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

Optimal. Leaf size=83 \[ i c \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2+x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2-2 b c \log \left (\frac {2 c}{c+i x}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )+i b^2 c \text {Li}_2\left (1-\frac {2 c}{c+i x}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [B]  time = 0.44, antiderivative size = 478, normalized size of antiderivative = 5.76, number of steps used = 31, number of rules used = 14, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.167, Rules used = {5029, 2448, 263, 31, 2449, 2391, 2556, 12, 2462, 260, 2416, 2394, 2393, 2315} \[ -\frac {1}{2} i b^2 c \text {PolyLog}\left (2,\frac {c-i x}{2 c}\right )+\frac {1}{2} i b^2 c \text {PolyLog}\left (2,\frac {c+i x}{2 c}\right )-\frac {1}{2} i b^2 c \text {PolyLog}\left (2,-\frac {i c}{x}\right )+\frac {1}{2} i b^2 c \text {PolyLog}\left (2,\frac {i c}{x}\right )+\frac {1}{2} i b^2 c \text {PolyLog}\left (2,1-\frac {i x}{c}\right )-\frac {1}{2} i b^2 c \text {PolyLog}\left (2,1+\frac {i x}{c}\right )+a^2 x+i a b x \log \left (1-\frac {i c}{x}\right )-i a b x \log \left (1+\frac {i c}{x}\right )+a b c \log (c-i x)+a b c \log (c+i x)+\frac {1}{4} b^2 (-x+i c) \log ^2\left (1-\frac {i c}{x}\right )-\frac {1}{4} b^2 (x+i c) \log ^2\left (1+\frac {i c}{x}\right )+\frac {1}{2} b^2 x \log \left (1-\frac {i c}{x}\right ) \log \left (1+\frac {i c}{x}\right )-\frac {1}{2} i b^2 c \log \left (1+\frac {i c}{x}\right ) \log (-c-i x)+\frac {1}{2} i b^2 c \log (-c-i x) \log \left (\frac {c-i x}{2 c}\right )+\frac {1}{2} i b^2 c \log \left (1-\frac {i c}{x}\right ) \log (-c+i x)-\frac {1}{2} i b^2 c \log (-c+i x) \log \left (\frac {c+i x}{2 c}\right )-\frac {1}{2} i b^2 c \log (-c-i x) \log \left (-\frac {i x}{c}\right )+\frac {1}{2} i b^2 c \log (-c+i x) \log \left (\frac {i x}{c}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcTan[c/x])^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2315

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

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 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 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 2448

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

Rule 2449

Int[((a_.) + Log[(c_.)*((d_) + (e_.)/(x_))^(p_.)]*(b_.))^(q_), x_Symbol] :> Simp[((e + d*x)*(a + b*Log[c*(d +
e/x)^p])^q)/d, x] + Dist[(b*e*p*q)/d, Int[(a + b*Log[c*(d + e/x)^p])^(q - 1)/x, x], x] /; FreeQ[{a, b, c, d, e
, p}, x] && IGtQ[q, 0]

Rule 2462

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

Rule 2556

Int[Log[v_]*Log[w_], x_Symbol] :> Simp[x*Log[v]*Log[w], x] + (-Int[SimplifyIntegrand[(x*Log[w]*D[v, x])/v, x],
x] - Int[SimplifyIntegrand[(x*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFreeQ[v, x] && InverseFunctionFree
Q[w, x]

Rule 5029

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(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, n}, x] && IGtQ[p, 0] && IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 105, normalized size = 1.27 \[ a \left (a x+b c \log \left (\frac {c^2}{x^2}+1\right )-2 b c \log \left (\frac {c}{x}\right )\right )+2 b \tan ^{-1}\left (\frac {c}{x}\right ) \left (a x-b c \log \left (1-e^{2 i \tan ^{-1}\left (\frac {c}{x}\right )}\right )\right )+i b^2 c \text {Li}_2\left (e^{2 i \tan ^{-1}\left (\frac {c}{x}\right )}\right )+b^2 (x+i c) \tan ^{-1}\left (\frac {c}{x}\right )^2 \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c/x])^2,x]

[Out]

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

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} \arctan \left (\frac {c}{x}\right )^{2} + 2 \, a b \arctan \left (\frac {c}{x}\right ) + a^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c/x) + a)^2, x)

