\(\int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx\) [63]

Optimal result

Integrand size = 28, antiderivative size = 132 \[ \int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 i \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b} \]

[Out]

-2*I*arctan(b*x+a)*arctan((1+I*(b*x+a))^(1/2)/(1-I*(b*x+a))^(1/2))/b+I*polylog(2,-I*(1+I*(b*x+a))^(1/2)/(1-I*(
b*x+a))^(1/2))/b-I*polylog(2,I*(1+I*(b*x+a))^(1/2)/(1-I*(b*x+a))^(1/2))/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {5163, 5006} \[ \int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 i \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b} \]

[In]

Int[ArcTan[a + b*x]/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((-2*I)*ArcTan[a + b*x]*ArcTan[Sqrt[1 + I*(a + b*x)]/Sqrt[1 - I*(a + b*x)]])/b + (I*PolyLog[2, ((-I)*Sqrt[1 +
I*(a + b*x)])/Sqrt[1 - I*(a + b*x)]])/b - (I*PolyLog[2, (I*Sqrt[1 + I*(a + b*x)])/Sqrt[1 - I*(a + b*x)]])/b

Rule 5006

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[-2*I*(a + b*ArcTan[c*x])*(
ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]]/(c*Sqrt[d])), x] + (Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 + I*c*x]/Sqrt[1
- I*c*x])]/(c*Sqrt[d])), x] - Simp[I*b*(PolyLog[2, I*(Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x])]/(c*Sqrt[d])), x]) /; F
reeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rule 5163

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(q_.), x_Symbol] :> Di
st[1/d, Subst[Int[(C/d^2 + (C/d^2)*x^2)^q*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B,
C, p, q}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\arctan (x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b} \\ & = -\frac {2 i \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.51 \[ \int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {i \left (2 \arctan \left (e^{i \arctan (a+b x)}\right ) \arctan (a+b x)-\operatorname {PolyLog}\left (2,-i e^{i \arctan (a+b x)}\right )+\operatorname {PolyLog}\left (2,i e^{i \arctan (a+b x)}\right )\right )}{b} \]

[In]

Integrate[ArcTan[a + b*x]/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((-I)*(2*ArcTan[E^(I*ArcTan[a + b*x])]*ArcTan[a + b*x] - PolyLog[2, (-I)*E^(I*ArcTan[a + b*x])] + PolyLog[2, I
*E^(I*ArcTan[a + b*x])]))/b

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {-\arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+\arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+i \operatorname {dilog}\left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \operatorname {dilog}\left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{b}\)	\(135\)

[In]

int(arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/b*(-arctan(b*x+a)*ln(1+I*(1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))+arctan(b*x+a)*ln(1-I*(1+I*(b*x+a))/(1+(b*x+a)^2)
^(1/2))+I*dilog(1+I*(1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))-I*dilog(1-I*(1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2)))

Fricas [F]

\[ \int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \]

[In]

integrate(arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(arctan(b*x + a)/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1), x)

Sympy [F]

\[ \int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\operatorname {atan}{\left (a + b x \right )}}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \]

[In]

integrate(atan(b*x+a)/(b**2*x**2+2*a*b*x+a**2+1)**(1/2),x)

[Out]

Integral(atan(a + b*x)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1), x)

Maxima [F]

\[ \int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \]

[In]

integrate(arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(arctan(b*x + a)/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1), x)

Giac [F]

\[ \int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \]

[In]

integrate(arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )}{\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}} \,d x \]

[In]

int(atan(a + b*x)/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2),x)

[Out]

int(atan(a + b*x)/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2), x)