3.188 \(\int \cot (e+f x) (a+b \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=26 \[ \frac{a \log (\sin (e+f x))}{f}-\frac{b \log (\cos (e+f x))}{f} \]

[Out]

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

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Rubi [A]  time = 0.0292719, antiderivative size = 26, 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 = {3625, 3475} \[ \frac{a \log (\sin (e+f x))}{f}-\frac{b \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3625

Int[((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[A, Int[1/Tan[e + f*x],
x], x] + Dist[C, Int[Tan[e + f*x], x], x] /; FreeQ[{e, f, A, C}, x] && NeQ[A, C]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.039285, size = 34, normalized size = 1.31 \[ \frac{a (\log (\tan (e+f x))+\log (\cos (e+f x)))}{f}-\frac{b \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 27, normalized size = 1. \begin{align*} -{\frac{b\ln \left ( \cos \left ( fx+e \right ) \right ) }{f}}+{\frac{a\ln \left ( \sin \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*tan(f*x+e)^2),x)

[Out]

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

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Maxima [A]  time = 1.07368, size = 42, normalized size = 1.62 \begin{align*} -\frac{b \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - a \log \left (\sin \left (f x + e\right )^{2}\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.12898, size = 113, normalized size = 4.35 \begin{align*} \frac{a \log \left (\frac{\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - b \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.627207, size = 58, normalized size = 2.23 \begin{align*} \begin{cases} - \frac{a \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{a \log{\left (\tan{\left (e + f x \right )} \right )}}{f} + \frac{b \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} & \text{for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\left (e \right )}\right ) \cot{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*log(tan(e + f*x)**2 + 1)/(2*f) + a*log(tan(e + f*x))/f + b*log(tan(e + f*x)**2 + 1)/(2*f), Ne(f,
0)), (x*(a + b*tan(e)**2)*cot(e), True))

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Giac [A]  time = 1.22549, size = 47, normalized size = 1.81 \begin{align*} \frac{a \log \left (\sin \left (f x + e\right )^{2}\right ) - b \log \left (-\sin \left (f x + e\right )^{2} + 1\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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