\(\int \frac {(a+b \arctan (c x)) (d+e \log (1+c^2 x^2))}{x^3} \, dx\) [1291]

Optimal result

Integrand size = 26, antiderivative size = 154 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=b c^2 e \arctan (c x)+a c^2 e \log (x)-\frac {1}{2} a c^2 e \log \left (1+c^2 x^2\right )-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,i c x) \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=-\frac {a d+b c d x+b d \arctan (c x)+b c^2 d x^2 \arctan (c x)-2 b c^2 e x^2 \arctan (c x)-2 a c^2 e x^2 \log (x)+a e \log \left (1+c^2 x^2\right )+b c e x \log \left (1+c^2 x^2\right )+a c^2 e x^2 \log \left (1+c^2 x^2\right )+b e \arctan (c x) \log \left (1+c^2 x^2\right )+b c^2 e x^2 \arctan (c x) \log \left (1+c^2 x^2\right )-i b c^2 e x^2 \operatorname {PolyLog}(2,-i c x)+i b c^2 e x^2 \operatorname {PolyLog}(2,i c x)}{2 x^2} \] Input:

Integrate[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^3,x]
 

Output:

-1/2*(a*d + b*c*d*x + b*d*ArcTan[c*x] + b*c^2*d*x^2*ArcTan[c*x] - 2*b*c^2* 
e*x^2*ArcTan[c*x] - 2*a*c^2*e*x^2*Log[x] + a*e*Log[1 + c^2*x^2] + b*c*e*x* 
Log[1 + c^2*x^2] + a*c^2*e*x^2*Log[1 + c^2*x^2] + b*e*ArcTan[c*x]*Log[1 + 
c^2*x^2] + b*c^2*e*x^2*ArcTan[c*x]*Log[1 + c^2*x^2] - I*b*c^2*e*x^2*PolyLo 
g[2, (-I)*c*x] + I*b*c^2*e*x^2*PolyLog[2, I*c*x])/x^2
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {5556, 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.

\(\displaystyle \int \frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^3} \, dx\)

\(\Big \downarrow \) 5556

\(\displaystyle -2 c^2 e \int \left (-\frac {a+b c x}{2 x \left (c^2 x^2+1\right )}-\frac {b \arctan (c x)}{2 x}\right )dx-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{2 x^2}-\frac {1}{2} b c^2 \arctan (c x) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{2 x}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{2 x^2}-2 c^2 e \left (\frac {1}{4} a \log \left (c^2 x^2+1\right )-\frac {1}{2} a \log (x)-\frac {1}{2} b \arctan (c x)-\frac {1}{4} i b \operatorname {PolyLog}(2,-i c x)+\frac {1}{4} i b \operatorname {PolyLog}(2,i c x)\right )-\frac {1}{2} b c^2 \arctan (c x) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{2 x}\)

Input:

Int[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^3,x]
 

Output:

-1/2*(b*c*(d + e*Log[1 + c^2*x^2]))/x - (b*c^2*ArcTan[c*x]*(d + e*Log[1 + 
c^2*x^2]))/2 - ((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/(2*x^2) - 2* 
c^2*e*(-1/2*(b*ArcTan[c*x]) - (a*Log[x])/2 + (a*Log[1 + c^2*x^2])/4 - (I/4 
)*b*PolyLog[2, (-I)*c*x] + (I/4)*b*PolyLog[2, I*c*x])
 

Defintions of rubi rules used

rule 2009

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

rule 5556

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*( 
e_.))*(x_)^(m_.), x_Symbol] :> With[{u = IntHide[x^m*(a + b*ArcTan[c*x]), x 
]}, Simp[(d + e*Log[f + g*x^2])   u, x] - Simp[2*e*g   Int[ExpandIntegrand[ 
x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Intege 
rQ[m] && NeQ[m, -1]
 

Maple [F]

\[\int \frac {\left (a +b \arctan \left (c x \right )\right ) \left (d +e \ln \left (c^{2} x^{2}+1\right )\right )}{x^{3}}d x\]

Input:

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^3,x)
 

Output:

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^3,x)
 

Fricas [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}} \,d x } \] Input:

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^3,x, algorithm="fricas" 
)
 

Output:

integral((b*d*arctan(c*x) + a*d + (b*e*arctan(c*x) + a*e)*log(c^2*x^2 + 1) 
)/x^3, x)
 

Sympy [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e \log {\left (c^{2} x^{2} + 1 \right )}\right )}{x^{3}}\, dx \] Input:

integrate((a+b*atan(c*x))*(d+e*ln(c**2*x**2+1))/x**3,x)
 

Output:

Integral((a + b*atan(c*x))*(d + e*log(c**2*x**2 + 1))/x**3, x)
 

Maxima [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}} \,d x } \] Input:

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^3,x, algorithm="maxima" 
)
 

Output:

-1/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*d - 1/2*(c^2*(log(c^2*x 
^2 + 1) - log(x^2)) + log(c^2*x^2 + 1)/x^2)*a*e + 1/2*(4*c^4*x^2*integrate 
(1/2*x*arctan(c*x)/(c^2*x^2 + 1), x) + 2*c^2*x^2*arctan(c*x) + 4*c^2*x^2*i 
ntegrate(1/2*arctan(c*x)/(c^2*x^3 + x), x) - (c*x + (c^2*x^2 + 1)*arctan(c 
*x))*log(c^2*x^2 + 1))*b*e/x^2 - 1/2*a*d/x^2
 

Giac [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}} \,d x } \] Input:

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^3,x, algorithm="giac")
 

Output:

integrate((b*arctan(c*x) + a)*(e*log(c^2*x^2 + 1) + d)/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (c^2\,x^2+1\right )\right )}{x^3} \,d x \] Input:

int(((a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)))/x^3,x)
 

Output:

int(((a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)))/x^3, x)
 

Reduce [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\frac {-\mathit {atan} \left (c x \right ) b \,c^{2} d \,x^{2}-\mathit {atan} \left (c x \right ) b d +2 \left (\int \frac {\mathit {atan} \left (c x \right ) \mathrm {log}\left (c^{2} x^{2}+1\right )}{x^{3}}d x \right ) b e \,x^{2}-\mathrm {log}\left (c^{2} x^{2}+1\right ) a \,c^{2} e \,x^{2}-\mathrm {log}\left (c^{2} x^{2}+1\right ) a e +2 \,\mathrm {log}\left (x \right ) a \,c^{2} e \,x^{2}-a d -b c d x}{2 x^{2}} \] Input:

int((a+b*atan(c*x))*(d+e*log(c^2*x^2+1))/x^3,x)
 

Output:

( - atan(c*x)*b*c**2*d*x**2 - atan(c*x)*b*d + 2*int((atan(c*x)*log(c**2*x* 
*2 + 1))/x**3,x)*b*e*x**2 - log(c**2*x**2 + 1)*a*c**2*e*x**2 - log(c**2*x* 
*2 + 1)*a*e + 2*log(x)*a*c**2*e*x**2 - a*d - b*c*d*x)/(2*x**2)