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

Optimal. Leaf size=156 \[ -\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d}+\frac {a x}{c^2 d}+\frac {b \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c^3 d}-\frac {i b \tan ^{-1}(c x)}{2 c^3 d}+\frac {i b x}{2 c^2 d}+\frac {b x \tan ^{-1}(c x)}{c^2 d}-\frac {b \log \left (c^2 x^2+1\right )}{2 c^3 d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4866, 4852, 321, 203, 4846, 260, 4854, 2402, 2315} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d}+\frac {a x}{c^2 d}-\frac {b \log \left (c^2 x^2+1\right )}{2 c^3 d}+\frac {i b x}{2 c^2 d}+\frac {b x \tan ^{-1}(c x)}{c^2 d}-\frac {i b \tan ^{-1}(c x)}{2 c^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 321

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

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 4852

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

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 4866

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 132, normalized size = 0.85 \[ -\frac {i \tan ^{-1}(c x) \left (-2 i a+b c^2 x^2+2 i b c x+2 b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+b\right )+i a c^2 x^2-i a \log \left (c^2 x^2+1\right )-2 a c x+b \log \left (c^2 x^2+1\right )+b \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-i b c x+2 b \tan ^{-1}(c x)^2}{2 c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a x^{2}}{2 \, c d x - 2 i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2*log(-(c*x + I)/(c*x - I)) - 2*I*a*x^2)/(2*c*d*x - 2*I*d), 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),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.08, size = 308, normalized size = 1.97 \[ \frac {a x}{c^{2} d}-\frac {i b \arctan \left (c x \right ) x^{2}}{2 c d}+\frac {i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{c^{3} d}-\frac {a \arctan \left (c x \right )}{c^{3} d}+\frac {b x \arctan \left (c x \right )}{c^{2} d}+\frac {i b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{8 c^{3} d}-\frac {i a \,x^{2}}{2 c d}+\frac {b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 c^{3} d}+\frac {b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 c^{3} d}-\frac {b \ln \left (c x -i\right )^{2}}{4 c^{3} d}+\frac {i b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{4 c^{3} d}+\frac {b}{2 c^{3} d}-\frac {b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{16 c^{3} d}-\frac {3 i b \arctan \left (c x \right )}{4 c^{3} d}+\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 c^{3} d}-\frac {i b \arctan \left (\frac {c x}{2}\right )}{8 c^{3} d}-\frac {3 b \ln \left (c^{2} x^{2}+1\right )}{8 c^{3} d}+\frac {i b x}{2 c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {i \, c x^{2} - 2 \, x}{c^{2} d} - \frac {2 i \, \log \left (i \, c x + 1\right )}{c^{3} d}\right )} - \frac {\frac {1}{2} \, {\left ({\left (2 \, {\left (\frac {x^{2}}{c^{4} d} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{6} d}\right )} \log \left (c^{2} x^{2} + 1\right ) - \frac {2 \, c^{2} x^{2} - \log \left (c^{2} x^{2} + 1\right )^{2} - 2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6} d}\right )} c^{6} d + 8 i \, c^{6} d \int \frac {x^{3} \arctan \left (c x\right )}{c^{4} d x^{2} + c^{2} d}\,{d x} - 4 \, {\left (2 \, {\left (\frac {x}{c^{4} d} - \frac {\arctan \left (c x\right )}{c^{5} d}\right )} \arctan \left (c x\right ) + \frac {\arctan \left (c x\right )^{2} - \log \left (c^{2} x^{2} + 1\right )}{c^{5} d}\right )} c^{5} d + 4 i \, c^{5} d \int \frac {x^{2} \log \left (c^{2} x^{2} + 1\right )}{c^{4} d x^{2} + c^{2} d}\,{d x} - 8 i \, c^{4} d \int \frac {x \arctan \left (c x\right )}{c^{4} d x^{2} + c^{2} d}\,{d x} + 4 i \, c^{3} d \int \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4} d x^{2} + c^{2} d}\,{d x} + 2 \, c^{2} x^{2} + 4 i \, c x + 2 \, {\left (2 i \, c^{2} x^{2} - 4 \, c x - 2 i\right )} \arctan \left (c x\right ) + 4 \, \arctan \left (c x\right )^{2} - 2 \, {\left (c^{2} x^{2} + 2 i \, c x + 1\right )} \log \left (c^{2} x^{2} + 1\right ) + \log \left (c^{2} x^{2} + 1\right )^{2} + 4 \, \log \left (8 \, c^{4} d x^{2} + 8 \, c^{2} d\right )\right )} b}{8 \, c^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*((I*c*x^2 - 2*x)/(c^2*d) - 2*I*log(I*c*x + 1)/(c^3*d)) - 1/8*(32*I*c^6*d*integrate(1/8*x^3*arctan(c*x)/
(c^4*d*x^2 + c^2*d), x) + 16*c^6*d*integrate(1/8*x^3*log(c^2*x^2 + 1)/(c^4*d*x^2 + c^2*d), x) - 32*c^5*d*integ
rate(1/8*x^2*arctan(c*x)/(c^4*d*x^2 + c^2*d), x) + 16*I*c^5*d*integrate(1/8*x^2*log(c^2*x^2 + 1)/(c^4*d*x^2 +
c^2*d), x) - 32*I*c^4*d*integrate(1/8*x*arctan(c*x)/(c^4*d*x^2 + c^2*d), x) - 16*c^4*d*integrate(1/8*x*log(c^2
*x^2 + 1)/(c^4*d*x^2 + c^2*d), x) + 16*I*c^3*d*integrate(1/8*log(c^2*x^2 + 1)/(c^4*d*x^2 + c^2*d), x) + c^2*x^
2 + 2*I*c*x + (2*I*c^2*x^2 - 4*c*x - 2*I)*arctan(c*x) + 2*arctan(c*x)^2 - (c^2*x^2 + 2*I*c*x + 1)*log(c^2*x^2
+ 1) + log(c^2*x^2 + 1)^2 + 2*log(8*c^4*d*x^2 + 8*c^2*d))*b/(c^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \left (\int \frac {2 b \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac {4 a c^{3} x^{3}}{c^{2} x^{2} + 1}\, dx + \int \frac {b c^{2} x^{2}}{c^{2} x^{2} + 1}\, dx + \int \frac {4 i a c^{2} x^{2}}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {2 i b c x}{c^{2} x^{2} + 1}\right )\, dx + \int \left (- \frac {i b c^{3} x^{3}}{c^{2} x^{2} + 1}\right )\, dx + \int \frac {2 b c^{2} x^{2} \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac {2 i b c x \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {2 i b c^{3} x^{3} \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx\right )}{4 c^{2} d} + \frac {\left (b c^{2} x^{2} + 2 i b c x - 2 b \log {\left (i c x + 1 \right )}\right ) \log {\left (- i c x + 1 \right )}}{4 c^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(Integral(2*b*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(4*a*c**3*x**3/(c**2*x**2 + 1), x) + Integral(b*
c**2*x**2/(c**2*x**2 + 1), x) + Integral(4*I*a*c**2*x**2/(c**2*x**2 + 1), x) + Integral(-2*I*b*c*x/(c**2*x**2
+ 1), x) + Integral(-I*b*c**3*x**3/(c**2*x**2 + 1), x) + Integral(2*b*c**2*x**2*log(I*c*x + 1)/(c**2*x**2 + 1)
, x) + Integral(2*I*b*c*x*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(-2*I*b*c**3*x**3*log(I*c*x + 1)/(c**2*
x**2 + 1), x))/(4*c**2*d) + (b*c**2*x**2 + 2*I*b*c*x - 2*b*log(I*c*x + 1))*log(-I*c*x + 1)/(4*c**3*d)

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