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

Optimal. Leaf size=53 \[ \frac{(a+b)^2 \log (\sin (e+f x))}{f}-\frac{b (2 a+b) \log (\cos (e+f x))}{f}+\frac{b^2 \sec ^2(e+f x)}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.0738816, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4138, 446, 88} \[ \frac{(a+b)^2 \log (\sin (e+f x))}{f}-\frac{b (2 a+b) \log (\cos (e+f x))}{f}+\frac{b^2 \sec ^2(e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.22654, size = 84, normalized size = 1.58 \[ \frac{2 (a \cos (e+f x)+b \sec (e+f x))^2 \left (2 \cos ^2(e+f x) \left ((a+b)^2 \log (\sin (e+f x))-b (2 a+b) \log (\cos (e+f x))\right )+b^2\right )}{f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.057, size = 60, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}\ln \left ( \sin \left ( fx+e \right ) \right ) }{f}}+2\,{\frac{ab\ln \left ( \tan \left ( fx+e \right ) \right ) }{f}}+{\frac{{b}^{2}}{2\,f \left ( \cos \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)*(a+b*sec(f*x+e)^2)^2,x)

[Out]

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

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Maxima [A]  time = 0.984089, size = 86, normalized size = 1.62 \begin{align*} -\frac{{\left (2 \, a b + b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) -{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac{b^{2}}{\sin \left (f x + e\right )^{2} - 1}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.535818, size = 203, normalized size = 3.83 \begin{align*} -\frac{{\left (2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (\cos \left (f x + e\right )^{2}\right ) -{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (-\frac{1}{4} \, \cos \left (f x + e\right )^{2} + \frac{1}{4}\right ) - b^{2}}{2 \, f \cos \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.39487, size = 358, normalized size = 6.75 \begin{align*} -\frac{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 ) +{\left (2 \, a b + b^{2}\right )} \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 ) - \frac{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 b - 2 \, b^{2}}{\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}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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