2*(a^5 - 5*a^4*b + 10*a^3*b^2 - 14*a^2*b^3 + 11*a*b^4 - 3*b^5)*sin
(f*x + e)^4 - (4*a^5 - 16*a^4*b + 24*a^3*b^2 - 24*a^2*b^3 + 9*a*b^4)*sin(f*x + e)^2)/((a^8 - 5*a^7*b + 10*a^6*
b^2 - 10*a^5*b^3 + 5*a^4*b^4 - a^3*b^5)*sin(f*x + e)^6 - 2*(a^8 - 4*a^7*b + 6*a^6*b^2 - 4*a^5*b^3 + a^4*b^4)*s
in(f*x + e)^4 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*sin(f*x + e)^2) + 2*(a + 3*b)*log(sin(f*x + e)^2)/a^4)/f

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Fricas [B]  time = 2.05262, size = 1175, normalized size = 6.49 \begin{align*} -\frac{{\left (2 \, a^{4} b^{2} - 6 \, a^{3} b^{3} + 13 \, a^{2} b^{4} - 6 \, a b^{5}\right )} \tan \left (f x + e\right )^{6} + 2 \, a^{6} - 6 \, a^{5} b + 6 \, a^{4} b^{2} - 2 \, a^{3} b^{3} + 2 \,{\left (2 \, a^{5} b - 5 \, a^{4} b^{2} + 7 \, a^{3} b^{3} + 2 \, a^{2} b^{4} - 3 \, a b^{5}\right )} \tan \left (f x + e\right )^{4} +{\left (2 \, a^{6} - 2 \, a^{5} b - 6 \, a^{4} b^{2} + 18 \, a^{3} b^{3} - 9 \, a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2} + 2 \,{\left ({\left (a^{4} b^{2} - 6 \, a^{2} b^{4} + 8 \, a b^{5} - 3 \, b^{6}\right )} \tan \left (f x + e\right )^{6} + 2 \,{\left (a^{5} b - 6 \, a^{3} b^{3} + 8 \, a^{2} b^{4} - 3 \, a b^{5}\right )} \tan \left (f x + e\right )^{4} +{\left (a^{6} - 6 \, a^{4} b^{2} + 8 \, a^{3} b^{3} - 3 \, a^{2} b^{4}\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 ) + 2 \,{\left ({\left (6 \, a^{2} b^{4} - 8 \, a b^{5} + 3 \, b^{6}\right )} \tan \left (f x + e\right )^{6} + 2 \,{\left (6 \, a^{3} b^{3} - 8 \, a^{2} b^{4} + 3 \, a b^{5}\right )} \tan \left (f x + e\right )^{4} +{\left (6 \, a^{4} b^{2} - 8 \, a^{3} b^{3} + 3 \, a^{2} b^{4}\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 )}{4 \,{\left ({\left (a^{7} b^{2} - 3 \, a^{6} b^{3} + 3 \, a^{5} b^{4} - a^{4} b^{5}\right )} f \tan \left (f x + e\right )^{6} + 2 \,{\left (a^{8} b - 3 \, a^{7} b^{2} + 3 \, a^{6} b^{3} - a^{5} b^{4}\right )} f \tan \left (f x + e\right )^{4} +{\left (a^{9} - 3 \, a^{8} b + 3 \, a^{7} b^{2} - a^{6} b^{3}\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)^3,x, algorithm="fricas")

[Out]

-1/4*((2*a^4*b^2 - 6*a^3*b^3 + 13*a^2*b^4 - 6*a*b^5)*tan(f*x + e)^6 + 2*a^6 - 6*a^5*b + 6*a^4*b^2 - 2*a^3*b^3
+ 2*(2*a^5*b - 5*a^4*b^2 + 7*a^3*b^3 + 2*a^2*b^4 - 3*a*b^5)*tan(f*x + e)^4 + (2*a^6 - 2*a^5*b - 6*a^4*b^2 + 18
*a^3*b^3 - 9*a^2*b^4)*tan(f*x + e)^2 + 2*((a^4*b^2 - 6*a^2*b^4 + 8*a*b^5 - 3*b^6)*tan(f*x + e)^6 + 2*(a^5*b -
6*a^3*b^3 + 8*a^2*b^4 - 3*a*b^5)*tan(f*x + e)^4 + (a^6 - 6*a^4*b^2 + 8*a^3*b^3 - 3*a^2*b^4)*tan(f*x + e)^2)*lo
g(tan(f*x + e)^2/(tan(f*x + e)^2 + 1)) + 2*((6*a^2*b^4 - 8*a*b^5 + 3*b^6)*tan(f*x + e)^6 + 2*(6*a^3*b^3 - 8*a^
2*b^4 + 3*a*b^5)*tan(f*x + e)^4 + (6*a^4*b^2 - 8*a^3*b^3 + 3*a^2*b^4)*tan(f*x + e)^2)*log((b*tan(f*x + e)^2 +
a)/(tan(f*x + e)^2 + 1)))/((a^7*b^2 - 3*a^6*b^3 + 3*a^5*b^4 - a^4*b^5)*f*tan(f*x + e)^6 + 2*(a^8*b - 3*a^7*b^2
+ 3*a^6*b^3 - a^5*b^4)*f*tan(f*x + e)^4 + (a^9 - 3*a^8*b + 3*a^7*b^2 - a^6*b^3)*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)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.71084, size = 1218, normalized size = 6.73 \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)^3,x, algorithm="giac")

[Out]

-1/8*(4*(6*a^2*b^2 - 8*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^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3) - 8
*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 2*(18*a^4*b^2 - 24*a^3*b^3
+ 9*a^2*b^4 + 72*a^4*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 208*a^3*b^3*(cos(f*x + e) - 1)/(cos(f*x + e)
+ 1) + 172*a^2*b^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 48*a*b^5*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 10
8*a^4*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 368*a^3*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 +
502*a^2*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 288*a*b^5*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 +
64*b^6*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 72*a^4*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 208*
a^3*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 172*a^2*b^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 48
*a*b^5*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 18*a^4*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 - 24*a
^3*b^3*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 9*a^2*b^4*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/((a^7
- 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*(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)^2) + 4*(a + 3*b)*log(-(cos(f*x + e) - 1)/(cos(f*x
+ e) + 1))/a^4 - (a + 4*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 12*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))
*(cos(f*x + e) + 1)/(a^4*(cos(f*x + e) - 1)) - (cos(f*x + e) - 1)/(a^3*(cos(f*x + e) + 1)))/f