3.387 \(\int \frac {x^4 \tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx\)

Optimal. Leaf size=217 \[ -\frac {2 \text {Li}_3\left (1-\frac {2}{i a x+1}\right )}{a^5 c}-\frac {4 i \text {Li}_2\left (1-\frac {2}{i a x+1}\right ) \tan ^{-1}(a x)}{a^5 c}+\frac {\tan ^{-1}(a x)^4}{4 a^5 c}-\frac {4 i \tan ^{-1}(a x)^3}{3 a^5 c}-\frac {\tan ^{-1}(a x)^2}{2 a^5 c}-\frac {4 \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)^2}{a^5 c}-\frac {x \tan ^{-1}(a x)^3}{a^4 c}+\frac {x \tan ^{-1}(a x)}{a^4 c}-\frac {x^2 \tan ^{-1}(a x)^2}{2 a^3 c}+\frac {x^3 \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\log \left (a^2 x^2+1\right )}{2 a^5 c} \]

[Out]

x*arctan(a*x)/a^4/c-1/2*arctan(a*x)^2/a^5/c-1/2*x^2*arctan(a*x)^2/a^3/c-4/3*I*arctan(a*x)^3/a^5/c-x*arctan(a*x
)^3/a^4/c+1/3*x^3*arctan(a*x)^3/a^2/c+1/4*arctan(a*x)^4/a^5/c-4*arctan(a*x)^2*ln(2/(1+I*a*x))/a^5/c-1/2*ln(a^2
*x^2+1)/a^5/c-4*I*arctan(a*x)*polylog(2,1-2/(1+I*a*x))/a^5/c-2*polylog(3,1-2/(1+I*a*x))/a^5/c

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Rubi [A]  time = 0.63, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {4916, 4852, 4846, 260, 4884, 4920, 4854, 4994, 6610} \[ -\frac {2 \text {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )}{a^5 c}-\frac {4 i \tan ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{a^5 c}-\frac {\log \left (a^2 x^2+1\right )}{2 a^5 c}+\frac {x^3 \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {x^2 \tan ^{-1}(a x)^2}{2 a^3 c}+\frac {\tan ^{-1}(a x)^4}{4 a^5 c}-\frac {x \tan ^{-1}(a x)^3}{a^4 c}-\frac {4 i \tan ^{-1}(a x)^3}{3 a^5 c}-\frac {\tan ^{-1}(a x)^2}{2 a^5 c}+\frac {x \tan ^{-1}(a x)}{a^4 c}-\frac {4 \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)^2}{a^5 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*ArcTan[a*x]^3)/(c + a^2*c*x^2),x]

[Out]

(x*ArcTan[a*x])/(a^4*c) - ArcTan[a*x]^2/(2*a^5*c) - (x^2*ArcTan[a*x]^2)/(2*a^3*c) - (((4*I)/3)*ArcTan[a*x]^3)/
(a^5*c) - (x*ArcTan[a*x]^3)/(a^4*c) + (x^3*ArcTan[a*x]^3)/(3*a^2*c) + ArcTan[a*x]^4/(4*a^5*c) - (4*ArcTan[a*x]
^2*Log[2/(1 + I*a*x)])/(a^5*c) - Log[1 + a^2*x^2]/(2*a^5*c) - ((4*I)*ArcTan[a*x]*PolyLog[2, 1 - 2/(1 + I*a*x)]
)/(a^5*c) - (2*PolyLog[3, 1 - 2/(1 + I*a*x)])/(a^5*c)

Rule 260

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

Rule 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 4884

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

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]

