3.1.34 \(\int \frac {(a+b \text {ArcTan}(c+d x))^2}{e+f x} \, dx\) [34]
Optimal. Leaf size=261 \[ -\frac {(a+b \text {ArcTan}(c+d x))^2 \log \left (\frac {2}{1-i (c+d x)}\right )}{f}+\frac {(a+b \text {ArcTan}(c+d x))^2 \log \left (\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}+\frac {i b (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,1-\frac {2}{1-i (c+d x)}\right )}{f}-\frac {i b (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1-i (c+d x)}\right )}{2 f}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 f} \]
[Out]
-(a+b*arctan(d*x+c))^2*ln(2/(1-I*(d*x+c)))/f+(a+b*arctan(d*x+c))^2*ln(2*d*(f*x+e)/(d*e+I*f-c*f)/(1-I*(d*x+c)))
/f+I*b*(a+b*arctan(d*x+c))*polylog(2,1-2/(1-I*(d*x+c)))/f-I*b*(a+b*arctan(d*x+c))*polylog(2,1-2*d*(f*x+e)/(d*e
+I*f-c*f)/(1-I*(d*x+c)))/f-1/2*b^2*polylog(3,1-2/(1-I*(d*x+c)))/f+1/2*b^2*polylog(3,1-2*d*(f*x+e)/(d*e+I*f-c*f
)/(1-I*(d*x+c)))/f
________________________________________________________________________________________
Rubi [A]
time = 0.13, antiderivative size = 261, 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 = {5155, 4968}
\begin {gather*} -\frac {i b (a+b \text {ArcTan}(c+d x)) \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+i f) (1-i (c+d x))}\right )}{f}+\frac {(a+b \text {ArcTan}(c+d x))^2 \log \left (\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}+\frac {i b \text {Li}_2\left (1-\frac {2}{1-i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))}{f}-\frac {\log \left (\frac {2}{1-i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))^2}{f}+\frac {b^2 \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 {b^2 \text {Li}_3\left (1-\frac {2}{1-i (c+d x)}\right )}{2 f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTan[c + d*x])^2/(e + f*x),x]
[Out]
-(((a + b*ArcTan[c + d*x])^2*Log[2/(1 - I*(c + d*x))])/f) + ((a + b*ArcTan[c + d*x])^2*Log[(2*d*(e + f*x))/((d
*e + I*f - c*f)*(1 - I*(c + d*x)))])/f + (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, 1 - 2/(1 - I*(c + d*x))])/f -
(I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(c + d*x)))])/f - (b^2*
PolyLog[3, 1 - 2/(1 - I*(c + d*x))])/(2*f) + (b^2*PolyLog[3, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(c
+ d*x)))])/(2*f)
Rule 4968
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^2)*(Log[
2/(1 - I*c*x)]/e), x] + (Simp[(a + b*ArcTan[c*x])^2*(Log[2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/e), x] + S
imp[I*b*(a + b*ArcTan[c*x])*(PolyLog[2, 1 - 2/(1 - I*c*x)]/e), x] - Simp[I*b*(a + b*ArcTan[c*x])*(PolyLog[2, 1
- 2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/e), x] - Simp[b^2*(PolyLog[3, 1 - 2/(1 - I*c*x)]/(2*e)), x] + Si
mp[b^2*(PolyLog[3, 1 - 2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(2*e)), x]) /; FreeQ[{a, b, c, d, e}, x] &&
NeQ[c^2*d^2 + e^2, 0]
Rule 5155
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 )^2}{e+f x} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{\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 )^2 \log \left (\frac {2}{1-i (c+d x)}\right )}{f}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}+\frac {i b \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1-i (c+d x)}\right )}{f}-\frac {i b \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-i (c+d x)}\right )}{2 f}+\frac {b^2 \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}\\ \end {align*}
________________________________________________________________________________________
Mathematica [F]
time = 5.13, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b \text {ArcTan}(c+d x))^2}{e+f x} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
Integrate[(a + b*ArcTan[c + d*x])^2/(e + f*x),x]
[Out]
Integrate[(a + b*ArcTan[c + d*x])^2/(e + f*x), x]
________________________________________________________________________________________
Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 2.12, size = 2022, normalized size = 7.75
