3.328 \(\int \cot ^3(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=57 \[ -\frac{\left (a^2-b^2\right ) \log (\sin (e+f x))}{f}-\frac{(a+b)^2 \csc ^2(e+f x)}{2 f}-\frac{b^2 \log (\cos (e+f x))}{f} \]

[Out]

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

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Rubi [A]  time = 0.0805385, antiderivative size = 57, 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} \[ -\frac{\left (a^2-b^2\right ) \log (\sin (e+f x))}{f}-\frac{(a+b)^2 \csc ^2(e+f x)}{2 f}-\frac{b^2 \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.181189, size = 81, normalized size = 1.42 \[ -\frac{2 \left (a \cos ^2(e+f x)+b\right )^2 \left (2 \left (a^2-b^2\right ) \log (\sin (e+f x))+(a+b)^2 \csc ^2(e+f x)+2 b^2 \log (\cos (e+f x))\right )}{f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.063, size = 78, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2} \left ( \cot \left ( fx+e \right ) \right ) ^{2}}{2\,f}}-{\frac{{a}^{2}\ln \left ( \sin \left ( fx+e \right ) \right ) }{f}}-{\frac{ab}{f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}-{\frac{{b}^{2}}{2\,f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}+{\frac{{b}^{2}\ln \left ( \tan \left ( fx+e \right ) \right ) }{f}} \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)^2,x)

[Out]

-1/2/f*a^2*cot(f*x+e)^2-a^2*ln(sin(f*x+e))/f-1/f*a*b/sin(f*x+e)^2-1/2/f*b^2/sin(f*x+e)^2+1/f*b^2*ln(tan(f*x+e)
)

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Maxima [A]  time = 0.983106, size = 81, normalized size = 1.42 \begin{align*} -\frac{b^{2} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) +{\left (a^{2} - b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac{a^{2} + 2 \, a b + b^{2}}{\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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.527658, size = 231, normalized size = 4.05 \begin{align*} \frac{a^{2} + 2 \, a b + b^{2} -{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \log \left (\cos \left (f x + e\right )^{2}\right ) -{\left ({\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} + b^{2}\right )} \log \left (-\frac{1}{4} \, \cos \left (f x + e\right )^{2} + \frac{1}{4}\right )}{2 \,{\left (f \cos \left (f x + e\right )^{2} - 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)^2,x, algorithm="fricas")

[Out]

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

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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*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.40244, size = 339, normalized size = 5.95 \begin{align*} \frac{a^{2}{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, a b{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + b^{2}{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 4 \, a^{2} \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right ) - 4 \, b^{2} \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2\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)^2,x, algorithm="giac")

[Out]

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