3.130 \(\int \frac {(a+b \tan ^{-1}(c x))^3}{x (d+i c d x)} \, dx\)
Optimal. Leaf size=128 \[ \frac {3 b^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 d}+\frac {3 i b \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac {\log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {3 i b^3 \text {Li}_4\left (\frac {2}{i c x+1}-1\right )}{4 d} \]
[Out]
(a+b*arctan(c*x))^3*ln(2-2/(1+I*c*x))/d+3/2*I*b*(a+b*arctan(c*x))^2*polylog(2,-1+2/(1+I*c*x))/d+3/2*b^2*(a+b*a
rctan(c*x))*polylog(3,-1+2/(1+I*c*x))/d-3/4*I*b^3*polylog(4,-1+2/(1+I*c*x))/d
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Rubi [A] time = 0.23, antiderivative size = 128, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{4868, 4884, 4994, 4998, 6610} \[ \frac {3 b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 d}+\frac {3 i b \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac {3 i b^3 \text {PolyLog}\left (4,-1+\frac {2}{1+i c x}\right )}{4 d}+\frac {\log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTan[c*x])^3/(x*(d + I*c*d*x)),x]
[Out]
((a + b*ArcTan[c*x])^3*Log[2 - 2/(1 + I*c*x)])/d + (((3*I)/2)*b*(a + b*ArcTan[c*x])^2*PolyLog[2, -1 + 2/(1 + I
*c*x)])/d + (3*b^2*(a + b*ArcTan[c*x])*PolyLog[3, -1 + 2/(1 + I*c*x)])/(2*d) - (((3*I)/4)*b^3*PolyLog[4, -1 +
2/(1 + I*c*x)])/d
Rule 4868
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTan[c*x]
)^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2 - 2/(1 + (e*x)/d)
])/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]
Rule 4884
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]
Rule 4994
Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u
])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - c*x))^2, 0]
Rule 4998
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a
+ b*ArcTan[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] - Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[k
+ 1, u])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 -
(2*I)/(I - c*x))^2, 0]
Rule 6610
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
; !FalseQ[w]] /; FreeQ[n, x]
Rubi steps
\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{x (d+i c d x)} \, dx &=\frac {\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {(3 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=\frac {\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\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 d}-\frac {\left (3 i b^2 c\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=\frac {\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\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 d}+\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 d}-\frac {\left (3 b^3 c\right ) \int \frac {\text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 d}\\ &=\frac {\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\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 d}+\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 d}-\frac {3 i b^3 \text {Li}_4\left (-1+\frac {2}{1+i c x}\right )}{4 d}\\ \end {align*}
