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

Optimal. Leaf size=74 \[ -\frac{b^2 \log \left (a \cos ^2(e+f x)+b\right )}{2 a f (a+b)^2}-\frac{\csc ^2(e+f x)}{2 f (a+b)}-\frac{(a+2 b) \log (\sin (e+f x))}{f (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.110046, antiderivative size = 74, 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 = {4138, 446, 88} \[ -\frac{b^2 \log \left (a \cos ^2(e+f x)+b\right )}{2 a f (a+b)^2}-\frac{\csc ^2(e+f x)}{2 f (a+b)}-\frac{(a+2 b) \log (\sin (e+f x))}{f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

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)}{a+b \sec ^2(e+f x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^5}{\left (1-x^2\right )^2 \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1-x)^2 (b+a x)} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(a+b) (-1+x)^2}+\frac{a+2 b}{(a+b)^2 (-1+x)}+\frac{b^2}{(a+b)^2 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\csc ^2(e+f x)}{2 (a+b) f}-\frac{b^2 \log \left (b+a \cos ^2(e+f x)\right )}{2 a (a+b)^2 f}-\frac{(a+2 b) \log (\sin (e+f x))}{(a+b)^2 f}\\ \end{align*}

Mathematica [A]  time = 0.234977, size = 100, normalized size = 1.35 \[ -\frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (b^2 \log \left (-a \sin ^2(e+f x)+a+b\right )+a (a+b) \csc ^2(e+f x)+2 a (a+2 b) \log (\sin (e+f x))\right )}{4 a f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(a*(a + b)*Csc[e + f*x]^2 + 2*a*(a + 2*b)*Log[Sin[e + f*x]] + b^2*Log[a + b -
a*Sin[e + f*x]^2])*Sec[e + f*x]^2)/(4*a*(a + b)^2*f*(a + b*Sec[e + f*x]^2))

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Maple [B]  time = 0.087, size = 158, normalized size = 2.1 \begin{align*} -{\frac{{b}^{2}\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,a \left ( a+b \right ) ^{2}f}}-{\frac{1}{f \left ( 4\,a+4\,b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) }}-{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ) a}{2\,f \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ) b}{f \left ( a+b \right ) ^{2}}}+{\frac{1}{f \left ( 4\,a+4\,b \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }}-{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ) a}{2\,f \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ) b}{f \left ( a+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00575, size = 117, normalized size = 1.58 \begin{align*} -\frac{\frac{b^{2} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{3} + 2 \, a^{2} b + a b^{2}} + \frac{{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{1}{{\left (a + b\right )} \sin \left (f x + e\right )^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.867789, size = 289, normalized size = 3.91 \begin{align*} \frac{a^{2} + a b -{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \,{\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b\right )} \log \left (\frac{1}{2} \, \sin \left (f x + e\right )\right )}{2 \,{\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{3}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**3/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(cot(e + f*x)**3/(a + b*sec(e + f*x)**2), x)

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Giac [B]  time = 1.40536, size = 419, normalized size = 5.66 \begin{align*} -\frac{\frac{4 \, b^{2} \log \left (a + b + \frac{2 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{2 \, 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}} + \frac{b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{a^{3} + 2 \, a^{2} b + a b^{2}} + \frac{4 \,{\left (a + 2 \, b\right )} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{{\left (a + b + \frac{4 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{8 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (\cos \left (f x + e\right ) - 1\right )}} - \frac{8 \, \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\cos \left (f x + e\right ) - 1}{{\left (a + b\right )}{\left (\cos \left (f x + e\right ) + 1\right )}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*b^2*log(a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1)
+ a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^3 + 2*a^2*b +
a*b^2) + 4*(a + 2*b)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a^2 + 2*a*b + b^2) - (a + b + 4*a*(cos(f*x +
e) - 1)/(cos(f*x + e) + 1) + 8*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))*(cos(f*x + e) + 1)/((a^2 + 2*a*b + b^
2)*(cos(f*x + e) - 1)) - 8*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)/a - (cos(f*x + e) - 1)/((a + b)*(co
s(f*x + e) + 1)))/f