Rule 4994

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {x^4 \tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx &=-\frac {\int \frac {x^2 \tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx}{a^2}+\frac {\int x^2 \tan ^{-1}(a x)^3 \, dx}{a^2 c}\\ &=\frac {x^3 \tan ^{-1}(a x)^3}{3 a^2 c}+\frac {\int \frac {\tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx}{a^4}-\frac {\int \tan ^{-1}(a x)^3 \, dx}{a^4 c}-\frac {\int \frac {x^3 \tan ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{a c}\\ &=-\frac {x \tan ^{-1}(a x)^3}{a^4 c}+\frac {x^3 \tan ^{-1}(a x)^3}{3 a^2 c}+\frac {\tan ^{-1}(a x)^4}{4 a^5 c}-\frac {\int x \tan ^{-1}(a x)^2 \, dx}{a^3 c}+\frac {\int \frac {x \tan ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{a^3 c}+\frac {3 \int \frac {x \tan ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{a^3 c}\\ &=-\frac {x^2 \tan ^{-1}(a x)^2}{2 a^3 c}-\frac {4 i \tan ^{-1}(a x)^3}{3 a^5 c}-\frac {x \tan ^{-1}(a x)^3}{a^4 c}+\frac {x^3 \tan ^{-1}(a x)^3}{3 a^2 c}+\frac {\tan ^{-1}(a x)^4}{4 a^5 c}-\frac {\int \frac {\tan ^{-1}(a x)^2}{i-a x} \, dx}{a^4 c}-\frac {3 \int \frac {\tan ^{-1}(a x)^2}{i-a x} \, dx}{a^4 c}+\frac {\int \frac {x^2 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{a^2 c}\\ &=-\frac {x^2 \tan ^{-1}(a x)^2}{2 a^3 c}-\frac {4 i \tan ^{-1}(a x)^3}{3 a^5 c}-\frac {x \tan ^{-1}(a x)^3}{a^4 c}+\frac {x^3 \tan ^{-1}(a x)^3}{3 a^2 c}+\frac {\tan ^{-1}(a x)^4}{4 a^5 c}-\frac {4 \tan ^{-1}(a x)^2 \log \left (\frac {2}{1+i a x}\right )}{a^5 c}+\frac {\int \tan ^{-1}(a x) \, dx}{a^4 c}-\frac {\int \frac {\tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{a^4 c}+\frac {2 \int \frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^4 c}+\frac {6 \int \frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^4 c}\\ &=\frac {x \tan ^{-1}(a x)}{a^4 c}-\frac {\tan ^{-1}(a x)^2}{2 a^5 c}-\frac {x^2 \tan ^{-1}(a x)^2}{2 a^3 c}-\frac {4 i \tan ^{-1}(a x)^3}{3 a^5 c}-\frac {x \tan ^{-1}(a x)^3}{a^4 c}+\frac {x^3 \tan ^{-1}(a x)^3}{3 a^2 c}+\frac {\tan ^{-1}(a x)^4}{4 a^5 c}-\frac {4 \tan ^{-1}(a x)^2 \log \left (\frac {2}{1+i a x}\right )}{a^5 c}-\frac {4 i \tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{a^5 c}+\frac {i \int \frac {\text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^4 c}+\frac {(3 i) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^4 c}-\frac {\int \frac {x}{1+a^2 x^2} \, dx}{a^3 c}\\ &=\frac {x \tan ^{-1}(a x)}{a^4 c}-\frac {\tan ^{-1}(a x)^2}{2 a^5 c}-\frac {x^2 \tan ^{-1}(a x)^2}{2 a^3 c}-\frac {4 i \tan ^{-1}(a x)^3}{3 a^5 c}-\frac {x \tan ^{-1}(a x)^3}{a^4 c}+\frac {x^3 \tan ^{-1}(a x)^3}{3 a^2 c}+\frac {\tan ^{-1}(a x)^4}{4 a^5 c}-\frac {4 \tan ^{-1}(a x)^2 \log \left (\frac {2}{1+i a x}\right )}{a^5 c}-\frac {\log \left (1+a^2 x^2\right )}{2 a^5 c}-\frac {4 i \tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{a^5 c}-\frac {2 \text {Li}_3\left (1-\frac {2}{1+i a x}\right )}{a^5 c}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 154, normalized size = 0.71 \[ \frac {4 a^3 x^3 \tan ^{-1}(a x)^3-6 \log \left (a^2 x^2+1\right )-6 a^2 x^2 \tan ^{-1}(a x)^2+48 i \tan ^{-1}(a x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(a x)}\right )-24 \text {Li}_3\left (-e^{2 i \tan ^{-1}(a x)}\right )+3 \tan ^{-1}(a x)^4-12 a x \tan ^{-1}(a x)^3+16 i \tan ^{-1}(a x)^3-6 \tan ^{-1}(a x)^2+12 a x \tan ^{-1}(a x)-48 \tan ^{-1}(a x)^2 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )}{12 a^5 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*ArcTan[a*x]^3)/(c + a^2*c*x^2),x]

[Out]

(12*a*x*ArcTan[a*x] - 6*ArcTan[a*x]^2 - 6*a^2*x^2*ArcTan[a*x]^2 + (16*I)*ArcTan[a*x]^3 - 12*a*x*ArcTan[a*x]^3
+ 4*a^3*x^3*ArcTan[a*x]^3 + 3*ArcTan[a*x]^4 - 48*ArcTan[a*x]^2*Log[1 + E^((2*I)*ArcTan[a*x])] - 6*Log[1 + a^2*
x^2] + (48*I)*ArcTan[a*x]*PolyLog[2, -E^((2*I)*ArcTan[a*x])] - 24*PolyLog[3, -E^((2*I)*ArcTan[a*x])])/(12*a^5*
c)

