\(\int \frac {(a+b \arctan (c x^n)) (d+e \log (f x^m))}{x} \, dx\) [153]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int \frac {\left (a+b \arctan \left (c x^n\right )\right ) \left (d+e \log \left (f x^m\right )\right )}{x} \, dx=a d \log (x)+\frac {a e \log ^2\left (f x^m\right )}{2 m}+\frac {i b d \operatorname {PolyLog}\left (2,-i c x^n\right )}{2 n}+\frac {i b e \log \left (f x^m\right ) \operatorname {PolyLog}\left (2,-i c x^n\right )}{2 n}-\frac {i b d \operatorname {PolyLog}\left (2,i c x^n\right )}{2 n}-\frac {i b e \log \left (f x^m\right ) \operatorname {PolyLog}\left (2,i c x^n\right )}{2 n}-\frac {i b e m \operatorname {PolyLog}\left (3,-i c x^n\right )}{2 n^2}+\frac {i b e m \operatorname {PolyLog}\left (3,i c x^n\right )}{2 n^2} \] Output:

a*d*ln(x)+1/2*a*e*ln(f*x^m)^2/m+1/2*I*b*d*polylog(2,-I*c*x^n)/n+1/2*I*b*e* 
ln(f*x^m)*polylog(2,-I*c*x^n)/n-1/2*I*b*d*polylog(2,I*c*x^n)/n-1/2*I*b*e*l 
n(f*x^m)*polylog(2,I*c*x^n)/n-1/2*I*b*e*m*polylog(3,-I*c*x^n)/n^2+1/2*I*b* 
e*m*polylog(3,I*c*x^n)/n^2
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int \frac {\left (a+b \arctan \left (c x^n\right )\right ) \left (d+e \log \left (f x^m\right )\right )}{x} \, dx=-\frac {b c e m x^n \, _4F_3\left (\frac {1}{2},\frac {1}{2},\frac {1}{2},1;\frac {3}{2},\frac {3}{2},\frac {3}{2};-c^2 x^{2 n}\right )}{n^2}+\frac {b c x^n \, _3F_2\left (\frac {1}{2},\frac {1}{2},1;\frac {3}{2},\frac {3}{2};-c^2 x^{2 n}\right ) \left (d+e \log \left (f x^m\right )\right )}{n}+\frac {1}{2} a \log (x) \left (2 d-e m \log (x)+2 e \log \left (f x^m\right )\right ) \] Input:

Integrate[((a + b*ArcTan[c*x^n])*(d + e*Log[f*x^m]))/x,x]
 

Output:

-((b*c*e*m*x^n*HypergeometricPFQ[{1/2, 1/2, 1/2, 1}, {3/2, 3/2, 3/2}, -(c^ 
2*x^(2*n))])/n^2) + (b*c*x^n*HypergeometricPFQ[{1/2, 1/2, 1}, {3/2, 3/2}, 
-(c^2*x^(2*n))]*(d + e*Log[f*x^m]))/n + (a*Log[x]*(2*d - e*m*Log[x] + 2*e* 
Log[f*x^m]))/2
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {7293, 2009}

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.

\(\displaystyle \int \frac {\left (a+b \arctan \left (c x^n\right )\right ) \left (d+e \log \left (f x^m\right )\right )}{x} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {d \left (a+b \arctan \left (c x^n\right )\right )}{x}+\frac {e \log \left (f x^m\right ) \left (a+b \arctan \left (c x^n\right )\right )}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle a d \log (x)+\frac {a e \log ^2\left (f x^m\right )}{2 m}+\frac {i b d \operatorname {PolyLog}\left (2,-i c x^n\right )}{2 n}-\frac {i b d \operatorname {PolyLog}\left (2,i c x^n\right )}{2 n}+\frac {i b e \operatorname {PolyLog}\left (2,-i c x^n\right ) \log \left (f x^m\right )}{2 n}-\frac {i b e \operatorname {PolyLog}\left (2,i c x^n\right ) \log \left (f x^m\right )}{2 n}-\frac {i b e m \operatorname {PolyLog}\left (3,-i c x^n\right )}{2 n^2}+\frac {i b e m \operatorname {PolyLog}\left (3,i c x^n\right )}{2 n^2}\)

