3.74 \(\int \frac {(d+i c d x) (a+b \tan ^{-1}(c x))^2}{x^3} \, dx\)

Optimal. Leaf size=159 \[ \frac {1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2+2 i b c^2 d \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x}-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{x}+b^2 c^2 d \text {Li}_2\left (\frac {2}{1-i c x}-1\right )-\frac {1}{2} b^2 c^2 d \log \left (c^2 x^2+1\right )+b^2 c^2 d \log (x) \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4876, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 2447} \[ b^2 c^2 d \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+\frac {1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2+2 i b c^2 d \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x}-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{2} b^2 c^2 d \log \left (c^2 x^2+1\right )+b^2 c^2 d \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)*(a + b*ArcTan[c*x])^2)/x^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 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 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])

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 4918

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

Rule 4924

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

Rubi steps

\begin {align*} \int \frac {(d+i c d x) \left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx &=\int \left (\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x^3}+\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x^2}\right ) \, dx\\ &=d \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx+(i c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx\\ &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+(b c d) \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx+\left (2 i b c^2 d\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx\\ &=c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+(b c d) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (2 b c^2 d\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx-\left (b c^3 d\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx\\ &=-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 i b c^2 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+\left (b^2 c^2 d\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx-\left (2 i b^2 c^3 d\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 i b c^2 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+b^2 c^2 d \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )+\frac {1}{2} \left (b^2 c^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 i b c^2 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+b^2 c^2 d \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )+\frac {1}{2} \left (b^2 c^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (b^2 c^4 d\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {1}{2} c^2 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+b^2 c^2 d \log (x)-\frac {1}{2} b^2 c^2 d \log \left (1+c^2 x^2\right )+2 i b c^2 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+b^2 c^2 d \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 190, normalized size = 1.19 \[ -\frac {d \left (2 i a^2 c x+a^2-4 i a b c^2 x^2 \log (c x)+2 i a b c^2 x^2 \log \left (c^2 x^2+1\right )+2 b \tan ^{-1}(c x) \left (a c^2 x^2+2 i a c x+a-2 i b c^2 x^2 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+b c x\right )+2 a b c x-2 b^2 c^2 x^2 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-2 b^2 c^2 x^2 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )-b^2 (c x-i)^2 \tan ^{-1}(c x)^2\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)*(a + b*ArcTan[c*x])^2)/x^3,x]

[Out]

-1/2*(d*(a^2 + (2*I)*a^2*c*x + 2*a*b*c*x - b^2*(-I + c*x)^2*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]*(a + (2*I)*a*c*x +
b*c*x + a*c^2*x^2 - (2*I)*b*c^2*x^2*Log[1 - E^((2*I)*ArcTan[c*x])]) - (4*I)*a*b*c^2*x^2*Log[c*x] - 2*b^2*c^2*
x^2*Log[(c*x)/Sqrt[1 + c^2*x^2]] + (2*I)*a*b*c^2*x^2*Log[1 + c^2*x^2] - 2*b^2*c^2*x^2*PolyLog[2, E^((2*I)*ArcT
an[c*x])]))/x^2

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ \frac {8 \, x^{2} {\rm integral}\left (\frac {2 i \, a^{2} c^{3} d x^{3} + 2 \, a^{2} c^{2} d x^{2} + 2 i \, a^{2} c d x + 2 \, a^{2} d - {\left (2 \, a b c^{3} d x^{3} + 2 \, {\left (-i \, a b + b^{2}\right )} c^{2} d x^{2} + {\left (2 \, a b - i \, b^{2}\right )} c d x - 2 i \, a b d\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, {\left (c^{2} x^{5} + x^{3}\right )}}, x\right ) + {\left (2 i \, b^{2} c d x + b^{2} d\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2}}{8 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^3,x, algorithm="fricas")

[Out]

