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

Optimal. Leaf size=193 \[ \frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.0572581, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {4890, 4886} \[ \frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4890

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/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 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 x)}{\sqrt{c+a^2 c x^2}} \, dx &=\frac{\sqrt{1+a^2 x^2} \int \frac{\tan ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{\sqrt{c+a^2 c x^2}}\\ &=-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{c+a^2 c x^2}}+\frac{i \sqrt{1+a^2 x^2} \text{Li}_2\left (-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{c+a^2 c x^2}}-\frac{i \sqrt{1+a^2 x^2} \text{Li}_2\left (\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.107153, size = 118, normalized size = 0.61 \[ \frac{\sqrt{c \left (a^2 x^2+1\right )} \left (i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )-i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )+\tan ^{-1}(a x) \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )\right )}{a c \sqrt{a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.352, size = 150, normalized size = 0.8 \begin{align*}{\frac{i}{ca} \left ( i\arctan \left ( ax \right ) \ln \left ( 1+{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -i\arctan \left ( ax \right ) \ln \left ( 1-{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) +{\it dilog} \left ( 1+{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -{\it dilog} \left ( 1-{i \left ( 1+iax \right ){\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)/(a^2*c*x^2+c)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(atan(a*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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*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)