3.2.17 \(\int \frac {(a+b \text {ArcTan}(c x^3))^2}{x} \, dx\) [117]

Optimal. Leaf size=154 \[ \frac {2}{3} \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^3}\right )-\frac {1}{3} i b \left (a+b \text {ArcTan}\left (c x^3\right )\right ) \text {PolyLog}\left (2,1-\frac {2}{1+i c x^3}\right )+\frac {1}{3} i b \left (a+b \text {ArcTan}\left (c x^3\right )\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+i c x^3}\right )-\frac {1}{6} b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x^3}\right )+\frac {1}{6} b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x^3}\right ) \]

[Out]

-2/3*(a+b*arctan(c*x^3))^2*arctanh(-1+2/(1+I*c*x^3))-1/3*I*b*(a+b*arctan(c*x^3))*polylog(2,1-2/(1+I*c*x^3))+1/
3*I*b*(a+b*arctan(c*x^3))*polylog(2,-1+2/(1+I*c*x^3))-1/6*b^2*polylog(3,1-2/(1+I*c*x^3))+1/6*b^2*polylog(3,-1+
2/(1+I*c*x^3))

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Rubi [A]
time = 0.21, antiderivative size = 154, 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 = {4944, 4942, 5108, 5004, 5114, 6745} \begin {gather*} -\frac {1}{3} i b \text {Li}_2\left (1-\frac {2}{i c x^3+1}\right ) \left (a+b \text {ArcTan}\left (c x^3\right )\right )+\frac {1}{3} i b \text {Li}_2\left (\frac {2}{i c x^3+1}-1\right ) \left (a+b \text {ArcTan}\left (c x^3\right )\right )+\frac {2}{3} \tanh ^{-1}\left (1-\frac {2}{1+i c x^3}\right ) \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2-\frac {1}{6} b^2 \text {Li}_3\left (1-\frac {2}{i c x^3+1}\right )+\frac {1}{6} b^2 \text {Li}_3\left (\frac {2}{i c x^3+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^3])^2/x,x]

[Out]

(2*(a + b*ArcTan[c*x^3])^2*ArcTanh[1 - 2/(1 + I*c*x^3)])/3 - (I/3)*b*(a + b*ArcTan[c*x^3])*PolyLog[2, 1 - 2/(1
+ I*c*x^3)] + (I/3)*b*(a + b*ArcTan[c*x^3])*PolyLog[2, -1 + 2/(1 + I*c*x^3)] - (b^2*PolyLog[3, 1 - 2/(1 + I*c
*x^3)])/6 + (b^2*PolyLog[3, -1 + 2/(1 + I*c*x^3)])/6

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 4944

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

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Mathematica [A]
time = 0.08, size = 167, normalized size = 1.08 \begin {gather*} \frac {1}{6} \left (4 \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2 \tanh ^{-1}\left (1+\frac {2 i}{-i+c x^3}\right )+b \left (2 i \left (a+b \text {ArcTan}\left (c x^3\right )\right ) \text {PolyLog}\left (2,\frac {i+c x^3}{i-c x^3}\right )-2 i \left (a+b \text {ArcTan}\left (c x^3\right )\right ) \text {PolyLog}\left (2,\frac {i+c x^3}{-i+c x^3}\right )+b \left (\text {PolyLog}\left (3,\frac {i+c x^3}{i-c x^3}\right )-\text {PolyLog}\left (3,\frac {i+c x^3}{-i+c x^3}\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^3])^2/x,x]

[Out]

(4*(a + b*ArcTan[c*x^3])^2*ArcTanh[1 + (2*I)/(-I + c*x^3)] + b*((2*I)*(a + b*ArcTan[c*x^3])*PolyLog[2, (I + c*
x^3)/(I - c*x^3)] - (2*I)*(a + b*ArcTan[c*x^3])*PolyLog[2, (I + c*x^3)/(-I + c*x^3)] + b*(PolyLog[3, (I + c*x^
3)/(I - c*x^3)] - PolyLog[3, (I + c*x^3)/(-I + c*x^3)])))/6

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctan \left (c \,x^{3}\right )\right )^{2}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^2/x,x, algorithm="maxima")

[Out]

a^2*log(x) + 1/16*integrate((12*b^2*arctan(c*x^3)^2 + b^2*log(c^2*x^6 + 1)^2 + 32*a*b*arctan(c*x^3))/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))^2/x,x, algorithm="fricas")

[Out]

integral((b^2*arctan(c*x^3)^2 + 2*a*b*arctan(c*x^3) + a^2)/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^{3} \right )}\right )^{2}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**3))**2/x,x)

[Out]

Integral((a + b*atan(c*x**3))**2/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))^2/x,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x^3\right )\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))^2/x,x)

[Out]

int((a + b*atan(c*x^3))^2/x, x)

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