3.5.63 \(\int \frac {\cot (c+d x)}{a+b \tan (c+d x)} \, dx\) [463]

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

[Out]

-b*x/(a^2+b^2)+ln(sin(d*x+c))/a/d-b^2*ln(a*cos(d*x+c)+b*sin(d*x+c))/a/(a^2+b^2)/d

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Rubi [A]
time = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 = {3652, 3611, 3556} \begin {gather*} -\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} \end {gather*}

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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[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] && EqQ[a*c + b*d, 0]

Rule 3652

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]

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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.21, size = 91, normalized size = 1.38 \begin {gather*} -\frac {\frac {\log (i-\tan (c+d x))}{a+i b}-\frac {2 \log (\tan (c+d x))}{a}+\frac {\log (i+\tan (c+d x))}{a-i b}+\frac {2 b^2 \log (a+b \tan (c+d x))}{a^3+a b^2}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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))/d

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Maple [A]
time = 0.27, size = 80, normalized size = 1.21

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tan \left (d x +c \right )\right )}{a}-\frac {b^{2} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a}+\frac {-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}-b \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(80\)
default	\(\frac {\frac {\ln \left (\tan \left (d x +c \right )\right )}{a}-\frac {b^{2} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a}+\frac {-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}-b \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(80\)
norman	\(-\frac {b x}{a^{2}+b^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{d a}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}-\frac {b^{2} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) d a}\)	\(86\)
risch	\(-\frac {i x}{i b -a}+\frac {2 i b^{2} x}{\left (a^{2}+b^{2}\right ) a}+\frac {2 i b^{2} c}{\left (a^{2}+b^{2}\right ) d a}-\frac {2 i x}{a}-\frac {2 i c}{d a}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{2}+b^{2}\right ) d a}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 84, normalized size = 1.27 \begin {gather*} -\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 {gather*}

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 = 1.02, size = 98, normalized size = 1.48 \begin {gather*} -\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 {gather*}

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 [C] Result contains complex when optimal does not.
time = 0.89, size = 626, normalized size = 9.48 \begin {gather*} \begin {cases} \frac {\tilde {\infty } x \cot {\left (c \right )}}{\tan {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {- \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {\log {\left (\tan {\left (c + d x \right )} \right )}}{d}}{a} & \text {for}\: b = 0 \\\frac {- x - \frac {1}{d \tan {\left (c + d x \right )}}}{b} & \text {for}\: a = 0 \\\frac {d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {2 i \log {\left (\tan {\left (c + d x \right )} \right )} \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {2 \log {\left (\tan {\left (c + d x \right )} \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {1}{2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = - i b \\\frac {d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {2 i \log {\left (\tan {\left (c + d x \right )} \right )} \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {2 \log {\left (\tan {\left (c + d x \right )} \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {1}{2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = i b \\\frac {x \cot {\left (c \right )}}{a + b \tan {\left (c \right )}} & \text {for}\: d = 0 \\- \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d + 2 a b^{2} d} + \frac {2 a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a b^{2} d} - \frac {2 a b d x}{2 a^{3} d + 2 a b^{2} d} - \frac {2 b^{2} \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a b^{2} d} + \frac {2 b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a b^{2} d} & \text {otherwise} \end {cases} \end {gather*}

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)), ((-log(tan(c + d*x)**2 + 1)/(2*d) + log(tan(c
+ d*x))/d)/a, Eq(b, 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*l
og(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)
), (-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 = 0.58, size = 88, normalized size = 1.33 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 4.13, size = 95, normalized size = 1.44 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {b^2\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,d\,\left (a^2+b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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