| | |
method |
result |
size |
| | |
derivativedivides |
\(\text {Expression too large to display}\) |
\(2022\) |
default |
\(\text {Expression too large to display}\) |
\(2022\) |
| | |
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctan(d*x+c))^2/(f*x+e),x,method=_RETURNVERBOSE)
[Out]
1/d*(a^2*d*ln(c*f-d*e-f*(d*x+c))/f+b^2*d*ln(c*f-d*e-f*(d*x+c))/f*arctan(d*x+c)^2-b^2*d/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)+1/2*I*b^2*d/f*arctan(d*x+c)^2*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+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^3+1/2*I*b^2*d/f*arctan(d*x
+c)^2*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))*Pi*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*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+(1+I*(d*x+c))^2/(1+(d*x+c)^2)
))+I*a*b*d*ln(c*f-d*e-f*(d*x+c))/f*ln((I*f-f*(d*x+c))/(d*e+I*f-c*f))-I*a*b*d*ln(c*f-d*e-f*(d*x+c))/f*ln((I*f+f
*(d*x+c))/(c*f-d*e+I*f))+1/2*I*b^2*d/(c*f-d*e+I*f)*polylog(3,(c*f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+
I*f-c*f))-1/2*b^2*d/f*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-1/2*I*b^2*d/f*arctan(d*x+c)^2*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))*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+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2+b^2*d/(c*f-d*e+I*f)*arctan(d*x+c)*polylog(2,(c*f-d*e+I*f)*(
1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+I*b^2*d/(c*f-d*e+I*f)*arctan(d*x+c)^2*ln(1-(c*f-d*e+I*f)*(1+I*(d*x
+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+b^2*d*c/(c*f-d*e+I*f)*arctan(d*x+c)^2*ln(1-(c*f-d*e+I*f)*(1+I*(d*x+c))^2/(
1+(d*x+c)^2)/(d*e+I*f-c*f))-I*a*b*d/f*dilog((I*f+f*(d*x+c))/(c*f-d*e+I*f))+1/2*b^2*d*c/(c*f-d*e+I*f)*polylog(3
,(c*f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-b^2*d^2/f*e/(c*f-d*e+I*f)*arctan(d*x+c)^2*ln(1-(c*
f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+2*I*b^2*d^2/f*e*arctan(d*x+c)*polylog(2,(c*f-d*e+I*f)*
(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)-1/2*b^2*d^2/f*e/(c*f-d*e+I*f)*polylog(3,(c*f-
d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+2*a*b*d*ln(c*f-d*e-f*(d*x+c))/f*arctan(d*x+c)+I*a*b*d/f*
dilog((I*f-f*(d*x+c))/(d*e+I*f-c*f))-1/2*I*b^2*d/f*arctan(d*x+c)^2*Pi*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))
)*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+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2+I*b^2*d/f*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x
+c)^2))-I*b^2*d*c/(c*f-d*e+I*f)*arctan(d*x+c)*polylog(2,(c*f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c
*f)))
________________________________________________________________________________________
Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))^2/(f*x+e),x, algorithm="maxima")
[Out]
a^2*log(f*x + e)/f + integrate(1/16*(12*b^2*arctan(d*x + c)^2 + b^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 + 32*a*
b*arctan(d*x + c))/(f*x + e), x)
________________________________________________________________________________________
Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))^2/(f*x+e),x, algorithm="fricas")
[Out]
integral((b^2*arctan(d*x + c)^2 + 2*a*b*arctan(d*x + c) + a^2)/(f*x + e), x)
________________________________________________________________________________________
Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atan(d*x+c))**2/(f*x+e),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))^2/(f*x+e),x, algorithm="giac")
[Out]
sage0*x
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^2}{e+f\,x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*atan(c + d*x))^2/(e + f*x),x)
[Out]
int((a + b*atan(c + d*x))^2/(e + f*x), x)
________________________________________________________________________________________