3.52 \(\int \frac {x^2 (a+b \tan ^{-1}(c x))}{(d+i c d x)^2} \, dx\)

Optimal. Leaf size=167 \[ \frac {a+b \tan ^{-1}(c x)}{c^3 d^2 (-c x+i)}+\frac {2 i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^2}-\frac {a x}{c^2 d^2}-\frac {b \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{c^3 d^2}-\frac {i b}{2 c^3 d^2 (-c x+i)}+\frac {i b \tan ^{-1}(c x)}{2 c^3 d^2}-\frac {b x \tan ^{-1}(c x)}{c^2 d^2}+\frac {b \log \left (c^2 x^2+1\right )}{2 c^3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4876, 4846, 260, 4862, 627, 44, 203, 4854, 2402, 2315} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3 d^2}+\frac {a+b \tan ^{-1}(c x)}{c^3 d^2 (-c x+i)}+\frac {2 i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^2}-\frac {a x}{c^2 d^2}+\frac {b \log \left (c^2 x^2+1\right )}{2 c^3 d^2}-\frac {i b}{2 c^3 d^2 (-c x+i)}-\frac {b x \tan ^{-1}(c x)}{c^2 d^2}+\frac {i b \tan ^{-1}(c x)}{2 c^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 2315

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

Rule 2402

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 4846

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[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 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 4876

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])

