Integrand size = 16, antiderivative size = 230 \[ \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^3}{x} \, dx=-2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3 \text {arctanh}\left (1-\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{2} i b \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{2} i b \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{2} b^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \operatorname {PolyLog}\left (3,1-\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{2} b^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \operatorname {PolyLog}\left (3,-1+\frac {2}{1+\frac {i c}{x}}\right )-\frac {3}{4} i b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+\frac {i c}{x}}\right )+\frac {3}{4} i b^3 \operatorname {PolyLog}\left (4,-1+\frac {2}{1+\frac {i c}{x}}\right ) \]
2*(a+b*arccot(x/c))^3*arctanh(-1+2/(1+I*c/x))+3/2*I*b*(a+b*arccot(x/c))^2* polylog(2,1-2/(1+I*c/x))-3/2*I*b*(a+b*arccot(x/c))^2*polylog(2,-1+2/(1+I*c /x))+3/2*b^2*(a+b*arccot(x/c))*polylog(3,1-2/(1+I*c/x))-3/2*b^2*(a+b*arcco t(x/c))*polylog(3,-1+2/(1+I*c/x))-3/4*I*b^3*polylog(4,1-2/(1+I*c/x))+3/4*I *b^3*polylog(4,-1+2/(1+I*c/x))
Time = 0.27 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.83 \[ \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^3}{x} \, dx=a^3 \log (x)-\frac {3}{2} i a^2 b \left (\operatorname {PolyLog}\left (2,-\frac {i c}{x}\right )-\operatorname {PolyLog}\left (2,\frac {i c}{x}\right )\right )+3 a b^2 \left (\frac {i \pi ^3}{24}-\frac {2}{3} i \arctan \left (\frac {c}{x}\right )^3-\arctan \left (\frac {c}{x}\right )^2 \log \left (1-e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )+\arctan \left (\frac {c}{x}\right )^2 \log \left (1+e^{2 i \arctan \left (\frac {c}{x}\right )}\right )-i \arctan \left (\frac {c}{x}\right ) \operatorname {PolyLog}\left (2,e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )-i \arctan \left (\frac {c}{x}\right ) \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (\frac {c}{x}\right )}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{2 i \arctan \left (\frac {c}{x}\right )}\right )\right )+\frac {1}{64} i b^3 \left (\pi ^4-32 \arctan \left (\frac {c}{x}\right )^4+64 i \arctan \left (\frac {c}{x}\right )^3 \log \left (1-e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )-64 i \arctan \left (\frac {c}{x}\right )^3 \log \left (1+e^{2 i \arctan \left (\frac {c}{x}\right )}\right )-96 \arctan \left (\frac {c}{x}\right )^2 \operatorname {PolyLog}\left (2,e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )-96 \arctan \left (\frac {c}{x}\right )^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (\frac {c}{x}\right )}\right )+96 i \arctan \left (\frac {c}{x}\right ) \operatorname {PolyLog}\left (3,e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )-96 i \arctan \left (\frac {c}{x}\right ) \operatorname {PolyLog}\left (3,-e^{2 i \arctan \left (\frac {c}{x}\right )}\right )+48 \operatorname {PolyLog}\left (4,e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )+48 \operatorname {PolyLog}\left (4,-e^{2 i \arctan \left (\frac {c}{x}\right )}\right )\right ) \]
a^3*Log[x] - ((3*I)/2)*a^2*b*(PolyLog[2, ((-I)*c)/x] - PolyLog[2, (I*c)/x] ) + 3*a*b^2*((I/24)*Pi^3 - ((2*I)/3)*ArcTan[c/x]^3 - ArcTan[c/x]^2*Log[1 - E^((-2*I)*ArcTan[c/x])] + ArcTan[c/x]^2*Log[1 + E^((2*I)*ArcTan[c/x])] - I*ArcTan[c/x]*PolyLog[2, E^((-2*I)*ArcTan[c/x])] - I*ArcTan[c/x]*PolyLog[2 , -E^((2*I)*ArcTan[c/x])] - PolyLog[3, E^((-2*I)*ArcTan[c/x])]/2 + PolyLog [3, -E^((2*I)*ArcTan[c/x])]/2) + (I/64)*b^3*(Pi^4 - 32*ArcTan[c/x]^4 + (64 *I)*ArcTan[c/x]^3*Log[1 - E^((-2*I)*ArcTan[c/x])] - (64*I)*ArcTan[c/x]^3*L og[1 + E^((2*I)*ArcTan[c/x])] - 96*ArcTan[c/x]^2*PolyLog[2, E^((-2*I)*ArcT an[c/x])] - 96*ArcTan[c/x]^2*PolyLog[2, -E^((2*I)*ArcTan[c/x])] + (96*I)*A rcTan[c/x]*PolyLog[3, E^((-2*I)*ArcTan[c/x])] - (96*I)*ArcTan[c/x]*PolyLog [3, -E^((2*I)*ArcTan[c/x])] + 48*PolyLog[4, E^((-2*I)*ArcTan[c/x])] + 48*P olyLog[4, -E^((2*I)*ArcTan[c/x])])
