3.213 \(\int \frac{\tan (e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=36 \[ -\frac{\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 f (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.0530929, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3670, 444, 36, 31} \[ -\frac{\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 f (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0283086, size = 37, normalized size = 1.03 \[ -\frac{\log \left (a+b \tan ^2(e+f x)\right )+2 \log (\cos (e+f x))}{2 f (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 50, normalized size = 1.4 \begin{align*} -{\frac{\ln \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)/(a+b*tan(f*x+e)^2),x)

[Out]

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

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Maxima [A]  time = 1.05478, size = 41, normalized size = 1.14 \begin{align*} -\frac{\log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{2 \,{\left (a - b\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.08921, size = 90, normalized size = 2.5 \begin{align*} -\frac{\log \left (\frac{b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \,{\left (a - b\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.34803, size = 143, normalized size = 3.97 \begin{align*} \begin{cases} \frac{\tilde{\infty } x}{\tan{\left (e \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge f = 0 \\- \frac{1}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text{for}\: a = b \\\frac{x \tan{\left (e \right )}}{a + b \tan ^{2}{\left (e \right )}} & \text{for}\: f = 0 \\\frac{\log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a f} & \text{for}\: b = 0 \\- \frac{\log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \tan{\left (e + f x \right )} \right )}}{2 a f - 2 b f} - \frac{\log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \tan{\left (e + f x \right )} \right )}}{2 a f - 2 b f} + \frac{\log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a f - 2 b f} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x/tan(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (-1/(2*b*f*tan(e + f*x)**2 + 2*b*f), Eq(a, b)), (x*t
an(e)/(a + b*tan(e)**2), Eq(f, 0)), (log(tan(e + f*x)**2 + 1)/(2*a*f), Eq(b, 0)), (-log(-I*sqrt(a)*sqrt(1/b) +
tan(e + f*x))/(2*a*f - 2*b*f) - log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(2*a*f - 2*b*f) + log(tan(e + f*x)**2
+ 1)/(2*a*f - 2*b*f), True))

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Giac [A]  time = 1.48737, size = 72, normalized size = 2. \begin{align*} \frac{\frac{2 \, \log \left (2\right )}{a - b} - \frac{\log \left ({\left | a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b \right |}\right )}{a - b}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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