3.1.18 \(\int \frac {(a+b \text {ArcTan}(c x))^3}{d+e x} \, dx\) [18]
Optimal. Leaf size=320 \[ -\frac {(a+b \text {ArcTan}(c x))^3 \log \left (\frac {2}{1-i c x}\right )}{e}+\frac {(a+b \text {ArcTan}(c x))^3 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e}+\frac {3 i b (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e}-\frac {3 i b (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\frac {3 b^2 (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e}+\frac {3 b^2 (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\frac {3 i b^3 \text {PolyLog}\left (4,1-\frac {2}{1-i c x}\right )}{4 e}+\frac {3 i b^3 \text {PolyLog}\left (4,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{4 e} \]
[Out]
-(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*arctan(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
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Rubi [A]
time = 0.04, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps
used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {4970}
\begin {gather*} \frac {3 b^2 (a+b \text {ArcTan}(c x)) \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))}{2 e}-\frac {3 i b (a+b \text {ArcTan}(c x))^2 \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}+\frac {(a+b \text {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 \text {Li}_2\left (1-\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))^2}{2 e}-\frac {\log \left (\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))^3}{e}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{4 e}-\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{1-i c x}\right )}{4 e} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTan[c*x])^3/(d + e*x),x]
[Out]
-(((a + b*ArcTan[c*x])^3*Log[2/(1 - I*c*x)])/e) + ((a + b*ArcTan[c*x])^3*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 -
I*c*x))])/e + (((3*I)/2)*b*(a + b*ArcTan[c*x])^2*PolyLog[2, 1 - 2/(1 - I*c*x)])/e - (((3*I)/2)*b*(a + b*ArcTa
n[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
Rule 4970
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] + S
imp[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]
Rubi steps
\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d+e x} \, dx &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1-i c x}\right )}{e}+\frac {\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e}+\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 e}-\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 e}+\frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{1-i c x}\right )}{4 e}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{4 e}\\ \end {align*}
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Mathematica [F]
time = 137.07, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b \text {ArcTan}(c x))^3}{d+e x} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
Integrate[(a + b*ArcTan[c*x])^3/(d + e*x),x]
[Out]
Integrate[(a + b*ArcTan[c*x])^3/(d + e*x), x]
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 2.78, size = 2664, normalized size = 8.32
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| method |
result |
size |
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| derivativedivides |
\(\text {Expression too large to display}\) |
\(2664\) |
| default |
\(\text {Expression too large to display}\) |
\(2664\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctan(c*x))^3/(e*x+d),x,method=_RETURNVERBOSE)
[Out]
1/c*(3*a^2*b*c*ln(c*e*x+c*d)/e*arctan(c*x)+a^3*c*ln(c*e*x+c*d)/e+1/2*I*b^3*c/e*arctan(c*x)^3*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+I*c*x)^2/(c^2*x^2+1)+1))^3*Pi+3/2*I*a^2*b*c*ln(c*
e*x+c*d)/e*ln((I*e-c*e*x)/(c*d+I*e))-3/2*I*a^2*b*c*ln(c*e*x+c*d)/e*ln((I*e+c*e*x)/(I*e-c*d))+3/2*a*b^2*c^2/e*d
/(-I*e+c*d)*polylog(3,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+3*I*a*b^2*c/e*arctan(c*x)*polylog(2,-(1+I*c
*x)^2/(c^2*x^2+1))-3*I*a*b^2*c*arctan(c*x)*polylog(2,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))/(e+I*c*d)-3/
2*a*b^2*c/e*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+3/2*a*b^2*c*polylog(3,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2
+1))/(e+I*c*d)+b^3*c*ln(c*e*x+c*d)/e*arctan(c*x)^3-b^3*c/e*arctan(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)-3/2*b^3*c/e*arctan(c*x)*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+b^3*c*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*c*d)+3/2*b^3*c*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*c*d)+3/4*I*b^3*c*polylog(4,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))/(e+I
*c*d)-3/4*I*b^3*c/e*polylog(4,-(1+I*c*x)^2/(c^2*x^2+1))+3/2*I*a^2*b*c/e*dilog((I*e-c*e*x)/(c*d+I*e))-3/2*I*a^2
*b*c/e*dilog((I*e+c*e*x)/(I*e-c*d))+3*a*b^2*c*ln(c*e*x+c*d)/e*arctan(c*x)^2-3*a*b^2*c/e*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)+3*a*b^2*c*arctan(c*x)^2*ln(1-(I*e-c*d)/(c*d+I*e)*
(1+I*c*x)^2/(c^2*x^2+1))/(e+I*c*d)-3/2*I*b^3*c*arctan(c*x)^2*polylog(2,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^
2+1))/(e+I*c*d)+3/2*I*b^3*c/e*arctan(c*x)^2*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+3/2*I*a*b^2*c/e*arctan(c*x)^2*
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+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I/
((1+I*c*x)^2/(c^2*x^2+1)+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))*Pi-1/2
*I*b^3*c/e*arctan(c*x)^3*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+I*c*x)^
2/(c^2*x^2+1)+1))^2*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))*Pi-3/2*I*b^3*c^
2/e*d/(-I*e+c*d)*arctan(c*x)^2*polylog(2,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))-1/2*I*b^3*c/e*arctan(c*x
)^3*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+I*c*x)^2/(c^2*x^2+1)+1))^2*c
sgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*Pi+3*a*b^2*c^2/e*d/(-I*e+c*d)*arctan(c*x)^2*ln(1-(I*e-c*d)/(c*d+I*e)*(1+I*c
*x)^2/(c^2*x^2+1))+3/2*I*a*b^2*c/e*arctan(c*x)^2*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+I*c*x)^2/(c^2*x^2+1)+1))^3*Pi+1/2*I*b^3*c/e*arctan(c*x)^3*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+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+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))*Pi-3/2*I*a*b^2*c/e*arctan(c*x)^2*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+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I/((1+I*
c*x)^2/(c^2*x^2+1)+1))*Pi-3/2*I*a*b^2*c/e*arctan(c*x)^2*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+I*c*x)^2/(c^2*x^2+1)+1))^2*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))*Pi-3*I*a*b^2*c^2/e*d/(-I*e+c*d)*arctan(c*x)*polylog(2,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2
+1))+b^3*c^2/e*d/(-I*e+c*d)*arctan(c*x)^3*ln(1-(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+3/2*b^3*c^2/e*d/(-
I*e+c*d)*arctan(c*x)*polylog(3,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+3/4*I*b^3*c^2/e*d/(-I*e+c*d)*polyl
og(4,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1)))
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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(c*x))^3/(e*x+d),x, algorithm="maxima")
[Out]
a^3*e^(-1)*log(x*e + d) + 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))/(x*e + d), x)
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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(c*x))^3/(e*x+d),x, algorithm="fricas")
[Out]
integral((b^3*arctan(c*x)^3 + 3*a*b^2*arctan(c*x)^2 + 3*a^2*b*arctan(c*x) + a^3)/(x*e + d), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3}}{d + e x}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atan(c*x))**3/(e*x+d),x)
[Out]
Integral((a + b*atan(c*x))**3/(d + e*x), x)
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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(c*x))^3/(e*x+d),x, algorithm="giac")
[Out]
sage0*x
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{d+e\,x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*atan(c*x))^3/(d + e*x),x)
[Out]
int((a + b*atan(c*x))^3/(d + e*x), x)
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