Integrand size = 26, antiderivative size = 100 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2} \, dx=\frac {c e (a+b \arctan (c x))^2}{b}-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x}+\frac {1}{2} b c \left (d+e \log \left (1+c^2 x^2\right )\right ) \log \left (1-\frac {1}{1+c^2 x^2}\right )-\frac {1}{2} b c e \operatorname {PolyLog}\left (2,\frac {1}{1+c^2 x^2}\right ) \]
c*e*(a+b*arctan(c*x))^2/b-(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x+1/2*b*c* (d+e*ln(c^2*x^2+1))*ln(1-1/(c^2*x^2+1))-1/2*b*c*e*polylog(2,1/(c^2*x^2+1))
Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2} \, dx=\frac {c e (a+b \arctan (c x))^2}{b}-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x}+b c \left (-\frac {\left (d+e \log \left (1+c^2 x^2\right )\right ) \left (d-2 e \log \left (-c^2 x^2\right )+e \log \left (1+c^2 x^2\right )\right )}{4 e}+\frac {1}{2} e \operatorname {PolyLog}\left (2,1+c^2 x^2\right )\right ) \]
(c*e*(a + b*ArcTan[c*x])^2)/b - ((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^ 2]))/x + b*c*(-1/4*((d + e*Log[1 + c^2*x^2])*(d - 2*e*Log[-(c^2*x^2)] + e* Log[1 + c^2*x^2]))/e + (e*PolyLog[2, 1 + c^2*x^2])/2)
Time = 0.64 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.82, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5552, 2925, 2858, 25, 27, 2779, 2838, 5419}
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.
\(\displaystyle \int \frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 5552 |
\(\displaystyle 2 c^2 e \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx+b c \int \frac {d+e \log \left (c^2 x^2+1\right )}{x \left (c^2 x^2+1\right )}dx-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle 2 c^2 e \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx+\frac {1}{2} b c \int \frac {d+e \log \left (c^2 x^2+1\right )}{x^2 \left (c^2 x^2+1\right )}dx^2-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}\) |
\(\Big \downarrow \) 2858 |
\(\displaystyle 2 c^2 e \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx+\frac {b \int \frac {d+e \log \left (c^2 x^2+1\right )}{x^4}d\left (c^2 x^2+1\right )}{2 c}-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 c^2 e \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx-\frac {b \int -\frac {d+e \log \left (c^2 x^2+1\right )}{x^4}d\left (c^2 x^2+1\right )}{2 c}-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 c^2 e \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx-\frac {1}{2} b c \int -\frac {d+e \log \left (c^2 x^2+1\right )}{c^2 x^4}d\left (c^2 x^2+1\right )-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}\) |
\(\Big \downarrow \) 2779 |
\(\displaystyle 2 c^2 e \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx-\frac {1}{2} b c \left (e \int \frac {\log \left (1-\frac {1}{x^2}\right )}{x^2}d\left (c^2 x^2+1\right )-\log \left (1-\frac {1}{x^2}\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )\right )-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle 2 c^2 e \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}-\frac {1}{2} b c \left (e \operatorname {PolyLog}\left (2,\frac {1}{x^2}\right )-\log \left (1-\frac {1}{x^2}\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )\right )\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle -\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}+\frac {c e (a+b \arctan (c x))^2}{b}-\frac {1}{2} b c \left (e \operatorname {PolyLog}\left (2,\frac {1}{x^2}\right )-\log \left (1-\frac {1}{x^2}\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )\right )\) |
(c*e*(a + b*ArcTan[c*x])^2)/b - ((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^ 2]))/x - (b*c*(-(Log[1 - x^(-2)]*(d + e*Log[1 + c^2*x^2])) + e*PolyLog[2, x^(-2)]))/2
3.13.92.3.1 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_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_ .)*(x_))^(q_.)*((h_.) + (i_.)*(x_))^(r_.), x_Symbol] :> Simp[1/e Subst[In t[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[r, 0]) && IntegerQ[2*r]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Si mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 ] || IGtQ[q, 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_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*( e_.))*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(d + e*Log[f + g*x^2])*((a + b*ArcTan[c*x])/(m + 1)), x] + (-Simp[b*(c/(m + 1)) Int[x^(m + 1)*((d + e* Log[f + g*x^2])/(1 + c^2*x^2)), x], x] - Simp[2*e*(g/(m + 1)) Int[x^(m + 2)*((a + b*ArcTan[c*x])/(f + g*x^2)), x], x]) /; FreeQ[{a, b, c, d, e, f, g }, x] && ILtQ[m/2, 0]
\[\int \frac {\left (a +b \arctan \left (c x \right )\right ) \left (d +e \ln \left (c^{2} x^{2}+1\right )\right )}{x^{2}}d x\]
\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{2}} \,d x } \]
Exception generated. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2} \, dx=\text {Exception raised: TypeError} \]
\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{2}} \,d x } \]
-1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d + (2*c*arctan (c*x) - log(c^2*x^2 + 1)/x)*a*e + b*e*integrate(arctan(c*x)*log(c^2*x^2 + 1)/x^2, x) - a*d/x
Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (c^2\,x^2+1\right )\right )}{x^2} \,d x \]