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

Optimal. Leaf size=250 \[ -\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{2 a^3 \sqrt{a^2 c x^2+c}}+\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{2 a^3 \sqrt{a^2 c x^2+c}}-\frac{\sqrt{a^2 c x^2+c}}{2 a^3 c}+\frac{i \sqrt{a^2 x^2+1} \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right ) \tan ^{-1}(a x)}{a^3 \sqrt{a^2 c x^2+c}}+\frac{x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{2 a^2 c} \]

[Out]

-Sqrt[c + a^2*c*x^2]/(2*a^3*c) + (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(2*a^2*c) + (I*Sqrt[1 + a^2*x^2]*ArcTan[a
*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a^3*Sqrt[c + a^2*c*x^2]) - ((I/2)*Sqrt[1 + a^2*x^2]*PolyLog[2, (
(-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^3*Sqrt[c + a^2*c*x^2]) + ((I/2)*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqr
t[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^3*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 0.146525, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4952, 261, 4890, 4886} \[ -\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{2 a^3 \sqrt{a^2 c x^2+c}}+\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{2 a^3 \sqrt{a^2 c x^2+c}}-\frac{\sqrt{a^2 c x^2+c}}{2 a^3 c}+\frac{i \sqrt{a^2 x^2+1} \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right ) \tan ^{-1}(a x)}{a^3 \sqrt{a^2 c x^2+c}}+\frac{x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{2 a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[c + a^2*c*x^2]/(2*a^3*c) + (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(2*a^2*c) + (I*Sqrt[1 + a^2*x^2]*ArcTan[a
*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a^3*Sqrt[c + a^2*c*x^2]) - ((I/2)*Sqrt[1 + a^2*x^2]*PolyLog[2, (
(-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^3*Sqrt[c + a^2*c*x^2]) + ((I/2)*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqr
t[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^3*Sqrt[c + a^2*c*x^2])

Rule 4952

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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{x^2 \tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx &=\frac{x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{2 a^2 c}-\frac{\int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{2 a^2}-\frac{\int \frac{x}{\sqrt{c+a^2 c x^2}} \, dx}{2 a}\\ &=-\frac{\sqrt{c+a^2 c x^2}}{2 a^3 c}+\frac{x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{2 a^2 c}-\frac{\sqrt{1+a^2 x^2} \int \frac{\tan ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{2 a^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{c+a^2 c x^2}}{2 a^3 c}+\frac{x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{2 a^2 c}+\frac{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^3 \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 )}{2 a^3 \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 )}{2 a^3 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.5769, size = 158, normalized size = 0.63 \[ -\frac{\sqrt{c \left (a^2 x^2+1\right )} \left (i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )-i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )+\sqrt{a^2 x^2+1}-a x \sqrt{a^2 x^2+1} \tan ^{-1}(a x)+\tan ^{-1}(a x) \log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )}{2 a^3 c \sqrt{a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Sqrt[c*(1 + a^2*x^2)]*(Sqrt[1 + a^2*x^2] - a*x*Sqrt[1 + a^2*x^2]*ArcTan[a*x] + ArcTan[a*x]*Log[1 - I*E^(I*Ar
cTan[a*x])] - 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])]))/(2*a^3*c*Sqrt[1 + a^2*x^2])

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Maple [A]  time = 0.895, size = 184, normalized size = 0.7 \begin{align*}{\frac{\arctan \left ( ax \right ) xa-1}{2\,c{a}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{{\frac{i}{2}}}{c{a}^{3}} \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(x^2*arctan(a*x)/(a^2*c*x^2+c)^(1/2),x)

[Out]

1/2*(arctan(a*x)*x*a-1)*(c*(a*x-I)*(a*x+I))^(1/2)/c/a^3-1/2*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)/a^3/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(x^2*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{x^{2} \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(x^2*arctan(a*x)/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral(x^2*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{x^{2} \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(x**2*atan(a*x)/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(x**2*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{x^{2} \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(x^2*arctan(a*x)/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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