∫ (a+bArcTan(cx))3 _______________________________

3.2.23 \(\int \frac {(a+b \text {ArcTan}(c x))^3}{d+i c d x} \, dx\) [123]

Optimal. Leaf size=139 \[ \frac {i (a+b \text {ArcTan}(c x))^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )}{4 c d} \]

[Out]

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

b*arctan(c*x))*polylog(3,1-2/(1+I*c*x))/c/d+3/4*b^3*polylog(4,1-2/(1+I*c*x))/c/d

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Rubi [A]
time = 0.17, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4964, 5004, 5114, 5118, 6745} \begin {gather*} \frac {3 i b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))}{2 c d}-\frac {3 b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))^2}{2 c d}+\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^3}{c d}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{i c x+1}\right )}{4 c d} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*ArcTan[c*x])^3/(d + I*c*d*x),x]

[Out]

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

/(2*c*d) + (((3*I)/2)*b^2*(a + b*ArcTan[c*x])*PolyLog[3, 1 - 2/(1 + I*c*x)])/(c*d) + (3*b^3*PolyLog[4, 1 - 2/(

1 + I*c*x)])/(4*c*d)

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 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 5118

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[I*(a +

b*ArcTan[c*x])^p*(PolyLog[k + 1, u]/(2*c*d)), x] - Dist[b*p*(I/2), Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[k

+ 1, u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 -

2*(I/(I - c*x)))^2, 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 \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d+i c d x} \, dx &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {(3 i b) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {\left (3 b^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c d}-\frac {\left (3 i b^3\right ) \int \frac {\text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 d}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1+i c x}\right )}{4 c d}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 133, normalized size = 0.96 \begin {gather*} \frac {i \left (4 (a+b \text {ArcTan}(c x))^3 \log \left (\frac {2 d}{d+i c d x}\right )+3 i b \left (2 (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )-b \left (2 i (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,\frac {i+c x}{-i+c x}\right )+b \text {PolyLog}\left (4,\frac {i+c x}{-i+c x}\right )\right )\right )\right )}{4 c d} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((I/4)*(4*(a + b*ArcTan[c*x])^3*Log[(2*d)/(d + I*c*d*x)] + (3*I)*b*(2*(a + b*ArcTan[c*x])^2*PolyLog[2, (I + c*

x)/(-I + c*x)] - b*((2*I)*(a + b*ArcTan[c*x])*PolyLog[3, (I + c*x)/(-I + c*x)] + b*PolyLog[4, (I + c*x)/(-I +

c*x)]))))/(c*d)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.66, size = 1940, normalized size = 13.96


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(1940\)
default	\(\text {Expression too large to display}\)	\(1940\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a^3/d*arctan(c*x)+1/2*b^3/d*arctan(c*x)^3*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csg

n((1+I*c*x)^2/(c^2*x^2+1))*Pi+3/2*a*b^2/d*arctan(c*x)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+

1))^2*csgn((1+I*c*x)^2/(c^2*x^2+1))*Pi-3/2*a*b^2/d*arctan(c*x)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^

2*x^2+1)+1))^2*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*Pi+3/2*a*b^2/d*arctan(c*x)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/(

(1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*Pi-3/2*a*b^2/d*arcta

n(c*x)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2

/(c^2*x^2+1)+1))*Pi+1/2*b^3/d*arctan(c*x)^3*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+

I*c*x)^2/(c^2*x^2+1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*Pi-3/4*b^3/d*polylog(4,-(1+I*c*x)^2/(c^2*x^2+1))+1/2

*b^3/d*arctan(c*x)^4-1/2*b^3/d*arctan(c*x)^3*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*Pi-

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

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

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

rctan(c*x)^3*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-1/2*b^3/d*arctan(c*x)^3*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2

/(c^2*x^2+1)+1))^3*Pi+1/2*b^3/d*arctan(c*x)^3*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*Pi

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

)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn(I/((1+I*c*x)^

2/(c^2*x^2+1)+1))*Pi+3/2*a^2*b/d*dilog(1/2+1/2*I*c*x)+3/4*a^2*b/d*ln(1+I*c*x)^2+2*a*b^2/d*arctan(c*x)^3-1/2*b^

3/d*arctan(c*x)^3*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+

1))*Pi+1/2*b^3/d*arctan(c*x)^3*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(1+I*c*x)^2/(c

^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*Pi-1/2*b^3/d*arctan(c*x)^3*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/

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

c*x)-3/2*a*b^2/d*arctan(c*x)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*Pi+3/2*a*b^2/d*arct

an(c*x)^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*Pi-3/2*a*b^2/d*arctan(c*x)^2*csgn(I*(1

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

an(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-1/2*I*a^3/d*ln(c^2*x^2+1))

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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(c*x))^3/(d+I*c*d*x),x, algorithm="maxima")

[Out]

-I*a^3*log(I*c*d*x + d)/(c*d) + 1/128*(16*b^3*arctan(c*x)^4 + 16*I*b^3*arctan(c*x)^3*log(c^2*x^2 + 1) + 4*I*b^

3*arctan(c*x)*log(c^2*x^2 + 1)^3 - b^3*log(c^2*x^2 + 1)^4 + 16*(b^3*arctan(c*x)^4/(c*d) + 8*b^3*c*integrate(1/

16*x*log(c^2*x^2 + 1)^3/(c^2*d*x^2 + d), x) + 8*a*b^2*arctan(c*x)^3/(c*d) + 12*a^2*b*arctan(c*x)^2/(c*d))*c*d

- 128*I*c*d*integrate(1/32*(40*b^3*c*x*arctan(c*x)^3 + 6*b^3*c*x*arctan(c*x)*log(c^2*x^2 + 1)^2 + 96*a*b^2*c*x

*arctan(c*x)^2 + 96*a^2*b*c*x*arctan(c*x) + 12*b^3*arctan(c*x)^2*log(c^2*x^2 + 1) + b^3*log(c^2*x^2 + 1)^3)/(c

^2*d*x^2 + d), x))/(c*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(c*x))^3/(d+I*c*d*x),x, algorithm="fricas")

[Out]

integral(-1/8*(b^3*log(-(c*x + I)/(c*x - I))^3 - 6*I*a*b^2*log(-(c*x + I)/(c*x - I))^2 - 12*a^2*b*log(-(c*x +

I)/(c*x - I)) + 8*I*a^3)/(c*d*x - I*d), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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(c*x))^3/(d+I*c*d*x),x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^3/(d + c*d*x*1i),x)

[Out]

int((a + b*atan(c*x))^3/(d + c*d*x*1i), x)

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