Integrand size = 14, antiderivative size = 213 \[ \int \frac {(a+b \arctan (c x))^3}{x^4} \, dx=-\frac {b^2 c^2 (a+b \arctan (c x))}{x}-\frac {1}{2} b c^3 (a+b \arctan (c x))^2-\frac {b c (a+b \arctan (c x))^2}{2 x^2}+\frac {1}{3} i c^3 (a+b \arctan (c x))^3-\frac {(a+b \arctan (c x))^3}{3 x^3}+b^3 c^3 \log (x)-\frac {1}{2} b^3 c^3 \log \left (1+c^2 x^2\right )-b c^3 (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1-i c x}\right )+i b^2 c^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )-\frac {1}{2} b^3 c^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-i c x}\right ) \]
-b^2*c^2*(a+b*arctan(c*x))/x-1/2*b*c^3*(a+b*arctan(c*x))^2-1/2*b*c*(a+b*ar ctan(c*x))^2/x^2+1/3*I*c^3*(a+b*arctan(c*x))^3-1/3*(a+b*arctan(c*x))^3/x^3 +b^3*c^3*ln(x)-1/2*b^3*c^3*ln(c^2*x^2+1)-b*c^3*(a+b*arctan(c*x))^2*ln(2-2/ (1-I*c*x))+I*b^2*c^3*(a+b*arctan(c*x))*polylog(2,-1+2/(1-I*c*x))-1/2*b^3*c ^3*polylog(3,-1+2/(1-I*c*x))
Time = 1.38 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b \arctan (c x))^3}{x^4} \, dx=\frac {1}{6} \left (-\frac {2 a^3}{x^3}-\frac {3 a^2 b c}{x^2}-\frac {6 a^2 b \arctan (c x)}{x^3}-6 a^2 b c^3 \log (x)+3 a^2 b c^3 \log \left (1+c^2 x^2\right )+\frac {6 i a b^2 \left (i c^2 x^2+\left (i+c^3 x^3\right ) \arctan (c x)^2+i c x \arctan (c x) \left (1+c^2 x^2+2 c^2 x^2 \log \left (1-e^{2 i \arctan (c x)}\right )\right )+c^3 x^3 \operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )\right )}{x^3}+6 b^3 c^3 \left (-i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+\frac {1}{24} \left (i \pi ^3-\frac {24 \arctan (c x)}{c x}+\left (-8 i-\frac {8}{c^3 x^3}\right ) \arctan (c x)^3+\arctan (c x)^2 \left (-12-\frac {12}{c^2 x^2}-24 \log \left (1-e^{-2 i \arctan (c x)}\right )\right )+24 \log (c x)-12 \log \left (1+c^2 x^2\right )-12 \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )\right )\right )\right ) \]
((-2*a^3)/x^3 - (3*a^2*b*c)/x^2 - (6*a^2*b*ArcTan[c*x])/x^3 - 6*a^2*b*c^3* Log[x] + 3*a^2*b*c^3*Log[1 + c^2*x^2] + ((6*I)*a*b^2*(I*c^2*x^2 + (I + c^3 *x^3)*ArcTan[c*x]^2 + I*c*x*ArcTan[c*x]*(1 + c^2*x^2 + 2*c^2*x^2*Log[1 - E ^((2*I)*ArcTan[c*x])]) + c^3*x^3*PolyLog[2, E^((2*I)*ArcTan[c*x])]))/x^3 + 6*b^3*c^3*((-I)*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + (I*Pi^3 - (24*ArcTan[c*x])/(c*x) + (-8*I - 8/(c^3*x^3))*ArcTan[c*x]^3 + ArcTan[c*x ]^2*(-12 - 12/(c^2*x^2) - 24*Log[1 - E^((-2*I)*ArcTan[c*x])]) + 24*Log[c*x ] - 12*Log[1 + c^2*x^2] - 12*PolyLog[3, E^((-2*I)*ArcTan[c*x])])/24))/6
Time = 1.60 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5361, 5453, 5361, 5453, 5361, 243, 47, 14, 16, 5419, 5459, 5403, 5527, 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.