Input:

Int[((a + b*ArcTan[c*x^n])*(d + e*Log[f*x^m]))/x,x]
 

Output:

a*d*Log[x] + (a*e*Log[f*x^m]^2)/(2*m) + ((I/2)*b*d*PolyLog[2, (-I)*c*x^n]) 
/n + ((I/2)*b*e*Log[f*x^m]*PolyLog[2, (-I)*c*x^n])/n - ((I/2)*b*d*PolyLog[ 
2, I*c*x^n])/n - ((I/2)*b*e*Log[f*x^m]*PolyLog[2, I*c*x^n])/n - ((I/2)*b*e 
*m*PolyLog[3, (-I)*c*x^n])/n^2 + ((I/2)*b*e*m*PolyLog[3, I*c*x^n])/n^2
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [C] (warning: unable to verify)

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

Time = 159.48 (sec) , antiderivative size = 547, normalized size of antiderivative = 3.36

method	result	size
risch	\(\frac {\left (-\frac {i e \pi \,\operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x^{m}\right ) \operatorname {csgn}\left (i f \,x^{m}\right )}{4}+\frac {i e \pi \,\operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i f \,x^{m}\right )^{2}}{4}+\frac {i e \pi \,\operatorname {csgn}\left (i x^{m}\right ) \operatorname {csgn}\left (i f \,x^{m}\right )^{2}}{4}-\frac {i e \pi \operatorname {csgn}\left (i f \,x^{m}\right )^{3}}{4}+\frac {e \ln \left (f \right )}{2}+\frac {d}{2}\right ) \left (-i b \operatorname {dilog}\left (1-i c \,x^{n}\right )+2 \ln \left (x^{n}\right ) a +i b \operatorname {dilog}\left (1+i c \,x^{n}\right )\right )}{n}-\frac {i e b \ln \left (-i \left (-c \,x^{n}+i\right )\right ) \ln \left (-i c \,x^{n}\right ) \ln \left (x^{m}\right )}{2 n}-\frac {i b e m \operatorname {polylog}\left (3, -i c \,x^{n}\right )}{2 n^{2}}-\frac {i e b \ln \left (x \right ) \ln \left (-i \left (c \,x^{n}+i\right )\right ) \ln \left (x^{m}\right )}{2}+\frac {i e b \ln \left (x \right ) \ln \left (1-i c \,x^{n}\right ) \ln \left (x^{m}\right )}{2}+\frac {i e b \ln \left (x \right )^{2} \ln \left (-i \left (c \,x^{n}+i\right )\right ) m}{2}-\frac {i e b \ln \left (x \right )^{2} \ln \left (1-i c \,x^{n}\right ) m}{2}+\frac {i e b \operatorname {dilog}\left (-i \left (c \,x^{n}+i\right )\right ) m \ln \left (x \right )}{2 n}+\frac {i b e m \operatorname {polylog}\left (3, i c \,x^{n}\right )}{2 n^{2}}+\frac {e a \ln \left (x^{m}\right )^{2}}{2 m}+\frac {i e b \ln \left (x \right ) \ln \left (-i \left (-c \,x^{n}+i\right )\right ) \ln \left (x^{m}\right )}{2}+\frac {i e b m \ln \left (x \right ) \operatorname {polylog}\left (2, -i c \,x^{n}\right )}{2 n}+\frac {i e b \ln \left (x \right )^{2} \ln \left (1+i c \,x^{n}\right ) m}{2}-\frac {i e b m \ln \left (x \right ) \operatorname {polylog}\left (2, i c \,x^{n}\right )}{2 n}+\frac {i e b \ln \left (-i \left (-c \,x^{n}+i\right )\right ) \ln \left (-i c \,x^{n}\right ) m \ln \left (x \right )}{2 n}+\frac {i e b \operatorname {dilog}\left (-i c \,x^{n}\right ) m \ln \left (x \right )}{2 n}-\frac {i e b \ln \left (x \right ) \ln \left (1+i c \,x^{n}\right ) \ln \left (x^{m}\right )}{2}-\frac {i e b \operatorname {dilog}\left (-i c \,x^{n}\right ) \ln \left (x^{m}\right )}{2 n}-\frac {i e b \ln \left (x \right )^{2} \ln \left (-i \left (-c \,x^{n}+i\right )\right ) m}{2}-\frac {i e b \operatorname {dilog}\left (-i \left (c \,x^{n}+i\right )\right ) \ln \left (x^{m}\right )}{2 n}\)	\(547\)

