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

Optimal. Leaf size=151 \[ -\frac {1}{2} i b \text {Li}_2\left (1-\frac {2}{i c x^2+1}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac {1}{2} i b \text {Li}_2\left (\frac {2}{i c x^2+1}-1\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2-\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{i c x^2+1}\right )+\frac {1}{4} b^2 \text {Li}_3\left (\frac {2}{i c x^2+1}-1\right ) \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5031, 4850, 4988, 4884, 4994, 6610} \[ -\frac {1}{2} i b \text {PolyLog}\left (2,1-\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac {1}{2} i b \text {PolyLog}\left (2,-1+\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )-\frac {1}{4} b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x^2}\right )+\frac {1}{4} b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x^2}\right )+\tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^2])^2/x,x]

[Out]

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

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 5031

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTan[c*x])^p
/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 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}\left (c x^2\right )\right )^2}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right )-(2 b c) \operatorname {Subst}\left (\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,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right )+(b c) \operatorname {Subst}\left (\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,x,x^2\right )-(b c) \operatorname {Subst}\left (\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,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right )-\frac {1}{2} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (1-\frac {2}{1+i c x^2}\right )+\frac {1}{2} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x^2}\right )+\frac {1}{2} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )-\frac {1}{2} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right )-\frac {1}{2} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (1-\frac {2}{1+i c x^2}\right )+\frac {1}{2} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x^2}\right )-\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{1+i c x^2}\right )+\frac {1}{4} b^2 \text {Li}_3\left (-1+\frac {2}{1+i c x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 165, normalized size = 1.09 \[ \frac {1}{4} b \left (2 i \text {Li}_2\left (\frac {c x^2+i}{i-c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )-2 i \text {Li}_2\left (\frac {c x^2+i}{c x^2-i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+b \left (\text {Li}_3\left (\frac {c x^2+i}{i-c x^2}\right )-\text {Li}_3\left (\frac {c x^2+i}{c x^2-i}\right )\right )\right )+\tanh ^{-1}\left (1+\frac {2 i}{c x^2-i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arctan \left (c x^{2}\right )^{2} + 2 \, a b \arctan \left (c x^{2}\right ) + a^{2}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctan \left (c \,x^{2}\right )\right )^{2}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \log \relax (x) + \frac {1}{16} \, \int \frac {12 \, b^{2} \arctan \left (c x^{2}\right )^{2} + b^{2} \log \left (c^{2} x^{4} + 1\right )^{2} + 32 \, a b \arctan \left (c x^{2}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(x) + 1/16*integrate((12*b^2*arctan(c*x^2)^2 + b^2*log(c^2*x^4 + 1)^2 + 32*a*b*arctan(c*x^2))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^2))^2/x,x)

[Out]

int((a + b*atan(c*x^2))^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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