1/8*(8*x^2*integral(1/2*(2*I*a^2*c^3*d*x^3 + 2*a^2*c^2*d*x^2 + 2*I*a^2*c*d*x + 2*a^2*d - (2*a*b*c^3*d*x^3 + 2*
(-I*a*b + b^2)*c^2*d*x^2 + (2*a*b - I*b^2)*c*d*x - 2*I*a*b*d)*log(-(c*x + I)/(c*x - I)))/(c^2*x^5 + x^3), x) +
(2*I*b^2*c*d*x + b^2*d)*log(-(c*x + I)/(c*x - I))^2)/x^2

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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((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^3,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.13, size = 487, normalized size = 3.06 \[ 2 i c^{2} d a b \ln \left (c x \right )-i c^{2} d a b \ln \left (c^{2} x^{2}+1\right )-\frac {i c d \,b^{2} \arctan \left (c x \right )^{2}}{x}-i c^{2} d \,b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )+2 i c^{2} d \,b^{2} \arctan \left (c x \right ) \ln \left (c x \right )-\frac {b^{2} c^{2} d \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {d \,a^{2}}{2 x^{2}}-c^{2} d \,b^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )+c^{2} d \,b^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )-\frac {i c d \,a^{2}}{x}-\frac {c^{2} d \,b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-\frac {c^{2} d \,b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {c^{2} d \,b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{2}+\frac {c^{2} d \,b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}-c^{2} d a b \arctan \left (c x \right )-\frac {c d \,b^{2} \arctan \left (c x \right )}{x}-\frac {c d a b}{x}-\frac {d a b \arctan \left (c x \right )}{x^{2}}-\frac {2 i c d a b \arctan \left (c x \right )}{x}+\frac {c^{2} d \,b^{2} \ln \left (c x +i\right )^{2}}{4}-\frac {c^{2} d \,b^{2} \ln \left (c x -i\right )^{2}}{4}-c^{2} d \,b^{2} \dilog \left (i c x +1\right )+c^{2} d \,b^{2} \dilog \left (-i c x +1\right )-\frac {c^{2} d \,b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {c^{2} d \,b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{2}-\frac {c^{2} d \,b^{2} \arctan \left (c x \right )^{2}}{2}+c^{2} d \,b^{2} \ln \left (c x \right )-\frac {d \,b^{2} \arctan \left (c x \right )^{2}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^3,x)

[Out]

-I*c^2*d*a*b*ln(c^2*x^2+1)-I*c*d*b^2*arctan(c*x)^2/x-I*c^2*d*b^2*arctan(c*x)*ln(c^2*x^2+1)+2*I*c^2*d*b^2*arcta
n(c*x)*ln(c*x)+2*I*c^2*d*a*b*ln(c*x)-2*I*c*d*a*b*arctan(c*x)/x-1/2*b^2*c^2*d*ln(c^2*x^2+1)-1/2*d*a^2/x^2-c^2*d
*b^2*ln(c*x)*ln(1+I*c*x)+c^2*d*b^2*ln(c*x)*ln(1-I*c*x)-c^2*d*a*b*arctan(c*x)-c*d*b^2*arctan(c*x)/x-I*c*d*a^2/x
-c*d*a*b/x-d*a*b*arctan(c*x)/x^2-1/2*c^2*d*b^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))-1/2*c^2*d*b^2*ln(I+c*x)*ln(c^2*x^2
+1)+1/2*c^2*d*b^2*ln(I+c*x)*ln(1/2*I*(c*x-I))+1/2*c^2*d*b^2*ln(c*x-I)*ln(c^2*x^2+1)-1/2*c^2*d*b^2*arctan(c*x)^
2-c^2*d*b^2*dilog(1+I*c*x)+c^2*d*b^2*dilog(1-I*c*x)-1/2*c^2*d*b^2*dilog(-1/2*I*(I+c*x))+1/4*c^2*d*b^2*ln(I+c*x
)^2+1/2*c^2*d*b^2*dilog(1/2*I*(c*x-I))+c^2*d*b^2*ln(c*x)-1/4*c^2*d*b^2*ln(c*x-I)^2-1/2*d*b^2*arctan(c*x)^2/x^2

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maxima [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((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^3,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x^3,x)

[Out]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x^3, 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((d+I*c*d*x)*(a+b*atan(c*x))**2/x**3,x)

[Out]

Timed out

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