3.1.33 \(\int (a+b \text {ArcTan}(c+d x))^2 \, dx\) [33]

Optimal. Leaf size=102 \[ \frac {i (a+b \text {ArcTan}(c+d x))^2}{d}+\frac {(c+d x) (a+b \text {ArcTan}(c+d x))^2}{d}+\frac {2 b (a+b \text {ArcTan}(c+d x)) \log \left (\frac {2}{1+i (c+d x)}\right )}{d}+\frac {i b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5147, 4930, 5040, 4964, 2449, 2352} \begin {gather*} \frac {(c+d x) (a+b \text {ArcTan}(c+d x))^2}{d}+\frac {i (a+b \text {ArcTan}(c+d x))^2}{d}+\frac {2 b \log \left (\frac {2}{1+i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))}{d}+\frac {i b^2 \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^2,x]

[Out]

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

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]

Rule 5147

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcTan[x])^p, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 104, normalized size = 1.02 \begin {gather*} \frac {a^2 (c+d x)}{d}+\frac {b \left (b (-i+c+d x) \text {ArcTan}(c+d x)^2+2 \text {ArcTan}(c+d x) \left (a (c+d x)+b \log \left (1+e^{2 i \text {ArcTan}(c+d x)}\right )\right )-a \log \left (1+(c+d x)^2\right )-i b \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c+d x)}\right )\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.14, size = 146, normalized size = 1.43

method	result	size
derivativedivides	\(\frac {\left (d x +c \right ) a^{2}-i \arctan \left (d x +c \right )^{2} b^{2}+\arctan \left (d x +c \right )^{2} b^{2} \left (d x +c \right )-i \polylog \left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) b^{2}+2 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) b^{2}+2 a b \left (d x +c \right ) \arctan \left (d x +c \right )-a b \ln \left (1+\left (d x +c \right )^{2}\right )}{d}\)	\(146\)
default	\(\frac {\left (d x +c \right ) a^{2}-i \arctan \left (d x +c \right )^{2} b^{2}+\arctan \left (d x +c \right )^{2} b^{2} \left (d x +c \right )-i \polylog \left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) b^{2}+2 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) b^{2}+2 a b \left (d x +c \right ) \arctan \left (d x +c \right )-a b \ln \left (1+\left (d x +c \right )^{2}\right )}{d}\)	\(146\)
risch	\(-\frac {\ln \left (-i d x -i c +1\right )^{2} x \,b^{2}}{4}+\frac {\ln \left (-i d x -i c +1\right ) x \,b^{2}}{2}-\frac {i b^{2} \ln \left (-i d x -i c +1\right ) \left (-i d x -i c +1\right )}{2 d}+\frac {i b^{2} \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{4 d}+\frac {i b^{2} \dilog \left (\frac {1}{2}-\frac {1}{2} i d x -\frac {1}{2} i c \right )}{d}-\frac {b^{2} \left (d x +c -i\right ) \ln \left (1+i \left (d x +c \right )\right )^{2}}{4 d}-\frac {\ln \left (-i d x -i c +1\right )^{2} b^{2} c}{4 d}+\left (\frac {b^{2} x \ln \left (1-i \left (d x +c \right )\right )}{2}-\frac {i b \left (2 a d x -b \ln \left (1-i \left (d x +c \right )\right )+i \ln \left (1-i \left (d x +c \right )\right ) b c \right )}{2 d}\right ) \ln \left (1+i \left (d x +c \right )\right )+\frac {a^{2} c}{d}-\frac {i b^{2} \ln \left (\frac {1}{2} i d x +\frac {1}{2} i c +\frac {1}{2}\right ) \ln \left (-i d x -i c +1\right )}{d}+\frac {b^{2} \arctan \left (d x +c \right )}{2 d}+\frac {2 a b c \arctan \left (d x +c \right )}{d}+a^{2} x -\frac {i b^{2} c \arctan \left (d x +c \right )}{2 d}+\frac {i a^{2}}{d}+i \ln \left (-i d x -i c +1\right ) x a b -\frac {i \ln \left (-i d x -i c +1\right )^{2} b^{2}}{4 d}+\frac {i b^{2} \ln \left (\frac {1}{2} i d x +\frac {1}{2} i c +\frac {1}{2}\right ) \ln \left (\frac {1}{2}-\frac {1}{2} i d x -\frac {1}{2} i c \right )}{d}-\frac {a b \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{d}+\frac {b^{2} c \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{4 d}\)	\(469\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(d*x+c))^2,x, algorithm="maxima")

[Out]

1/16*(12*c^2*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 4*(3*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d -
arctan((d^2*x + c*d)/d)^3/d)*c^2 + 4*x*arctan(d*x + c)^2 + 192*d^2*integrate(1/16*x^2*arctan(d*x + c)^2/(d^2*x
^2 + 2*c*d*x + c^2 + 1), x) + 16*d^2*integrate(1/16*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x
+ c^2 + 1), x) + 384*c*d*integrate(1/16*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 64*d^2*integra
te(1/16*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 32*c*d*integrate(1/16*x*log(d
^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 64*c*d*integrate(1/16*x*log(d^2*x^2 + 2*c*d*
x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 16*c^2*integrate(1/16*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2
*x^2 + 2*c*d*x + c^2 + 1), x) - x*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 + 12*arctan(d*x + c)^2*arctan((d^2*x + c*
d)/d)/d - 12*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d + 4*arctan((d^2*x + c*d)/d)^3/d - 128*d*integrate(1/1
6*x*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 16*integrate(1/16*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(
d^2*x^2 + 2*c*d*x + c^2 + 1), x))*b^2 + a^2*x + (2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*a*b/d

________________________________________________________________________________________

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(d*x+c))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \operatorname {atan}{\left (c + d x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c + d*x))**2, x)

________________________________________________________________________________________

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(d*x+c))^2,x, algorithm="giac")

[Out]

sage0*x

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))^2,x)

[Out]

int((a + b*atan(c + d*x))^2, x)

________________________________________________________________________________________