3.1.50 \(\int \text {ArcTan}(c+d \tan (a+b x)) \, dx\) [50]

Optimal. Leaf size=198 \[ x \text {ArcTan}(c+d \tan (a+b x))+\frac {1}{2} i x \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )-\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )+\frac {\text {PolyLog}\left (2,-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}-\frac {\text {PolyLog}\left (2,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b} \]

[Out]

x*arctan(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
xp(2*I*a+2*I*b*x)/(c+I*(1+d)))/b

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Rubi [A]
time = 0.18, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5275, 2221, 2317, 2438} \begin {gather*} x \text {ArcTan}(d \tan (a+b x)+c)+\frac {\text {Li}_2\left (-\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{4 b}-\frac {\text {Li}_2\left (-\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[c + d*Tan[a + b*x]],x]

[Out]

x*ArcTan[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)))]
/(4*b)

Rule 2221

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]

Rule 2317

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]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 5275

Int[ArcTan[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcTan[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]) /; Fre
eQ[{a, b, c, d}, x] && NeQ[(c + I*d)^2, -1]

Rubi steps

\begin {align*} \int \tan ^{-1}(c+d \tan (a+b x)) \, dx &=x \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 {Li}_2\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}-\frac {\text {Li}_2\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(555\) vs. \(2(198)=396\).
time = 5.02, size = 555, normalized size = 2.80 \begin {gather*} x \text {ArcTan}(c+d \tan (a+b x))+\frac {x \left (-4 a d \text {ArcTan}(c+d \tan (a+b x))-i \sqrt {-d^2} \left (\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 )+\text {PolyLog}\left (2,\frac {d^2 (1-i \tan (a+b x))}{i c d+d^2-i \sqrt {-d^2}}\right )\right )+i \sqrt {-d^2} \left (\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 )+\text {PolyLog}\left (2,\frac {d^2 (1-i \tan (a+b x))}{i c d+d^2+i \sqrt {-d^2}}\right )\right )+i \sqrt {-d^2} \left (\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 )+\text {PolyLog}\left (2,\frac {d^2 (1+i \tan (a+b x))}{-i c d+d^2+i \sqrt {-d^2}}\right )\right )-i \sqrt {-d^2} \left (\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 )+\text {PolyLog}\left (2,\frac {d^2 (1+i \tan (a+b x))}{d^2-i \left (c d+\sqrt {-d^2}\right )}\right )\right )\right )}{2 d (2 a-i \log (1-i \tan (a+b x))+i \log (1+i \tan (a+b x)))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[c + d*Tan[a + b*x]],x]

[Out]

