3.2.0 ∫ cot−1(c + d tan(a + bx)) dx [160]

Integrand size = 11, antiderivative size = 198 \[ \int \cot ^{-1}(c+d \tan (a+b x)) \, dx=x \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{2} i x \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {\operatorname {PolyLog}\left (2,-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (2,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b} \]

x*arccot(c+d*tan(b*x+a))-1/2*I*x*ln(1+(1+I*c+d)*exp(2*I*a+2*I*b*x)/(1+I*c-d))+1/2*I*x*ln(1+(c+I*(1-d))*exp(2*I

*a+2*I*b*x)/(c+I*(1+d)))-1/4*polylog(2,-(1+I*c+d)*exp(2*I*a+2*I*b*x)/(1+I*c-d))/b+1/4*polylog(2,-(c+I*(1-d))*e

Rubi [A] (veriﬁed)

Time = 0.17 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5276, 2221, 2317, 2438} \[ \int \cot ^{-1}(c+d \tan (a+b x)) \, dx=-\frac {\operatorname {PolyLog}\left (2,-\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (2,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{4 b}-\frac {1}{2} i x \log \left (1+\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )+\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )+x \cot ^{-1}(d \tan (a+b x)+c) \]

x*ArcCot[c + d*Tan[a + b*x]] - (I/2)*x*Log[1 + ((1 + I*c + d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c - d)] + (I/2)*

x*Log[1 + ((c + I*(1 - d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 + d))] - PolyLog[2, -(((1 + I*c + d)*E^((2*I)*a

+ (2*I)*b*x))/(1 + I*c - d))]/(4*b) + PolyLog[2, -(((c + I*(1 - d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 + d)))]

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +

(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]

- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,

a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int

[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,

Int[ArcCot[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcCot[c + d*Tan[a + b*x]], x] + (-Dist

[b*(1 - I*c - d), Int[x*(E^(2*I*a + 2*I*b*x)/(1 - I*c + d + (1 - I*c - d)*E^(2*I*a + 2*I*b*x))), x], x] + Dist

[b*(1 + I*c + d), Int[x*(E^(2*I*a + 2*I*b*x)/(1 + I*c - d + (1 + I*c + d)*E^(2*I*a + 2*I*b*x))), x], x]) /; Fr

eeQ[{a, b, c, d}, x] && NeQ[(c + I*d)^2, -1]

Rubi steps \begin{align*} \text {integral}& = x \cot ^{-1}(c+d \tan (a+b x))-(b (1-i c-d)) \int \frac {e^{2 i a+2 i b x} x}{1-i c+d+(1-i c-d) e^{2 i a+2 i b x}} \, dx+(b (1+i c+d)) \int \frac {e^{2 i a+2 i b x} x}{1+i c-d+(1+i c+d) e^{2 i a+2 i b x}} \, dx \\ & = x \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{2} i x \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {1}{2} i \int \log \left (1+\frac {(1-i c-d) e^{2 i a+2 i b x}}{1-i c+d}\right ) \, dx+\frac {1}{2} i \int \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right ) \, dx \\ & = x \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{2} i x \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {\text {Subst}\left (\int \frac {\log \left (1+\frac {(1-i c-d) x}{1-i c+d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{4 b}+\frac {\text {Subst}\left (\int \frac {\log \left (1+\frac {(1+i c+d) x}{1+i c-d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{4 b} \\ & = x \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{2} i x \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {\operatorname {PolyLog}\left (2,-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (2,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(549\) vs. \(2(198)=396\).