Input:

int((a+b*arctan(c*x^n))*(d+e*ln(f*x^m))/x,x,method=_RETURNVERBOSE)
 

Output:

(-1/4*I*e*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/4*I*e*Pi*csgn(I*f)*csgn 
(I*f*x^m)^2+1/4*I*e*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/4*I*e*Pi*csgn(I*f*x^m 
)^3+1/2*e*ln(f)+1/2*d)/n*(-I*b*dilog(1-I*c*x^n)+2*ln(x^n)*a+I*b*dilog(1+I* 
c*x^n))-1/2*I*e*b/n*ln(-I*(-c*x^n+I))*ln(-I*c*x^n)*ln(x^m)-1/2*I*b*e*m*pol 
ylog(3,-I*c*x^n)/n^2-1/2*I*e*b*ln(x)*ln(-I*(c*x^n+I))*ln(x^m)+1/2*I*e*b*ln 
(x)*ln(1-I*c*x^n)*ln(x^m)+1/2*I*e*b*ln(x)^2*ln(-I*(c*x^n+I))*m-1/2*I*e*b*l 
n(x)^2*ln(1-I*c*x^n)*m+1/2*I*e*b/n*dilog(-I*(c*x^n+I))*m*ln(x)+1/2*I*b*e*m 
*polylog(3,I*c*x^n)/n^2+1/2*e*a/m*ln(x^m)^2+1/2*I*e*b*ln(x)*ln(-I*(-c*x^n+ 
I))*ln(x^m)+1/2*I*e*b*m/n*ln(x)*polylog(2,-I*c*x^n)+1/2*I*e*b*ln(x)^2*ln(1 
+I*c*x^n)*m-1/2*I*e*b*m/n*ln(x)*polylog(2,I*c*x^n)+1/2*I*e*b/n*ln(-I*(-c*x 
^n+I))*ln(-I*c*x^n)*m*ln(x)+1/2*I*e*b/n*dilog(-I*c*x^n)*m*ln(x)-1/2*I*e*b* 
ln(x)*ln(1+I*c*x^n)*ln(x^m)-1/2*I*e*b/n*dilog(-I*c*x^n)*ln(x^m)-1/2*I*e*b* 
ln(x)^2*ln(-I*(-c*x^n+I))*m-1/2*I*e*b/n*dilog(-I*(c*x^n+I))*ln(x^m)
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (121) = 242\).

Time = 0.13 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a+b \arctan \left (c x^n\right )\right ) \left (d+e \log \left (f x^m\right )\right )}{x} \, dx=\frac {2 \, a e m n^{2} \log \left (x\right )^{2} + 2 i \, b e m {\rm polylog}\left (3, i \, c x^{n}\right ) - 2 i \, b e m {\rm polylog}\left (3, -i \, c x^{n}\right ) + 2 \, {\left (b e m n^{2} \log \left (x\right )^{2} + 2 \, {\left (b e n^{2} \log \left (f\right ) + b d n^{2}\right )} \log \left (x\right )\right )} \arctan \left (c x^{n}\right ) - 2 \, {\left (i \, b e m n \log \left (x\right ) + i \, b e n \log \left (f\right ) + i \, b d n\right )} {\rm Li}_2\left (i \, c x^{n}\right ) - 2 \, {\left (-i \, b e m n \log \left (x\right ) - i \, b e n \log \left (f\right ) - i \, b d n\right )} {\rm Li}_2\left (-i \, c x^{n}\right ) + {\left (i \, b e m n^{2} \log \left (x\right )^{2} - 2 \, {\left (-i \, b e n^{2} \log \left (f\right ) - i \, b d n^{2}\right )} \log \left (x\right )\right )} \log \left (i \, c x^{n} + 1\right ) + {\left (-i \, b e m n^{2} \log \left (x\right )^{2} - 2 \, {\left (i \, b e n^{2} \log \left (f\right ) + i \, b d n^{2}\right )} \log \left (x\right )\right )} \log \left (-i \, c x^{n} + 1\right ) + 4 \, {\left (a e n^{2} \log \left (f\right ) + a d n^{2}\right )} \log \left (x\right )}{4 \, n^{2}} \] Input:

