3.1.18 \(\int \frac {(a+b \arctan (c x))^3}{d+e x} \, dx\) [18]

3.1.18.1 Optimal result

Integrand size = 18, antiderivative size = 320 \[ \int \frac {(a+b \arctan (c x))^3}{d+e x} \, dx=-\frac {(a+b \arctan (c x))^3 \log \left (\frac {2}{1-i c x}\right )}{e}+\frac {(a+b \arctan (c x))^3 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e}+\frac {3 i b (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e}-\frac {3 i b (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\frac {3 b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e}+\frac {3 b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\frac {3 i b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1-i c x}\right )}{4 e}+\frac {3 i b^3 \operatorname {PolyLog}\left (4,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{4 e} \]

output

-(a+b*arctan(c*x))^3*ln(2/(1-I*c*x))/e+(a+b*arctan(c*x))^3*ln(2*c*(e*x+d)/ 
(c*d+I*e)/(1-I*c*x))/e+3/2*I*b*(a+b*arctan(c*x))^2*polylog(2,1-2/(1-I*c*x) 
)/e-3/2*I*b*(a+b*arctan(c*x))^2*polylog(2,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x 
))/e-3/2*b^2*(a+b*arctan(c*x))*polylog(3,1-2/(1-I*c*x))/e+3/2*b^2*(a+b*arc 
tan(c*x))*polylog(3,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/e-3/4*I*b^3*polylog 
(4,1-2/(1-I*c*x))/e+3/4*I*b^3*polylog(4,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x)) 
/e
 

3.1.18.2 Mathematica [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x))^3}{d+e x} \, dx=\text {\$Aborted} \]

input

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

output

$Aborted
 

3.1.18.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {5385}

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.

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

\(\Big \downarrow \) 5385

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

input

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

output

-(((a + b*ArcTan[c*x])^3*Log[2/(1 - I*c*x)])/e) + ((a + b*ArcTan[c*x])^3*L 
og[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e + (((3*I)/2)*b*(a + b*Arc 
Tan[c*x])^2*PolyLog[2, 1 - 2/(1 - I*c*x)])/e - (((3*I)/2)*b*(a + b*ArcTan[ 
c*x])^2*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e - (3* 
b^2*(a + b*ArcTan[c*x])*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*e) + (3*b^2*(a + 
 b*ArcTan[c*x])*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))]) 
/(2*e) - (((3*I)/4)*b^3*PolyLog[4, 1 - 2/(1 - I*c*x)])/e + (((3*I)/4)*b^3* 
PolyLog[4, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e
 

3.1.18.3.1 Defintions of rubi rules used

rule 5385

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^3/((d_) + (e_.)*(x_)), x_Symbol] :> 
Simp[(-(a + b*ArcTan[c*x])^3)*(Log[2/(1 - I*c*x)]/e), x] + (Simp[(a + b*Arc 
Tan[c*x])^3*(Log[2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/e), x] + Simp[3 
*I*b*(a + b*ArcTan[c*x])^2*(PolyLog[2, 1 - 2/(1 - I*c*x)]/(2*e)), x] - Simp 
[3*I*b*(a + b*ArcTan[c*x])^2*(PolyLog[2, 1 - 2*c*((d + e*x)/((c*d + I*e)*(1 
 - I*c*x)))]/(2*e)), x] - Simp[3*b^2*(a + b*ArcTan[c*x])*(PolyLog[3, 1 - 2/ 
(1 - I*c*x)]/(2*e)), x] + Simp[3*b^2*(a + b*ArcTan[c*x])*(PolyLog[3, 1 - 2* 
c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(2*e)), x] - Simp[3*I*b^3*(PolyLog 
[4, 1 - 2/(1 - I*c*x)]/(4*e)), x] + Simp[3*I*b^3*(PolyLog[4, 1 - 2*c*((d + 
e*x)/((c*d + I*e)*(1 - I*c*x)))]/(4*e)), x]) /; FreeQ[{a, b, c, d, e}, x] & 
& NeQ[c^2*d^2 + e^2, 0]
 

3.1.18.4 Maple [C] (warning: unable to verify)

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

Time = 18.20 (sec) , antiderivative size = 2398, normalized size of antiderivative = 7.49

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2398\)
default	\(\text {Expression too large to display}\)	\(2398\)
parts	\(\text {Expression too large to display}\)	\(2405\)

input

int((a+b*arctan(c*x))^3/(e*x+d),x,method=_RETURNVERBOSE)
 

output

1/c*(a^3*c*ln(c*e*x+c*d)/e+b^3*c*(ln(c*e*x+c*d)/e*arctan(c*x)^3-3/e*(1/3*a 
rctan(c*x)^3*ln(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I 
*e+c*d)-1/6*I*Pi*csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2 
*x^2+1)+I*e+c*d)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*(csgn(I*(-I*e*(1+I*c*x)^2/(c 
^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+c*d))*csgn(I/(1+(1+I*c*x)^2/(c^2 
*x^2+1)))-csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1) 
+I*e+c*d)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1))) 
-csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+c*d) 
)*csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+c*d 
)/(1+(1+I*c*x)^2/(c^2*x^2+1)))+csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1 
+I*c*x)^2/(c^2*x^2+1)+I*e+c*d)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2)*arctan(c*x) 
^3-1/2*I*arctan(c*x)^2*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+1/2*arctan(c*x) 
*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+1/4*I*polylog(4,-(1+I*c*x)^2/(c^2*x^2 
+1))-1/3*c*d/(c*d-I*e)*arctan(c*x)^3*ln(1-(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/ 
(c^2*x^2+1))-1/2*c*d/(c*d-I*e)*arctan(c*x)*polylog(3,(I*e-c*d)/(c*d+I*e)*( 
1+I*c*x)^2/(c^2*x^2+1))+1/2*I*c*d/(c*d-I*e)*arctan(c*x)^2*polylog(2,(I*e-c 
*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))-1/4*I*c*d/(c*d-I*e)*polylog(4,(I*e- 
c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))-1/3*e*arctan(c*x)^3*ln(1-(I*e-c*d) 
/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))/(e+I*d*c)-1/2*e*arctan(c*x)*polylog(3, 
(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))/(e+I*d*c)+1/2*I*e*arctan(c...
 

3.1.18.5 Fricas [F]

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

input

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

output

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

3.1.18.6 Sympy [F]

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

input

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

output

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

3.1.18.7 Maxima [F]

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

input

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

output

a^3*log(e*x + d)/e + integrate(1/32*(28*b^3*arctan(c*x)^3 + 3*b^3*arctan(c 
*x)*log(c^2*x^2 + 1)^2 + 96*a*b^2*arctan(c*x)^2 + 96*a^2*b*arctan(c*x))/(e 
*x + d), x)
 

3.1.18.8 Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x))^3}{d+e x} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.1.18.9 Mupad [F(-1)]

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

input

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

output

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