Integrand size = 16, antiderivative size = 100 \[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^4} \, dx=-\frac {1}{3} i c \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^3}+\frac {2}{3} b c \left (a+b \arctan \left (c x^3\right )\right ) \log \left (2-\frac {2}{1-i c x^3}\right )-\frac {1}{3} i b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x^3}\right ) \]
-1/3*I*c*(a+b*arctan(c*x^3))^2-1/3*(a+b*arctan(c*x^3))^2/x^3+2/3*b*c*(a+b* arctan(c*x^3))*ln(2-2/(1-I*c*x^3))-1/3*I*b^2*c*polylog(2,-1+2/(1-I*c*x^3))
Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^4} \, dx=\frac {b^2 \left (-1-i c x^3\right ) \arctan \left (c x^3\right )^2+2 b \arctan \left (c x^3\right ) \left (-a+b c x^3 \log \left (1-e^{2 i \arctan \left (c x^3\right )}\right )\right )-a \left (a-2 b c x^3 \log \left (c x^3\right )+b c x^3 \log \left (1+c^2 x^6\right )\right )-i b^2 c x^3 \operatorname {PolyLog}\left (2,e^{2 i \arctan \left (c x^3\right )}\right )}{3 x^3} \]
(b^2*(-1 - I*c*x^3)*ArcTan[c*x^3]^2 + 2*b*ArcTan[c*x^3]*(-a + b*c*x^3*Log[ 1 - E^((2*I)*ArcTan[c*x^3])]) - a*(a - 2*b*c*x^3*Log[c*x^3] + b*c*x^3*Log[ 1 + c^2*x^6]) - I*b^2*c*x^3*PolyLog[2, E^((2*I)*ArcTan[c*x^3])])/(3*x^3)
Time = 0.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5363, 5361, 5459, 5403, 2897}
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.
\(\displaystyle \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^4} \, dx\) |
\(\Big \downarrow \) 5363 |
\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^6}dx^3\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {1}{3} \left (2 b c \int \frac {a+b \arctan \left (c x^3\right )}{x^3 \left (c^2 x^6+1\right )}dx^3-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^3}\right )\) |
\(\Big \downarrow \) 5459 |
\(\displaystyle \frac {1}{3} \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^3}+2 b c \left (i \int \frac {a+b \arctan \left (c x^3\right )}{x^3 \left (c x^3+i\right )}dx^3-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b}\right )\right )\) |
\(\Big \downarrow \) 5403 |
\(\displaystyle \frac {1}{3} \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^3}+2 b c \left (i \left (i b c \int \frac {\log \left (2-\frac {2}{1-i c x^3}\right )}{c^2 x^6+1}dx^3-i \log \left (2-\frac {2}{1-i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )\right )-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b}\right )\right )\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle \frac {1}{3} \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^3}+2 b c \left (i \left (-i \log \left (2-\frac {2}{1-i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{1-i c x^3}-1\right )\right )-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b}\right )\right )\) |
(-((a + b*ArcTan[c*x^3])^2/x^3) + 2*b*c*(((-1/2*I)*(a + b*ArcTan[c*x^3])^2 )/b + I*((-I)*(a + b*ArcTan[c*x^3])*Log[2 - 2/(1 - I*c*x^3)] - (b*PolyLog[ 2, -1 + 2/(1 - I*c*x^3)])/2)))/3
3.2.18.3.1 Defintions of rubi rules used
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
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_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTan[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simplif y[(m + 1)/n]]
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_.)/((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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 4.78 (sec) , antiderivative size = 11455, normalized size of antiderivative = 114.55
method | result | size |
default | \(\text {Expression too large to display}\) | \(11455\) |
parts | \(\text {Expression too large to display}\) | \(11455\) |
\[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \]
\[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^4} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x^{3} \right )}\right )^{2}}{x^{4}}\, dx \]
\[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \]
-1/3*(c*(log(c^2*x^6 + 1) - log(x^6)) + 2*arctan(c*x^3)/x^3)*a*b + 1/48*(4 8*x^3*integrate(-1/16*(4*c^2*x^6*log(c^2*x^6 + 1) - 8*c*x^3*arctan(c*x^3) - 12*(c^2*x^6 + 1)*arctan(c*x^3)^2 - (c^2*x^6 + 1)*log(c^2*x^6 + 1)^2)/(c^ 2*x^10 + x^4), x) - 4*arctan(c*x^3)^2 + log(c^2*x^6 + 1)^2)*b^2/x^3 - 1/3* a^2/x^3
\[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x^3\right )\right )}^2}{x^4} \,d x \]