integrate((a+b*arctan(c*x^n))*(d+e*log(f*x^m))/x,x, algorithm="fricas")
 

Output:

1/4*(2*a*e*m*n^2*log(x)^2 + 2*I*b*e*m*polylog(3, I*c*x^n) - 2*I*b*e*m*poly 
log(3, -I*c*x^n) + 2*(b*e*m*n^2*log(x)^2 + 2*(b*e*n^2*log(f) + b*d*n^2)*lo 
g(x))*arctan(c*x^n) - 2*(I*b*e*m*n*log(x) + I*b*e*n*log(f) + I*b*d*n)*dilo 
g(I*c*x^n) - 2*(-I*b*e*m*n*log(x) - I*b*e*n*log(f) - I*b*d*n)*dilog(-I*c*x 
^n) + (I*b*e*m*n^2*log(x)^2 - 2*(-I*b*e*n^2*log(f) - I*b*d*n^2)*log(x))*lo 
g(I*c*x^n + 1) + (-I*b*e*m*n^2*log(x)^2 - 2*(I*b*e*n^2*log(f) + I*b*d*n^2) 
*log(x))*log(-I*c*x^n + 1) + 4*(a*e*n^2*log(f) + a*d*n^2)*log(x))/n^2
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \arctan \left (c x^n\right )\right ) \left (d+e \log \left (f x^m\right )\right )}{x} \, dx=\text {Timed out} \] Input:

integrate((a+b*atan(c*x**n))*(d+e*ln(f*x**m))/x,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*arctan(c*x^n))*(d+e*log(f*x^m))/x,x, algorithm="maxima")
 

Output:

1/2*a*e*log(f*x^m)^2/m + a*d*log(x) - 1/2*(b*e*m*log(x)^2 - 2*b*e*log(x)*l 
og(x^m) - 2*(b*e*log(f) + b*d)*log(x))*arctan(c*x^n) - integrate(-1/2*(b*c 
*e*m*n*x^n*log(x)^2 - 2*b*c*e*n*x^n*log(x)*log(x^m) - 2*(b*c*e*log(f) + b* 
c*d)*n*x^n*log(x))/(c^2*x*x^(2*n) + x), x)
 

Giac [F]

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

integrate((a+b*arctan(c*x^n))*(d+e*log(f*x^m))/x,x, algorithm="giac")
 

Output:

integrate((b*arctan(c*x^n) + a)*(e*log(f*x^m) + d)/x, x)
 

Mupad [F(-1)]

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

int(((a + b*atan(c*x^n))*(d + e*log(f*x^m)))/x,x)
 

Output:

int(((a + b*atan(c*x^n))*(d + e*log(f*x^m)))/x, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \arctan \left (c x^n\right )\right ) \left (d+e \log \left (f x^m\right )\right )}{x} \, dx=\frac {2 \left (\int \frac {\mathit {atan} \left (x^{n} c \right )}{x}d x \right ) b d m +2 \left (\int \frac {\mathit {atan} \left (x^{n} c \right ) \mathrm {log}\left (x^{m} f \right )}{x}d x \right ) b e m +\mathrm {log}\left (x^{m} f \right )^{2} a e +2 \,\mathrm {log}\left (x \right ) a d m}{2 m} \] Input:

int((a+b*atan(c*x^n))*(d+e*log(f*x^m))/x,x)
 

Output:

(2*int(atan(x**n*c)/x,x)*b*d*m + 2*int((atan(x**n*c)*log(x**m*f))/x,x)*b*e 
*m + log(x**m*f)**2*a*e + 2*log(x)*a*d*m)/(2*m)