\(\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx\) [123]
Optimal result
Integrand size = 22, antiderivative size = 139 \[
\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\frac {i (a+b \arctan (c x))^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )}{4 c d}
\]
[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
Rubi [A] (verified)
Time = 0.17 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of
steps used = 4, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4964, 5004, 5114, 5118, 6745}
\[
\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\frac {3 i b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c d}-\frac {3 b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))^2}{2 c d}+\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c d}+\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{i c x+1}\right )}{4 c d}
\]
[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*}
\text {integral}& = \frac {i (a+b \arctan (c x))^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {(3 i b) \int \frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = \frac {i (a+b \arctan (c x))^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {\left (3 b^2\right ) \int \frac {(a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = \frac {i (a+b \arctan (c x))^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c d}-\frac {\left (3 i b^3\right ) \int \frac {\operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 d} \\ & = \frac {i (a+b \arctan (c x))^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )}{4 c d} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.06 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96
\[
\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\frac {i \left (4 (a+b \arctan (c x))^3 \log \left (\frac {2 d}{d+i c d x}\right )+3 i b \left (2 (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )-b \left (2 i (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,\frac {i+c x}{-i+c x}\right )+b \operatorname {PolyLog}\left (4,\frac {i+c x}{-i+c x}\right )\right )\right )\right )}{4 c d}
\]
[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)
Maple [C] (warning: unable to verify)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.54 (sec) , antiderivative size = 1629, normalized size of antiderivative =
11.72
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| method | result | size |
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| derivativedivides |
\(\text {Expression too large to display}\) |
\(1629\) |
| default |
\(\text {Expression too large to display}\) |
\(1629\) |
| parts |
\(\text {Expression too large to display}\) |
\(1640\) |
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[In]
int((a+b*arctan(c*x))^3/(d+I*c*d*x),x,method=_RETURNVERBOSE)
[Out]
1/c*(-1/2*I*a^3/d*ln(c^2*x^2+1)+a^3/d*arctan(c*x)+b^3/d*(-I*ln(1+I*c*x)*arctan(c*x)^3+3*I*(1/3*arctan(c*x)^3*l
n(2*I*(1+I*c*x)^2/(c^2*x^2+1))+1/6*I*Pi*(csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3+csgn(I/((
1+I*c*x)^2/(c^2*x^2+1)+1))*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))*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)/((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-csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1
))*csgn((1+I*c*x)^2/(c^2*x^2+1))*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))^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))+csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))
^2-1)*arctan(c*x)^3-1/2*I*arctan(c*x)^2*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+1/2*arctan(c*x)*polylog(3,-(1+I*c*
x)^2/(c^2*x^2+1))+1/4*I*polylog(4,-(1+I*c*x)^2/(c^2*x^2+1))-1/6*I*arctan(c*x)^4))+3*a*b^2/d*(-I*ln(1+I*c*x)*ar
ctan(c*x)^2+2*I*(1/2*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))+1/4*I*Pi*(csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+
I*c*x)^2/(c^2*x^2+1)+1))^3+csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*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))*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)/((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-csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*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))^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))+csgn(I*(1+I*c*x)^2/(c^2
*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2-1)*arctan(c*x)^2-1/2*I*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+
1/4*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-1/3*I*arctan(c*x)^3))+3*a^2*b/d*(-I*ln(1+I*c*x)*arctan(c*x)-1/2*(ln(1+
I*c*x)-ln(1/2+1/2*I*c*x))*ln(1/2-1/2*I*c*x)+1/2*dilog(1/2+1/2*I*c*x)+1/4*ln(1+I*c*x)^2))
Fricas [F]
\[
\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{i \, c d x + d} \,d x }
\]
[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)
Sympy [F(-1)]
Timed out. \[
\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\text {Timed out}
\]
[In]
integrate((a+b*atan(c*x))**3/(d+I*c*d*x),x)
[Out]
Timed out
Maxima [F]
\[
\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{i \, c d x + d} \,d x }
\]
[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)
Giac [F]
\[
\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{i \, c d x + d} \,d x }
\]
[In]
integrate((a+b*arctan(c*x))^3/(d+I*c*d*x),x, algorithm="giac")
[Out]
sage0*x
Mupad [F(-1)]
Timed out. \[
\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{d+c\,d\,x\,1{}\mathrm {i}} \,d x
\]
[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)