\(\int (a+b \arctan (c x))^2 \, dx\) [18]

Optimal result

Integrand size = 10, antiderivative size = 83 \[ \int (a+b \arctan (c x))^2 \, dx=\frac {i (a+b \arctan (c x))^2}{c}+x (a+b \arctan (c x))^2+\frac {2 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c}+\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4930, 5040, 4964, 2449, 2352} \[ \int (a+b \arctan (c x))^2 \, dx=x (a+b \arctan (c x))^2+\frac {i (a+b \arctan (c x))^2}{c}+\frac {2 b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}+\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c} \]

[In]

Int[(a + b*ArcTan[c*x])^2,x]

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4930

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

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = x (a+b \arctan (c x))^2-(2 b c) \int \frac {x (a+b \arctan (c x))}{1+c^2 x^2} \, dx \\ & = \frac {i (a+b \arctan (c x))^2}{c}+x (a+b \arctan (c x))^2+(2 b) \int \frac {a+b \arctan (c x)}{i-c x} \, dx \\ & = \frac {i (a+b \arctan (c x))^2}{c}+x (a+b \arctan (c x))^2+\frac {2 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c}-\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx \\ & = \frac {i (a+b \arctan (c x))^2}{c}+x (a+b \arctan (c x))^2+\frac {2 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c}+\frac {\left (2 i b^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c} \\ & = \frac {i (a+b \arctan (c x))^2}{c}+x (a+b \arctan (c x))^2+\frac {2 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c}+\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.08 \[ \int (a+b \arctan (c x))^2 \, dx=\frac {b^2 (-i+c x) \arctan (c x)^2+2 b \arctan (c x) \left (a c x+b \log \left (1+e^{2 i \arctan (c x)}\right )\right )+a \left (a c x-b \log \left (1+c^2 x^2\right )\right )-i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )}{c} \]

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

Maple [A] (verified)

Time = 2.19 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.48

method	result	size
derivativedivides	\(\frac {c x \,a^{2}-i \arctan \left (c x \right )^{2} b^{2}+\arctan \left (c x \right )^{2} b^{2} c x +2 \arctan \left (c x \right ) \ln \left (1+\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right ) b^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right ) b^{2}+2 a b c x \arctan \left (c x \right )-a b \ln \left (c^{2} x^{2}+1\right )}{c}\)	\(123\)
default	\(\frac {c x \,a^{2}-i \arctan \left (c x \right )^{2} b^{2}+\arctan \left (c x \right )^{2} b^{2} c x +2 \arctan \left (c x \right ) \ln \left (1+\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right ) b^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right ) b^{2}+2 a b c x \arctan \left (c x \right )-a b \ln \left (c^{2} x^{2}+1\right )}{c}\)	\(123\)
parts	\(a^{2} x +b^{2} \arctan \left (c x \right )^{2} x -\frac {i b^{2} \arctan \left (c x \right )^{2}}{c}-\frac {i b^{2} \operatorname {polylog}\left (2, -\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right )}{c}+\frac {2 b^{2} \arctan \left (c x \right ) \ln \left (1+\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right )}{c}+2 a b x \arctan \left (c x \right )-\frac {a b \ln \left (c^{2} x^{2}+1\right )}{c}\)	\(128\)
risch	\(\frac {i b^{2} \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{c}-\frac {b a \ln \left (i c x +1\right )}{c}+\frac {b^{2} \ln \left (i c x +1\right ) \ln \left (-i c x +1\right ) x}{2}+\frac {i b^{2} \ln \left (i c x +1\right ) \ln \left (-i c x +1\right )}{2 c}+\frac {2 a b}{c}-\frac {b^{2} \ln \left (i c x +1\right )^{2} x}{4}-\frac {i \ln \left (-i c x +1\right )^{2} b^{2}}{4 c}+\frac {i b^{2} \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{c}+\frac {i b^{2} \ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {i b^{2} \ln \left (i c x +1\right )}{c}-i b a \ln \left (i c x +1\right ) x +a^{2} x -\frac {\ln \left (-i c x +1\right ) a b}{c}-\frac {i b^{2} \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{c}+i \ln \left (-i c x +1\right ) a b x -\frac {\ln \left (-i c x +1\right )^{2} b^{2} x}{4}+\frac {i a^{2}}{c}+\frac {i b^{2} \ln \left (i c x +1\right )^{2}}{4 c}+\frac {i b^{2}}{c}-\frac {b^{2} \arctan \left (c x \right )}{c}\)	\(322\)

[In]

int((a+b*arctan(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int (a+b \arctan (c x))^2 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{2} \,d x } \]

[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)

Sympy [F]

\[ \int (a+b \arctan (c x))^2 \, dx=\int \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}\, dx \]

[In]

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

[Out]

Integral((a + b*atan(c*x))**2, x)

Maxima [F]

\[ \int (a+b \arctan (c x))^2 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (a+b \arctan (c x))^2 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{2} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int (a+b \arctan (c x))^2 \, dx=\int {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2 \,d x \]

[In]

int((a + b*atan(c*x))^2,x)

[Out]

int((a + b*atan(c*x))^2, x)