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maple [B]  time = 0.10, size = 357, normalized size = 4.30 \[ a^{2} x +b^{2} x \arctan \left (\frac {c}{x}\right )^{2}-2 c \,b^{2} \ln \left (\frac {c}{x}\right ) \arctan \left (\frac {c}{x}\right )+c \,b^{2} \arctan \left (\frac {c}{x}\right ) \ln \left (1+\frac {c^{2}}{x^{2}}\right )+i c \,b^{2} \dilog \left (1-\frac {i c}{x}\right )+i c \,b^{2} \ln \left (\frac {c}{x}\right ) \ln \left (1-\frac {i c}{x}\right )+\frac {i c \,b^{2} \ln \left (i+\frac {c}{x}\right ) \ln \left (\frac {i \left (\frac {c}{x}-i\right )}{2}\right )}{2}-\frac {i c \,b^{2} \ln \left (\frac {c}{x}-i\right ) \ln \left (-\frac {i \left (i+\frac {c}{x}\right )}{2}\right )}{2}-\frac {i c \,b^{2} \ln \left (i+\frac {c}{x}\right ) \ln \left (1+\frac {c^{2}}{x^{2}}\right )}{2}+\frac {i c \,b^{2} \ln \left (\frac {c}{x}-i\right ) \ln \left (1+\frac {c^{2}}{x^{2}}\right )}{2}-i c \,b^{2} \ln \left (\frac {c}{x}\right ) \ln \left (1+\frac {i c}{x}\right )+\frac {i c \,b^{2} \ln \left (i+\frac {c}{x}\right )^{2}}{4}+\frac {i c \,b^{2} \dilog \left (\frac {i \left (\frac {c}{x}-i\right )}{2}\right )}{2}-i c \,b^{2} \dilog \left (1+\frac {i c}{x}\right )-\frac {i c \,b^{2} \ln \left (\frac {c}{x}-i\right )^{2}}{4}-\frac {i c \,b^{2} \dilog \left (-\frac {i \left (i+\frac {c}{x}\right )}{2}\right )}{2}+2 a b x \arctan \left (\frac {c}{x}\right )-2 c a b \ln \left (\frac {c}{x}\right )+c a b \ln \left (1+\frac {c^{2}}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x+b^2*x*arctan(c/x)^2-2*c*b^2*ln(c/x)*arctan(c/x)+c*b^2*arctan(c/x)*ln(1+c^2/x^2)-1/2*I*c*b^2*dilog(-1/2*I
*(I+c/x))+I*c*b^2*ln(c/x)*ln(1-I*c/x)+1/2*I*c*b^2*dilog(1/2*I*(c/x-I))+1/2*I*c*b^2*ln(I+c/x)*ln(1/2*I*(c/x-I))
-1/2*I*c*b^2*ln(c/x-I)*ln(-1/2*I*(I+c/x))-1/2*I*c*b^2*ln(I+c/x)*ln(1+c^2/x^2)+1/2*I*c*b^2*ln(c/x-I)*ln(1+c^2/x
^2)-I*c*b^2*ln(c/x)*ln(1+I*c/x)+1/4*I*c*b^2*ln(I+c/x)^2-I*c*b^2*dilog(1+I*c/x)-1/4*I*c*b^2*ln(c/x-I)^2+I*c*b^2
*dilog(1-I*c/x)+2*a*b*x*arctan(c/x)-2*c*a*b*ln(c/x)+c*a*b*ln(1+c^2/x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (2 \, x \arctan \left (\frac {c}{x}\right ) + c \log \left (c^{2} + x^{2}\right )\right )} a b + \frac {1}{16} \, {\left (12 \, c \arctan \left (\frac {c}{x}\right )^{2} \arctan \left (\frac {x}{c}\right ) + 4 \, {\left (\frac {3 \, \arctan \left (\frac {c}{x}\right ) \arctan \left (\frac {x}{c}\right )^{2}}{c} + \frac {\arctan \left (\frac {x}{c}\right )^{3}}{c}\right )} c^{2} + 4 \, x \arctan \left (c, x\right )^{2} + 16 \, c^{2} \int \frac {\log \left (c^{2} + x^{2}\right )^{2}}{16 \, {\left (c^{2} + x^{2}\right )}}\,{d x} - x \log \left (c^{2} + x^{2}\right )^{2} + 128 \, c \int \frac {x \arctan \left (\frac {c}{x}\right )}{16 \, {\left (c^{2} + x^{2}\right )}}\,{d x} + 192 \, \int \frac {x^{2} \arctan \left (\frac {c}{x}\right )^{2}}{16 \, {\left (c^{2} + x^{2}\right )}}\,{d x} + 16 \, \int \frac {x^{2} \log \left (c^{2} + x^{2}\right )^{2}}{16 \, {\left (c^{2} + x^{2}\right )}}\,{d x} + 64 \, \int \frac {x^{2} \log \left (c^{2} + x^{2}\right )}{16 \, {\left (c^{2} + x^{2}\right )}}\,{d x}\right )} b^{2} + a^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*arctan(c/x) + c*log(c^2 + x^2))*a*b + 1/16*(12*c*arctan(c/x)^2*arctan(x/c) + 4*(3*arctan(c/x)*arctan(x/c)
^2/c + arctan(x/c)^3/c)*c^2 + 4*x*arctan2(c, x)^2 + 16*c^2*integrate(1/16*log(c^2 + x^2)^2/(c^2 + x^2), x) - x
*log(c^2 + x^2)^2 + 128*c*integrate(1/16*x*arctan(c/x)/(c^2 + x^2), x) + 192*integrate(1/16*x^2*arctan(c/x)^2/
(c^2 + x^2), x) + 16*integrate(1/16*x^2*log(c^2 + x^2)^2/(c^2 + x^2), x) + 64*integrate(1/16*x^2*log(c^2 + x^2
)/(c^2 + x^2), x))*b^2 + a^2*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c/x))^2,x)

[Out]

int((a + b*atan(c/x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c/x))**2, x)

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