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Mathematica [B] time = 0.18, size = 311, normalized size = 2.43 \[ \frac {4 a^3 \log \left (\frac {2 i}{-c x+i}\right )+8 a^3 \tanh ^{-1}\left (\frac {c x+i}{c x-i}\right )+12 a^2 b \log \left (\frac {2 i}{-c x+i}\right ) \tan ^{-1}(c x)+24 a^2 b \tan ^{-1}(c x) \tanh ^{-1}\left (\frac {c x+i}{c x-i}\right )+6 b^2 \text {Li}_3\left (\frac {c x+i}{i-c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+12 a b^2 \log \left (\frac {2 i}{-c x+i}\right ) \tan ^{-1}(c x)^2+24 a b^2 \tan ^{-1}(c x)^2 \tanh ^{-1}\left (\frac {c x+i}{c x-i}\right )+6 i b \text {Li}_2\left (\frac {c x+i}{i-c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-3 i b^3 \text {Li}_4\left (\frac {c x+i}{i-c x}\right )+4 b^3 \log \left (\frac {2 i}{-c x+i}\right ) \tan ^{-1}(c x)^3+8 b^3 \tan ^{-1}(c x)^3 \tanh ^{-1}\left (\frac {c x+i}{c x-i}\right )}{4 d} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*ArcTan[c*x])^3/(x*(d + I*c*d*x)),x]
[Out]
(8*a^3*ArcTanh[(I + c*x)/(-I + c*x)] + 24*a^2*b*ArcTan[c*x]*ArcTanh[(I + c*x)/(-I + c*x)] + 24*a*b^2*ArcTan[c*
x]^2*ArcTanh[(I + c*x)/(-I + c*x)] + 8*b^3*ArcTan[c*x]^3*ArcTanh[(I + c*x)/(-I + c*x)] + 4*a^3*Log[(2*I)/(I -
c*x)] + 12*a^2*b*ArcTan[c*x]*Log[(2*I)/(I - c*x)] + 12*a*b^2*ArcTan[c*x]^2*Log[(2*I)/(I - c*x)] + 4*b^3*ArcTan
[c*x]^3*Log[(2*I)/(I - c*x)] + (6*I)*b*(a + b*ArcTan[c*x])^2*PolyLog[2, (I + c*x)/(I - c*x)] + 6*b^2*(a + b*Ar
cTan[c*x])*PolyLog[3, (I + c*x)/(I - c*x)] - (3*I)*b^3*PolyLog[4, (I + c*x)/(I - c*x)])/(4*d)
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fricas [F] time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{3} \log \left (-\frac {c x + i}{c x - i}\right )^{3} - 6 i \, a b^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 12 \, a^{2} b \log \left (-\frac {c x + i}{c x - i}\right ) + 8 i \, a^{3}}{8 \, c d x^{2} - 8 i \, d x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x))^3/x/(d+I*c*d*x),x, algorithm="fricas")
[Out]
integral(-(b^3*log(-(c*x + I)/(c*x - I))^3 - 6*I*a*b^2*log(-(c*x + I)/(c*x - I))^2 - 12*a^2*b*log(-(c*x + I)/(
c*x - I)) + 8*I*a^3)/(8*c*d*x^2 - 8*I*d*x), x)
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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(c*x))^3/x/(d+I*c*d*x),x, algorithm="giac")
[Out]
Timed out
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maple [C] time = 0.46, size = 3393, normalized size = 26.51 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctan(c*x))^3/x/(d+I*c*d*x),x)
[Out]
3/2*I*a*b^2/d*Pi*csgn(I*((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*((1+I*c*x)^2/(
c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-3/2*I*a*b^2/d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*
csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-1/2*I*b^
3/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*a
rctan(c*x)^3-1/2*I*b^3/d*Pi*arctan(c*x)^3*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x
)^2/(c^2*x^2+1)+1))^2-1/2*I*b^3/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+
I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^3-1/2*I*b^3/d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x
^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^3+3/2*I*a*b^2/d*Pi*csg
n(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-3/2*I*a*b^2/d*Pi*csgn(((1+I*c*x)^
2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+3/2*I*a*b^2/d*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1
)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-3/2*I*a*b^2/d*Pi*arctan(c*x)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1
+I*c*x)^2/(c^2*x^2+1)+1))^3-3*I*a*b^2/d*Pi*arctan(c*x)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)
