3.1194 \(\int \frac{c+d \tan (e+f x)}{a+b \tan (e+f x)} \, dx\)

Optimal. Leaf size=58 \[ \frac{(b c-a d) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )}+\frac{x (a c+b d)}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.0725751, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3531, 3530} \[ \frac{(b c-a d) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )}+\frac{x (a c+b d)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])/(a + b*Tan[e + f*x]),x]

[Out]

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

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.108863, size = 66, normalized size = 1.14 \[ \frac{2 (a c+b d) \tan ^{-1}(\tan (e+f x))-(b c-a d) \left (\log \left (\sec ^2(e+f x)\right )-2 \log (a+b \tan (e+f x))\right )}{2 f \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])/(a + b*Tan[e + f*x]),x]

[Out]

(2*(a*c + b*d)*ArcTan[Tan[e + f*x]] - (b*c - a*d)*(Log[Sec[e + f*x]^2] - 2*Log[a + b*Tan[e + f*x]]))/(2*(a^2 +
b^2)*f)

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Maple [B]  time = 0.022, size = 153, normalized size = 2.6 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ad}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) bc}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) ac}{f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) bd}{f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ) ad}{f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ) bc}{f \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))/(a+b*tan(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 1.82217, size = 120, normalized size = 2.07 \begin{align*} \frac{\frac{2 \,{\left (a c + b d\right )}{\left (f x + e\right )}}{a^{2} + b^{2}} + \frac{2 \,{\left (b c - a d\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{2} + b^{2}} - \frac{{\left (b c - a d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.33039, size = 174, normalized size = 3. \begin{align*} \frac{2 \,{\left (a c + b d\right )} f x +{\left (b c - a d\right )} \log \left (\frac{b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \,{\left (a^{2} + b^{2}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.61183, size = 524, normalized size = 9.03 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \left (c + d \tan{\left (e \right )}\right )}{\tan{\left (e \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge f = 0 \\- \frac{i c f x \tan{\left (e + f x \right )}}{- 2 b f \tan{\left (e + f x \right )} + 2 i b f} - \frac{c f x}{- 2 b f \tan{\left (e + f x \right )} + 2 i b f} - \frac{i c}{- 2 b f \tan{\left (e + f x \right )} + 2 i b f} - \frac{d f x \tan{\left (e + f x \right )}}{- 2 b f \tan{\left (e + f x \right )} + 2 i b f} + \frac{i d f x}{- 2 b f \tan{\left (e + f x \right )} + 2 i b f} + \frac{d}{- 2 b f \tan{\left (e + f x \right )} + 2 i b f} & \text{for}\: a = - i b \\- \frac{i c f x \tan{\left (e + f x \right )}}{2 b f \tan{\left (e + f x \right )} + 2 i b f} + \frac{c f x}{2 b f \tan{\left (e + f x \right )} + 2 i b f} - \frac{i c}{2 b f \tan{\left (e + f x \right )} + 2 i b f} + \frac{d f x \tan{\left (e + f x \right )}}{2 b f \tan{\left (e + f x \right )} + 2 i b f} + \frac{i d f x}{2 b f \tan{\left (e + f x \right )} + 2 i b f} - \frac{d}{2 b f \tan{\left (e + f x \right )} + 2 i b f} & \text{for}\: a = i b \\\frac{x \left (c + d \tan{\left (e \right )}\right )}{a + b \tan{\left (e \right )}} & \text{for}\: f = 0 \\\frac{c x + \frac{d \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f}}{a} & \text{for}\: b = 0 \\\frac{2 a c f x}{2 a^{2} f + 2 b^{2} f} - \frac{2 a d \log{\left (\frac{a}{b} + \tan{\left (e + f x \right )} \right )}}{2 a^{2} f + 2 b^{2} f} + \frac{a d \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a^{2} f + 2 b^{2} f} + \frac{2 b c \log{\left (\frac{a}{b} + \tan{\left (e + f x \right )} \right )}}{2 a^{2} f + 2 b^{2} f} - \frac{b c \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a^{2} f + 2 b^{2} f} + \frac{2 b d f x}{2 a^{2} f + 2 b^{2} f} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.3624, size = 132, normalized size = 2.28 \begin{align*} \frac{\frac{2 \,{\left (a c + b d\right )}{\left (f x + e\right )}}{a^{2} + b^{2}} - \frac{{\left (b c - a d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (b^{2} c - a b d\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a*c + b*d)*(f*x + e)/(a^2 + b^2) - (b*c - a*d)*log(tan(f*x + e)^2 + 1)/(a^2 + b^2) + 2*(b^2*c - a*b*d)
*log(abs(b*tan(f*x + e) + a))/(a^2*b + b^3))/f