3.1.30 \(\int \frac {(a+b \text {ArcTan}(c x))^3}{x} \, dx\) [30]

Optimal. Leaf size=206 \[ 2 (a+b \text {ArcTan}(c x))^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-\frac {3}{2} i b (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+\frac {3}{2} i b (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {3}{2} b^2 (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+\frac {3}{4} i b^3 \text {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )-\frac {3}{4} i b^3 \text {PolyLog}\left (4,-1+\frac {2}{1+i c x}\right ) \]

[Out]

-2*(a+b*arctan(c*x))^3*arctanh(-1+2/(1+I*c*x))-3/2*I*b*(a+b*arctan(c*x))^2*polylog(2,1-2/(1+I*c*x))+3/2*I*b*(a
+b*arctan(c*x))^2*polylog(2,-1+2/(1+I*c*x))-3/2*b^2*(a+b*arctan(c*x))*polylog(3,1-2/(1+I*c*x))+3/2*b^2*(a+b*ar
ctan(c*x))*polylog(3,-1+2/(1+I*c*x))+3/4*I*b^3*polylog(4,1-2/(1+I*c*x))-3/4*I*b^3*polylog(4,-1+2/(1+I*c*x))

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Rubi [A]
time = 0.29, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4942, 5108, 5004, 5114, 5118, 6745} \begin {gather*} -\frac {3}{2} b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))+\frac {3}{2} b^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))-\frac {3}{2} i b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))^2+\frac {3}{2} i b \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))^2+2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^3+\frac {3}{4} i b^3 \text {Li}_4\left (1-\frac {2}{i c x+1}\right )-\frac {3}{4} i b^3 \text {Li}_4\left (\frac {2}{i c x+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^3/x,x]

[Out]

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

Rule 4942

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

Rule 5004

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 5108

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

Rule 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 5118

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

Rule 6745

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} \, dx &=2 \left (a+b \tan ^{-1}(c x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-(6 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=2 \left (a+b \tan ^{-1}(c x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+(3 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(3 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=2 \left (a+b \tan ^{-1}(c x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-\frac {3}{2} i b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+\frac {3}{2} i b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\left (3 i b^2 c\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (3 i b^2 c\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=2 \left (a+b \tan ^{-1}(c x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-\frac {3}{2} i b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+\frac {3}{2} i b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {3}{2} b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{2} \left (3 b^3 c\right ) \int \frac {\text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac {1}{2} \left (3 b^3 c\right ) \int \frac {\text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=2 \left (a+b \tan ^{-1}(c x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-\frac {3}{2} i b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+\frac {3}{2} i b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {3}{2} b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\frac {3}{4} i b^3 \text {Li}_4\left (1-\frac {2}{1+i c x}\right )-\frac {3}{4} i b^3 \text {Li}_4\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 212, normalized size = 1.03 \begin {gather*} 2 (a+b \text {ArcTan}(c x))^3 \tanh ^{-1}\left (\frac {i+c x}{-i+c x}\right )+\frac {3}{4} i b \left (2 (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,\frac {i+c x}{i-c x}\right )-2 (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )+b \left (-2 i (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,\frac {i+c x}{i-c x}\right )+2 i (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,\frac {i+c x}{-i+c x}\right )+b \left (-\text {PolyLog}\left (4,\frac {i+c x}{i-c x}\right )+\text {PolyLog}\left (4,\frac {i+c x}{-i+c x}\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])^3/x,x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 4.77, size = 2309, normalized size = 11.21

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2309\)
default	\(\text {Expression too large to display}\)	\(2309\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))^3/x,x,method=_RETURNVERBOSE)

[Out]

3/2*I*a*b^2*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)^2+3/2*I*a^2*b*ln(c*
x)*ln(1+I*c*x)-3/2*I*a^2*b*ln(c*x)*ln(1-I*c*x)+1/2*I*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/
(c^2*x^2+1)))^3*arctan(c*x)^3-1/2*I*b^3*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arc
tan(c*x)^3+1/2*I*b^3*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)^3+3*I*a*b^
2*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-6*I*a*b^2*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2)
)+3/2*I*a*b^2*Pi*arctan(c*x)^2-6*I*a*b^2*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+a^3*ln(c*x)+b^3*ln
(c*x)*arctan(c*x)^3+3/2*I*a*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csg
n(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*arctan(c*x)^2+3/2*I*b^3*arctan(c*x)^2*polylog(2,-
(1+I*c*x)^2/(c^2*x^2+1))-3/2*I*a*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((
(1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2+3/2*I*a*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^
2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*arctan(
c*x)^2-3/2*I*a*b^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)/(1+(1+I*c*x)^2/(c
^2*x^2+1)))^2*arctan(c*x)^2-3/2*I*a*b^2*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)
-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2+1/2*I*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I/(1+(1
+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*arctan(c*x)^3+3*a*b^2*
ln(c*x)*arctan(c*x)^2-3*a*b^2*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+3*a*b^2*arctan(c*x)^2*ln(1-(1+I*c*x)
/(c^2*x^2+1)^(1/2))+3*a*b^2*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+3*a^2*b*ln(c*x)*arctan(c*x)+3/2*I*
a^2*b*dilog(1+I*c*x)-3/2*I*a^2*b*dilog(1-I*c*x)+1/2*I*b^3*Pi*arctan(c*x)^3-3*I*b^3*arctan(c*x)^2*polylog(2,-(1
+I*c*x)/(c^2*x^2+1)^(1/2))-3*I*b^3*arctan(c*x)^2*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*a*b^2*polylog(3,-(
1+I*c*x)^2/(c^2*x^2+1))+6*a*b^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*a*b^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1
)^(1/2))-b^3*arctan(c*x)^3*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+b^3*arctan(c*x)^3*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+6
*b^3*arctan(c*x)*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^3*arctan(c*x)^3*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*
b^3*arctan(c*x)*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*b^3*arctan(c*x)*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1)
)+6*I*b^3*polylog(4,(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*b^3*polylog(4,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/4*I*b^3*pol
ylog(4,-(1+I*c*x)^2/(c^2*x^2+1))+3/2*I*a*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1))
)^3*arctan(c*x)^2-3/2*I*a*b^2*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2
+1/2*I*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)
/(1+(1+I*c*x)^2/(c^2*x^2+1)))*arctan(c*x)^3-1/2*I*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^
2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^3-1/2*I*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+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^3-1/
2*I*b^3*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))
^2*arctan(c*x)^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*log(x) + 1/32*integrate((28*b^3*arctan(c*x)^3 + 3*b^3*arctan(c*x)*log(c^2*x^2 + 1)^2 + 96*a*b^2*arctan(c*x
)^2 + 96*a^2*b*arctan(c*x))/x, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^3/x,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, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c*x))**3/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^3/x,x)

[Out]

int((a + b*atan(c*x))^3/x, x)

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