3.462 \(\int \frac{1}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=45 \[ \frac{b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac{a x}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.043348, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3484, 3530} \[ \frac{b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac{a x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-1),x]

[Out]

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

Rule 3484

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2),
Int[(b - a*Tan[c + d*x])/(a + b*Tan[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 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{1}{a+b \tan (c+d x)} \, dx &=\frac{a x}{a^2+b^2}+\frac{b \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=\frac{a x}{a^2+b^2}+\frac{b \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 0.0564538, size = 76, normalized size = 1.69 \[ \frac{(-b-i a) \log (-\tan (c+d x)+i)+i (a+i b) \log (\tan (c+d x)+i)+2 b \log (a+b \tan (c+d x))}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(-1),x]

[Out]

(((-I)*a - b)*Log[I - Tan[c + d*x]] + I*(a + I*b)*Log[I + Tan[c + d*x]] + 2*b*Log[a + b*Tan[c + d*x]])/(2*(a^2
+ b^2)*d)

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Maple [A]  time = 0.017, size = 74, normalized size = 1.6 \begin{align*} -{\frac{b\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{a\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{b\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.64799, size = 93, normalized size = 2.07 \begin{align*} \frac{\frac{2 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac{2 \, b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} + b^{2}} - \frac{b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.98028, size = 147, normalized size = 3.27 \begin{align*} \frac{2 \, a d x + b \log \left (\frac{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \,{\left (a^{2} + b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.42587, size = 243, normalized size = 5.4 \begin{align*} \begin{cases} \frac{\tilde{\infty } x}{\tan{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\- \frac{i d x \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{d x}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{i}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = - i b \\- \frac{i d x \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{d x}{2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{i}{2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = i b \\\frac{x}{a + b \tan{\left (c \right )}} & \text{for}\: d = 0 \\\frac{x}{a} & \text{for}\: b = 0 \\\frac{2 a d x}{2 a^{2} d + 2 b^{2} d} + \frac{2 b \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} - \frac{b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.23145, size = 100, normalized size = 2.22 \begin{align*} \frac{\frac{2 \, b^{2} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}} + \frac{2 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac{b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

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