3.1.72 \(\int \frac {(d+i c d x) (a+b \text {ArcTan}(c x))^2}{x} \, dx\) [72]

Optimal. Leaf size=216 \[ -d (a+b \text {ArcTan}(c x))^2+i c d x (a+b \text {ArcTan}(c x))^2+2 d (a+b \text {ArcTan}(c x))^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 i b d (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )-b^2 d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-i b d (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.30, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4996, 4930, 5040, 4964, 2449, 2352, 4942, 5108, 5004, 5114, 6745} \begin {gather*} -i b d \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))+i b d \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))-d (a+b \text {ArcTan}(c x))^2+i c d x (a+b \text {ArcTan}(c x))^2+2 i b d \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))+2 d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2+b^2 (-d) \text {Li}_2\left (1-\frac {2}{i c x+1}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(a + b*ArcTan[c*x])^2) + I*c*d*x*(a + b*ArcTan[c*x])^2 + 2*d*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*
x)] + (2*I)*b*d*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)] - b^2*d*PolyLog[2, 1 - 2/(1 + I*c*x)] - I*b*d*(a + b*Ar
cTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*d*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] - (b^2*d*P
olyLog[3, 1 - 2/(1 + I*c*x)])/2 + (b^2*d*PolyLog[3, -1 + 2/(1 + I*c*x)])/2

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 4942

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 +
I*c*x)], x] - Dist[2*b*c*p, Int[(a + b*ArcTan[c*x])^(p - 1)*(ArcTanh[1 - 2/(1 + I*c*x)]/(1 + c^2*x^2)), x], x
] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[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 4996

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 5004

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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 5108

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[L
og[1 + u]*((a + b*ArcTan[c*x])^p/(d + e*x^2)), x], x] - Dist[1/2, Int[Log[1 - u]*((a + b*ArcTan[c*x])^p/(d + e
*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 - 2*(I/(I - c*x)))^
2, 0]

Rule 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 6745

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

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Mathematica [A]
time = 0.26, size = 272, normalized size = 1.26 \begin {gather*} d \left (i a^2 c x+a^2 \log (c x)+i a b \left (2 c x \text {ArcTan}(c x)-\log \left (1+c^2 x^2\right )\right )+b^2 \left (\text {ArcTan}(c x) \left ((1+i c x) \text {ArcTan}(c x)+2 i \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )\right )+i a b (\text {PolyLog}(2,-i c x)-\text {PolyLog}(2,i c x))+b^2 \left (-\frac {i \pi ^3}{24}+\frac {2}{3} i \text {ArcTan}(c x)^3+\text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )-\text {ArcTan}(c x)^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )+i \text {ArcTan}(c x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )+i \text {ArcTan}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

d*(I*a^2*c*x + a^2*Log[c*x] + I*a*b*(2*c*x*ArcTan[c*x] - Log[1 + c^2*x^2]) + b^2*(ArcTan[c*x]*((1 + I*c*x)*Arc
Tan[c*x] + (2*I)*Log[1 + E^((2*I)*ArcTan[c*x])]) + PolyLog[2, -E^((2*I)*ArcTan[c*x])]) + I*a*b*(PolyLog[2, (-I
)*c*x] - PolyLog[2, I*c*x]) + b^2*((-1/24*I)*Pi^3 + ((2*I)/3)*ArcTan[c*x]^3 + ArcTan[c*x]^2*Log[1 - E^((-2*I)*
ArcTan[c*x])] - ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])
] + I*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2 - PolyLog[3, -E^((
2*I)*ArcTan[c*x])]/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 2.57, size = 7034, normalized size = 32.56

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(7034\)
default	\(\text {Expression too large to display}\)	\(7034\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

1/4*I*b^2*c*d*x*arctan(c*x)^2 + 12*I*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)^2/(c^2*x^3 + x), x) + 4*b^2*c^3*
d*integrate(1/16*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + I*b^2*c^3*d*integrate(1/16*x^3*log(c^2*x
^2 + 1)^2/(c^2*x^3 + x), x) + 8*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)/(c^2*x^3 + x), x) + 4*I*b^2*c^3*d*int
egrate(1/16*x^3*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) - 1/4*b^2*c*d*x*arctan(c*x)*log(c^2*x^2 + 1) - 1/16*I*b^2*c
*d*x*log(c^2*x^2 + 1)^2 + 1/4*I*b^2*d*arctan(c*x)^3 + 12*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^3 +
x), x) - 4*I*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 32*a*b*c^2*d*integ
rate(1/16*x^2*arctan(c*x)/(c^2*x^3 + x), x) - 8*I*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)/(c^2*x^3 + x), x) +
1/96*b^2*d*log(c^2*x^2 + 1)^3 + I*a^2*c*d*x + 4*b^2*c*d*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^
3 + x), x) + I*b^2*c*d*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 1/16*b^2*d*log(c^2*x^2 + 1)^2 +
I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d + 12*b^2*d*integrate(1/16*arctan(c*x)^2/(c^2*x^3 + x), x) - 4*
I*b^2*d*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + b^2*d*integrate(1/16*log(c^2*x^2 + 1)^
2/(c^2*x^3 + x), x) + 32*a*b*d*integrate(1/16*arctan(c*x)/(c^2*x^3 + x), x) + a^2*d*log(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((d+I*c*d*x)*(a+b*arctan(c*x))^2/x,x, algorithm="fricas")

[Out]

integral(1/4*(4*I*a^2*c*d*x + 4*a^2*d + (-I*b^2*c*d*x - b^2*d)*log(-(c*x + I)/(c*x - I))^2 - 4*(a*b*c*d*x - I*
a*b*d)*log(-(c*x + I)/(c*x - I)))/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i d \left (\int a^{2} c\, dx + \int \left (- \frac {i a^{2}}{x}\right )\, dx + \int b^{2} c \operatorname {atan}^{2}{\left (c x \right )}\, dx + \int \left (- \frac {i b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{x}\right )\, dx + \int 2 a b c \operatorname {atan}{\left (c x \right )}\, dx + \int \left (- \frac {2 i a b \operatorname {atan}{\left (c x \right )}}{x}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*d*(Integral(a**2*c, x) + Integral(-I*a**2/x, x) + Integral(b**2*c*atan(c*x)**2, x) + Integral(-I*b**2*atan(c
*x)**2/x, x) + Integral(2*a*b*c*atan(c*x), x) + Integral(-2*I*a*b*atan(c*x)/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((d+I*c*d*x)*(a+b*arctan(c*x))^2/x,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 )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x,x)

[Out]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x, x)

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