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

Optimal. Leaf size=96 \[ -\frac{i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+\frac{i c}{x}}\right )}{c}-\frac{i \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2}{c}-\frac{\left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2}{x}-\frac{2 b \log \left (\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )}{c} \]

[Out]

((-I)*(a + b*ArcCot[x/c])^2)/c - (a + b*ArcCot[x/c])^2/x - (2*b*(a + b*ArcCot[x/c])*Log[2/(1 + (I*c)/x)])/c -
(I*b^2*PolyLog[2, 1 - 2/(1 + (I*c)/x)])/c

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Rubi [B]  time = 0.527944, antiderivative size = 259, normalized size of antiderivative = 2.7, number of steps used = 28, number of rules used = 12, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5035, 2454, 2389, 2296, 2295, 6715, 2430, 43, 2416, 2394, 2393, 2391} \[ \frac{i b^2 \text{PolyLog}\left (2,-\frac{-x+i c}{2 x}\right )}{2 c}-\frac{i b^2 \text{PolyLog}\left (2,\frac{x+i c}{2 x}\right )}{2 c}+\frac{b \log \left (1+\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )}{2 x}-\frac{i b \log \left (\frac{x+i c}{2 x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )}{2 c}-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c}-\frac{i b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c}-\frac{i b^2 \log \left (1+\frac{i c}{x}\right ) \log \left (-\frac{-x+i c}{2 x}\right )}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcTan[c/x])^2/x^2,x]

[Out]

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

Rule 5035

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^
m*(a + (I*b*Log[1 - I*c*x^n])/2 - (I*b*Log[1 + I*c*x^n])/2)^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[
p, 0] && IntegerQ[m] && IntegerQ[n]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 2430

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.)), x_Symbol] :> Simp[x*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]), x] + (-Dist[g*j*m, Int[(x
*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[b*e*n*p, Int[(x*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f
+ g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.111112, size = 107, normalized size = 1.11 \[ -\frac{-i b^2 x \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}\left (\frac{c}{x}\right )}\right )+a \left (a c+2 b x \log \left (\frac{1}{\sqrt{\frac{c^2}{x^2}+1}}\right )\right )+2 b \tan ^{-1}\left (\frac{c}{x}\right ) \left (a c+b x \log \left (1+e^{2 i \tan ^{-1}\left (\frac{c}{x}\right )}\right )\right )+b^2 (c-i x) \tan ^{-1}\left (\frac{c}{x}\right )^2}{c x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((b^2*(c - I*x)*ArcTan[c/x]^2 + 2*b*ArcTan[c/x]*(a*c + b*x*Log[1 + E^((2*I)*ArcTan[c/x])]) + a*(a*c + 2*b*x*L
og[1/Sqrt[1 + c^2/x^2]]) - I*b^2*x*PolyLog[2, -E^((2*I)*ArcTan[c/x])])/(c*x))

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Maple [A]  time = 0.088, size = 147, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2}}{x}}+{\frac{i{b}^{2}}{c} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{2}}-{\frac{{b}^{2}}{x} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{2}}+{\frac{i{b}^{2}}{c}{\it polylog} \left ( 2,-{ \left ( 1+{\frac{ic}{x}} \right ) ^{2} \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) ^{-1}} \right ) }-2\,{\frac{{b}^{2}}{c}\arctan \left ({\frac{c}{x}} \right ) \ln \left ({ \left ( 1+{\frac{ic}{x}} \right ) ^{2} \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) ^{-1}}+1 \right ) }-2\,{\frac{ab}{x}\arctan \left ({\frac{c}{x}} \right ) }+{\frac{ab}{c}\ln \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (\frac{c}{x}\right )^{2} + 2 \, a b \arctan \left (\frac{c}{x}\right ) + a^{2}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (\frac{c}{x} \right )}\right )^{2}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (\frac{c}{x}\right ) + a\right )}^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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