3.19 \(\int \frac {(a+b \tan ^{-1}(c x))^2}{x} \, dx\)

Optimal. Leaf size=132 \[ -i b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )+2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{2} b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4850, 4988, 4884, 4994, 6610} \[ -i b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{2} b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^2/x,x]

[Out]

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

Rule 4850

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 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 4988

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

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Mathematica [A]  time = 0.08, size = 144, normalized size = 1.09 \[ \frac {1}{2} b \left (2 i \text {Li}_2\left (\frac {c x+i}{i-c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 i \text {Li}_2\left (\frac {c x+i}{c x-i}\right ) \left (a+b \tan ^{-1}(c x)\right )+b \left (\text {Li}_3\left (\frac {c x+i}{i-c x}\right )-\text {Li}_3\left (\frac {c x+i}{c x-i}\right )\right )\right )+2 \tanh ^{-1}\left (\frac {c x+i}{c x-i}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])^2/x,x]

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + a^{2}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [C]  time = 0.51, size = 1128, normalized size = 8.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \log \relax (x) + \frac {1}{16} \, \int \frac {12 \, b^{2} \arctan \left (c x\right )^{2} + b^{2} \log \left (c^{2} x^{2} + 1\right )^{2} + 32 \, a b \arctan \left (c x\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(x) + 1/16*integrate((12*b^2*arctan(c*x)^2 + b^2*log(c^2*x^2 + 1)^2 + 32*a*b*arctan(c*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2/x,x)

[Out]

int((a + b*atan(c*x))^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c*x))**2/x, x)

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