\(\int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx\) [146]

Optimal result

Integrand size = 24, antiderivative size = 101 \[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\frac {i \log (d (a+b x)) \operatorname {PolyLog}(2,-i c (a+b x))}{2 b}-\frac {i \log (d (a+b x)) \operatorname {PolyLog}(2,i c (a+b x))}{2 b}-\frac {i \operatorname {PolyLog}(3,-i c (a+b x))}{2 b}+\frac {i \operatorname {PolyLog}(3,i c (a+b x))}{2 b} \] Output:

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
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78 \[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\frac {i (\log (d (a+b x)) \operatorname {PolyLog}(2,-i c (a+b x))-\log (d (a+b x)) \operatorname {PolyLog}(2,i c (a+b x))-\operatorname {PolyLog}(3,-i c (a+b x))+\operatorname {PolyLog}(3,i c (a+b x)))}{2 b} \] Input:

Integrate[(ArcTan[c*(a + b*x)]*Log[d*(a + b*x)])/(a + b*x),x]
 

Output:

((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
 

Rubi [A] (verified)

Time = 0.68 (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

\(\displaystyle \frac {i (\operatorname {PolyLog}(3,i c (a+b x))-\operatorname {PolyLog}(2,i c (a+b x)) \log (d (a+b x)))}{2 b}-\frac {i (\operatorname {PolyLog}(3,-i c (a+b x))-\operatorname {PolyLog}(2,-i c (a+b x)) \log (d (a+b x)))}{2 b}\)

Input:

Int[(ArcTan[c*(a + b*x)]*Log[d*(a + b*x)])/(a + b*x),x]
 

Output:

((-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
 

Defintions of rubi rules used

rule 2821

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]
 

rule 2881

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]
 

rule 2894

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]])
 

rule 5732

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]
 

rule 7143

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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 6.99 (sec) , antiderivative size = 1087, normalized size of antiderivative = 10.76

method	result	size
risch	\(\text {Expression too large to display}\)	\(1087\)

Input:

int(arctan(c*(b*x+a))*ln(d*(b*x+a))/(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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)*csgn(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)-1/2*I*polylog(3,-I 
*c*(b*x+a))/b+1/2*I*polylog(3,I*c*(b*x+a))/b-1/4/b*ln(-I*(-c*(b*x+a)+I))*l 
n(-I*c*(b*x+a))*Pi*csgn(I*d)*csgn(I*(b*x+a))*csgn(I*d*(b*x+a))-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*l 
n(-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+1/2*I/b*ln(b*x+a)*polylog(2,-I*c*(b*x+a))-1/2*I/b*dilog(-I*c*(b*x+a))*l 
n(d)-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))-1/4/b*dilog(-I*c*(b*x+a))*Pi*csgn(I*d)*csgn(I*(b*x+a))*c 
sgn(I*d*(b*x+a))+1/4*I/b*ln(-I*(I+c*(b*x+a)))*ln(b*x+a)^2-1/4*I/b*ln(b*x+a 
)^2*ln(1-I*c*(b*x+a))-1/2*I/b*ln(d)*dilog(-I*(I+c*(b*x+a)))-1/2*I/b*ln(b*x 
+a)*polylog(2,I*c*(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...
 

Fricas [F]

\[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int { \frac {\arctan \left ({\left (b x + a\right )} c\right ) \log \left ({\left (b x + a\right )} d\right )}{b x + a} \,d x } \] Input:

integrate(arctan(c*(b*x+a))*log(d*(b*x+a))/(b*x+a),x, algorithm="fricas")
 

Output:

integral(arctan(b*c*x + a*c)*log(b*d*x + a*d)/(b*x + a), x)
 

Sympy [F]

\[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int \frac {\log {\left (a d + b d x \right )} \operatorname {atan}{\left (a c + b c x \right )}}{a + b x}\, dx \] Input:

integrate(atan(c*(b*x+a))*ln(d*(b*x+a))/(b*x+a),x)
 

Output:

Integral(log(a*d + b*d*x)*atan(a*c + b*c*x)/(a + b*x), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(arctan(c*(b*x+a))*log(d*(b*x+a))/(b*x+a),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [F]

\[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int { \frac {\arctan \left ({\left (b x + a\right )} c\right ) \log \left ({\left (b x + a\right )} d\right )}{b x + a} \,d x } \] Input:

integrate(arctan(c*(b*x+a))*log(d*(b*x+a))/(b*x+a),x, algorithm="giac")
 

Output:

integrate(arctan((b*x + a)*c)*log((b*x + a)*d)/(b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int \frac {\mathrm {atan}\left (c\,\left (a+b\,x\right )\right )\,\ln \left (d\,\left (a+b\,x\right )\right )}{a+b\,x} \,d x \] Input:

int((atan(c*(a + b*x))*log(d*(a + b*x)))/(a + b*x),x)
 

Output:

int((atan(c*(a + b*x))*log(d*(a + b*x)))/(a + b*x), x)
 

Reduce [F]

\[ \int \frac {\arctan (c (a+b x)) \log (d (a+b x))}{a+b x} \, dx=\int \frac {\mathit {atan} \left (b c x +a c \right ) \mathrm {log}\left (b d x +a d \right )}{b x +a}d x \] Input:

int(atan(c*(b*x+a))*log(d*(b*x+a))/(b*x+a),x)
 

Output:

int((atan(a*c + b*c*x)*log(a*d + b*d*x))/(a + b*x),x)