Time = 1.03 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5359, 5357, 5523, 5529, 5533, 7164}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^3}{x} \, dx\) |
| \(\Big \downarrow \) 5359 |
| \(\displaystyle -\int x \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3d\frac {1}{x}\) |
| \(\Big \downarrow \) 5357 |
| \(\displaystyle 6 b c \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \text {arctanh}\left (1-\frac {2}{\frac {i c}{x}+1}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}-2 \text {arctanh}\left (1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3\) |
| \(\Big \downarrow \) 5523 |
| \(\displaystyle 6 b c \left (\frac {1}{2} \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \log \left (2-\frac {2}{\frac {i c}{x}+1}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}-\frac {1}{2} \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \log \left (\frac {2}{\frac {i c}{x}+1}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )-2 \text {arctanh}\left (1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3\) |
| \(\Big \downarrow \) 5529 |
| \(\displaystyle 6 b c \left (\frac {1}{2} \left (\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{\frac {i c}{x}+1}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2}{2 c}-i b \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{\frac {i c}{x}+1}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )+\frac {1}{2} \left (i b \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right ) \operatorname {PolyLog}\left (2,\frac {2}{\frac {i c}{x}+1}-1\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{\frac {i c}{x}+1}-1\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2}{2 c}\right )\right )-2 \text {arctanh}\left (1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3\) |
| \(\Big \downarrow \) 5533 |
| \(\displaystyle 6 b c \left (\frac {1}{2} \left (\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{\frac {i c}{x}+1}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2}{2 c}-i b \left (\frac {i \operatorname {PolyLog}\left (3,1-\frac {2}{\frac {i c}{x}+1}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{2 c}-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (3,1-\frac {2}{\frac {i c}{x}+1}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )\right )+\frac {1}{2} \left (i b \left (\frac {i \operatorname {PolyLog}\left (3,\frac {2}{\frac {i c}{x}+1}-1\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{2 c}-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (3,\frac {2}{\frac {i c}{x}+1}-1\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{\frac {i c}{x}+1}-1\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2}{2 c}\right )\right )-2 \text {arctanh}\left (1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3\) |
| \(\Big \downarrow \) 7164 |
| \(\displaystyle 6 b c \left (\frac {1}{2} \left (\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{\frac {i c}{x}+1}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2}{2 c}-i b \left (\frac {i \operatorname {PolyLog}\left (3,1-\frac {2}{\frac {i c}{x}+1}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{2 c}+\frac {b \operatorname {PolyLog}\left (4,1-\frac {2}{\frac {i c}{x}+1}\right )}{4 c}\right )\right )+\frac {1}{2} \left (i b \left (\frac {i \operatorname {PolyLog}\left (3,\frac {2}{\frac {i c}{x}+1}-1\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{2 c}+\frac {b \operatorname {PolyLog}\left (4,\frac {2}{\frac {i c}{x}+1}-1\right )}{4 c}\right )-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{\frac {i c}{x}+1}-1\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2}{2 c}\right )\right )-2 \text {arctanh}\left (1-\frac {2}{1+\frac {i c}{x}}\right ) \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3\) |
-2*(a + b*ArcTan[c/x])^3*ArcTanh[1 - 2/(1 + (I*c)/x)] + 6*b*c*((((I/2)*(a + b*ArcTan[c/x])^2*PolyLog[2, 1 - 2/(1 + (I*c)/x)])/c - I*b*(((I/2)*(a + b *ArcTan[c/x])*PolyLog[3, 1 - 2/(1 + (I*c)/x)])/c + (b*PolyLog[4, 1 - 2/(1 + (I*c)/x)])/(4*c)))/2 + (((-1/2*I)*(a + b*ArcTan[c/x])^2*PolyLog[2, -1 + 2/(1 + (I*c)/x)])/c + I*b*(((I/2)*(a + b*ArcTan[c/x])*PolyLog[3, -1 + 2/(1 + (I*c)/x)])/c + (b*PolyLog[4, -1 + 2/(1 + (I*c)/x)])/(4*c)))/2)
