[next] [prev] [prev-tail] [tail] [up]

3.39 \(\int \frac {(a+b \tan ^{-1}(c+d x))^3}{e+f x} \, dx\)

Optimal. Leaf size=372 \[ \frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2 d (e+f x)}{(d e-c f+i f) (1-i (c+d x))}\right )}{2 f}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+i f) (1-i (c+d x))}\right )}{2 f}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3 \log \left (\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}+\frac {3 i b \text {Li}_2\left (1-\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}-\frac {\log \left (\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^3}{f}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2 d (e+f x)}{(d e-c f+i f) (1-i (c+d x))}\right )}{4 f}-\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{1-i (c+d x)}\right )}{4 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 372, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5047, 4860} \[ \frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {PolyLog}\left (3,1-\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{2 f}-\frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{2 f}+\frac {3 i b \text {PolyLog}\left (2,1-\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}+\frac {3 i b^3 \text {PolyLog}\left (4,1-\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{4 f}-\frac {3 i b^3 \text {PolyLog}\left (4,1-\frac {2}{1-i (c+d x)}\right )}{4 f}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3 \log \left (\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}-\frac {\log \left (\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^3}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b*ArcTan[c + d*x])^3*Log[2/(1 - I*(c + d*x))])/f) + ((a + b*ArcTan[c + d*x])^3*Log[(2*d*(e + f*x))/((d
*e + I*f - c*f)*(1 - I*(c + d*x)))])/f + (((3*I)/2)*b*(a + b*ArcTan[c + d*x])^2*PolyLog[2, 1 - 2/(1 - I*(c + d
*x))])/f - (((3*I)/2)*b*(a + b*ArcTan[c + d*x])^2*PolyLog[2, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(c
+ d*x)))])/f - (3*b^2*(a + b*ArcTan[c + d*x])*PolyLog[3, 1 - 2/(1 - I*(c + d*x))])/(2*f) + (3*b^2*(a + b*ArcTa
n[c + d*x])*PolyLog[3, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(c + d*x)))])/(2*f) - (((3*I)/4)*b^3*Poly
Log[4, 1 - 2/(1 - I*(c + d*x))])/f + (((3*I)/4)*b^3*PolyLog[4, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(
c + d*x)))])/f

Rule 4860

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*ArcTan[c*x])^3*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] + Sim
p[(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*Po
lyLog[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
]

