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

Optimal. Leaf size=54 \[ \frac{(a+b) \log (1-\cos (e+f x))}{2 f}+\frac{(a-b) \log (\cos (e+f x)+1)}{2 f}+\frac{b \sec (e+f x)}{f} \]

[Out]

((a + b)*Log[1 - Cos[e + f*x]])/(2*f) + ((a - b)*Log[1 + Cos[e + f*x]])/(2*f) + (b*Sec[e + f*x])/f

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Rubi [A]  time = 0.069843, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {4138, 1802} \[ \frac{(a+b) \log (1-\cos (e+f x))}{2 f}+\frac{(a-b) \log (\cos (e+f x)+1)}{2 f}+\frac{b \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)*Log[1 - Cos[e + f*x]])/(2*f) + ((a - b)*Log[1 + Cos[e + f*x]])/(2*f) + (b*Sec[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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{b+a x^3}{x^2 \left (1-x^2\right )} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{-a-b}{2 (-1+x)}+\frac{b}{x^2}+\frac{-a+b}{2 (1+x)}\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{(a+b) \log (1-\cos (e+f x))}{2 f}+\frac{(a-b) \log (1+\cos (e+f x))}{2 f}+\frac{b \sec (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.0627475, size = 65, normalized size = 1.2 \[ \frac{a (\log (\tan (e+f x))+\log (\cos (e+f x)))}{f}+\frac{b \sec (e+f x)}{f}+\frac{b \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}-\frac{b \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 48, normalized size = 0.9 \begin{align*}{\frac{a\ln \left ( \sin \left ( fx+e \right ) \right ) }{f}}+{\frac{b}{f\cos \left ( fx+e \right ) }}+{\frac{b\ln \left ( \csc \left ( fx+e \right ) -\cot \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)^3),x)

[Out]

a*ln(sin(f*x+e))/f+1/f*b/cos(f*x+e)+1/f*b*ln(csc(f*x+e)-cot(f*x+e))

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Maxima [A]  time = 0.998987, size = 61, normalized size = 1.13 \begin{align*} \frac{{\left (a - b\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) +{\left (a + b\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) + \frac{2 \, b}{\cos \left (f x + e\right )}}{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)^3),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.513617, size = 177, normalized size = 3.28 \begin{align*} \frac{{\left (a - b\right )} \cos \left (f x + e\right ) \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) +{\left (a + b\right )} \cos \left (f x + e\right ) \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) + 2 \, b}{2 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.22243, size = 122, normalized size = 2.26 \begin{align*} -\frac{2 \, a \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right ) -{\left (a + b\right )} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) - \frac{4 \, b}{\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 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)^3),x, algorithm="giac")

[Out]

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