Integrand size = 12, antiderivative size = 119 \[ \int \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3 \, dx=i c \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3+x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3-3 b c \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2 \log \left (\frac {2 c}{c+i x}\right )+3 i b^2 c \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c}{c+i x}\right )-\frac {3}{2} b^3 c \operatorname {PolyLog}\left (3,1-\frac {2 c}{c+i x}\right ) \]
I*c*(a+b*arccot(x/c))^3+x*(a+b*arccot(x/c))^3-3*b*c*(a+b*arccot(x/c))^2*ln (2*c/(c+I*x))+3*I*b^2*c*(a+b*arccot(x/c))*polylog(2,1-2*c/(c+I*x))-3/2*b^3 *c*polylog(3,1-2*c/(c+I*x))
Time = 0.28 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.81 \[ \int \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3 \, dx=a^3 x+3 a^2 b x \arctan \left (\frac {c}{x}\right )+\frac {3}{2} a^2 b c \log \left (c^2+x^2\right )-3 a b^2 \left (-\left ((i c+x) \arctan \left (\frac {c}{x}\right )^2\right )+2 c \arctan \left (\frac {c}{x}\right ) \log \left (1-e^{2 i \arctan \left (\frac {c}{x}\right )}\right )-i c \operatorname {PolyLog}\left (2,e^{2 i \arctan \left (\frac {c}{x}\right )}\right )\right )-\frac {1}{8} b^3 \left (-i c \pi ^3+8 i c \arctan \left (\frac {c}{x}\right )^3-8 x \arctan \left (\frac {c}{x}\right )^3+24 c \arctan \left (\frac {c}{x}\right )^2 \log \left (1-e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )+24 i c \arctan \left (\frac {c}{x}\right ) \operatorname {PolyLog}\left (2,e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )+12 c \operatorname {PolyLog}\left (3,e^{-2 i \arctan \left (\frac {c}{x}\right )}\right )\right ) \]
a^3*x + 3*a^2*b*x*ArcTan[c/x] + (3*a^2*b*c*Log[c^2 + x^2])/2 - 3*a*b^2*(-( (I*c + x)*ArcTan[c/x]^2) + 2*c*ArcTan[c/x]*Log[1 - E^((2*I)*ArcTan[c/x])] - I*c*PolyLog[2, E^((2*I)*ArcTan[c/x])]) - (b^3*((-I)*c*Pi^3 + (8*I)*c*Arc Tan[c/x]^3 - 8*x*ArcTan[c/x]^3 + 24*c*ArcTan[c/x]^2*Log[1 - E^((-2*I)*ArcT an[c/x])] + (24*I)*c*ArcTan[c/x]*PolyLog[2, E^((-2*I)*ArcTan[c/x])] + 12*c *PolyLog[3, E^((-2*I)*ArcTan[c/x])]))/8
Time = 0.66 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5349, 5346, 27, 5456, 27, 5380, 27, 5530, 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.
\(\displaystyle \int \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3 \, dx\) |
\(\Big \downarrow \) 5349 |
\(\displaystyle \int \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3dx\) |
\(\Big \downarrow \) 5346 |
\(\displaystyle \frac {3 b \int \frac {c^2 x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2}{c^2+x^2}dx}{c}+x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 b c \int \frac {x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2}{c^2+x^2}dx+x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3\) |
\(\Big \downarrow \) 5456 |
\(\displaystyle x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3+3 b c \left (\frac {i \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3}{3 b}-\frac {\int \frac {c \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2}{i c-x}dx}{c}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3+3 b c \left (\frac {i \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3}{3 b}-\int \frac {\left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2}{i c-x}dx\right )\) |
\(\Big \downarrow \) 5380 |
\(\displaystyle x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3+3 b c \left (-\frac {2 b \int \frac {c^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \log \left (\frac {2 c}{c+i x}\right )}{c^2+x^2}dx}{c}+\frac {i \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3}{3 b}-\log \left (\frac {2 c}{c+i x}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3+3 b c \left (-2 b c \int \frac {\left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right ) \log \left (\frac {2 c}{c+i x}\right )}{c^2+x^2}dx+\frac {i \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3}{3 b}-\log \left (\frac {2 c}{c+i x}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2\right )\) |
\(\Big \downarrow \) 5530 |
\(\displaystyle x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3+3 b c \left (-2 b c \left (-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2 c}{c+i x}\right )}{c^2+x^2}dx-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 c}{c+i x}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )}{2 c}\right )+\frac {i \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3}{3 b}-\log \left (\frac {2 c}{c+i x}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2\right )\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3+3 b c \left (-2 b c \left (\frac {b \operatorname {PolyLog}\left (3,1-\frac {2 c}{c+i x}\right )}{4 c}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 c}{c+i x}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )}{2 c}\right )+\frac {i \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^3}{3 b}-\log \left (\frac {2 c}{c+i x}\right ) \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2\right )\) |
