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

Optimal. Leaf size=189 \[ a^3 c \text {Li}_3\left (\frac {2}{1-i a x}-1\right )-2 i a^3 c \text {Li}_2\left (\frac {2}{1-i a x}-1\right ) \tan ^{-1}(a x)+a^3 c \log (x)-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^3-\frac {1}{2} a^3 c \tan ^{-1}(a x)^2+2 a^3 c \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^2-\frac {a^2 c \tan ^{-1}(a x)^3}{x}-\frac {a^2 c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^3 c \log \left (a^2 x^2+1\right )-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a c \tan ^{-1}(a x)^2}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.58, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4950, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 4992, 6610} \[ a^3 c \text {PolyLog}\left (3,-1+\frac {2}{1-i a x}\right )-2 i a^3 c \tan ^{-1}(a x) \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )-\frac {1}{2} a^3 c \log \left (a^2 x^2+1\right )+a^3 c \log (x)-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^3-\frac {1}{2} a^3 c \tan ^{-1}(a x)^2-\frac {a^2 c \tan ^{-1}(a x)^3}{x}-\frac {a^2 c \tan ^{-1}(a x)}{x}+2 a^3 c \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^2-\frac {a c \tan ^{-1}(a x)^2}{2 x^2}-\frac {c \tan ^{-1}(a x)^3}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4868

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTan[c*x]
)^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2 - 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 4918

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

Rule 4924

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

Rule 4950

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

Rule 4992

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcT
an[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 {\left (c+a^2 c x^2\right ) \tan ^{-1}(a x)^3}{x^4} \, dx &=c \int \frac {\tan ^{-1}(a x)^3}{x^4} \, dx+\left (a^2 c\right ) \int \frac {\tan ^{-1}(a x)^3}{x^2} \, dx\\ &=-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)^3}{x}+(a c) \int \frac {\tan ^{-1}(a x)^2}{x^3 \left (1+a^2 x^2\right )} \, dx+\left (3 a^3 c\right ) \int \frac {\tan ^{-1}(a x)^2}{x \left (1+a^2 x^2\right )} \, dx\\ &=-i a^3 c \tan ^{-1}(a x)^3-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)^3}{x}+(a c) \int \frac {\tan ^{-1}(a x)^2}{x^3} \, dx+\left (3 i a^3 c\right ) \int \frac {\tan ^{-1}(a x)^2}{x (i+a x)} \, dx-\left (a^3 c\right ) \int \frac {\tan ^{-1}(a x)^2}{x \left (1+a^2 x^2\right )} \, dx\\ &=-\frac {a c \tan ^{-1}(a x)^2}{2 x^2}-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^3-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)^3}{x}+3 a^3 c \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )+\left (a^2 c\right ) \int \frac {\tan ^{-1}(a x)}{x^2 \left (1+a^2 x^2\right )} \, dx-\left (i a^3 c\right ) \int \frac {\tan ^{-1}(a x)^2}{x (i+a x)} \, dx-\left (6 a^4 c\right ) \int \frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {a c \tan ^{-1}(a x)^2}{2 x^2}-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^3-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)^3}{x}+2 a^3 c \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )-3 i a^3 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+\left (a^2 c\right ) \int \frac {\tan ^{-1}(a x)}{x^2} \, dx+\left (3 i a^4 c\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx-\left (a^4 c\right ) \int \frac {\tan ^{-1}(a x)}{1+a^2 x^2} \, dx+\left (2 a^4 c\right ) \int \frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {a^2 c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^3 c \tan ^{-1}(a x)^2-\frac {a c \tan ^{-1}(a x)^2}{2 x^2}-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^3-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)^3}{x}+2 a^3 c \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )-2 i a^3 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+\frac {3}{2} a^3 c \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )+\left (a^3 c\right ) \int \frac {1}{x \left (1+a^2 x^2\right )} \, dx-\left (i a^4 c\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {a^2 c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^3 c \tan ^{-1}(a x)^2-\frac {a c \tan ^{-1}(a x)^2}{2 x^2}-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^3-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)^3}{x}+2 a^3 c \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )-2 i a^3 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+a^3 c \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )+\frac {1}{2} \left (a^3 c\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {a^2 c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^3 c \tan ^{-1}(a x)^2-\frac {a c \tan ^{-1}(a x)^2}{2 x^2}-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^3-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)^3}{x}+2 a^3 c \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )-2 i a^3 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+a^3 c \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )+\frac {1}{2} \left (a^3 c\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (a^5 c\right ) \operatorname {Subst}\left (\int \frac {1}{1+a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a^2 c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^3 c \tan ^{-1}(a x)^2-\frac {a c \tan ^{-1}(a x)^2}{2 x^2}-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^3-\frac {c \tan ^{-1}(a x)^3}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)^3}{x}+a^3 c \log (x)-\frac {1}{2} a^3 c \log \left (1+a^2 x^2\right )+2 a^3 c \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )-2 i a^3 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+a^3 c \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 177, normalized size = 0.94 \[ \frac {1}{12} c \left (24 i a^3 \tan ^{-1}(a x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(a x)}\right )+12 a^3 \text {Li}_3\left (e^{-2 i \tan ^{-1}(a x)}\right )+8 i a^3 \tan ^{-1}(a x)^3-6 a^3 \tan ^{-1}(a x)^2+24 a^3 \tan ^{-1}(a x)^2 \log \left (1-e^{-2 i \tan ^{-1}(a x)}\right )-i \pi ^3 a^3-\frac {12 a^2 \tan ^{-1}(a x)^3}{x}-\frac {12 a^2 \tan ^{-1}(a x)}{x}+12 a^3 \log \left (\frac {a x}{\sqrt {a^2 x^2+1}}\right )-\frac {4 \tan ^{-1}(a x)^3}{x^3}-\frac {6 a \tan ^{-1}(a x)^2}{x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*((-I)*a^3*Pi^3 - (12*a^2*ArcTan[a*x])/x - 6*a^3*ArcTan[a*x]^2 - (6*a*ArcTan[a*x]^2)/x^2 + (8*I)*a^3*ArcTan[
a*x]^3 - (4*ArcTan[a*x]^3)/x^3 - (12*a^2*ArcTan[a*x]^3)/x + 24*a^3*ArcTan[a*x]^2*Log[1 - E^((-2*I)*ArcTan[a*x]
)] + 12*a^3*Log[(a*x)/Sqrt[1 + a^2*x^2]] + (24*I)*a^3*ArcTan[a*x]*PolyLog[2, E^((-2*I)*ArcTan[a*x])] + 12*a^3*
PolyLog[3, E^((-2*I)*ArcTan[a*x])]))/12

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )^{3}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [C]  time = 3.71, size = 5426, normalized size = 28.71 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^3\,\left (c\,a^2\,x^2+c\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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