Time = 0.54 (sec) , antiderivative size = 549, normalized size of antiderivative = 2.77 \[ \int \cot ^{-1}(c+d \tan (a+b x)) \, dx=x \cot ^{-1}(c+d \tan (a+b x))+\frac {x \left (4 a \sqrt {-d^2} \arctan (c+d \tan (a+b x))-i d \log (1-i \tan (a+b x)) \log \left (\frac {-c d+\sqrt {-d^2}-d^2 \tan (a+b x)}{-c d+i d^2+\sqrt {-d^2}}\right )+i d \log (1+i \tan (a+b x)) \log \left (\frac {c d-\sqrt {-d^2}+d^2 \tan (a+b x)}{c d+i d^2-\sqrt {-d^2}}\right )+i d \log (1-i \tan (a+b x)) \log \left (\frac {c d+\sqrt {-d^2}+d^2 \tan (a+b x)}{c d-i d^2+\sqrt {-d^2}}\right )-i d \log (1+i \tan (a+b x)) \log \left (\frac {c d+\sqrt {-d^2}+d^2 \tan (a+b x)}{c d+i d^2+\sqrt {-d^2}}\right )-i d \operatorname {PolyLog}\left (2,\frac {d^2 (1-i \tan (a+b x))}{i c d+d^2-i \sqrt {-d^2}}\right )+i d \operatorname {PolyLog}\left (2,\frac {d^2 (1-i \tan (a+b x))}{i c d+d^2+i \sqrt {-d^2}}\right )+i d \operatorname {PolyLog}\left (2,\frac {d^2 (1+i \tan (a+b x))}{-i c d+d^2+i \sqrt {-d^2}}\right )-i d \operatorname {PolyLog}\left (2,\frac {d^2 (1+i \tan (a+b x))}{d^2-i \left (c d+\sqrt {-d^2}\right )}\right )\right )}{2 \sqrt {-d^2} (2 a-i \log (1-i \tan (a+b x))+i \log (1+i \tan (a+b x)))} \]

x*ArcCot[c + d*Tan[a + b*x]] + (x*(4*a*Sqrt[-d^2]*ArcTan[c + d*Tan[a + b*x]] - I*d*Log[1 - I*Tan[a + b*x]]*Log

[(-(c*d) + Sqrt[-d^2] - d^2*Tan[a + b*x])/(-(c*d) + I*d^2 + Sqrt[-d^2])] + I*d*Log[1 + I*Tan[a + b*x]]*Log[(c*

d - Sqrt[-d^2] + d^2*Tan[a + b*x])/(c*d + I*d^2 - Sqrt[-d^2])] + I*d*Log[1 - I*Tan[a + b*x]]*Log[(c*d + Sqrt[-

d^2] + d^2*Tan[a + b*x])/(c*d - I*d^2 + Sqrt[-d^2])] - I*d*Log[1 + I*Tan[a + b*x]]*Log[(c*d + Sqrt[-d^2] + d^2

*Tan[a + b*x])/(c*d + I*d^2 + Sqrt[-d^2])] - I*d*PolyLog[2, (d^2*(1 - I*Tan[a + b*x]))/(I*c*d + d^2 - I*Sqrt[-

d^2])] + I*d*PolyLog[2, (d^2*(1 - I*Tan[a + b*x]))/(I*c*d + d^2 + I*Sqrt[-d^2])] + I*d*PolyLog[2, (d^2*(1 + I*

Tan[a + b*x]))/((-I)*c*d + d^2 + I*Sqrt[-d^2])] - I*d*PolyLog[2, (d^2*(1 + I*Tan[a + b*x]))/(d^2 - I*(c*d + Sq

rt[-d^2]))]))/(2*Sqrt[-d^2]*(2*a - I*Log[1 - I*Tan[a + b*x]] + I*Log[1 + I*Tan[a + b*x]]))

Maple [B] (veriﬁed)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1137 vs. \(2 (168 ) = 336\).

Time = 4.35 (sec) , antiderivative size = 1138, normalized size of antiderivative = 5.75


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(1138\)
default	\(\text {Expression too large to display}\)	\(1138\)
risch	\(\text {Expression too large to display}\)	\(4969\)

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

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

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

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

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

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

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

1+I*c+d)+1/4*d^2*polylog(2,(c-I*d+I)*(1+I*(d*((c+d*tan(b*x+a))/d-c/d)+c))^2/((d*((c+d*tan(b*x+a))/d-c/d)+c)^2+

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

d*tan(b*x+a))/d-c/d)+c)^2+1/4*d*polylog(2,(c-I*d+I)*(1+I*(d*((c+d*tan(b*x+a))/d-c/d)+c))^2/((d*((c+d*tan(b*x+a

))/d-c/d)+c)^2+1)/(I*d+I-c))/(1+I*c+d)+1/4*d/(c-I*d-I)*polylog(2,(c-I*d+I)*(1+I*(d*((c+d*tan(b*x+a))/d-c/d)+c)

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

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

+d*tan(b*x+a))/d-c/d)+c)^2-1/4*d*polylog(2,(I+c+I*d)*(1+I*(d*((c+d*tan(b*x+a))/d-c/d)+c))^2/((d*((c+d*tan(b*x+

Fricas [B] (veriﬁcation not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1101 vs. \(2 (141) = 282\).