+1))^2+1/2*I*b^3/d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1
)+1))^2*arctan(c*x)^3+1/2*I*b^3/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+
I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^3-b^3/d*ln(c*x-I)*arctan(c*x)^3+b^3/d*arctan(
c*x)^3*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))+6*a*b^2/d*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*a*b^2/d*polylog(3,-(
1+I*c*x)/(c^2*x^2+1)^(1/2))+b^3/d*arctan(c*x)^3*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^3/d*arctan(c*x)^3*ln(1+(1+
I*c*x)/(c^2*x^2+1)^(1/2))-b^3/d*arctan(c*x)^3*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+6*b^3/d*arctan(c*x)*polylog(3,(1+I
*c*x)/(c^2*x^2+1)^(1/2))+b^3/d*arctan(c*x)^3*ln(c*x)+6*b^3/d*arctan(c*x)*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2
))-1/2*I*b^3/d*arctan(c*x)^4+6*I*b^3/d*polylog(4,(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*b^3/d*polylog(4,-(1+I*c*x)/(
c^2*x^2+1)^(1/2))-I*a^3/d*arctan(c*x)-I*b^3/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*a
rctan(c*x)^3-1/2*I*b^3/d*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^3+3/2*
I*b*a^2/d*ln(c*x)*ln(1+I*c*x)-3/2*I*b*a^2/d*ln(c*x)*ln(1-I*c*x)+3/2*I*b*a^2/d*ln(c*x-I)*ln(-1/2*I*(I+c*x))-6*I
*a*b^2/d*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))-6*I*a*b^2/d*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*
x^2+1)^(1/2))+9/2*I*a*b^2/d*Pi*arctan(c*x)^2-3*I*b^3/d*arctan(c*x)^2*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-3
*I*b^3/d*arctan(c*x)^2*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*I*b^3/d*Pi*arctan(c*x)^3-3*b*a^2/d*ln(c*x-I)
*arctan(c*x)+3*a*b^2/d*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-3*a*b^2/d*ln(c*x-I)*arctan(c*x)^2+3*b*a^2
/d*arctan(c*x)*ln(c*x)+3/2*I*b*a^2/d*dilog(1+I*c*x)-3/2*I*b*a^2/d*dilog(1-I*c*x)+3/2*I*b*a^2/d*dilog(-1/2*I*(I
+c*x))-3/4*I*b*a^2/d*ln(c*x-I)^2+3*a*b^2/d*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+3*a*b^2/d*arctan(c*
x)^2*ln(c*x)-3*a*b^2/d*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+3*a*b^2/d*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2
*x^2+1)^(1/2))-2*I*a*b^2/d*arctan(c*x)^3-1/2*a^3/d*ln(c^2*x^2+1)+a^3/d*ln(c*x)-3/2*I*a*b^2/d*Pi*csgn(I*((1+I*c
*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1)
)^2*arctan(c*x)^2+3/2*I*a*b^2/d*Pi*arctan(c*x)^2*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2
+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2-3/2*I*a*b^2/d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c
^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-3/2*I*a*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn(
(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-3/2*I*a*b^2/d*Pi*csgn(I*((1+I*c*x)^2/(c^2
*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+3/2*I*a*b^2/d*Pi*c
sgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(
c^2*x^2+1)+1))*arctan(c*x)^2+1/2*I*b^3/d*Pi*csgn(I*((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*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^3-1/2*I*b^3/d*Pi*csgn(I/((1+
I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1
))*arctan(c*x)^3+1/2*I*b^3/d*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^3+
1/2*I*b^3/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^3-1/2*I*b^3/d*Pi*
