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

Optimal. Leaf size=213 \[ i b^2 c^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} b^3 c^3 \text{PolyLog}\left (3,-1+\frac{2}{1-i c x}\right )-\frac{b^2 c^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac{1}{3} i c^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{1}{2} b c^3 \left (a+b \tan ^{-1}(c x)\right )^2-b c^3 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b c \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{3 x^3}-\frac{1}{2} b^3 c^3 \log \left (c^2 x^2+1\right )+b^3 c^3 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.476335, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 4992, 6610} \[ i b^2 c^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} b^3 c^3 \text{PolyLog}\left (3,-1+\frac{2}{1-i c x}\right )-\frac{b^2 c^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac{1}{3} i c^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{1}{2} b c^3 \left (a+b \tan ^{-1}(c x)\right )^2-b c^3 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b c \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{3 x^3}-\frac{1}{2} b^3 c^3 \log \left (c^2 x^2+1\right )+b^3 c^3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^3/x^4,x]

[Out]

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

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

Mathematica [A]  time = 0.812281, size = 321, normalized size = 1.51 \[ \frac{i a b^2 \left (c^3 x^3 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+i c^2 x^2+\left (c^3 x^3+i\right ) \tan ^{-1}(c x)^2+i c x \tan ^{-1}(c x) \left (c^2 x^2+2 c^2 x^2 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+1\right )\right )}{x^3}+\frac{1}{24} b^3 c^3 \left (-24 i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )-12 \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )+24 \log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )-\frac{8 \tan ^{-1}(c x)^3}{c^3 x^3}-\frac{12 \tan ^{-1}(c x)^2}{c^2 x^2}-8 i \tan ^{-1}(c x)^3-12 \tan ^{-1}(c x)^2-\frac{24 \tan ^{-1}(c x)}{c x}-24 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+i \pi ^3\right )+\frac{1}{2} a^2 b c^3 \log \left (c^2 x^2+1\right )-a^2 b c^3 \log (x)-\frac{a^2 b c}{2 x^2}-\frac{a^2 b \tan ^{-1}(c x)}{x^3}-\frac{a^3}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 1.735, size = 5974, normalized size = 28.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \arctan \left (c x\right )^{3} + 3 \, a b^{2} \arctan \left (c x\right )^{2} + 3 \, a^{2} b \arctan \left (c x\right ) + a^{3}}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right )^{3}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c*x) + a)^3/x^4, x)