3.50 \(\int \tan ^{-1}(c+d \tan (a+b x)) \, dx\)

Optimal. Leaf size=198 \[ \frac {\text {Li}_2\left (-\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{4 b}-\frac {\text {Li}_2\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{4 b}+\frac {1}{2} i x \log \left (1+\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )-\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )+x \tan ^{-1}(d \tan (a+b x)+c) \]

[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 = 0.23, 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 = {5167, 2190, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,-\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{4 b}-\frac {\text {PolyLog}\left (2,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{4 b}+\frac {1}{2} i x \log \left (1+\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )-\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )+x \tan ^{-1}(d \tan (a+b x)+c) \]

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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 5167

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 {\operatorname {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 {\operatorname {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]  time = 7.92, size = 555, normalized size = 2.80 \[ x \tan ^{-1}(d \tan (a+b x)+c)+\frac {x \left (-i \sqrt {-d^2} \left (\text {Li}_2\left (\frac {d^2 (1-i \tan (a+b x))}{d^2+i c d-i \sqrt {-d^2}}\right )+\log (1-i \tan (a+b x)) \log \left (\frac {d^2 (-\tan (a+b x))-c d+\sqrt {-d^2}}{-c d+i d^2+\sqrt {-d^2}}\right )\right )+i \sqrt {-d^2} \left (\text {Li}_2\left (\frac {d^2 (1-i \tan (a+b x))}{d^2+i c d+i \sqrt {-d^2}}\right )+\log (1-i \tan (a+b x)) \log \left (\frac {d^2 \tan (a+b x)+c d+\sqrt {-d^2}}{c d-i d^2+\sqrt {-d^2}}\right )\right )+i \sqrt {-d^2} \left (\text {Li}_2\left (\frac {d^2 (i \tan (a+b x)+1)}{d^2-i c d+i \sqrt {-d^2}}\right )+\log (1+i \tan (a+b x)) \log \left (\frac {d^2 \tan (a+b x)+c d-\sqrt {-d^2}}{c d+i d^2-\sqrt {-d^2}}\right )\right )-i \sqrt {-d^2} \left (\text {Li}_2\left (\frac {d^2 (i \tan (a+b x)+1)}{d^2-i \left (c d+\sqrt {-d^2}\right )}\right )+\log (1+i \tan (a+b x)) \log \left (\frac {d^2 \tan (a+b x)+c d+\sqrt {-d^2}}{c d+i d^2+\sqrt {-d^2}}\right )\right )-4 a d \tan ^{-1}(d \tan (a+b x)+c)\right )}{2 d (-i \log (1-i \tan (a+b x))+i \log (1+i \tan (a+b x))+2 a)} \]

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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fricas [B]  time = 0.70, size = 1117, normalized size = 5.64 \[ \text {result too large to display} \]

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 - 2*I*a)*log(-(2*(I*c*d - d^2 + d)*tan(b*x + a)^2 - 2*c^2 -
2*I*c*d - (-2*I*c^2 + 4*c*d + 2*I*d^2 - 2*I)*tan(b*x + a) + 2*d - 2)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c
^2 + d^2 - 2*d + 1)) + (2*I*b*x + 2*I*a)*log(-(2*(I*c*d - d^2 - d)*tan(b*x + a)^2 - 2*c^2 - 2*I*c*d - (-2*I*c^
2 + 4*c*d + 2*I*d^2 - 2*I)*tan(b*x + a) - 2*d - 2)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1
)) + (2*I*b*x + 2*I*a)*log(-(2*(-I*c*d - d^2 + d)*tan(b*x + a)^2 - 2*c^2 + 2*I*c*d - (2*I*c^2 + 4*c*d - 2*I*d^
2 + 2*I)*tan(b*x + a) + 2*d - 2)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*d + 1)) + (-2*I*b*x - 2
*I*a)*log(-(2*(-I*c*d - d^2 - d)*tan(b*x + a)^2 - 2*c^2 + 2*I*c*d - (2*I*c^2 + 4*c*d - 2*I*d^2 + 2*I)*tan(b*x
+ a) - 2*d - 2)/((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)*t
an(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 -
2*c^2 - 2*I*c*d - (-2*I*c^2 + 4*c*d + 2*I*d^2 - 2*I)*tan(b*x + a) + 2*d - 2)/((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 - 2*c^2 - 2*I*c*d - (-2*I*c^2 + 4
*c*d + 2*I*d^2 - 2*I)*tan(b*x + a) - 2*d - 2)/((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 - 2*c^2 + 2*I*c*d - (2*I*c^2 + 4*c*d - 2*I*d^2 + 2*I)*tan(b*x +
a) + 2*d - 2)/((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 - 2*c^2 + 2*I*c*d - (2*I*c^2 + 4*c*d - 2*I*d^2 + 2*I)*tan(b*x + a) - 2*d - 2)/((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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \arctan \left (d \tan \left (b x + a\right ) + c\right )\,{d x} \]