arctan(c*x)^3*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -a^{3} {\left (\frac {\log \left (i \, c x + 1\right )}{d} - \frac {\log \relax (x)}{d}\right )} + \frac {-64 i \, b^{3} \arctan \left (c x\right )^{4} + 64 \, b^{3} \arctan \left (c x\right )^{3} \log \left (c^{2} x^{2} + 1\right ) + 4 i \, b^{3} \log \left (c^{2} x^{2} + 1\right )^{4} - i \, {\left (\frac {64 \, b^{3} \arctan \left (c x\right )^{4}}{d} + 96 \, b^{3} c^{2} \int \frac {x^{2} \arctan \left (c x\right )^{2} \log \left (c^{2} x^{2} + 1\right )}{c^{2} d x^{3} + d x}\,{d x} + \frac {3 \, b^{3} \log \left (c^{2} x^{2} + 1\right )^{4}}{d} + \frac {512 \, a b^{2} \arctan \left (c x\right )^{3}}{d} + {\left (\frac {\log \left (c^{2} x^{2} + 1\right )^{4}}{d} - \frac {4 \, {\left (\log \left (c^{2} x^{2} + 1\right )^{3} \log \left (-c^{2} x^{2}\right ) + 3 \, {\rm Li}_2\left (c^{2} x^{2} + 1\right ) \log \left (c^{2} x^{2} + 1\right )^{2} - 6 \, \log \left (c^{2} x^{2} + 1\right ) {\rm Li}_{3}(c^{2} x^{2} + 1) + 6 \, {\rm Li}_{4}(c^{2} x^{2} + 1)\right )}}{d}\right )} b^{3} + \frac {768 \, a^{2} b \arctan \left (c x\right )^{2}}{d} + 96 \, b^{3} \int \frac {\arctan \left (c x\right )^{2} \log \left (c^{2} x^{2} + 1\right )}{c^{2} d x^{3} + d x}\,{d x}\right )} d + 16 \, d \int -\frac {12 \, b^{3} c x \arctan \left (c x\right )^{2} \log \left (c^{2} x^{2} + 1\right ) - 96 \, a b^{2} \arctan \left (c x\right )^{2} - 96 \, a^{2} b \arctan \left (c x\right ) + 4 \, {\left (3 \, b^{3} c^{2} x^{2} - 7 \, b^{3}\right )} \arctan \left (c x\right )^{3} - 3 \, {\left (b^{3} c^{2} x^{2} + b^{3}\right )} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2}}{c^{2} d x^{3} + d x}\,{d x}}{512 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x))^3/x/(d+I*c*d*x),x, algorithm="maxima")
[Out]
-a^3*(log(I*c*x + 1)/d - log(x)/d) + 1/512*(-64*I*b^3*arctan(c*x)^4 + 64*b^3*arctan(c*x)^3*log(c^2*x^2 + 1) +
16*b^3*arctan(c*x)*log(c^2*x^2 + 1)^3 + 4*I*b^3*log(c^2*x^2 + 1)^4 - I*(64*b^3*arctan(c*x)^4/d + 6144*b^3*c^2*
integrate(1/64*x^2*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2*d*x^3 + d*x), x) + 3*b^3*log(c^2*x^2 + 1)^4/d + 512*a*b
^2*arctan(c*x)^3/d + 768*a^2*b*arctan(c*x)^2/d + 6144*b^3*integrate(1/64*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2*d
*x^3 + d*x), x) - 512*b^3*integrate(1/64*log(c^2*x^2 + 1)^3/(c^2*d*x^3 + d*x), x))*d - 512*d*integrate(1/32*(1
2*b^3*c*x*arctan(c*x)^2*log(c^2*x^2 + 1) + b^3*c*x*log(c^2*x^2 + 1)^3 - 96*a*b^2*arctan(c*x)^2 - 96*a^2*b*arct
an(c*x) + 4*(3*b^3*c^2*x^2 - 7*b^3)*arctan(c*x)^3 + 3*(b^3*c^2*x^2 - b^3)*arctan(c*x)*log(c^2*x^2 + 1)^2)/(c^2
*d*x^3 + d*x), x))/d
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{x\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*atan(c*x))^3/(x*(d + c*d*x*1i)),x)
[Out]
int((a + b*atan(c*x))^3/(x*(d + c*d*x*1i)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \left (\int \frac {a^{3}}{c x^{2} - i x}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c x \right )}}{c x^{2} - i x}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c x^{2} - i x}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c x \right )}}{c x^{2} - i x}\, dx\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atan(c*x))**3/x/(d+I*c*d*x),x)
[Out]
-I*(Integral(a**3/(c*x**2 - I*x), x) + Integral(b**3*atan(c*x)**3/(c*x**2 - I*x), x) + Integral(3*a*b**2*atan(
c*x)**2/(c*x**2 - I*x), x) + Integral(3*a**2*b*atan(c*x)/(c*x**2 - I*x), x))/d
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