\(\displaystyle \int \frac {(a+b \arctan (c x))^3}{x^4} \, dx\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle b c \int \frac {(a+b \arctan (c x))^2}{x^3 \left (c^2 x^2+1\right )}dx-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 5453 |
\(\displaystyle b c \left (\int \frac {(a+b \arctan (c x))^2}{x^3}dx-c^2 \int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle b c \left (c^2 \left (-\int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )+b c \int \frac {a+b \arctan (c x)}{x^2 \left (c^2 x^2+1\right )}dx-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 5453 |
\(\displaystyle b c \left (c^2 \left (-\int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )+b c \left (\int \frac {a+b \arctan (c x)}{x^2}dx-c^2 \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle b c \left (c^2 \left (-\int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )+b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+b c \int \frac {1}{x \left (c^2 x^2+1\right )}dx-\frac {a+b \arctan (c x)}{x}\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle b c \left (c^2 \left (-\int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )+b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \int \frac {1}{x^2 \left (c^2 x^2+1\right )}dx^2-\frac {a+b \arctan (c x)}{x}\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle b c \left (c^2 \left (-\int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )+b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \left (\int \frac {1}{x^2}dx^2-c^2 \int \frac {1}{c^2 x^2+1}dx^2\right )-\frac {a+b \arctan (c x)}{x}\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle b c \left (c^2 \left (-\int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )+b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \left (\log \left (x^2\right )-c^2 \int \frac {1}{c^2 x^2+1}dx^2\right )-\frac {a+b \arctan (c x)}{x}\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle b c \left (c^2 \left (-\int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )+b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle b c \left (c^2 \left (-\int \frac {(a+b \arctan (c x))^2}{x \left (c^2 x^2+1\right )}dx\right )+b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )-\frac {(a+b \arctan (c x))^3}{3 x^3}\) |
\(\Big \downarrow \) 5459 |
\(\displaystyle -\frac {(a+b \arctan (c x))^3}{3 x^3}+b c \left (-\left (c^2 \left (i \int \frac {(a+b \arctan (c x))^2}{x (c x+i)}dx-\frac {i (a+b \arctan (c x))^3}{3 b}\right )\right )+b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )\) |
\(\Big \downarrow \) 5403 |
\(\displaystyle -\frac {(a+b \arctan (c x))^3}{3 x^3}+b c \left (-\left (c^2 \left (i \left (2 i b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{c^2 x^2+1}dx-i \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2\right )-\frac {i (a+b \arctan (c x))^3}{3 b}\right )\right )+b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )\) |
\(\Big \downarrow \) 5527 |
\(\displaystyle -\frac {(a+b \arctan (c x))^3}{3 x^3}+b c \left (-\left (c^2 \left (i \left (2 i b c \left (\frac {i \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right ) (a+b \arctan (c x))}{2 c}-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )}{c^2 x^2+1}dx\right )-i \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2\right )-\frac {i (a+b \arctan (c x))^3}{3 b}\right )\right )+b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle -\frac {(a+b \arctan (c x))^3}{3 x^3}+b c \left (-\left (c^2 \left (i \left (2 i b c \left (\frac {i \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right ) (a+b \arctan (c x))}{2 c}-\frac {b \operatorname {PolyLog}\left (3,\frac {2}{1-i c x}-1\right )}{4 c}\right )-i \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2\right )-\frac {i (a+b \arctan (c x))^3}{3 b}\right )\right )+b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}\right )\) |