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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maple [B]  time = 1.22, size = 1002, normalized size = 5.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*arctan(tan(b*x+a))*arctan(c+d*tan(b*x+a))+1/2*I/b*arctan((c+d*tan(b*x+a))/d-c/d)*ln(1-(-I*d-I+c)*(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/b*arctan((c+d*tan(b*x+a))/d-c/d)^2+1/4/
b*polylog(2,(-I*d-I+c)*(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/b/(I+
I*d+c)*ln(1-(-I*d+I+c)*(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/b/(I+I*d+c)*ln(1-(-I*d+I+c)*(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/b/(I+I*d+c)*ln(1-(-I*d+I+c)*(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)*c-1/2*I*d/b/(I+I*d+c)*arctan((
c+d*tan(b*x+a))/d-c/d)^2-1/4*I*d/b/(I+I*d+c)*polylog(2,(-I*d+I+c)*(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/b/(I+I*d+c)*arctan((c+d*tan(b*x+a))/d-c/d)^2-1/2/b/(I+I*d+c)*arctan((c+d
*tan(b*x+a))/d-c/d)^2*c-1/4*I/b/(I+I*d+c)*polylog(2,(-I*d+I+c)*(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/b/(I+I*d+c)*polylog(2,(-I*d+I+c)*(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]  time = 0.52, size = 433, normalized size = 2.19 \[ \frac {d {\left (\frac {8 \, {\left (b x + a\right )} \arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + c d}{d}\right )}{d} - \frac {4 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (d^{2} + d\right )} \tan \left (b x + a\right )}{c^{2} + d^{2} + 2 \, d + 1}, \frac {c d \tan \left (b x + a\right ) + c^{2} + d + 1}{c^{2} + d^{2} + 2 \, d + 1}\right ) - 4 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (d^{2} - d\right )} \tan \left (b x + a\right )}{c^{2} + d^{2} - 2 \, d + 1}, \frac {c d \tan \left (b x + a\right ) + c^{2} - d + 1}{c^{2} + d^{2} - 2 \, d + 1}\right ) + \log \left (\tan \left (b x + a\right )^{2} + 1\right ) \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, c d \tan \left (b x + a\right ) + c^{2} + 1}{c^{2} + d^{2} + 2 \, d + 1}\right ) - \log \left (\tan \left (b x + a\right )^{2} + 1\right ) \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, c d \tan \left (b x + a\right ) + c^{2} + 1}{c^{2} + d^{2} - 2 \, d + 1}\right ) + 2 \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d + 1}\right ) - 2 \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d - 1}\right ) + 2 \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d + 1}\right ) - 2 \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d - 1}\right )}{d}\right )} + 8 \, {\left (b x + a\right )} \arctan \left (d \tan \left (b x + a\right ) + c\right ) - 8 \, {\left (b x + a\right )} \arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + c d}{d}\right )}{8 \, b} \]

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

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

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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