3.196 \(\int \frac {\tan ^{-1}(a x)}{x (c+a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=159 \[ -\frac {11 a x}{32 c^3 \left (a^2 x^2+1\right )}-\frac {a x}{16 c^3 \left (a^2 x^2+1\right )^2}+\frac {\tan ^{-1}(a x)}{2 c^3 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {i \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{2 c^3}-\frac {i \tan ^{-1}(a x)^2}{2 c^3}-\frac {11 \tan ^{-1}(a x)}{32 c^3}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^3} \]

[Out]

-1/16*a*x/c^3/(a^2*x^2+1)^2-11/32*a*x/c^3/(a^2*x^2+1)-11/32*arctan(a*x)/c^3+1/4*arctan(a*x)/c^3/(a^2*x^2+1)^2+
1/2*arctan(a*x)/c^3/(a^2*x^2+1)-1/2*I*arctan(a*x)^2/c^3+arctan(a*x)*ln(2-2/(1-I*a*x))/c^3-1/2*I*polylog(2,-1+2
/(1-I*a*x))/c^3

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Rubi [A]  time = 0.28, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4966, 4924, 4868, 2447, 4930, 199, 205} \[ -\frac {i \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c^3}-\frac {11 a x}{32 c^3 \left (a^2 x^2+1\right )}-\frac {a x}{16 c^3 \left (a^2 x^2+1\right )^2}+\frac {\tan ^{-1}(a x)}{2 c^3 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {i \tan ^{-1}(a x)^2}{2 c^3}-\frac {11 \tan ^{-1}(a x)}{32 c^3}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x)/(16*c^3*(1 + a^2*x^2)^2) - (11*a*x)/(32*c^3*(1 + a^2*x^2)) - (11*ArcTan[a*x])/(32*c^3) + ArcTan[a*x]/(4
*c^3*(1 + a^2*x^2)^2) + ArcTan[a*x]/(2*c^3*(1 + a^2*x^2)) - ((I/2)*ArcTan[a*x]^2)/c^3 + (ArcTan[a*x]*Log[2 - 2
/(1 - I*a*x)])/c^3 - ((I/2)*PolyLog[2, -1 + 2/(1 - I*a*x)])/c^3

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

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 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 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]

Rule 4930

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

Rule 4966

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

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )^3} \, dx &=-\left (a^2 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^3} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=\frac {\tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {1}{4} a \int \frac {1}{\left (c+a^2 c x^2\right )^3} \, dx+\frac {\int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )} \, dx}{c^2}-\frac {a^2 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=-\frac {a x}{16 c^3 \left (1+a^2 x^2\right )^2}+\frac {\tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {\tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac {i \tan ^{-1}(a x)^2}{2 c^3}+\frac {i \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c^3}-\frac {(3 a) \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{16 c}-\frac {a \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{2 c}\\ &=-\frac {a x}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac {11 a x}{32 c^3 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {\tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac {i \tan ^{-1}(a x)^2}{2 c^3}+\frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c^3}-\frac {a \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c^3}-\frac {(3 a) \int \frac {1}{c+a^2 c x^2} \, dx}{32 c^2}-\frac {a \int \frac {1}{c+a^2 c x^2} \, dx}{4 c^2}\\ &=-\frac {a x}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac {11 a x}{32 c^3 \left (1+a^2 x^2\right )}-\frac {11 \tan ^{-1}(a x)}{32 c^3}+\frac {\tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {\tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac {i \tan ^{-1}(a x)^2}{2 c^3}+\frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c^3}-\frac {i \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 90, normalized size = 0.57 \[ -\frac {64 i \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )+64 i \tan ^{-1}(a x)^2+24 \sin \left (2 \tan ^{-1}(a x)\right )+\sin \left (4 \tan ^{-1}(a x)\right )-4 \tan ^{-1}(a x) \left (32 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )+12 \cos \left (2 \tan ^{-1}(a x)\right )+\cos \left (4 \tan ^{-1}(a x)\right )\right )}{128 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/128*((64*I)*ArcTan[a*x]^2 - 4*ArcTan[a*x]*(12*Cos[2*ArcTan[a*x]] + Cos[4*ArcTan[a*x]] + 32*Log[1 - E^((2*I)
*ArcTan[a*x])]) + (64*I)*PolyLog[2, E^((2*I)*ArcTan[a*x])] + 24*Sin[2*ArcTan[a*x]] + Sin[4*ArcTan[a*x]])/c^3

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )}{a^{6} c^{3} x^{7} + 3 \, a^{4} c^{3} x^{5} + 3 \, a^{2} c^{3} x^{3} + c^{3} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(a*x)/(a^6*c^3*x^7 + 3*a^4*c^3*x^5 + 3*a^2*c^3*x^3 + c^3*x), 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(arctan(a*x)/x/(a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.11, size = 340, normalized size = 2.14 \[ \frac {\arctan \left (a x \right ) \ln \left (a x \right )}{c^{3}}-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 c^{3}}+\frac {\arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\arctan \left (a x \right )}{2 c^{3} \left (a^{2} x^{2}+1\right )}-\frac {11 x^{3} a^{3}}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {13 a x}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {11 \arctan \left (a x \right )}{32 c^{3}}+\frac {i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c^{3}}-\frac {i \dilog \left (-i a x +1\right )}{2 c^{3}}-\frac {i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c^{3}}+\frac {i \ln \left (a x -i\right )^{2}}{8 c^{3}}-\frac {i \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2 c^{3}}+\frac {i \dilog \left (i a x +1\right )}{2 c^{3}}-\frac {i \ln \left (a x +i\right )^{2}}{8 c^{3}}+\frac {i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c^{3}}-\frac {i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c^{3}}+\frac {i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c^{3}}+\frac {i \ln \left (a x \right ) \ln \left (i a x +1\right )}{2 c^{3}}-\frac {i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*arctan(a*x)*ln(a*x)-1/2/c^3*arctan(a*x)*ln(a^2*x^2+1)+1/4*arctan(a*x)/c^3/(a^2*x^2+1)^2+1/2*arctan(a*x)/
c^3/(a^2*x^2+1)-11/32/c^3/(a^2*x^2+1)^2*x^3*a^3-13/32*a*x/c^3/(a^2*x^2+1)^2-11/32*arctan(a*x)/c^3+1/4*I/c^3*ln
(I+a*x)*ln(a^2*x^2+1)-1/2*I/c^3*dilog(1-I*a*x)-1/4*I/c^3*ln(a*x-I)*ln(a^2*x^2+1)+1/8*I/c^3*ln(a*x-I)^2-1/2*I/c
^3*ln(a*x)*ln(1-I*a*x)-1/8*I/c^3*ln(I+a*x)^2+1/4*I/c^3*ln(a*x-I)*ln(-1/2*I*(I+a*x))+1/4*I/c^3*dilog(-1/2*I*(I+
a*x))+1/2*I/c^3*ln(a*x)*ln(1+I*a*x)-1/4*I/c^3*dilog(1/2*I*(a*x-I))-1/4*I/c^3*ln(I+a*x)*ln(1/2*I*(a*x-I))+1/2*I
/c^3*dilog(1+I*a*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x/(a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

integrate(arctan(a*x)/((a^2*c*x^2 + c)^3*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)/(x*(c + a^2*c*x^2)^3),x)

[Out]

int(atan(a*x)/(x*(c + a^2*c*x^2)^3), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RecursionError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)/x/(a**2*c*x**2+c)**3,x)

[Out]

Exception raised: RecursionError

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