Rule 5047

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x]
&& IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{e+f x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{\frac {d e-c f}{d}+\frac {f x}{d}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3 \log \left (\frac {2}{1-i (c+d x)}\right )}{f}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3 \log \left (\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}+\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1-i (c+d x)}\right )}{2 f}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 f}-\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2}{1-i (c+d x)}\right )}{2 f}+\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 f}-\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{1-i (c+d x)}\right )}{4 f}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{4 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 8.95, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{e+f x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arctan \left (d x + c\right )^{3} + 3 \, a b^{2} \arctan \left (d x + c\right )^{2} + 3 \, a^{2} b \arctan \left (d x + c\right ) + a^{3}}{f x + e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [C]  time = 0.69, size = 4389, normalized size = 11.80 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*I*a*b^2*c/(I*f+c*f-d*e)*arctan(d*x+c)*polylog(2,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-
3/2*I*a*b^2/f*Pi*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))
^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^3*arctan(d*x+c)^2-3*I*d*b^3/f*e*polylog(4,(I*
f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))/(4*I*f+4*c*f-4*d*e)-3*d*a*b^2/f*e/(I*f+c*f-d*e)*arctan
(d*x+c)^2*ln(1-(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-3/2*I*a*b^2/f*Pi*csgn(I/((1+I*(d*x+c
))^2/(1+(d*x+c)^2)+1))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d
*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(d*x+c)^2-3/2*I*a*b^2/f*Pi*csgn
(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+
c*f-d*e))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(
d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(d*x+c)^2+1/2*I*b^3/f*Pi*csgn(I/((1+I*(d*x+c
))^2/(1+(d*x+c)^2)+1))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d
*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^
2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*arctan(d*x+c)^3+3*I*d*b^3
/f*e*arctan(d*x+c)^2*polylog(2,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))/(2*I*f+2*c*f-2*d*e)+
3*a^2*b*ln(f*(d*x+c)-c*f+d*e)/f*arctan(d*x+c)+3*a*b^2*ln(f*(d*x+c)-c*f+d*e)/f*arctan(d*x+c)^2-3*a*b^2/f*arctan
(d*x+c)^2*ln(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c
)^2)-I*f+c*f-d*e)-3/4*b^3/(I*f+c*f-d*e)*polylog(4,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+a
^3*ln(f*(d*x+c)-c*f+d*e)/f-b^3/f*arctan(d*x+c)^3*ln(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(
d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)-3/2*b^3/f*arctan(d*x+c)*polylog(3,-(1+I*(d*x+c))^2/(1
+(d*x+c)^2))+3/2*b^3/(I*f+c*f-d*e)*arctan(d*x+c)^2*polylog(2,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+
I*f-c*f))+b^3*ln(f*(d*x+c)-c*f+d*e)/f*arctan(d*x+c)^3-3/2*a*b^2/f*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-3/
4*I*b^3/f*polylog(4,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-3/2*d*a*b^2/f*e/(I*f+c*f-d*e)*polylog(3,(I*f+c*f-d*e)*(1+I
*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-d*b^3/f*e/(I*f+c*f-d*e)*arctan(d*x+c)^3*ln(1-(I*f+c*f-d*e)*(1+I*(d*x+
c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-3/2*d*b^3/f*e/(I*f+c*f-d*e)*arctan(d*x+c)*polylog(3,(I*f+c*f-d*e)*(1+I*(d*x
+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-1/2*I*b^3/f*Pi*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))
^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f
*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1
))^2*arctan(d*x+c)^3-1/2*I*b^3/f*Pi*csgn(I/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(
d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1
+(d*x+c)^2)+1))^2*arctan(d*x+c)^3+3*I*a*b^2/f*Pi*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2
/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(d*x+
c)^2+3*a*b^2*c/(I*f+c*f-d*e)*arctan(d*x+c)^2*ln(1-(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-3
/2*I*b^3*c/(I*f+c*f-d*e)*arctan(d*x+c)^2*polylog(2,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+
3/2*I*a^2*b*ln(f*(d*x+c)-c*f+d*e)/f*ln((I*f-f*(d*x+c))/(d*e+I*f-c*f))-3/2*I*a^2*b*ln(f*(d*x+c)-c*f+d*e)/f*ln((
I*f+f*(d*x+c))/(I*f+c*f-d*e))+3*I*a*b^2/(I*f+c*f-d*e)*arctan(d*x+c)^2*ln(1-(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d
*x+c)^2)/(d*e+I*f-c*f))+3*I*a*b^2/f*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-3*I*a*b^2/f*Pi*arc
tan(d*x+c)^2-1/2*I*b^3/f*Pi*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1
+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^3*arctan(d*x+c)^3+6*I*d*a*b^2/f*e*
arctan(d*x+c)*polylog(2,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))/(2*I*f+2*c*f-2*d*e)+3/2*I*a
*b^2/f*Pi*csgn(I/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c)
)^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*
f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+
1))*arctan(d*x+c)^2-3/2*I*a^2*b/f*dilog((I*f+f*(d*x+c))/(I*f+c*f-d*e))+3/4*I*b^3*c/(I*f+c*f-d*e)*polylog(4,(I*
f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-I*b^3/f*Pi*arctan(d*x+c)^3+3/2*I*a*b^2/(I*f+c*f-d*e)*p
olylog(3,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+b^3*c/(I*f+c*f-d*e)*arctan(d*x+c)^3*ln(1-(
I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+3/2*b^3*c/(I*f+c*f-d*e)*arctan(d*x+c)*polylog(3,(I*f
+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+I*b^3/f*Pi*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*
f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+
1))^2*arctan(d*x+c)^3+3*a*b^2/(I*f+c*f-d*e)*arctan(d*x+c)*polylog(2,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2
)/(d*e+I*f-c*f))+I*b^3/(I*f+c*f-d*e)*arctan(d*x+c)^3*ln(1-(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f
-c*f))+3/2*a*b^2*c/(I*f+c*f-d*e)*polylog(3,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+3/2*I*b^
3/f*arctan(d*x+c)^2*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+3/2*I*b^3/(I*f+c*f-d*e)*arctan(d*x+c)*polylog(3,
(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+3/2*I*a^2*b/f*dilog((I*f-f*(d*x+c))/(d*e+I*f-c*f))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{3} \log \left (f x + e\right )}{f} + \int \frac {28 \, b^{3} \arctan \left (d x + c\right )^{3} + 3 \, b^{3} \arctan \left (d x + c\right ) \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2} + 96 \, a b^{2} \arctan \left (d x + c\right )^{2} + 96 \, a^{2} b \arctan \left (d x + c\right )}{32 \, {\left (f x + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{e+f\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atan}{\left (c + d x \right )}\right )^{3}}{e + f x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

[next] [prev] [prev-tail] [front] [up]