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{4} \arctan \left (a x\right )^{3}}{a^{2} c x^{2} + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

integral(x^4*arctan(a*x)^3/(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^4*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 3.07, size = 1740, normalized size = 8.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arctan(a*x)^3/(a^2*c*x^2+c),x)

[Out]

-1/2*arctan(a*x)^2/a^5/c+1/4*arctan(a*x)^4/a^5/c-1/2*x^2*arctan(a*x)^2/a^3/c-x*arctan(a*x)^3/a^4/c+1/3*x^3*arc
tan(a*x)^3/a^2/c+I/a^5/c*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))^2*arctan(a*x)^2*P
i+1/a^5/c*ln((1+I*a*x)^2/(a^2*x^2+1)+1)-2/a^5/c*polylog(3,-(1+I*a*x)^2/(a^2*x^2+1))-1/a^4/c*csgn(I*(1+I*a*x)^4
/(a^2*x^2+1)^2+2*I*(1+I*a*x)^2/(a^2*x^2+1)+I)^2*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)+I)*arctan(a*x)^2*Pi*x-I/a^5/c*a
rctan(a*x)+4/3*I/a^5/c*arctan(a*x)^3-4/a^5/c*ln(2)*arctan(a*x)^2+2/a^5/c*arctan(a*x)^2*ln(a^2*x^2+1)-4/a^5/c*a
rctan(a*x)^2*ln((1+I*a*x)/(a^2*x^2+1)^(1/2))+1/a^4/c*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2*csgn(I*((1+I*a*x)
^2/(a^2*x^2+1)+1))*arctan(a*x)^2*Pi*x-1/2/a^4/c*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)*csgn(I*((1+I*a*x)^2/(a^2
*x^2+1)+1))^2*arctan(a*x)^2*Pi*x+I/a^5/c*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2*csgn(I*((1+I*a*x)^2/(a^2*x^2+
1)+1))*arctan(a*x)^2*Pi+I/a^5/c*csgn(I*(1+I*a*x)^4/(a^2*x^2+1)^2+2*I*(1+I*a*x)^2/(a^2*x^2+1)+I)^2*csgn(I*(1+I*
a*x)^2/(a^2*x^2+1)+I)*arctan(a*x)^2*Pi+4*I/a^5/c*arctan(a*x)*polylog(2,-(1+I*a*x)^2/(a^2*x^2+1))+1/2/a^4/c*csg
n(I*(1+I*a*x)^4/(a^2*x^2+1)^2+2*I*(1+I*a*x)^2/(a^2*x^2+1)+I)*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)+I)^2*arctan(a*x)^2
*Pi*x-2*I/a^5/c*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^2*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*arctan(a*x)^2*Pi-I/a^5/c
*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2*arctan(a*x)^2
*Pi-I/a^5/c*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1)
^2)*arctan(a*x)^2*Pi-1/2*I/a^5/c*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1))^2*a
rctan(a*x)^2*Pi-1/2*I/a^5/c*csgn(I*(1+I*a*x)^4/(a^2*x^2+1)^2+2*I*(1+I*a*x)^2/(a^2*x^2+1)+I)*csgn(I*(1+I*a*x)^2
/(a^2*x^2+1)+I)^2*arctan(a*x)^2*Pi+x*arctan(a*x)/a^4/c+1/2/a^4/c*csgn(I*(1+I*a*x)^4/(a^2*x^2+1)^2+2*I*(1+I*a*x
)^2/(a^2*x^2+1)+I)^3*arctan(a*x)^2*Pi*x-1/2/a^4/c*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3*arctan(a*x)^2*Pi*x+I
/a^5/c*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^3*arctan(a*x)^2*Pi+I/a^5/c*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/
(a^2*x^2+1)+1)^2)^3*arctan(a*x)^2*Pi-1/2*I/a^5/c*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3*arctan(a*x)^2*Pi-1/2*
I/a^5/c*csgn(I*(1+I*a*x)^4/(a^2*x^2+1)^2+2*I*(1+I*a*x)^2/(a^2*x^2+1)+I)^3*arctan(a*x)^2*Pi+I/a^5/c*csgn(I*(1+I
*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)*csgn(I/((1+I*a*x)^2/(a^2*x^
2+1)+1)^2)*arctan(a*x)^2*Pi

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\mathrm {atan}\left (a\,x\right )}^3}{c\,a^2\,x^2+c} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*atan(a*x)^3)/(c + a^2*c*x^2),x)

[Out]

int((x^4*atan(a*x)^3)/(c + a^2*c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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