3.1.38 \(\int (a+b \text {ArcTan}(c+d x))^3 \, dx\) [38]

Optimal. Leaf size=143 \[ \frac {i (a+b \text {ArcTan}(c+d x))^3}{d}+\frac {(c+d x) (a+b \text {ArcTan}(c+d x))^3}{d}+\frac {3 b (a+b \text {ArcTan}(c+d x))^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}+\frac {3 i b^2 (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5147, 4930, 5040, 4964, 5004, 5114, 6745} \begin {gather*} \frac {3 i b^2 \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right ) (a+b \text {ArcTan}(c+d x))}{d}+\frac {(c+d x) (a+b \text {ArcTan}(c+d x))^3}{d}+\frac {i (a+b \text {ArcTan}(c+d x))^3}{d}+\frac {3 b \log \left (\frac {2}{1+i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))^2}{d}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{i (c+d x)+1}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^3,x]

[Out]

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

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 5004

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 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 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 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]

Rule 6745

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 \left (a+b \tan ^{-1}(c+d x)\right )^3 \, dx &=\frac {\text {Subst}\left (\int \left (a+b \tan ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}-\frac {(3 b) \text {Subst}\left (\int \frac {x \left (a+b \tan ^{-1}(x)\right )^2}{1+x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac {(3 b) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{i-x} \, dx,x,c+d x\right )}{d}\\ &=\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac {3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}-\frac {\left (6 b^2\right ) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \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 )^3}{d}+\frac {(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac {3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{d}-\frac {\left (3 i b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (1-\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 )^3}{d}+\frac {(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac {3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{d}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 190, normalized size = 1.33 \begin {gather*} \frac {2 b^3 (-i+c+d x) \text {ArcTan}(c+d x)^3+6 b^2 \text {ArcTan}(c+d x)^2 \left (a (-i+c+d x)+b \log \left (1+e^{2 i \text {ArcTan}(c+d x)}\right )\right )+6 a b \text {ArcTan}(c+d x) \left (a (c+d x)+2 b \log \left (1+e^{2 i \text {ArcTan}(c+d x)}\right )\right )+2 a^2 \left (a (c+d x)+3 b \log \left (\frac {1}{\sqrt {1+(c+d x)^2}}\right )\right )-6 i b^2 (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c+d x)}\right )+3 b^3 \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c+d x)}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c + d*x])^3,x]

[Out]

(2*b^3*(-I + c + d*x)*ArcTan[c + d*x]^3 + 6*b^2*ArcTan[c + d*x]^2*(a*(-I + c + d*x) + b*Log[1 + E^((2*I)*ArcTa
n[c + d*x])]) + 6*a*b*ArcTan[c + d*x]*(a*(c + d*x) + 2*b*Log[1 + E^((2*I)*ArcTan[c + d*x])]) + 2*a^2*(a*(c + d
*x) + 3*b*Log[1/Sqrt[1 + (c + d*x)^2]]) - (6*I)*b^2*(a + b*ArcTan[c + d*x])*PolyLog[2, -E^((2*I)*ArcTan[c + d*
x])] + 3*b^3*PolyLog[3, -E^((2*I)*ArcTan[c + d*x])])/(2*d)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (136 ) = 272\).
time = 0.46, size = 297, normalized size = 2.08

