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

Optimal. Leaf size=251 \[ -\frac {x \tan ^{-1}(a x)}{a^2 c \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {i \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {1}{a^3 c \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right ) \tan ^{-1}(a x)}{a^3 c \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 251, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.25, size = 155, normalized size = 0.62 \[ -\frac {\sqrt {a^2 x^2+1} \left (\frac {1}{\sqrt {a^2 x^2+1}}+\frac {a x \tan ^{-1}(a x)}{\sqrt {a^2 x^2+1}}-i \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+i \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )+\tan ^{-1}(a x) \left (-\log \left (1-i e^{i \tan ^{-1}(a x)}\right )\right )+\tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )}{a^3 c \sqrt {c \left (a^2 x^2+1\right )}} \]

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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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} x^{2} \arctan \left (a x\right )}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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]

sage0*x

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maple [A]  time = 3.00, size = 247, normalized size = 0.98 \[ -\frac {\left (i+\arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) c^{2} a^{3}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )-i\right )}{2 \left (a^{2} x^{2}+1\right ) c^{2} a^{3}}+\frac {i \left (i \arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+\dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{2} a^{3}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,\mathrm {atan}\left (a\,x\right )}{{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*atan(a*x))/(c + a^2*c*x^2)^(3/2),x)

[Out]

int((x^2*atan(a*x))/(c + a^2*c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {atan}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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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