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

Optimal. Leaf size=161 \[ -\frac {c^2 \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {a+b \tan ^{-1}(c x)}{2 d x^2}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {i b c^2 \text {Li}_2\left (\frac {2}{i c x+1}-1\right )}{2 d}+\frac {i b c^2 \log \left (c^2 x^2+1\right )}{2 d}-\frac {i b c^2 \log (x)}{d}-\frac {b c^2 \tan ^{-1}(c x)}{2 d}-\frac {b c}{2 d x} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4870, 4852, 325, 203, 266, 36, 29, 31, 4868, 2447} \[ -\frac {i b c^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {c^2 \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {a+b \tan ^{-1}(c x)}{2 d x^2}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac {i b c^2 \log \left (c^2 x^2+1\right )}{2 d}-\frac {i b c^2 \log (x)}{d}-\frac {b c^2 \tan ^{-1}(c x)}{2 d}-\frac {b c}{2 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(2*d*x) - (b*c^2*ArcTan[c*x])/(2*d) - (a + b*ArcTan[c*x])/(2*d*x^2) + (I*c*(a + b*ArcTan[c*x]))/(d*x) -
(I*b*c^2*Log[x])/d + ((I/2)*b*c^2*Log[1 + c^2*x^2])/d - (c^2*(a + b*ArcTan[c*x])*Log[2 - 2/(1 + I*c*x)])/d -
((I/2)*b*c^2*PolyLog[2, -1 + 2/(1 + I*c*x)])/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 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 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 325

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

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

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Mathematica [C]  time = 0.19, size = 178, normalized size = 1.11 \[ -\frac {2 c^2 \log \left (\frac {2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {a+b \tan ^{-1}(c x)}{x^2}-\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )}{x}+2 a c^2 \log (x)+\frac {b c \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )}{x}+i b c^2 \text {Li}_2(-i c x)-i b c^2 \text {Li}_2(i c x)+i b c^2 \text {Li}_2\left (\frac {c x+i}{c x-i}\right )+i b c^2 \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*((a + b*ArcTan[c*x])/x^2 - ((2*I)*c*(a + b*ArcTan[c*x]))/x + (b*c*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x
^2)])/x + 2*a*c^2*Log[x] + 2*c^2*(a + b*ArcTan[c*x])*Log[(2*I)/(I - c*x)] + I*b*c^2*(2*Log[x] - Log[1 + c^2*x^
2]) + I*b*c^2*PolyLog[2, (-I)*c*x] - I*b*c^2*PolyLog[2, I*c*x] + I*b*c^2*PolyLog[2, (I + c*x)/(-I + c*x)])/d

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c d x^{4} - 2 i \, d x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^3/(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^4 - 2*I*d*x^3), 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^3/(d+I*c*d*x),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.07, size = 335, normalized size = 2.08 \[ -\frac {a}{2 d \,x^{2}}-\frac {i c^{2} b \dilog \left (i c x +1\right )}{2 d}-\frac {c^{2} a \ln \left (c x \right )}{d}+\frac {c^{2} a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {i c^{2} a \arctan \left (c x \right )}{d}-\frac {b \arctan \left (c x \right )}{2 d \,x^{2}}+\frac {i c^{2} b \dilog \left (-i c x +1\right )}{2 d}-\frac {c^{2} b \ln \left (c x \right ) \arctan \left (c x \right )}{d}+\frac {c^{2} b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d}+\frac {i b \,c^{2} \ln \left (c^{2} x^{2}+1\right )}{2 d}-\frac {b c}{2 d x}+\frac {i c b \arctan \left (c x \right )}{d x}-\frac {b \,c^{2} \arctan \left (c x \right )}{2 d}-\frac {i c^{2} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 d}-\frac {i c^{2} b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}-\frac {i c^{2} b \ln \left (c x \right )}{d}-\frac {i c^{2} b \ln \left (-\frac {i \left (c x +i\right )}{2}\right ) \ln \left (c x -i\right )}{2 d}+\frac {i c^{2} b \ln \left (c x -i\right )^{2}}{4 d}+\frac {i c a}{d x}+\frac {i c^{2} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c^2*log(I*c*x + 1)/d - 2*c^2*log(x)/d + (2*I*c*x - 1)/(d*x^2))*a + (-I*c*integrate(arctan(c*x)/(c^2*d*x
^4 + d*x^2), x) + integrate(arctan(c*x)/(c^2*d*x^5 + d*x^3), x))*b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))/x**3/(d+I*c*d*x),x)

[Out]

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

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