3.1.22 \(\int \frac {(a+b \arctan (c x))^2}{x^4} \, dx\) [22]

3.1.22.1 Optimal result

Integrand size = 14, antiderivative size = 140 \[ \int \frac {(a+b \arctan (c x))^2}{x^4} \, dx=-\frac {b^2 c^2}{3 x}-\frac {1}{3} b^2 c^3 \arctan (c x)-\frac {b c (a+b \arctan (c x))}{3 x^2}+\frac {1}{3} i c^3 (a+b \arctan (c x))^2-\frac {(a+b \arctan (c x))^2}{3 x^3}-\frac {2}{3} b c^3 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )+\frac {1}{3} i b^2 c^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right ) \]

output

-1/3*b^2*c^2/x-1/3*b^2*c^3*arctan(c*x)-1/3*b*c*(a+b*arctan(c*x))/x^2+1/3*I 
*c^3*(a+b*arctan(c*x))^2-1/3*(a+b*arctan(c*x))^2/x^3-2/3*b*c^3*(a+b*arctan 
(c*x))*ln(2-2/(1-I*c*x))+1/3*I*b^2*c^3*polylog(2,-1+2/(1-I*c*x))
 

3.1.22.2 Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b \arctan (c x))^2}{x^4} \, dx=-\frac {a^2+a b c x+b^2 c^2 x^2+b^2 \left (1-i c^3 x^3\right ) \arctan (c x)^2+b \arctan (c x) \left (2 a+b c x+b c^3 x^3+2 b c^3 x^3 \log \left (1-e^{2 i \arctan (c x)}\right )\right )+2 a b c^3 x^3 \log (c x)-a b c^3 x^3 \log \left (1+c^2 x^2\right )-i b^2 c^3 x^3 \operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )}{3 x^3} \]

input

Integrate[(a + b*ArcTan[c*x])^2/x^4,x]
 

output

-1/3*(a^2 + a*b*c*x + b^2*c^2*x^2 + b^2*(1 - I*c^3*x^3)*ArcTan[c*x]^2 + b* 
ArcTan[c*x]*(2*a + b*c*x + b*c^3*x^3 + 2*b*c^3*x^3*Log[1 - E^((2*I)*ArcTan 
[c*x])]) + 2*a*b*c^3*x^3*Log[c*x] - a*b*c^3*x^3*Log[1 + c^2*x^2] - I*b^2*c 
^3*x^3*PolyLog[2, E^((2*I)*ArcTan[c*x])])/x^3
 

3.1.22.3 Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5361, 5453, 5361, 264, 216, 5459, 5403, 2897}

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.

\(\displaystyle \int \frac {(a+b \arctan (c x))^2}{x^4} \, dx\)

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5453

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 5459

\(\displaystyle -\frac {(a+b \arctan (c x))^2}{3 x^3}+\frac {2}{3} b c \left (-\left (c^2 \left (i \int \frac {a+b \arctan (c x)}{x (c x+i)}dx-\frac {i (a+b \arctan (c x))^2}{2 b}\right )\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )\)

\(\Big \downarrow \) 5403

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

\(\Big \downarrow \) 2897

\(\displaystyle -\frac {(a+b \arctan (c x))^2}{3 x^3}+\frac {2}{3} b c \left (-\left (c^2 \left (i \left (-i \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )\right )-\frac {i (a+b \arctan (c x))^2}{2 b}\right )\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )\)

input

Int[(a + b*ArcTan[c*x])^2/x^4,x]
 

output

-1/3*(a + b*ArcTan[c*x])^2/x^3 + (2*b*c*(-1/2*(a + b*ArcTan[c*x])/x^2 + (b 
*c*(-x^(-1) - c*ArcTan[c*x]))/2 - c^2*(((-1/2*I)*(a + b*ArcTan[c*x])^2)/b 
+ I*((-I)*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)] - (b*PolyLog[2, -1 + 
2/(1 - I*c*x)])/2))))/3
 

3.1.22.3.1 Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 2897

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ 
D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && 
PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

rule 5361

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> 
 Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 
1))   Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], 
x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & 
& IntegerQ[m])) && NeQ[m, -1]
 

rule 5403

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_ 
Symbol] :> Simp[(a + b*ArcTan[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Si 
mp[b*c*(p/d)   Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))]/(1 
 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2* 
d^2 + e^2, 0]
 

rule 5453

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], 
 x] - Simp[e/(d*f^2)   Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/(d + e*x^2) 
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]
 

rule 5459

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), 
x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*d*(p + 1))), x] + Si 
mp[I/d   Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b, c, 
 d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]
 

3.1.22.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (122 ) = 244\).