Time = 0.34 (sec) , antiderivative size = 1101, normalized size of antiderivative = 5.56 \[ \int \cot ^{-1}(c+d \tan (a+b x)) \, dx=\text {Too large to display} \]

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

1/8*(8*b*x*arccot(d*tan(b*x + a) + c) - 2*(-I*b*x - I*a)*log(-2*((I*c*d - d^2 + d)*tan(b*x + a)^2 - c^2 - I*c*

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

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

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

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

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

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

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

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

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

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

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

2 + 1)) - dilog(2*((I*c*d - d^2 + d)*tan(b*x + a)^2 - c^2 - I*c*d + (I*c^2 - 2*c*d - I*d^2 + I)*tan(b*x + a) +

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

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

2 + c^2 + d^2 + 2*d + 1) + 1) - dilog(2*((-I*c*d - d^2 + d)*tan(b*x + a)^2 - c^2 + I*c*d + (-I*c^2 - 2*c*d + I

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

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

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

Sympy [F]

\[ \int \cot ^{-1}(c+d \tan (a+b x)) \, dx=\int \operatorname {acot}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \]

Maxima [B] (veriﬁcation not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (141) = 282\).

Time = 0.32 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.19 \[ \int \cot ^{-1}(c+d \tan (a+b x)) \, dx=-\frac {d {\left (\frac {8 \, {\left (b x + a\right )} \arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + c d}{d}\right )}{d} - \frac {4 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (d^{2} + d\right )} \tan \left (b x + a\right )}{c^{2} + d^{2} + 2 \, d + 1}, \frac {c d \tan \left (b x + a\right ) + c^{2} + d + 1}{c^{2} + d^{2} + 2 \, d + 1}\right ) - 4 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (d^{2} - d\right )} \tan \left (b x + a\right )}{c^{2} + d^{2} - 2 \, d + 1}, \frac {c d \tan \left (b x + a\right ) + c^{2} - d + 1}{c^{2} + d^{2} - 2 \, d + 1}\right ) + \log \left (\tan \left (b x + a\right )^{2} + 1\right ) \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, c d \tan \left (b x + a\right ) + c^{2} + 1}{c^{2} + d^{2} + 2 \, d + 1}\right ) - \log \left (\tan \left (b x + a\right )^{2} + 1\right ) \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, c d \tan \left (b x + a\right ) + c^{2} + 1}{c^{2} + d^{2} - 2 \, d + 1}\right ) + 2 \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d + 1}\right ) - 2 \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d - 1}\right ) + 2 \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d + 1}\right ) - 2 \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d - 1}\right )}{d}\right )} - 8 \, {\left (b x + a\right )} \operatorname {arccot}\left (d \tan \left (b x + a\right ) + c\right ) - 8 \, {\left (b x + a\right )} \arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + c d}{d}\right )}{8 \, b} \]

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

-1/8*(d*(8*(b*x + a)*arctan((d^2*tan(b*x + a) + c*d)/d)/d - (4*(b*x + a)*arctan2((c*d + (d^2 + d)*tan(b*x + a)

)/(c^2 + d^2 + 2*d + 1), (c*d*tan(b*x + a) + c^2 + d + 1)/(c^2 + d^2 + 2*d + 1)) - 4*(b*x + a)*arctan2((c*d +

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

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

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

an(b*x + a) - d)/(I*c + d + 1)) - 2*dilog(-(I*d*tan(b*x + a) - d)/(I*c + d - 1)) + 2*dilog((I*d*tan(b*x + a) +

d)/(-I*c + d + 1)) - 2*dilog((I*d*tan(b*x + a) + d)/(-I*c + d - 1)))/d) - 8*(b*x + a)*arccot(d*tan(b*x + a) +

c) - 8*(b*x + a)*arctan((d^2*tan(b*x + a) + c*d)/d))/b

Giac [F]

\[ \int \cot ^{-1}(c+d \tan (a+b x)) \, dx=\int { \operatorname {arccot}\left (d \tan \left (b x + a\right ) + c\right ) \,d x } \]

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

Mupad [F(-1)]

Timed out. \[ \int \cot ^{-1}(c+d \tan (a+b x)) \, dx=\int \mathrm {acot}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \]

\(\int \cot ^{-1}(c+d \tan (a+b x)) \, dx\) [160]

Optimal result