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

Optimal. Leaf size=177 \[ \frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{\sqrt{a^2 c x^2+c}}-\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{\sqrt{a^2 c x^2+c}}-\frac{2 \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{\sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.134127, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {4958, 4954} \[ \frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{\sqrt{a^2 c x^2+c}}-\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{\sqrt{a^2 c x^2+c}}-\frac{2 \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{\sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]/(x*Sqrt[c + a^2*c*x^2]),x]

[Out]

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

Rule 4958

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + c^2*
x^2]/Sqrt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/(x*Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[e, c^2*d] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rule 4954

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

Rubi steps

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

Mathematica [A]  time = 0.14381, size = 100, normalized size = 0.56 \[ \frac{\sqrt{a^2 x^2+1} \left (i \text{PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )-i \text{PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )+\tan ^{-1}(a x) \left (\log \left (1-e^{i \tan ^{-1}(a x)}\right )-\log \left (1+e^{i \tan ^{-1}(a x)}\right )\right )\right )}{\sqrt{c \left (a^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[a*x]/(x*Sqrt[c + a^2*c*x^2]),x]

[Out]

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

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Maple [A]  time = 0.393, size = 139, normalized size = 0.8 \begin{align*}{\frac{-i}{c} \left ( i\arctan \left ( ax \right ) \ln \left ( 1-{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -i\arctan \left ( ax \right ) \ln \left ( 1+{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) +{\it polylog} \left ( 2,{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -{\it polylog} \left ( 2,-{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) \right ) \sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(a*x)/x/(a^2*c*x^2+c)^(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{\sqrt{a^{2} c x^{2} + c} \arctan \left (a x\right )}{a^{2} c x^{3} + c x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*arctan(a*x)/(a^2*c*x^3 + c*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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