3.270 \(\int \frac{A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0678102, 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} \[ \frac{(A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac{x (a A+b B)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(a + b*Tan[c + d*x]),x]

[Out]

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

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

Mathematica [A]  time = 0.103249, size = 66, normalized size = 1.14 \[ \frac{2 (a A+b B) \tan ^{-1}(\tan (c+d x))-(A b-a B) \left (\log \left (\sec ^2(c+d x)\right )-2 \log (a+b \tan (c+d x))\right )}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(a + b*Tan[c + d*x]),x]

[Out]

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

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Maple [B]  time = 0.031, size = 153, normalized size = 2.6 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Ab}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) aB}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) Ab}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) aB}{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.48983, size = 119, normalized size = 2.05 \begin{align*} \frac{\frac{2 \,{\left (A a + B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} - \frac{2 \,{\left (B a - A b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} + b^{2}} + \frac{{\left (B a - A b\right )} \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((A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.64692, size = 174, normalized size = 3. \begin{align*} \frac{2 \,{\left (A a + B b\right )} d x -{\left (B a - A b\right )} \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((A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*(A*a + B*b)*d*x - (B*a - A*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 = 2.56906, size = 524, normalized size = 9.03 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \left (A + B \tan{\left (c \right )}\right )}{\tan{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{A x + \frac{B \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d}}{a} & \text{for}\: b = 0 \\- \frac{i A d x \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{A d x}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{i A}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{B d x \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{i B d x}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{B}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = - i b \\- \frac{i A d x \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{A d x}{2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{i A}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{B d x \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{i B d x}{2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{B}{2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = i b \\\frac{x \left (A + B \tan{\left (c \right )}\right )}{a + b \tan{\left (c \right )}} & \text{for}\: d = 0 \\\frac{2 A a d x}{2 a^{2} d + 2 b^{2} d} + \frac{2 A b \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} - \frac{A b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} - \frac{2 B a \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} + \frac{B a \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} + \frac{2 B b d x}{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((A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.22887, size = 127, normalized size = 2.19 \begin{align*} \frac{\frac{2 \,{\left (A a + B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} + \frac{{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \,{\left (B a b - A b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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