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

Optimal. Leaf size=100 \[ -\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {i c \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d}+\frac {b c \text {Li}_2\left (\frac {2}{i c x+1}-1\right )}{2 d}+\frac {b c \log (x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4870, 4852, 266, 36, 29, 31, 4868, 2447} \[ \frac {b c \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {i c \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d}+\frac {b c \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4870

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 149, normalized size = 1.49 \[ -\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {i c \log \left (\frac {2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {i a c \log (x)}{d}+\frac {b c \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )}{2 d}+\frac {b c \text {Li}_2(-i c x)}{2 d}-\frac {b c \text {Li}_2(i c x)}{2 d}+\frac {b c \text {Li}_2\left (-\frac {c x+i}{i-c x}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c d x^{3} - 2 i \, d x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [B]  time = 0.07, size = 252, normalized size = 2.52 \[ -\frac {a}{d x}-\frac {i c a \ln \left (c x \right )}{d}-\frac {i c b \arctan \left (c x \right ) \ln \left (c x \right )}{d}-\frac {c a \arctan \left (c x \right )}{d}-\frac {b \arctan \left (c x \right )}{d x}+\frac {i c a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {i c b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d}+\frac {c b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 d}-\frac {c b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 d}+\frac {c b \dilog \left (i c x +1\right )}{2 d}-\frac {c b \dilog \left (-i c x +1\right )}{2 d}+\frac {c b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}+\frac {c b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}-\frac {c b \ln \left (c x -i\right )^{2}}{4 d}+\frac {c b \ln \left (c x \right )}{d}-\frac {b c \ln \left (c^{2} x^{2}+1\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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