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

Optimal. Leaf size=113 \[ \frac {i \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{2 a^4 c}+\frac {i \tan ^{-1}(a x)^2}{2 a^4 c}+\frac {\tan ^{-1}(a x)}{2 a^4 c}+\frac {\log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^4 c}-\frac {x}{2 a^3 c}+\frac {x^2 \tan ^{-1}(a x)}{2 a^2 c} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 113, 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 = {4916, 4852, 321, 203, 4920, 4854, 2402, 2315} \[ \frac {i \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^4 c}+\frac {x^2 \tan ^{-1}(a x)}{2 a^2 c}-\frac {x}{2 a^3 c}+\frac {i \tan ^{-1}(a x)^2}{2 a^4 c}+\frac {\tan ^{-1}(a x)}{2 a^4 c}+\frac {\log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 4852

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

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 4916

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 120, normalized size = 1.06 \[ \frac {i \text {Li}_2\left (-\frac {a x+i}{i-a x}\right )}{2 a^4 c}+\frac {i \tan ^{-1}(a x)^2}{2 a^4 c}+\frac {\tan ^{-1}(a x)}{2 a^4 c}+\frac {\log \left (\frac {2 i}{-a x+i}\right ) \tan ^{-1}(a x)}{a^4 c}-\frac {x}{2 a^3 c}+\frac {x^2 \tan ^{-1}(a x)}{2 a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \arctan \left (a x\right )}{a^{2} c x^{2} + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [B]  time = 0.12, size = 238, normalized size = 2.11 \[ \frac {x^{2} \arctan \left (a x \right )}{2 a^{2} c}-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 a^{4} c}-\frac {x}{2 a^{3} c}+\frac {\arctan \left (a x \right )}{2 a^{4} c}-\frac {i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 a^{4} c}+\frac {i \ln \left (a x -i\right )^{2}}{8 a^{4} c}+\frac {i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 a^{4} c}+\frac {i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 a^{4} c}+\frac {i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 a^{4} c}-\frac {i \ln \left (a x +i\right )^{2}}{8 a^{4} c}-\frac {i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{4 a^{4} c}-\frac {i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{4 a^{4} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(a*x)/a^2/c-1/2/a^4/c*arctan(a*x)*ln(a^2*x^2+1)-1/2*x/a^3/c+1/2*arctan(a*x)/a^4/c-1/4*I/a^4/c*ln
(a*x-I)*ln(a^2*x^2+1)+1/8*I/a^4/c*ln(a*x-I)^2+1/4*I/a^4/c*dilog(-1/2*I*(I+a*x))+1/4*I/a^4/c*ln(a*x-I)*ln(-1/2*
I*(I+a*x))+1/4*I/a^4/c*ln(I+a*x)*ln(a^2*x^2+1)-1/8*I/a^4/c*ln(I+a*x)^2-1/4*I/a^4/c*dilog(1/2*I*(a*x-I))-1/4*I/
a^4/c*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 \[ \int \frac {x^{3} \arctan \left (a x\right )}{a^{2} c x^{2} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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