Time = 2.91 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.21

method	result	size
parts	\(-\frac {a^{2}}{3 x^{3}}+b^{2} c^{3} \left (-\frac {\arctan \left (c x \right )^{2}}{3 c^{3} x^{3}}+\frac {\arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {\arctan \left (c x \right )}{3 c^{2} x^{2}}-\frac {2 \ln \left (c x \right ) \arctan \left (c x \right )}{3}+\frac {i \left (\ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )-\frac {\ln \left (c x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )-\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{6}-\frac {i \left (\ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )-\frac {\ln \left (c x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (c x -i\right )}{2}\right )-\ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )\right )}{6}-\frac {1}{3 c x}-\frac {\arctan \left (c x \right )}{3}-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{3}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{3}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{3}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{3}\right )+2 a b \,c^{3} \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {1}{6 c^{2} x^{2}}-\frac {\ln \left (c x \right )}{3}\right )\)	\(310\)
derivativedivides	\(c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}+b^{2} \left (-\frac {\arctan \left (c x \right )^{2}}{3 c^{3} x^{3}}+\frac {\arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {\arctan \left (c x \right )}{3 c^{2} x^{2}}-\frac {2 \ln \left (c x \right ) \arctan \left (c x \right )}{3}+\frac {i \left (\ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )-\frac {\ln \left (c x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )-\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{6}-\frac {i \left (\ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )-\frac {\ln \left (c x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (c x -i\right )}{2}\right )-\ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )\right )}{6}-\frac {1}{3 c x}-\frac {\arctan \left (c x \right )}{3}-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{3}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{3}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{3}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{3}\right )+2 a b \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {1}{6 c^{2} x^{2}}-\frac {\ln \left (c x \right )}{3}\right )\right )\)	\(311\)
default	\(c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}+b^{2} \left (-\frac {\arctan \left (c x \right )^{2}}{3 c^{3} x^{3}}+\frac {\arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {\arctan \left (c x \right )}{3 c^{2} x^{2}}-\frac {2 \ln \left (c x \right ) \arctan \left (c x \right )}{3}+\frac {i \left (\ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )-\frac {\ln \left (c x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )-\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{6}-\frac {i \left (\ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )-\frac {\ln \left (c x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (c x -i\right )}{2}\right )-\ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )\right )}{6}-\frac {1}{3 c x}-\frac {\arctan \left (c x \right )}{3}-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{3}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{3}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{3}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{3}\right )+2 a b \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {1}{6 c^{2} x^{2}}-\frac {\ln \left (c x \right )}{3}\right )\right )\)	\(311\)

input

int((a+b*arctan(c*x))^2/x^4,x,method=_RETURNVERBOSE)
 

output

-1/3*a^2/x^3+b^2*c^3*(-1/3/c^3/x^3*arctan(c*x)^2+1/3*arctan(c*x)*ln(c^2*x^ 
2+1)-1/3/c^2/x^2*arctan(c*x)-2/3*ln(c*x)*arctan(c*x)+1/6*I*(ln(c*x-I)*ln(c 
^2*x^2+1)-1/2*ln(c*x-I)^2-dilog(-1/2*I*(c*x+I))-ln(c*x-I)*ln(-1/2*I*(c*x+I 
)))-1/6*I*(ln(c*x+I)*ln(c^2*x^2+1)-1/2*ln(c*x+I)^2-dilog(1/2*I*(c*x-I))-ln 
(c*x+I)*ln(1/2*I*(c*x-I)))-1/3/c/x-1/3*arctan(c*x)-1/3*I*ln(c*x)*ln(1+I*c* 
x)+1/3*I*ln(c*x)*ln(1-I*c*x)-1/3*I*dilog(1+I*c*x)+1/3*I*dilog(1-I*c*x))+2* 
a*b*c^3*(-1/3/c^3/x^3*arctan(c*x)+1/6*ln(c^2*x^2+1)-1/6/c^2/x^2-1/3*ln(c*x 
))
 

3.1.22.5 Fricas [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^4} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{4}} \,d x } \]

input

integrate((a+b*arctan(c*x))^2/x^4,x, algorithm="fricas")
 

output

integral((b^2*arctan(c*x)^2 + 2*a*b*arctan(c*x) + a^2)/x^4, x)
 

3.1.22.6 Sympy [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^4} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{x^{4}}\, dx \]

input

integrate((a+b*atan(c*x))**2/x**4,x)
 

output

Integral((a + b*atan(c*x))**2/x**4, x)
 

3.1.22.7 Maxima [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^4} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{4}} \,d x } \]

input

integrate((a+b*arctan(c*x))^2/x^4,x, algorithm="maxima")
 

output

1/3*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)* 
a*b + 1/48*(48*x^3*integrate(-1/48*(4*c^2*x^2*log(c^2*x^2 + 1) - 8*c*x*arc 
tan(c*x) - 36*(c^2*x^2 + 1)*arctan(c*x)^2 - 3*(c^2*x^2 + 1)*log(c^2*x^2 + 
1)^2)/(c^2*x^6 + x^4), x) - 4*arctan(c*x)^2 + log(c^2*x^2 + 1)^2)*b^2/x^3 
- 1/3*a^2/x^3
 

3.1.22.8 Giac [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^4} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{4}} \,d x } \]

input

integrate((a+b*arctan(c*x))^2/x^4,x, algorithm="giac")
 

output

sage0*x
 

3.1.22.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x^4} \,d x \]

input

int((a + b*atan(c*x))^2/x^4,x)
 

output

int((a + b*atan(c*x))^2/x^4, x)