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

Optimal. Leaf size=66 \[ -\frac{b^2 \log (a \cos (c+d x)+b \sin (c+d x))}{a d \left (a^2+b^2\right )}-\frac{b x}{a^2+b^2}+\frac{\log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.0762826, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3571, 3530, 3475} \[ -\frac{b^2 \log (a \cos (c+d x)+b \sin (c+d x))}{a d \left (a^2+b^2\right )}-\frac{b x}{a^2+b^2}+\frac{\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3571

Int[1/(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[((a*
c - b*d)*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[b^2/((b*c - a*d)*(a^2 + b^2)), Int[(b - a*Tan[e + f*x])/(a +
b*Tan[e + f*x]), x], x] - Dist[d^2/((b*c - a*d)*(c^2 + d^2)), Int[(d - c*Tan[e + f*x])/(c + d*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[c^2 + d^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]

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

Mathematica [C]  time = 0.156277, size = 91, normalized size = 1.38 \[ -\frac{\frac{2 b^2 \log (a+b \tan (c+d x))}{a^3+a b^2}+\frac{\log (-\tan (c+d x)+i)}{a+i b}+\frac{\log (\tan (c+d x)+i)}{a-i b}-\frac{2 \log (\tan (c+d x))}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[I - Tan[c + d*x]]/(a + I*b) - (2*Log[Tan[c + d*x]])/a + Log[I + Tan[c + d*x]]/(a - I*b) + (2*b^2*Log[a +
b*Tan[c + d*x]])/(a^3 + a*b^2))/(2*d)

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Maple [A]  time = 0.067, size = 95, normalized size = 1.4 \begin{align*} -{\frac{a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{b\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}}-{\frac{{b}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.57049, size = 113, normalized size = 1.71 \begin{align*} -\frac{\frac{2 \, b^{2} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{3} + a b^{2}} + \frac{2 \,{\left (d x + c\right )} b}{a^{2} + b^{2}} + \frac{a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, \log \left (\tan \left (d x + c\right )\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.14481, size = 231, normalized size = 3.5 \begin{align*} -\frac{2 \, a b d x + b^{2} \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 ) -{\left (a^{2} + b^{2}\right )} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \,{\left (a^{3} + a b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 10.9664, size = 626, normalized size = 9.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*cot(c)/tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-x - 1/(d*tan(c + d*x)))/b, Eq(a, 0)), (-d*
x*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) + I*d*x/(-2*b*d*tan(c + d*x) + 2*I*b*d) + I*log(tan(c + d*x)**2
+ 1)*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) + log(tan(c + d*x)**2 + 1)/(-2*b*d*tan(c + d*x) + 2*I*b*d)
- 2*I*log(tan(c + d*x))*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - 2*log(tan(c + d*x))/(-2*b*d*tan(c + d*x
) + 2*I*b*d) - 1/(-2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, -I*b)), (d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*
d) + I*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*
b*d) - log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d*x) + 2*I*b*d) - 2*I*log(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan
(c + d*x) + 2*I*b*d) + 2*log(tan(c + d*x))/(2*b*d*tan(c + d*x) + 2*I*b*d) + 1/(2*b*d*tan(c + d*x) + 2*I*b*d),
Eq(a, I*b)), (x*cot(c)/(a + b*tan(c)), Eq(d, 0)), ((-log(tan(c + d*x)**2 + 1)/(2*d) + log(tan(c + d*x))/d)/a,
Eq(b, 0)), (-a**2*log(tan(c + d*x)**2 + 1)/(2*a**3*d + 2*a*b**2*d) + 2*a**2*log(tan(c + d*x))/(2*a**3*d + 2*a*
b**2*d) - 2*a*b*d*x/(2*a**3*d + 2*a*b**2*d) - 2*b**2*log(a/b + tan(c + d*x))/(2*a**3*d + 2*a*b**2*d) + 2*b**2*
log(tan(c + d*x))/(2*a**3*d + 2*a*b**2*d), True))

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Giac [A]  time = 1.27943, size = 119, normalized size = 1.8 \begin{align*} -\frac{\frac{2 \, b^{3} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{3} b + a b^{3}} + \frac{2 \,{\left (d x + c\right )} b}{a^{2} + b^{2}} + \frac{a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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