\(\int \frac {(a+b \arctan (c x))^2}{x^2 (d+e x)} \, dx\) [146]

Optimal result

Integrand size = 21, antiderivative size = 473 \[ \int \frac {(a+b \arctan (c x))^2}{x^2 (d+e x)} \, dx=-\frac {i c (a+b \arctan (c x))^2}{d}-\frac {(a+b \arctan (c x))^2}{d x}-\frac {2 e (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )}{d^2}-\frac {e (a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{d^2}+\frac {e (a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^2}+\frac {2 b c (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}+\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{d^2}-\frac {i b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d}+\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{d^2}-\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d^2}-\frac {i b e (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^2}-\frac {b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 d^2}+\frac {b^2 e \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 d^2}-\frac {b^2 e \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d^2}+\frac {b^2 e \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^2} \] Output:

-I*c*(a+b*arctan(c*x))^2/d-(a+b*arctan(c*x))^2/d/x+2*e*(a+b*arctan(c*x))^2 
*arctanh(-1+2/(1+I*c*x))/d^2-e*(a+b*arctan(c*x))^2*ln(2/(1-I*c*x))/d^2+e*( 
a+b*arctan(c*x))^2*ln(2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/d^2+2*b*c*(a+b*arct 
an(c*x))*ln(2-2/(1-I*c*x))/d+I*b*e*(a+b*arctan(c*x))*polylog(2,1-2/(1-I*c* 
x))/d^2-I*b^2*c*polylog(2,-1+2/(1-I*c*x))/d+I*b*e*(a+b*arctan(c*x))*polylo 
g(2,1-2/(1+I*c*x))/d^2-I*b*e*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))/d 
^2-I*b*e*(a+b*arctan(c*x))*polylog(2,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/d^ 
2-1/2*b^2*e*polylog(3,1-2/(1-I*c*x))/d^2+1/2*b^2*e*polylog(3,1-2/(1+I*c*x) 
)/d^2-1/2*b^2*e*polylog(3,-1+2/(1+I*c*x))/d^2+1/2*b^2*e*polylog(3,1-2*c*(e 
*x+d)/(c*d+I*e)/(1-I*c*x))/d^2
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 473, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5411, 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.

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

\(\Big \downarrow \) 5411

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 5411

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f* 
x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] & 
& IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
 

Maple [C] (warning: unable to verify)

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

Time = 52.06 (sec) , antiderivative size = 38040, normalized size of antiderivative = 80.42

method	result	size
parts	\(\text {Expression too large to display}\)	\(38040\)
derivativedivides	\(\text {Expression too large to display}\)	\(38050\)
default	\(\text {Expression too large to display}\)	\(38050\)

Input:

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

Output:

result too large to display
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x^2 (d+e x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

a^2*(e*log(e*x + d)/d^2 - e*log(x)/d^2 - 1/(d*x)) - 1/16*(4*b^2*arctan(c*x 
)^2 - b^2*log(c^2*x^2 + 1)^2 - 16*d*x*integrate(1/16*(12*(b^2*c^2*d*x^2 + 
b^2*d)*arctan(c*x)^2 + (b^2*c^2*d*x^2 + b^2*d)*log(c^2*x^2 + 1)^2 + 8*(b^2 
*c*d*x + 4*a*b*d + (4*a*b*c^2*d + b^2*c*e)*x^2)*arctan(c*x) - 4*(b^2*c^2*e 
*x^3 + b^2*c^2*d*x^2)*log(c^2*x^2 + 1))/(c^2*d*e*x^5 + c^2*d^2*x^4 + d*e*x 
^3 + d^2*x^2), x))/(d*x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^2 (d+e x)} \, dx=\frac {2 \left (\int \frac {\mathit {atan} \left (c x \right )}{c^{2} e \,x^{5}+c^{2} d \,x^{4}+e \,x^{3}+d \,x^{2}}d x \right ) a b \,d^{2} x +2 \left (\int \frac {\mathit {atan} \left (c x \right )}{c^{2} e \,x^{3}+c^{2} d \,x^{2}+e x +d}d x \right ) a b \,c^{2} d^{2} x +\left (\int \frac {\mathit {atan} \left (c x \right )^{2}}{c^{2} e \,x^{5}+c^{2} d \,x^{4}+e \,x^{3}+d \,x^{2}}d x \right ) b^{2} d^{2} x +\left (\int \frac {\mathit {atan} \left (c x \right )^{2}}{c^{2} e \,x^{3}+c^{2} d \,x^{2}+e x +d}d x \right ) b^{2} c^{2} d^{2} x +\mathrm {log}\left (e x +d \right ) a^{2} e x -\mathrm {log}\left (x \right ) a^{2} e x -a^{2} d}{d^{2} x} \] Input:

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

Output:

(2*int(atan(c*x)/(c**2*d*x**4 + c**2*e*x**5 + d*x**2 + e*x**3),x)*a*b*d**2 
*x + 2*int(atan(c*x)/(c**2*d*x**2 + c**2*e*x**3 + d + e*x),x)*a*b*c**2*d** 
2*x + int(atan(c*x)**2/(c**2*d*x**4 + c**2*e*x**5 + d*x**2 + e*x**3),x)*b* 
*2*d**2*x + int(atan(c*x)**2/(c**2*d*x**2 + c**2*e*x**3 + d + e*x),x)*b**2 
*c**2*d**2*x + log(d + e*x)*a**2*e*x - log(x)*a**2*e*x - a**2*d)/(d**2*x)