Rubi steps

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

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Mathematica [A]  time = 0.81, size = 153, normalized size = 0.92 \[ -\frac {4 i a \log \left (c^2 x^2+1\right )+4 a c x+\frac {4 a}{c x-i}-8 a \tan ^{-1}(c x)+b \left (-2 \log \left (c^2 x^2+1\right )-4 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-8 \tan ^{-1}(c x)^2-i \sin \left (2 \tan ^{-1}(c x)\right )+\cos \left (2 \tan ^{-1}(c x)\right )+2 \tan ^{-1}(c x) \left (2 c x-4 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+\sin \left (2 \tan ^{-1}(c x)\right )+i \cos \left (2 \tan ^{-1}(c x)\right )\right )\right )}{4 c^3 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*(4*a*c*x + (4*a)/(-I + c*x) - 8*a*ArcTan[c*x] + (4*I)*a*Log[1 + c^2*x^2] + b*(-8*ArcTan[c*x]^2 + Cos[2*Ar
cTan[c*x]] - 2*Log[1 + c^2*x^2] - 4*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - I*Sin[2*ArcTan[c*x]] + 2*ArcTan[c*x]*
(2*c*x + I*Cos[2*ArcTan[c*x]] - (4*I)*Log[1 + E^((2*I)*ArcTan[c*x])] + Sin[2*ArcTan[c*x]])))/(c^3*d^2)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {-i \, b x^{2} \log \left (-\frac {c x + i}{c x - i}\right ) - 2 \, a x^{2}}{2 \, {\left (c^{2} d^{2} x^{2} - 2 i \, c d^{2} x - d^{2}\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.08, size = 316, normalized size = 1.89 \[ -\frac {a x}{c^{2} d^{2}}-\frac {a}{c^{3} d^{2} \left (c x -i\right )}+\frac {2 a \arctan \left (c x \right )}{c^{3} d^{2}}-\frac {i b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{8 c^{3} d^{2}}-\frac {b x \arctan \left (c x \right )}{c^{2} d^{2}}-\frac {b \arctan \left (c x \right )}{c^{3} d^{2} \left (c x -i\right )}-\frac {2 i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{c^{3} d^{2}}-\frac {b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{c^{3} d^{2}}-\frac {b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{c^{3} d^{2}}+\frac {b \ln \left (c x -i\right )^{2}}{2 c^{3} d^{2}}+\frac {b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{16 c^{3} d^{2}}+\frac {i b \arctan \left (\frac {c x}{2}\right )}{8 c^{3} d^{2}}-\frac {i b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{4 c^{3} d^{2}}+\frac {i b}{2 c^{3} d^{2} \left (c x -i\right )}-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{c^{3} d^{2}}+\frac {3 b \ln \left (c^{2} x^{2}+1\right )}{8 c^{3} d^{2}}+\frac {3 i b \arctan \left (c x \right )}{4 c^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*x/c^2/d^2-1/c^3*a/d^2/(c*x-I)+2/c^3*a/d^2*arctan(c*x)-1/8*I/c^3*b/d^2*arctan(1/6*c^3*x^3+7/6*c*x)-b*x*arcta
n(c*x)/c^2/d^2-1/c^3*b/d^2*arctan(c*x)/(c*x-I)-2*I/c^3*b/d^2*arctan(c*x)*ln(c*x-I)-1/c^3*b/d^2*ln(c*x-I)*ln(-1
/2*I*(I+c*x))-1/c^3*b/d^2*dilog(-1/2*I*(I+c*x))+1/2/c^3*b/d^2*ln(c*x-I)^2+1/16/c^3*b/d^2*ln(c^4*x^4+10*c^2*x^2
+9)+1/8*I/c^3*b/d^2*arctan(1/2*c*x)-1/4*I/c^3*b/d^2*arctan(1/2*c*x-1/2*I)+1/2*I/c^3*b/d^2/(c*x-I)-I/c^3*a/d^2*
ln(c^2*x^2+1)+3/8*b*ln(c^2*x^2+1)/c^3/d^2+3/4*I/c^3*b/d^2*arctan(c*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(1/(c^4*d^2*x - I*c^3*d^2) + x/(c^2*d^2) + 2*I*log(c*x - I)/(c^3*d^2)) - 1/32*(-16*I*c^2*x^2 - 8*(4*c*x - 4
*I)*arctan(c*x)^2 - 8*(c*x - I)*log(c^2*x^2 + 1)^2 - (8*I*c^4*d^2*x + 8*c^3*d^2)*((c*(x/(c^6*d^2*x^2 + c^4*d^2
) + arctan(c*x)/(c^5*d^2)) - 2*arctan(c*x)/(c^6*d^2*x^2 + c^4*d^2))*c + 8*integrate(1/4*log(c^2*x^2 + 1)/(c^6*
d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x)) + 8*(c^4*d^2*x - I*c^3*d^2)*(c*(c^2/(c^8*d^2*x^2 + c^6*d^2) + log(c^2*
x^2 + 1)/(c^6*d^2*x^2 + c^4*d^2)) + 16*integrate(1/4*arctan(c*x)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x))
+ (16*I*c^5*d^2*x + 16*c^4*d^2)*(c*(x/(c^6*d^2*x^2 + c^4*d^2) + arctan(c*x)/(c^5*d^2)) - 8*c*integrate(1/4*x^2
*log(c^2*x^2 + 1)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - 2*arctan(c*x)/(c^6*d^2*x^2 + c^4*d^2)) + 16*(c
^5*d^2*x - I*c^4*d^2)*(16*c*integrate(1/4*x^2*arctan(c*x)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - c^2/(c
^8*d^2*x^2 + c^6*d^2) - log(c^2*x^2 + 1)/(c^6*d^2*x^2 + c^4*d^2)) - 16*c*x + 16*(c^2*x^2 - I*c*x + 1)*arctan(c
*x) + 32*(c^7*d^2*x - I*c^6*d^2)*integrate(1/4*(2*c*x^4*arctan(c*x) + x^3*log(c^2*x^2 + 1))/(c^6*d^2*x^4 + 2*c
^4*d^2*x^2 + c^2*d^2), x) - 8*(4*I*c^7*d^2*x + 4*c^6*d^2)*integrate(1/4*(c*x^4*log(c^2*x^2 + 1) - 2*x^3*arctan
(c*x))/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - 8*(-12*I*c^6*d^2*x - 12*c^5*d^2)*integrate(1/4*(2*c*x^3*a
rctan(c*x) + x^2*log(c^2*x^2 + 1))/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 96*(c^6*d^2*x - I*c^5*d^2)*in
tegrate(1/4*(c*x^3*log(c^2*x^2 + 1) - 2*x^2*arctan(c*x))/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - 8*(-I*c
^2*x^2 - 2*I)*log(c^2*x^2 + 1))*b/(c^4*d^2*x - I*c^3*d^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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