Integrand size = 24, antiderivative size = 101 \[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\frac {i \log (d (a+b x)) \operatorname {PolyLog}(2,-i c (a+b x))}{2 b}-\frac {i \log (d (a+b x)) \operatorname {PolyLog}(2,i c (a+b x))}{2 b}-\frac {i \operatorname {PolyLog}(3,-i c (a+b x))}{2 b}+\frac {i \operatorname {PolyLog}(3,i c (a+b x))}{2 b} \]
1/2*I*ln(d*(b*x+a))*polylog(2,-I*c*(b*x+a))/b-1/2*I*ln(d*(b*x+a))*polylog( 2,I*c*(b*x+a))/b-1/2*I*polylog(3,-I*c*(b*x+a))/b+1/2*I*polylog(3,I*c*(b*x+ a))/b
Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78 \[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\frac {i (\log (d (a+b x)) \operatorname {PolyLog}(2,-i c (a+b x))-\log (d (a+b x)) \operatorname {PolyLog}(2,i c (a+b x))-\operatorname {PolyLog}(3,-i c (a+b x))+\operatorname {PolyLog}(3,i c (a+b x)))}{2 b} \]
((I/2)*(Log[d*(a + b*x)]*PolyLog[2, (-I)*c*(a + b*x)] - Log[d*(a + b*x)]*P olyLog[2, I*c*(a + b*x)] - PolyLog[3, (-I)*c*(a + b*x)] + PolyLog[3, I*c*( a + b*x)]))/b
Time = 0.61 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5732, 2894, 2881, 2821, 7143}
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.
\(\displaystyle \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx\) |
\(\Big \downarrow \) 5732 |
\(\displaystyle \frac {1}{2} i \int \frac {\log (d (a+b x)) \log (1-i c (a+b x))}{a+b x}dx-\frac {1}{2} i \int \frac {\log (d (a+b x)) \log (i c (a+b x)+1)}{a+b x}dx\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle \frac {1}{2} i \int \frac {\log (d (a+b x)) \log (-i a c-i b x c+1)}{a+b x}dx-\frac {1}{2} i \int \frac {\log (d (a+b x)) \log (i a c+i b x c+1)}{a+b x}dx\) |
\(\Big \downarrow \) 2881 |
\(\displaystyle \frac {i \int \frac {\log (d (a+b x)) \log (1-i c (a+b x))}{a+b x}d(a+b x)}{2 b}-\frac {i \int \frac {\log (d (a+b x)) \log (i c (a+b x)+1)}{a+b x}d(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2821 |
\(\displaystyle \frac {i \left (\int \frac {\operatorname {PolyLog}(2,i c (a+b x))}{a+b x}d(a+b x)-\operatorname {PolyLog}(2,i c (a+b x)) \log (d (a+b x))\right )}{2 b}-\frac {i \left (\int \frac {\operatorname {PolyLog}(2,-i c (a+b x))}{a+b x}d(a+b x)-\operatorname {PolyLog}(2,-i c (a+b x)) \log (d (a+b x))\right )}{2 b}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {i (\operatorname {PolyLog}(3,i c (a+b x))-\operatorname {PolyLog}(2,i c (a+b x)) \log (d (a+b x)))}{2 b}-\frac {i (\operatorname {PolyLog}(3,-i c (a+b x))-\operatorname {PolyLog}(2,-i c (a+b x)) \log (d (a+b x)))}{2 b}\) |
((-1/2*I)*(-(Log[d*(a + b*x)]*PolyLog[2, (-I)*c*(a + b*x)]) + PolyLog[3, ( -I)*c*(a + b*x)]))/b + ((I/2)*(-(Log[d*(a + b*x)]*PolyLog[2, I*c*(a + b*x) ]) + PolyLog[3, I*c*(a + b*x)]))/b
3.2.46.3.1 Defintions of rubi rules used
Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b _.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c *x^n])^p/m), x] + Simp[b*n*(p/m) Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c *x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log [(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Sym bol] :> Simp[1/e Subst[Int[(k*(x/d))^r*(a + b*Log[c*x^n])^p*(f + g*Log[h* ((e*i - d*j)/e + j*(x/e))^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r}, x] && EqQ[e*k - d*l, 0]
Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*( a + b*Log[c*ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p}, x] && Lin earQ[v, x] && !LinearMatchQ[v, x] && !(EqQ[n, 1] && MatchQ[c*v, (e_.)*((f _) + (g_.)*x) /; FreeQ[{e, f, g}, x]])
Int[(ArcTan[v_]*Log[w_])/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[I/2 Int[ Log[1 - I*v]*(Log[w]/(a + b*x)), x], x] - Simp[I/2 Int[Log[1 + I*v]*(Log[ w]/(a + b*x)), x], x] /; FreeQ[{a, b}, x] && LinearQ[v, x] && LinearQ[w, x] && EqQ[Simplify[D[v/(a + b*x), x]], 0] && EqQ[Simplify[D[w/(a + b*x), x]], 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 6.95 (sec) , antiderivative size = 1087, normalized size of antiderivative = 10.76
1/4/b*Pi*csgn(I*d)*csgn(I*d*(b*x+a))^2*dilog(-I*(c*(b*x+a)+I))+1/4/b*Pi*cs gn(I*(b*x+a))*csgn(I*d*(b*x+a))^2*dilog(-I*(c*(b*x+a)+I))-1/4/b*Pi*csgn(I* d)*csgn(I*(b*x+a))*csgn(I*d*(b*x+a))*dilog(-I*(c*(b*x+a)+I))-1/4/b*ln(-I*( -c*(b*x+a)+I))*ln(b*x+a)*Pi*csgn(I*d)*csgn(I*d*(b*x+a))^2-1/4/b*ln(-I*(-c* (b*x+a)+I))*ln(b*x+a)*Pi*csgn(I*(b*x+a))*csgn(I*d*(b*x+a))^2+1/4/b*ln(-I*( -c*(b*x+a)+I))*ln(-I*c*(b*x+a))*Pi*csgn(I*d)*csgn(I*d*(b*x+a))^2+1/4/b*ln( -I*(-c*(b*x+a)+I))*ln(-I*c*(b*x+a))*Pi*csgn(I*(b*x+a))*csgn(I*d*(b*x+a))^2 -1/4/b*dilog(-I*c*(b*x+a))*Pi*csgn(I*d)*csgn(I*(b*x+a))*csgn(I*d*(b*x+a))- 1/4*I*(-I*Pi*ln(b*x+a)*csgn(I*d)*csgn(I*(b*x+a))*csgn(I*d*(b*x+a))+I*Pi*ln (b*x+a)*csgn(I*d)*csgn(I*d*(b*x+a))^2+I*Pi*ln(b*x+a)*csgn(I*(b*x+a))*csgn( I*d*(b*x+a))^2-I*Pi*ln(b*x+a)*csgn(I*d*(b*x+a))^3+2*ln(d)*ln(b*x+a)+ln(b*x +a)^2)/b*ln(1+I*c*(b*x+a))-1/4/b*Pi*csgn(I*d*(b*x+a))^3*dilog(-I*(c*(b*x+a )+I))+1/2*I*polylog(3,I*c*(b*x+a))/b-1/4/b*dilog(-I*c*(b*x+a))*Pi*csgn(I*d *(b*x+a))^3+1/4*I/b*ln(b*x+a)^2*ln(1+I*c*(b*x+a))+1/2*I/b*ln(b*x+a)*polylo g(2,-I*c*(b*x+a))-1/2*I/b*dilog(-I*c*(b*x+a))*ln(d)+1/4/b*ln(-I*(-c*(b*x+a )+I))*ln(b*x+a)*Pi*csgn(I*d*(b*x+a))^3-1/4/b*ln(-I*(-c*(b*x+a)+I))*ln(-I*c *(b*x+a))*Pi*csgn(I*d*(b*x+a))^3+1/4/b*dilog(-I*c*(b*x+a))*Pi*csgn(I*d)*cs gn(I*d*(b*x+a))^2+1/4/b*dilog(-I*c*(b*x+a))*Pi*csgn(I*(b*x+a))*csgn(I*d*(b *x+a))^2+1/2*I/b*ln(-I*(-c*(b*x+a)+I))*ln(b*x+a)*ln(d)-1/2*I/b*ln(-I*(-c*( b*x+a)+I))*ln(-I*c*(b*x+a))*ln(d)-1/4*I/b*ln(b*x+a)^2*ln(1-I*c*(b*x+a))...
\[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int { \frac {\arctan \left ({\left (b x + a\right )} c\right ) \log \left ({\left (b x + a\right )} d\right )}{b x + a} \,d x } \]
\[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int \frac {\log {\left (a d + b d x \right )} \operatorname {atan}{\left (a c + b c x \right )}}{a + b x}\, dx \]
Exception generated. \[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int { \frac {\arctan \left ({\left (b x + a\right )} c\right ) \log \left ({\left (b x + a\right )} d\right )}{b x + a} \,d x } \]
Timed out. \[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int \frac {\mathrm {atan}\left (c\,\left (a+b\,x\right )\right )\,\ln \left (d\,\left (a+b\,x\right )\right )}{a+b\,x} \,d x \]