method	result	size
derivativedivides	\(\frac {\left (d x +c \right ) a^{3}-i b^{3} \arctan \left (d x +c \right )^{3}+b^{3} \arctan \left (d x +c \right )^{3} \left (d x +c \right )+3 b^{3} \arctan \left (d x +c \right )^{2} \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-3 i b^{3} \arctan \left (d x +c \right ) \polylog \left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )+\frac {3 b^{3} \polylog \left (3, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )}{2}-3 i \arctan \left (d x +c \right )^{2} a \,b^{2}+3 \arctan \left (d x +c \right )^{2} a \,b^{2} \left (d x +c \right )-3 i \polylog \left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) a \,b^{2}+6 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) a \,b^{2}+3 a^{2} b \left (d x +c \right ) \arctan \left (d x +c \right )-\frac {3 a^{2} b \ln \left (1+\left (d x +c \right )^{2}\right )}{2}}{d}\)	\(297\)
default	\(\frac {\left (d x +c \right ) a^{3}-i b^{3} \arctan \left (d x +c \right )^{3}+b^{3} \arctan \left (d x +c \right )^{3} \left (d x +c \right )+3 b^{3} \arctan \left (d x +c \right )^{2} \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-3 i b^{3} \arctan \left (d x +c \right ) \polylog \left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )+\frac {3 b^{3} \polylog \left (3, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )}{2}-3 i \arctan \left (d x +c \right )^{2} a \,b^{2}+3 \arctan \left (d x +c \right )^{2} a \,b^{2} \left (d x +c \right )-3 i \polylog \left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) a \,b^{2}+6 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) a \,b^{2}+3 a^{2} b \left (d x +c \right ) \arctan \left (d x +c \right )-\frac {3 a^{2} b \ln \left (1+\left (d x +c \right )^{2}\right )}{2}}{d}\)	\(297\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*((d*x+c)*a^3-I*b^3*arctan(d*x+c)^3+b^3*arctan(d*x+c)^3*(d*x+c)+3*b^3*arctan(d*x+c)^2*ln(1+(1+I*(d*x+c))^2/
(1+(d*x+c)^2))-3*I*b^3*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+3/2*b^3*polylog(3,-(1+I*(d*x+c)
)^2/(1+(d*x+c)^2))-3*I*arctan(d*x+c)^2*a*b^2+3*arctan(d*x+c)^2*a*b^2*(d*x+c)-3*I*polylog(2,-(1+I*(d*x+c))^2/(1
+(d*x+c)^2))*a*b^2+6*arctan(d*x+c)*ln(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))*a*b^2+3*a^2*b*(d*x+c)*arctan(d*x+c)-3/2
*a^2*b*ln(1+(d*x+c)^2))

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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))^3,x, algorithm="maxima")

[Out]

7/8*b^3*c^2*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 1/8*b^3*x*arctan(d*x + c)^3 + 3*a*b^2*c^2*arctan(d*x
+ c)^2*arctan((d^2*x + c*d)/d)/d - 3/32*b^3*x*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 - (3*arctan(
d*x + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2*c^2 - 7/32*(6*arctan(d*x + c)^2*arct
an((d^2*x + c*d)/d)^2/d - 4*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^3/d + arctan((d^2*x + c*d)/d)^4/d)*b^3*c^2
+ 7/8*b^3*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 28*b^3*d^2*integrate(1/32*x^2*arctan(d*x + c)^3/(d^2*
x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*d^2*integrate(1/32*x^2*arctan(d*x + c)*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) + 96*a*b^2*d^2*integrate(1/32*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^
2 + 1), x) + 56*b^3*c*d*integrate(1/32*x*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3*d^2*inte
grate(1/32*x^2*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 6*b^3*c*d*
integrate(1/32*x*arctan(d*x + c)*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) + 192*a*
b^2*c*d*integrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3*c*d*integrate(1/32*x*arc
tan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*c^2*integrate(1/32*arc
tan(d*x + c)*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) + 3*a*b^2*arctan(d*x + c)^2*
arctan((d^2*x + c*d)/d)/d - 12*b^3*d*integrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*
b^3*d*integrate(1/32*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) - (3*arctan(d*x +
c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2 - 7/32*(6*arctan(d*x + c)^2*arctan((d^2*x
+ c*d)/d)^2/d - 4*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^3/d + arctan((d^2*x + c*d)/d)^4/d)*b^3 + a^3*x + 3*b
^3*integrate(1/32*arctan(d*x + c)*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) + 3/2*(
2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*a^2*b/d

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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))^3,x, algorithm="fricas")

[Out]

integral(b^3*arctan(d*x + c)^3 + 3*a*b^2*arctan(d*x + c)^2 + 3*a^2*b*arctan(d*x + c) + a^3, x)

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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 )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(d*x+c))**3,x)

[Out]

Integral((a + b*atan(c + d*x))**3, x)

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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))^3,x, algorithm="giac")

[Out]

sage0*x

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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 )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))^3,x)

[Out]

int((a + b*atan(c + d*x))^3, x)

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