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

Optimal. Leaf size=228 \[ b c d \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )-b c d \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )-i c d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c d \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 i c d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-i b^2 c d \text {Li}_2\left (\frac {2}{1-i c x}-1\right )-\frac {1}{2} i b^2 c d \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} i b^2 c d \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4876, 4852, 4924, 4868, 2447, 4850, 4988, 4884, 4994, 6610} \[ b c d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-b c d \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-i b^2 c d \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )-\frac {1}{2} i b^2 c d \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c d \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )-i c d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c d \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 i c d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)*(a + b*ArcTan[c*x])^2)/x^2,x]

[Out]

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

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 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 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 4876

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

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

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Mathematica [A]  time = 0.44, size = 289, normalized size = 1.27 \[ \frac {i d \left (a^2 c x \log (x)+i a^2+i a b \left (c x \left (\log \left (c^2 x^2+1\right )-2 \log (c x)\right )+2 \tan ^{-1}(c x)\right )+i a b c x (\text {Li}_2(-i c x)-\text {Li}_2(i c x))+i b^2 \left (i c x \left (\tan ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )\right )+\tan ^{-1}(c x)^2-2 c x \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )+\frac {1}{24} b^2 c x \left (24 i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+24 i \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+12 \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-12 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+16 i \tan ^{-1}(c x)^3+24 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-24 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-i \pi ^3\right )\right )}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*d*(I*a^2 + a^2*c*x*Log[x] + I*a*b*(2*ArcTan[c*x] + c*x*(-2*Log[c*x] + Log[1 + c^2*x^2])) + I*b^2*(ArcTan[c*
x]^2 - 2*c*x*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + I*c*x*(ArcTan[c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*
x])])) + I*a*b*c*x*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x]) + (b^2*c*x*((-I)*Pi^3 + (16*I)*ArcTan[c*x]^3 + 2
4*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - 24*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*Arc
Tan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 12*PolyL
og[3, E^((-2*I)*ArcTan[c*x])] - 12*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/24))/x

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {4 i \, a^{2} c d x + 4 \, a^{2} d + {\left (-i \, b^{2} c d x - b^{2} d\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - {\left (4 \, a b c d x - 4 i \, a b d\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{4 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/4*(4*I*a^2*c*d*x + 4*a^2*d + (-I*b^2*c*d*x - b^2*d)*log(-(c*x + I)/(c*x - I))^2 - (4*a*b*c*d*x - 4*
I*a*b*d)*log(-(c*x + I)/(c*x - I)))/x^2, 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((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^2,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 2.07, size = 5963, normalized size = 26.15 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x^2,x)

[Out]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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