x*ArcTan[c + d*Tan[a + b*x]] + (x*(-4*a*d*ArcTan[c + d*Tan[a + b*x]] - I*Sqrt[-d^2]*(Log[1 - I*Tan[a + b*x]]*L
og[(-(c*d) + Sqrt[-d^2] - d^2*Tan[a + b*x])/(-(c*d) + I*d^2 + Sqrt[-d^2])] + PolyLog[2, (d^2*(1 - I*Tan[a + b*
x]))/(I*c*d + d^2 - I*Sqrt[-d^2])]) + I*Sqrt[-d^2]*(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])] + PolyLog[2, (d^2*(1 - I*Tan[a + b*x]))/(I*c*d + d^2 + I*Sqrt[-d^2])]) + I
*Sqrt[-d^2]*(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])] + P
olyLog[2, (d^2*(1 + I*Tan[a + b*x]))/((-I)*c*d + d^2 + I*Sqrt[-d^2])]) - I*Sqrt[-d^2]*(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])] + PolyLog[2, (d^2*(1 + I*Tan[a + b*x]))
/(d^2 - I*(c*d + Sqrt[-d^2]))])))/(2*d*(2*a - I*Log[1 - I*Tan[a + b*x]] + I*Log[1 + I*Tan[a + b*x]]))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1000 vs. \(2 (168 ) = 336\).
time = 0.78, size = 1001, normalized size = 5.06 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] 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 = 0.53, size = 433, normalized size = 2.19 \begin {gather*} \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 )} \arctan \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] 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 = 0.65, size = 1101, normalized size = 5.56 \begin {gather*} \frac {8 \, b x \arctan \left (d \tan \left (b x + a\right ) + c\right ) - 2 \, {\left (i \, b x + i \, a\right )} \log \left (-\frac {2 \, {\left ({\left (i \, c d - d^{2} + d\right )} \tan \left (b x + a\right )^{2} - c^{2} - i \, c d + {\left (i \, c^{2} - 2 \, c d - i \, d^{2} + i\right )} \tan \left (b x + a\right ) + d - 1\right )}}{{\left (c^{2} + d^{2} - 2 \, d + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + d^{2} - 2 \, d + 1}\right ) - 2 \, {\left (-i \, b x - i \, a\right )} \log \left (-\frac {2 \, {\left ({\left (i \, c d - d^{2} - d\right )} \tan \left (b x + a\right )^{2} - c^{2} - i \, c d + {\left (i \, c^{2} - 2 \, c d - i \, d^{2} + i\right )} \tan \left (b x + a\right ) - d - 1\right )}}{{\left (c^{2} + d^{2} + 2 \, d + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + d^{2} + 2 \, d + 1}\right ) - 2 \, {\left (-i \, b x - i \, a\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, c d - d^{2} + d\right )} \tan \left (b x + a\right )^{2} - c^{2} + i \, c d + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2} - i\right )} \tan \left (b x + a\right ) + d - 1\right )}}{{\left (c^{2} + d^{2} - 2 \, d + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + d^{2} - 2 \, d + 1}\right ) - 2 \, {\left (i \, b x + i \, a\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, c d - d^{2} - d\right )} \tan \left (b x + a\right )^{2} - c^{2} + i \, c d + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2} - i\right )} \tan \left (b x + a\right ) - d - 1\right )}}{{\left (c^{2} + d^{2} + 2 \, d + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + d^{2} + 2 \, d + 1}\right ) + 2 i \, a \log \left (\frac {{\left (i \, c d + d^{2} + d\right )} \tan \left (b x + a\right )^{2} - c^{2} + i \, c d + {\left (i \, c^{2} + i \, d^{2} + 2 i \, d + i\right )} \tan \left (b x + a\right ) - d - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 i \, a \log \left (\frac {{\left (i \, c d + d^{2} - d\right )} \tan \left (b x + a\right )^{2} - c^{2} + i \, c d + {\left (i \, c^{2} + i \, d^{2} - 2 i \, d + i\right )} \tan \left (b x + a\right ) + d - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 i \, a \log \left (\frac {{\left (i \, c d - d^{2} + d\right )} \tan \left (b x + a\right )^{2} + c^{2} + i \, c d + {\left (i \, c^{2} + i \, d^{2} - 2 i \, d + i\right )} \tan \left (b x + a\right ) - d + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 i \, a \log \left (\frac {{\left (i \, c d - d^{2} - d\right )} \tan \left (b x + a\right )^{2} + c^{2} + i \, c d + {\left (i \, c^{2} + i \, d^{2} + 2 i \, d + i\right )} \tan \left (b x + a\right ) + d + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + {\rm Li}_2\left (\frac {2 \, {\left ({\left (i \, c d - d^{2} + d\right )} \tan \left (b x + a\right )^{2} - c^{2} - i \, c d + {\left (i \, c^{2} - 2 \, c d - i \, d^{2} + i\right )} \tan \left (b x + a\right ) + d - 1\right )}}{{\left (c^{2} + d^{2} - 2 \, d + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + d^{2} - 2 \, d + 1} + 1\right ) - {\rm Li}_2\left (\frac {2 \, {\left ({\left (i \, c d - d^{2} - d\right )} \tan \left (b x + a\right )^{2} - c^{2} - i \, c d + {\left (i \, c^{2} - 2 \, c d - i \, d^{2} + i\right )} \tan \left (b x + a\right ) - d - 1\right )}}{{\left (c^{2} + d^{2} + 2 \, d + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + d^{2} + 2 \, d + 1} + 1\right ) + {\rm Li}_2\left (\frac {2 \, {\left ({\left (-i \, c d - d^{2} + d\right )} \tan \left (b x + a\right )^{2} - c^{2} + i \, c d + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2} - i\right )} \tan \left (b x + a\right ) + d - 1\right )}}{{\left (c^{2} + d^{2} - 2 \, d + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + d^{2} - 2 \, d + 1} + 1\right ) - {\rm Li}_2\left (\frac {2 \, {\left ({\left (-i \, c d - d^{2} - d\right )} \tan \left (b x + a\right )^{2} - c^{2} + i \, c d + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2} - i\right )} \tan \left (b x + a\right ) - d - 1\right )}}{{\left (c^{2} + d^{2} + 2 \, d + 1\right )} \tan \left (b x + a\right )^{2} + c^{2} + d^{2} + 2 \, d + 1} + 1\right )}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*b*x*arctan(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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \operatorname {atan}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(c + d*tan(a + b*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {atan}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(c + d*tan(a + b*x)),x)

[Out]

int(atan(c + d*tan(a + b*x)), x)

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