3.13.95 \(\int \frac {(a+b \arctan (c x)) (d+e \log (1+c^2 x^2))}{x^5} \, dx\) [1295]

3.13.95.1 Optimal result

Integrand size = 26, antiderivative size = 225 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=-\frac {a c^2 e}{4 x^2}-\frac {5 b c^3 e}{12 x}-\frac {11}{12} b c^4 e \arctan (c x)-\frac {b c^2 e \arctan (c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{4} a c^4 e \log \left (1+c^2 x^2\right )-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,-i c x)+\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,i c x) \]

output

-1/4*a*c^2*e/x^2-5/12*b*c^3*e/x-11/12*b*c^4*e*arctan(c*x)-1/4*b*c^2*e*arct 
an(c*x)/x^2-1/2*a*c^4*e*ln(x)+1/4*a*c^4*e*ln(c^2*x^2+1)-1/12*b*c*(d+e*ln(c 
^2*x^2+1))/x^3+1/4*b*c^3*(d+e*ln(c^2*x^2+1))/x+1/4*b*c^4*arctan(c*x)*(d+e* 
ln(c^2*x^2+1))-1/4*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^4-1/4*I*b*c^4*e 
*polylog(2,-I*c*x)+1/4*I*b*c^4*e*polylog(2,I*c*x)
 

3.13.95.2 Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=-\frac {3 a d+b c d x+3 a c^2 e x^2-3 b c^3 d x^3+5 b c^3 e x^3+3 b d \arctan (c x)+3 b c^2 e x^2 \arctan (c x)-3 b c^4 d x^4 \arctan (c x)+11 b c^4 e x^4 \arctan (c x)+6 a c^4 e x^4 \log (x)+3 a e \log \left (1+c^2 x^2\right )+b c e x \log \left (1+c^2 x^2\right )-3 b c^3 e x^3 \log \left (1+c^2 x^2\right )-3 a c^4 e x^4 \log \left (1+c^2 x^2\right )+3 b e \arctan (c x) \log \left (1+c^2 x^2\right )-3 b c^4 e x^4 \arctan (c x) \log \left (1+c^2 x^2\right )+3 i b c^4 e x^4 \operatorname {PolyLog}(2,-i c x)-3 i b c^4 e x^4 \operatorname {PolyLog}(2,i c x)}{12 x^4} \]

input

Integrate[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^5,x]
 

output

-1/12*(3*a*d + b*c*d*x + 3*a*c^2*e*x^2 - 3*b*c^3*d*x^3 + 5*b*c^3*e*x^3 + 3 
*b*d*ArcTan[c*x] + 3*b*c^2*e*x^2*ArcTan[c*x] - 3*b*c^4*d*x^4*ArcTan[c*x] + 
 11*b*c^4*e*x^4*ArcTan[c*x] + 6*a*c^4*e*x^4*Log[x] + 3*a*e*Log[1 + c^2*x^2 
] + b*c*e*x*Log[1 + c^2*x^2] - 3*b*c^3*e*x^3*Log[1 + c^2*x^2] - 3*a*c^4*e* 
x^4*Log[1 + c^2*x^2] + 3*b*e*ArcTan[c*x]*Log[1 + c^2*x^2] - 3*b*c^4*e*x^4* 
ArcTan[c*x]*Log[1 + c^2*x^2] + (3*I)*b*c^4*e*x^4*PolyLog[2, (-I)*c*x] - (3 
*I)*b*c^4*e*x^4*PolyLog[2, I*c*x])/x^4
 

3.13.95.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.96, 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.

\(\displaystyle \int \frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^5} \, dx\)

\(\Big \downarrow \) 5556

\(\displaystyle -2 c^2 e \int \left (-\frac {-3 b c^3 x^3+b c x+3 a}{12 x^3 \left (c^2 x^2+1\right )}-\frac {b \left (1-c^2 x^2\right ) \arctan (c x)}{4 x^3}\right )dx-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x^4}+\frac {1}{4} b c^4 \arctan (c x) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{12 x^3}+\frac {b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x^4}-2 c^2 e \left (-\frac {1}{8} a c^2 \log \left (c^2 x^2+1\right )+\frac {1}{4} a c^2 \log (x)+\frac {a}{8 x^2}+\frac {11}{24} b c^2 \arctan (c x)+\frac {b \arctan (c x)}{8 x^2}+\frac {1}{8} i b c^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{8} i b c^2 \operatorname {PolyLog}(2,i c x)+\frac {5 b c}{24 x}\right )+\frac {1}{4} b c^4 \arctan (c x) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{12 x^3}+\frac {b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x}\)

input

Int[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^5,x]
 

output

-1/12*(b*c*(d + e*Log[1 + c^2*x^2]))/x^3 + (b*c^3*(d + e*Log[1 + c^2*x^2]) 
)/(4*x) + (b*c^4*ArcTan[c*x]*(d + e*Log[1 + c^2*x^2]))/4 - ((a + b*ArcTan[ 
c*x])*(d + e*Log[1 + c^2*x^2]))/(4*x^4) - 2*c^2*e*(a/(8*x^2) + (5*b*c)/(24 
*x) + (11*b*c^2*ArcTan[c*x])/24 + (b*ArcTan[c*x])/(8*x^2) + (a*c^2*Log[x]) 
/4 - (a*c^2*Log[1 + c^2*x^2])/8 + (I/8)*b*c^2*PolyLog[2, (-I)*c*x] - (I/8) 
*b*c^2*PolyLog[2, I*c*x])
 

3.13.95.3.1 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]
 

3.13.95.4 Maple [F]

\[\int \frac {\left (a +b \arctan \left (c x \right )\right ) \left (d +e \ln \left (c^{2} x^{2}+1\right )\right )}{x^{5}}d x\]

input

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^5,x)
 

output

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^5,x)
 

3.13.95.5 Fricas [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \]

input

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^5,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^5, x)
 

3.13.95.6 Sympy [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e \log {\left (c^{2} x^{2} + 1 \right )}\right )}{x^{5}}\, dx \]

input

integrate((a+b*atan(c*x))*(d+e*ln(c**2*x**2+1))/x**5,x)
 

output

Integral((a + b*atan(c*x))*(d + e*log(c**2*x**2 + 1))/x**5, x)
 

3.13.95.7 Maxima [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \]

input

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^5,x, algorithm="maxima" 
)
 

output

1/12*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d 
 + 1/4*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c^2 - log(c^2*x^2 + 
1)/x^4)*a*e - 1/12*(72*c^6*x^4*integrate(1/12*x*arctan(c*x)/(c^2*x^2 + 1), 
 x) + 8*c^4*x^4*arctan(c*x) - 72*c^2*x^4*integrate(1/12*arctan(c*x)/(c^2*x 
^5 + x^3), x) + 2*c^3*x^3 - (3*c^3*x^3 - c*x + 3*(c^4*x^4 - 1)*arctan(c*x) 
)*log(c^2*x^2 + 1))*b*e/x^4 - 1/4*a*d/x^4
 

3.13.95.8 Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\text {Timed out} \]

input

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^5,x, algorithm="giac")
 

output

Timed out
 

3.13.95.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (c^2\,x^2+1\right )\right )}{x^5} \,d x \]

input

int(((a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)))/x^5,x)
 

output

int(((a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)))/x^5, x)