x*(a + b*ArcCot[x/c])^3 + 3*b*c*(((I/3)*(a + b*ArcCot[x/c])^3)/b - (a + b* ArcCot[x/c])^2*Log[(2*c)/(c + I*x)] - 2*b*c*(((-1/2*I)*(a + b*ArcCot[x/c]) *PolyLog[2, 1 - (2*c)/(c + I*x)])/c + (b*PolyLog[3, 1 - (2*c)/(c + I*x)])/ (4*c)))
3.2.50.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCot[c*x^n])^p, x] + Simp[b*c*n*p Int[x^n*((a + b*ArcCot[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])
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_), x_Symbol] :> Int[(a + b*A rcCot[1/(x^n*c)])^p, x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1] && ILtQ[n, 0]
Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcCot[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] - Simp[b*c*( p/e) Int[(a + b*ArcCot[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 ]
Int[(((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[I*((a + b*ArcCot[c*x])^(p + 1)/(b*e*(p + 1))), x] - Simp[ 1/(c*d) Int[(a + b*ArcCot[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]
Int[(Log[u_]*((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 ), x_Symbol] :> Simp[(-I)*(a + b*ArcCot[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)) , x] - Simp[b*p*(I/2) Int[(a + b*ArcCot[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[(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 = 102.88 (sec) , antiderivative size = 2028, normalized size of antiderivative = 17.04
method | result | size |
parts | \(\text {Expression too large to display}\) | \(2028\) |
derivativedivides | \(\text {Expression too large to display}\) | \(2031\) |
default | \(\text {Expression too large to display}\) | \(2031\) |
x*a^3-b^3*c*(-1/c*x*arctan(c/x)^3+3*ln(c/x)*arctan(c/x)^2-3/2*arctan(c/x)^ 2*ln(1+c^2/x^2)+3*arctan(c/x)^2*ln((1+I*c/x)/(1+c^2/x^2)^(1/2))-3*arctan(c /x)^2*ln((1+I*c/x)^2/(1+c^2/x^2)-1)-I*arctan(c/x)^3+3/4*(-I*Pi*csgn(I/((1+ I*c/x)^2/(1+c^2/x^2)+1)^2)*csgn(I*(1+I*c/x)^2/(1+c^2/x^2))*csgn(I*(1+I*c/x )^2/(1+c^2/x^2)/((1+I*c/x)^2/(1+c^2/x^2)+1)^2)+2*I*Pi*csgn(((1+I*c/x)^2/(1 +c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^3+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))^3+2*I*Pi*csgn(I*(1+I*c/x)/(1+c^2/ x^2)^(1/2))*csgn(I*(1+I*c/x)^2/(1+c^2/x^2))^2-2*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))^2-2*I*Pi*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-I*Pi*c sgn(I*(1+I*c/x)/(1+c^2/x^2)^(1/2))^2*csgn(I*(1+I*c/x)^2/(1+c^2/x^2))+I*Pi* csgn(I*((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)^2 )+I*Pi*csgn(I*((1+I*c/x)^2/(1+c^2/x^2)+1)^2)^3-2*I*Pi*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)^2)^2-2*I*Pi*csgn(((1+I* c/x)^2/(1+c^2/x^2)-1)/((1+I*c/x)^2/(1+c^2/x^2)+1))^2+2*I*Pi*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))*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*I*Pi+I*Pi*csgn(I*(1+I*c/ x)^2/(1+c^2/x^2))*csgn(I*(1+I*c/x)^2/(1+c^2/x^2)/((1+I*c/x)^2/(1+c^2/x^2)+ 1)^2)^2-I*Pi*csgn(I*(1+I*c/x)^2/(1+c^2/x^2))^3+2*I*Pi*csgn(I*((1+I*c/x)...
\[ \int \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3 \, dx=\int { {\left (b \arctan \left (\frac {c}{x}\right ) + a\right )}^{3} \,d x } \]
\[ \int \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3 \, dx=\int \left (a + b \operatorname {atan}{\left (\frac {c}{x} \right )}\right )^{3}\, dx \]
\[ \int \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3 \, dx=\int { {\left (b \arctan \left (\frac {c}{x}\right ) + a\right )}^{3} \,d x } \]
7/8*b^3*c*arctan(c/x)^3*arctan(x/c) + 3*a*b^2*c*arctan(c/x)^2*arctan(x/c) + 1/8*b^3*x*arctan2(c, x)^3 - 3/32*b^3*x*arctan2(c, x)*log(c^2 + x^2)^2 + (3*arctan(c/x)*arctan(x/c)^2/c + arctan(x/c)^3/c)*a*b^2*c^2 + 7/32*(6*arct an(c/x)^2*arctan(x/c)^2/c + 4*arctan(c/x)*arctan(x/c)^3/c + arctan(x/c)^4/ c)*b^3*c^2 + 3*b^3*c^2*integrate(1/32*arctan(c/x)*log(c^2 + x^2)^2/(c^2 + x^2), x) + 12*b^3*c*integrate(1/32*x*arctan(c/x)^2/(c^2 + x^2), x) - 3*b^3 *c*integrate(1/32*x*log(c^2 + x^2)^2/(c^2 + x^2), x) + 3/2*(2*x*arctan(c/x ) + c*log(c^2 + x^2))*a^2*b + a^3*x + 28*b^3*integrate(1/32*x^2*arctan(c/x )^3/(c^2 + x^2), x) + 3*b^3*integrate(1/32*x^2*arctan(c/x)*log(c^2 + x^2)^ 2/(c^2 + x^2), x) + 96*a*b^2*integrate(1/32*x^2*arctan(c/x)^2/(c^2 + x^2), x) + 12*b^3*integrate(1/32*x^2*arctan(c/x)*log(c^2 + x^2)/(c^2 + x^2), x)
\[ \int \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3 \, dx=\int { {\left (b \arctan \left (\frac {c}{x}\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int \left (a+b \arctan \left (\frac {c}{x}\right )\right )^3 \, dx=\int {\left (a+b\,\mathrm {atan}\left (\frac {c}{x}\right )\right )}^3 \,d x \]