3.2.51.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 + I*c*x)], x] - Simp[2*b*c*p Int[(a + b *ArcTan[c*x])^(p - 1)*(ArcTanh[1 - 2/(1 + I*c*x)]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[1 /n Subst[Int[(a + b*ArcTan[c*x])^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]
Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x _)^2), x_Symbol] :> Simp[1/2 Int[Log[1 + u]*((a + b*ArcTan[c*x])^p/(d + e *x^2)), x], x] - Simp[1/2 Int[Log[1 - u]*((a + b*ArcTan[c*x])^p/(d + e*x^ 2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 - 2*(I/(I - c*x)))^2, 0]
Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 ), x_Symbol] :> Simp[(-I)*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)) , x] + Simp[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]
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] - Simp[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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 26.33 (sec) , antiderivative size = 2225, normalized size of antiderivative = 9.67
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(2225\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(2226\) |
| default | \(\text {Expression too large to display}\) | \(2226\) |
a^3*ln(x)+b^3*(-ln(c/x)*arctan(c/x)^3+arctan(c/x)^3*ln((1+I*c/x)^2/(1+c^2/ x^2)-1)-arctan(c/x)^3*ln(1-(1+I*c/x)/(1+c^2/x^2)^(1/2))+3*I*arctan(c/x)^2* polylog(2,(1+I*c/x)/(1+c^2/x^2)^(1/2))-6*arctan(c/x)*polylog(3,(1+I*c/x)/( 1+c^2/x^2)^(1/2))-6*I*polylog(4,(1+I*c/x)/(1+c^2/x^2)^(1/2))-arctan(c/x)^3 *ln((1+I*c/x)/(1+c^2/x^2)^(1/2)+1)+3*I*arctan(c/x)^2*polylog(2,-(1+I*c/x)/ (1+c^2/x^2)^(1/2))-6*arctan(c/x)*polylog(3,-(1+I*c/x)/(1+c^2/x^2)^(1/2))-6 *I*polylog(4,-(1+I*c/x)/(1+c^2/x^2)^(1/2))-1/2*I*Pi*(csgn(I*((1+I*c/x)^2/( 1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))*csgn(((1+I*c/x)^2/(1+c^2/x^2)-1 )/((1+I*c/x)^2/(1+c^2/x^2)+1))-csgn(((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x) ^2/(1+c^2/x^2)+1))^2-csgn(I/((1+I*c/x)^2/(1+c^2/x^2)+1))*csgn(I*((1+I*c/x) ^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^2+csgn(I/((1+I*c/x)^2/(1+c^ 2/x^2)+1))*csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1)) *csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1))+csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1)/(( 1+I*c/x)^2/(1+c^2/x^2)+1))^3-csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x) ^2/(1+c^2/x^2)+1))^2*csgn(I*((1+I*c/x)^2/(1+c^2/x^2)-1))-csgn(I*((1+I*c/x) ^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))*csgn(((1+I*c/x)^2/(1+c^2/x^ 2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^2+csgn(((1+I*c/x)^2/(1+c^2/x^2)-1)/((1+ I*c/x)^2/(1+c^2/x^2)+1))^3+1)*arctan(c/x)^3-3/2*I*arctan(c/x)^2*polylog(2, -(1+I*c/x)^2/(1+c^2/x^2))+3/2*arctan(c/x)*polylog(3,-(1+I*c/x)^2/(1+c^2/x^ 2))+3/4*I*polylog(4,-(1+I*c/x)^2/(1+c^2/x^2)))+3*a^2*b*(-ln(c/x)*arctan...
\[ \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^3}{x} \, dx=\int { \frac {{\left (b \arctan \left (\frac {c}{x}\right ) + a\right )}^{3}}{x} \,d x } \]
\[ \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^3}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (\frac {c}{x} \right )}\right )^{3}}{x}\, dx \]
\[ \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^3}{x} \, dx=\int { \frac {{\left (b \arctan \left (\frac {c}{x}\right ) + a\right )}^{3}}{x} \,d x } \]
a^3*log(x) + 1/32*integrate((28*b^3*arctan2(c, x)^3 + 3*b^3*arctan2(c, x)* log(c^2 + x^2)^2 + 96*a*b^2*arctan2(c, x)^2 + 96*a^2*b*arctan2(c, x))/x, x )
\[ \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^3}{x} \, dx=\int { \frac {{\left (b \arctan \left (\frac {c}{x}\right ) + a\right )}^{3}}{x} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \arctan \left (\frac {c}{x}\right )\right )^3}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (\frac {c}{x}\right )\right )}^3}{x} \,d x \]