\(\int \frac {(a+b \arctan (c x))^3}{(d+e x)^2} \, dx\) [19]

Optimal result

Integrand size = 18, antiderivative size = 499 \[ \int \frac {(a+b \arctan (c x))^3}{(d+e x)^2} \, dx=\frac {i c (a+b \arctan (c x))^3}{c^2 d^2+e^2}+\frac {c^2 d (a+b \arctan (c x))^3}{e \left (c^2 d^2+e^2\right )}-\frac {(a+b \arctan (c x))^3}{e (d+e x)}-\frac {3 b c (a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c (a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}-\frac {3 i b^2 c (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}-\frac {3 b^3 c \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5389, 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.

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

\(\Big \downarrow \) 5389

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 5389

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Sy 
mbol] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTan[c*x])^p/(e*(q + 1))), x] - S 
imp[b*c*(p/(e*(q + 1)))   Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p - 1), 
(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && 
 IGtQ[p, 1] && IntegerQ[q] && NeQ[q, -1]
 

Maple [C] (warning: unable to verify)

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

Time = 5.25 (sec) , antiderivative size = 2398, normalized size of antiderivative = 4.81

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2398\)
default	\(\text {Expression too large to display}\)	\(2398\)
parts	\(\text {Expression too large to display}\)	\(2406\)

Input:

int((a+b*arctan(c*x))^3/(e*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

3/2*((2*c*d*arctan(c*x)/(c^2*d^2*e + e^3) - log(c^2*x^2 + 1)/(c^2*d^2 + e^ 
2) + 2*log(e*x + d)/(c^2*d^2 + e^2))*c - 2*arctan(c*x)/(e^2*x + d*e))*a^2* 
b - a^3/(e^2*x + d*e) - 1/32*(4*b^3*arctan(c*x)^3 - 3*b^3*arctan(c*x)*log( 
c^2*x^2 + 1)^2 - 32*(e^2*x + d*e)*integrate(1/32*(28*(b^3*c^2*e*x^2 + b^3* 
e)*arctan(c*x)^3 + 12*(8*a*b^2*c^2*e*x^2 + b^3*c*e*x + b^3*c*d + 8*a*b^2*e 
)*arctan(c*x)^2 - 12*(b^3*c^2*e*x^2 + b^3*c^2*d*x)*arctan(c*x)*log(c^2*x^2 
 + 1) - 3*(b^3*c*e*x + b^3*c*d - (b^3*c^2*e*x^2 + b^3*e)*arctan(c*x))*log( 
c^2*x^2 + 1)^2)/(c^2*e^3*x^4 + 2*c^2*d*e^2*x^3 + 2*d*e^2*x + d^2*e + (c^2* 
d^2*e + e^3)*x^2), x))/(e^2*x + d*e)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \arctan (c x))^3}{(d+e x)^2} \, dx=\text {too large to display} \] Input:

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

Output:

(2*atan(c*x)**3*b**3*c**5*d**4*e*x + 2*atan(c*x)**3*b**3*c**3*d**3*e**2 + 
2*atan(c*x)**3*b**3*c**3*d**2*e**3*x + 2*atan(c*x)**3*b**3*c*d*e**4 - 3*at 
an(c*x)**2*a*b**2*c**5*d**5 + 3*atan(c*x)**2*a*b**2*c**5*d**4*e*x + 3*atan 
(c*x)**2*a*b**2*c*d*e**4 - 3*atan(c*x)**2*a*b**2*c*e**5*x + 3*atan(c*x)**2 
*b**3*c**4*d**4*e - 3*atan(c*x)**2*b**3*c**4*d**3*e**2*x + 3*atan(c*x)**2* 
b**3*c**2*d**2*e**3 - 3*atan(c*x)**2*b**3*c**2*d*e**4*x + 6*atan(c*x)*a**2 
*b*c**5*d**4*e*x - 6*atan(c*x)*a**2*b*c**3*d**3*e**2 - 6*atan(c*x)*a**2*b* 
c**3*d**2*e**3*x + 6*atan(c*x)*a**2*b*c*d*e**4 - 6*atan(c*x)*a*b**2*c**4*d 
**3*e**2*x + 6*atan(c*x)*a*b**2*c**2*d**2*e**3 + 6*atan(c*x)*a*b**2*c**2*d 
*e**4*x - 6*atan(c*x)*a*b**2*e**5 + 6*atan(c*x)*b**3*c**3*d**2*e**3*x - 6* 
atan(c*x)*b**3*c*d*e**4 + 6*int(atan(c*x)/(c**4*d**4*x**2 + 2*c**4*d**3*e* 
x**3 + c**4*d**2*e**2*x**4 + c**2*d**4 + 2*c**2*d**3*e*x - 2*c**2*d*e**3*x 
**3 - c**2*e**4*x**4 - d**2*e**2 - 2*d*e**3*x - e**4*x**2),x)*a*b**2*c**8* 
d**9 + 6*int(atan(c*x)/(c**4*d**4*x**2 + 2*c**4*d**3*e*x**3 + c**4*d**2*e* 
*2*x**4 + c**2*d**4 + 2*c**2*d**3*e*x - 2*c**2*d*e**3*x**3 - c**2*e**4*x** 
4 - d**2*e**2 - 2*d*e**3*x - e**4*x**2),x)*a*b**2*c**8*d**8*e*x - 12*int(a 
tan(c*x)/(c**4*d**4*x**2 + 2*c**4*d**3*e*x**3 + c**4*d**2*e**2*x**4 + c**2 
*d**4 + 2*c**2*d**3*e*x - 2*c**2*d*e**3*x**3 - c**2*e**4*x**4 - d**2*e**2 
- 2*d*e**3*x - e**4*x**2),x)*a*b**2*c**4*d**5*e**4 - 12*int(atan(c*x)/(c** 
4*d**4*x**2 + 2*c**4*d**3*e*x**3 + c**4*d**2*e**2*x**4 + c**2*d**4 + 2*...