3.63 \(\int \frac{\tan ^{-1}(a+b x)}{\sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=132 \[ \frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b}-\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b}-\frac{2 i \tan ^{-1}(a+b x) \tan ^{-1}\left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b} \]

[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

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Rubi [A]  time = 0.0962101, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {5055, 4886} \[ \frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b}-\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b}-\frac{2 i \tan ^{-1}(a+b x) \tan ^{-1}\left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b} \]

Antiderivative was successfully verified.

[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 5055

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*x^2)/d^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]

Rule 4886

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]

Rubi steps

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

Mathematica [A]  time = 0.101348, size = 97, normalized size = 0.73 \[ \frac{i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a+b x)}\right )-i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a+b x)}\right )+\tan ^{-1}(a+b x) \left (\log \left (1-i e^{i \tan ^{-1}(a+b x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a+b x)}\right )\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.397, size = 143, normalized size = 1.1 \begin{align*} -{\frac{\arctan \left ( bx+a \right ) }{b}\ln \left ( 1+{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) }+{\frac{\arctan \left ( bx+a \right ) }{b}\ln \left ( 1-{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) }+{\frac{i}{b}{\it dilog} \left ( 1+{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) }-{\frac{i}{b}{\it dilog} \left ( 1-{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (b x + a\right )}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atan}{\left (a + b x \right )}}{\sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (b x + a\right )}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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