3.233 \(\int \frac{x^2 \tan ^{-1}(a x)}{(c+a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=251 \[ \frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a^3 c \sqrt{a^2 c x^2+c}}-\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a^3 c \sqrt{a^2 c x^2+c}}-\frac{1}{a^3 c \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right ) \tan ^{-1}(a x)}{a^3 c \sqrt{a^2 c x^2+c}}-\frac{x \tan ^{-1}(a x)}{a^2 c \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.158616, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4934, 4890, 4886} \[ \frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a^3 c \sqrt{a^2 c x^2+c}}-\frac{i \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{a^3 c \sqrt{a^2 c x^2+c}}-\frac{1}{a^3 c \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right ) \tan ^{-1}(a x)}{a^3 c \sqrt{a^2 c x^2+c}}-\frac{x \tan ^{-1}(a x)}{a^2 c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcTan[a*x])/(c + a^2*c*x^2)^(3/2),x]

[Out]

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

Rule 4934

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q
+ 1))/(4*c^3*d*(q + 1)^2), x] + (-Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x],
x] + Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(2*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && E
qQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]

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

Mathematica [A]  time = 0.239759, size = 155, normalized size = 0.62 \[ -\frac{\sqrt{a^2 x^2+1} \left (-i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )+i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )+\frac{1}{\sqrt{a^2 x^2+1}}+\frac{a x \tan ^{-1}(a x)}{\sqrt{a^2 x^2+1}}+\tan ^{-1}(a x) \left (-\log \left (1-i e^{i \tan ^{-1}(a x)}\right )\right )+\tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )}{a^3 c \sqrt{c \left (a^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^2*ArcTan[a*x])/(c + a^2*c*x^2)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.766, size = 247, normalized size = 1. \begin{align*} -{\frac{ \left ( \arctan \left ( ax \right ) +i \right ) \left ( ax-i \right ) }{ \left ( 2\,{a}^{2}{x}^{2}+2 \right ){c}^{2}{a}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{ \left ( ax+i \right ) \left ( \arctan \left ( ax \right ) -i \right ) }{ \left ( 2\,{a}^{2}{x}^{2}+2 \right ){c}^{2}{a}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{i}{{c}^{2}{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)^(3/2),x)

[Out]

-1/2*(arctan(a*x)+I)*(a*x-I)*(c*(a*x-I)*(a*x+I))^(1/2)/(a^2*x^2+1)/c^2/a^3-1/2*(c*(a*x-I)*(a*x+I))^(1/2)*(a*x+
I)*(arctan(a*x)-I)/(a^2*x^2+1)/c^2/a^3+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^2/a^3

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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)^(3/2),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} x^{2} \arctan \left (a x\right )}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}, 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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*x^2*arctan(a*x)/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2), 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 )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

Integral(x**2*atan(a*x)/(c*(a**2*x**2 + 1))**(3/2), 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 )}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

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