-1/3*(a + b*ArcTan[c*x])^3/x^3 + b*c*(-1/2*(a + b*ArcTan[c*x])^2/x^2 + b*c *(-((a + b*ArcTan[c*x])/x) - (c*(a + b*ArcTan[c*x])^2)/(2*b) + (b*c*(Log[x ^2] - Log[1 + c^2*x^2]))/2) - c^2*(((-1/3*I)*(a + b*ArcTan[c*x])^3)/b + I* ((-I)*(a + b*ArcTan[c*x])^2*Log[2 - 2/(1 - I*c*x)] + (2*I)*b*c*(((I/2)*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 - I*c*x)])/c - (b*PolyLog[3, -1 + 2/ (1 - I*c*x)])/(4*c)))))
3.1.33.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & & IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_ Symbol] :> Simp[(a + b*ArcTan[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Si mp[b*c*(p/d) Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 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_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> 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]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Simp[e/(d*f^2) Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/(d + e*x^2) ), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*d*(p + 1))), x] + Si mp[I/d Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 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[(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 = 6.80 (sec) , antiderivative size = 2097, normalized size of antiderivative = 9.85
\[\text {Expression too large to display}\]
c^3*(-1/3/c^3/x^3*a^3+b^3*(-1/3/c^3/x^3*arctan(c*x)^3-1/2/c^2/x^2*arctan(c *x)^2-ln(c*x)*arctan(c*x)^2+1/2*arctan(c*x)^2*ln(c^2*x^2+1)-arctan(c*x)^2* ln((1+I*c*x)/(c^2*x^2+1)^(1/2))+arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1 )+1/12*arctan(c*x)*(3*I*arctan(c*x)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I *c*x)^2/(c^2*x^2+1))^2)*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1))^2)*csgn(I*(1+I* c*x)^2/(c^2*x^2+1))*Pi*c*x-6*I*arctan(c*x)*csgn(I*((1+I*c*x)^2/(c^2*x^2+1) -1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I* c*x)^2/(c^2*x^2+1)))*Pi*c*x-3*I*arctan(c*x)*csgn(I*(1+(1+I*c*x)^2/(c^2*x^2 +1))^2)^3*Pi*c*x-3*I*arctan(c*x)*csgn(I*(1+(1+I*c*x)^2/(c^2*x^2+1))^2)*csg n(I*(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*Pi*c*x-3*I*arctan(c*x)*csgn(I*(1+I*c*x) ^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1))^2)^2*csgn(I/(1+(1+I*c*x)^2/(c^2 *x^2+1))^2)*Pi*c*x+6*I*arctan(c*x)*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+( 1+I*c*x)^2/(c^2*x^2+1)))^2*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*Pi*c*x+3*I* arctan(c*x)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1))^2)^ 3*Pi*c*x-6*I*arctan(c*x)*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/( c^2*x^2+1)))^3*Pi*c*x-6*I*arctan(c*x)*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/( 1+(1+I*c*x)^2/(c^2*x^2+1)))^3*Pi*c*x-12*I*c*x-6*I*arctan(c*x)*Pi*c*x+6*I*a rctan(c*x)*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1))) ^2*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*Pi*c*x+3*I*arctan(c*x)*csgn(I*(1+I* c*x)^2/(c^2*x^2+1))^3*Pi*c*x-6*I*arctan(c*x)*csgn(I*((1+I*c*x)^2/(c^2*x...
\[ \int \frac {(a+b \arctan (c x))^3}{x^4} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{x^{4}} \,d x } \]
\[ \int \frac {(a+b \arctan (c x))^3}{x^4} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3}}{x^{4}}\, dx \]
\[ \int \frac {(a+b \arctan (c x))^3}{x^4} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{x^{4}} \,d x } \]
1/2*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)* a^2*b - 1/3*a^3/x^3 - 1/96*(4*b^3*arctan(c*x)^3 - 3*b^3*arctan(c*x)*log(c^ 2*x^2 + 1)^2 - 96*x^3*integrate(-1/32*(4*b^3*c^2*x^2*arctan(c*x)*log(c^2*x ^2 + 1) - 28*(b^3*c^2*x^2 + b^3)*arctan(c*x)^3 - 4*(24*a*b^2*c^2*x^2 + b^3 *c*x + 24*a*b^2)*arctan(c*x)^2 + (b^3*c*x - 3*(b^3*c^2*x^2 + b^3)*arctan(c *x))*log(c^2*x^2 + 1)^2)/(c^2*x^6 + x^4), x))/x^3
Timed out. \[ \int \frac {(a+b \arctan (c x))^3}{x^4} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \arctan (c x))^3}{x^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{x^4} \,d x \]