3.31 \(\int \frac{(a+b \tan ^{-1}(c x))^3}{x^2} \, dx\)
Optimal. Leaf size=116 \[ -3 i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{2} b^3 c \text{PolyLog}\left (3,-1+\frac{2}{1-i c x}\right )-i c \left (a+b \tan ^{-1}(c x)\right )^3-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x}+3 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]
[Out]
(-I)*c*(a + b*ArcTan[c*x])^3 - (a + b*ArcTan[c*x])^3/x + 3*b*c*(a + b*ArcTan[c*x])^2*Log[2 - 2/(1 - I*c*x)] -
(3*I)*b^2*c*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 - I*c*x)] + (3*b^3*c*PolyLog[3, -1 + 2/(1 - I*c*x)])/2
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Rubi [A] time = 0.267137, antiderivative size = 116, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used =
{4852, 4924, 4868, 4884, 4992, 6610} \[ -3 i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{2} b^3 c \text{PolyLog}\left (3,-1+\frac{2}{1-i c x}\right )-i c \left (a+b \tan ^{-1}(c x)\right )^3-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x}+3 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTan[c*x])^3/x^2,x]
[Out]
(-I)*c*(a + b*ArcTan[c*x])^3 - (a + b*ArcTan[c*x])^3/x + 3*b*c*(a + b*ArcTan[c*x])^2*Log[2 - 2/(1 - I*c*x)] -
(3*I)*b^2*c*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 - I*c*x)] + (3*b^3*c*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 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 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 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^2} \, dx &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x}+(3 b c) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x \left (1+c^2 x^2\right )} \, dx\\ &=-i c \left (a+b \tan ^{-1}(c x)\right )^3-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x}+(3 i b c) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x (i+c x)} \, dx\\ &=-i c \left (a+b \tan ^{-1}(c x)\right )^3-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x}+3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1-i c x}\right )-\left (6 b^2 c^2\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\\ &=-i c \left (a+b \tan ^{-1}(c x)\right )^3-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x}+3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1-i c x}\right )-3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )+\left (3 i b^3 c^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-i c \left (a+b \tan ^{-1}(c x)\right )^3-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x}+3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1-i c x}\right )-3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )+\frac{3}{2} b^3 c \text{Li}_3\left (-1+\frac{2}{1-i c x}\right )\\ \end{align*}
Mathematica [A] time = 0.368906, size = 214, normalized size = 1.84 \[ 3 a b^2 c \left (-i \left (\tan ^{-1}(c x)^2+\text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )\right )-\frac{\tan ^{-1}(c x)^2}{c x}+2 \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )+b^3 c \left (3 i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-\frac{\tan ^{-1}(c x)^3}{c x}+i \tan ^{-1}(c x)^3+3 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\frac{i \pi ^3}{8}\right )-\frac{3}{2} a^2 b c \log \left (c^2 x^2+1\right )+3 a^2 b c \log (x)-\frac{3 a^2 b \tan ^{-1}(c x)}{x}-\frac{a^3}{x} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*ArcTan[c*x])^3/x^2,x]
[Out]
-(a^3/x) - (3*a^2*b*ArcTan[c*x])/x + 3*a^2*b*c*Log[x] - (3*a^2*b*c*Log[1 + c^2*x^2])/2 + 3*a*b^2*c*(-(ArcTan[c
*x]^2/(c*x)) + 2*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] - I*(ArcTan[c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*
x])])) + b^3*c*((-I/8)*Pi^3 + I*ArcTan[c*x]^3 - ArcTan[c*x]^3/(c*x) + 3*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan
[c*x])] + (3*I)*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + (3*PolyLog[3, E^((-2*I)*ArcTan[c*x])])/2)
________________________________________________________________________________________
Maple [C] time = 0.341, size = 2159, normalized size = 18.6 \begin{align*} \text{result 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^2,x)
[Out]
3*c*b^3*ln(c*x)*arctan(c*x)^2+3*c*b^3*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/4*I*c*b^3*arctan(c*x)^
2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+
I*c*x)^2/(c^2*x^2+1)+1)^2)+3/2*I*c*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+
1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+3*I*c*a*b^2*dilog(1+I*c*x)-3
*I*c*a*b^2*dilog(1-I*c*x)+3/2*I*c*b^3*Pi*arctan(c*x)^2+6*c*a*b^2*ln(c*x)*arctan(c*x)-3*c*a*b^2*arctan(c*x)*ln(
c^2*x^2+1)-6*I*c*b^3*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))-6*I*c*b^3*arctan(c*x)*polylog(2,-(1+I*
c*x)/(c^2*x^2+1)^(1/2))+3/4*I*c*a*b^2*ln(c*x-I)^2+3/2*I*c*a*b^2*dilog(-1/2*I*(c*x+I))-3/4*I*c*a*b^2*ln(c*x+I)^
2-3/2*I*c*a*b^2*dilog(1/2*I*(c*x-I))+3/4*I*c*b^3*arctan(c*x)^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I
*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2+3/2*I*c*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)/(c^2*x
^2+1)^(1/2))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2+3/4*I*c*b^3*arctan(c*x)^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)
)^2*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)-3/2*I*c*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)
^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-3/2*I*c*b^3*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+
1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+3/2*I*c*b^3*Pi*csgn(I*((1+
I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)
+1))*arctan(c*x)^2+3*c*b^3*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-3*c*b^3*arctan(c*x)^2*ln((1+I*c*x)^
2/(c^2*x^2+1)-1)-3/2*c*b^3*arctan(c*x)^2*ln(c^2*x^2+1)+3*c*b^3*arctan(c*x)^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))+3
*c*a^2*b*ln(c*x)-3/2*c*a^2*b*ln(c^2*x^2+1)+3*c*b^3*arctan(c*x)^2*ln(2)-3*a^2*b/x*arctan(c*x)-3*a*b^2/x*arctan(
c*x)^2-I*c*b^3*arctan(c*x)^3-a^3/x-b^3/x*arctan(c*x)^3+6*c*b^3*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*c*b^3*
polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/4*I*c*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)
^2/(c^2*x^2+1)+1)^2)^3+3/4*I*c*b^3*arctan(c*x)^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3+3*I*c*a*b^2*ln(c*x
)*ln(1+I*c*x)-3*I*c*a*b^2*ln(c*x)*ln(1-I*c*x)-3/2*I*c*a*b^2*ln(c^2*x^2+1)*ln(c*x-I)+3/2*I*c*a*b^2*ln(c*x-I)*ln
(-1/2*I*(c*x+I))-3/2*I*c*a*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))+3/2*I*c*a*b^2*ln(c^2*x^2+1)*ln(c*x+I)+3/2*I*c*b^3*P
i*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2+3/2*I*c*b^3*Pi*csgn(((1+I*c*
x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-3/2*I*c*b^3*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-
1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-3/4*I*c*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3
+3/4*I*c*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2
*x^2+1)+1)^2)^2-3/4*I*c*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))^2*csgn(I*(1+I*c*x)^2/(c^2*x^2
+1))-3/2*I*c*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2
+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-3/2*I*c*b^3*arctan(c*x)^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+
1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2
________________________________________________________________________________________
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^2,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^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x))^3/x^2,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^2, x)
________________________________________________________________________________________
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^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atan(c*x))**3/x**2,x)
[Out]
Integral((a + b*atan(c*x))**3/x**2, x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x))^3/x^2,x, algorithm="giac")
[Out]
integrate((b*arctan(c*x) + a)^3/x^2, x)