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

Optimal. Leaf size=133 \[ -\frac {i \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{2 a^4 c^2}-\frac {i \tan ^{-1}(a x)^2}{2 a^4 c^2}-\frac {\tan ^{-1}(a x)}{4 a^4 c^2}-\frac {\log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^4 c^2}+\frac {\tan ^{-1}(a x)}{2 a^4 c^2 \left (a^2 x^2+1\right )}-\frac {x}{4 a^3 c^2 \left (a^2 x^2+1\right )} \]

[Out]

-1/4*x/a^3/c^2/(a^2*x^2+1)-1/4*arctan(a*x)/a^4/c^2+1/2*arctan(a*x)/a^4/c^2/(a^2*x^2+1)-1/2*I*arctan(a*x)^2/a^4
/c^2-arctan(a*x)*ln(2/(1+I*a*x))/a^4/c^2-1/2*I*polylog(2,1-2/(1+I*a*x))/a^4/c^2

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Rubi [A]  time = 0.16, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4964, 4920, 4854, 2402, 2315, 4930, 199, 205} \[ -\frac {i \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}-\frac {x}{4 a^3 c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{2 a^4 c^2 \left (a^2 x^2+1\right )}-\frac {i \tan ^{-1}(a x)^2}{2 a^4 c^2}-\frac {\tan ^{-1}(a x)}{4 a^4 c^2}-\frac {\log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(4*a^3*c^2*(1 + a^2*x^2)) - ArcTan[a*x]/(4*a^4*c^2) + ArcTan[a*x]/(2*a^4*c^2*(1 + a^2*x^2)) - ((I/2)*ArcTan
[a*x]^2)/(a^4*c^2) - (ArcTan[a*x]*Log[2/(1 + I*a*x)])/(a^4*c^2) - ((I/2)*PolyLog[2, 1 - 2/(1 + I*a*x)])/(a^4*c
^2)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4854

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

Rule 4920

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

Rule 4930

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

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int[
x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*Arc
Tan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m
, 1] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac {\int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{a^2}+\frac {\int \frac {x \tan ^{-1}(a x)}{c+a^2 c x^2} \, dx}{a^2 c}\\ &=\frac {\tan ^{-1}(a x)}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \tan ^{-1}(a x)^2}{2 a^4 c^2}-\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{2 a^3}-\frac {\int \frac {\tan ^{-1}(a x)}{i-a x} \, dx}{a^3 c^2}\\ &=-\frac {x}{4 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \tan ^{-1}(a x)^2}{2 a^4 c^2}-\frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}+\frac {\int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c^2}-\frac {\int \frac {1}{c+a^2 c x^2} \, dx}{4 a^3 c}\\ &=-\frac {x}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {\tan ^{-1}(a x)}{4 a^4 c^2}+\frac {\tan ^{-1}(a x)}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \tan ^{-1}(a x)^2}{2 a^4 c^2}-\frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {i \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )}{a^4 c^2}\\ &=-\frac {x}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {\tan ^{-1}(a x)}{4 a^4 c^2}+\frac {\tan ^{-1}(a x)}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \tan ^{-1}(a x)^2}{2 a^4 c^2}-\frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {i \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 77, normalized size = 0.58 \[ \frac {4 i \text {Li}_2\left (-e^{2 i \tan ^{-1}(a x)}\right )+4 i \tan ^{-1}(a x)^2-\sin \left (2 \tan ^{-1}(a x)\right )+2 \tan ^{-1}(a x) \left (\cos \left (2 \tan ^{-1}(a x)\right )-4 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )\right )}{8 a^4 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((4*I)*ArcTan[a*x]^2 + 2*ArcTan[a*x]*(Cos[2*ArcTan[a*x]] - 4*Log[1 + E^((2*I)*ArcTan[a*x])]) + (4*I)*PolyLog[2
, -E^((2*I)*ArcTan[a*x])] - Sin[2*ArcTan[a*x]])/(8*a^4*c^2)

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \arctan \left (a x\right )}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)/(a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

integral(x^3*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^3*arctan(a*x)/(a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.10, size = 257, normalized size = 1.93 \[ \frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 a^{4} c^{2}}+\frac {\arctan \left (a x \right )}{2 a^{4} c^{2} \left (a^{2} x^{2}+1\right )}-\frac {x}{4 a^{3} c^{2} \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )}{4 a^{4} c^{2}}+\frac {i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 a^{4} c^{2}}-\frac {i \ln \left (a x -i\right )^{2}}{8 a^{4} c^{2}}-\frac {i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 a^{4} c^{2}}-\frac {i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 a^{4} c^{2}}-\frac {i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 a^{4} c^{2}}+\frac {i \ln \left (a x +i\right )^{2}}{8 a^{4} c^{2}}+\frac {i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{4 a^{4} c^{2}}+\frac {i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{4 a^{4} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^4/c^2*arctan(a*x)*ln(a^2*x^2+1)+1/2*arctan(a*x)/a^4/c^2/(a^2*x^2+1)-1/4*x/a^3/c^2/(a^2*x^2+1)-1/4*arctan
(a*x)/a^4/c^2+1/4*I/a^4/c^2*ln(a*x-I)*ln(a^2*x^2+1)-1/8*I/a^4/c^2*ln(a*x-I)^2-1/4*I/a^4/c^2*dilog(-1/2*I*(I+a*
x))-1/4*I/a^4/c^2*ln(a*x-I)*ln(-1/2*I*(I+a*x))-1/4*I/a^4/c^2*ln(I+a*x)*ln(a^2*x^2+1)+1/8*I/a^4/c^2*ln(I+a*x)^2
+1/4*I/a^4/c^2*dilog(1/2*I*(a*x-I))+1/4*I/a^4/c^2*ln(I+a*x)*ln(1/2*I*(a*x-I))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)/(a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\mathrm {atan}\left (a\,x\right )}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{3} \operatorname {atan}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*atan(a*x)/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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