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

Optimal. Leaf size=205 \[ \frac {3 i a^2 \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{2 c^3}-\frac {a^2 \tan ^{-1}(a x)}{c^3 \left (a^2 x^2+1\right )}-\frac {a^2 \tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 i a^2 \tan ^{-1}(a x)^2}{2 c^3}+\frac {3 a^2 \tan ^{-1}(a x)}{32 c^3}-\frac {3 a^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^3}+\frac {19 a^3 x}{32 c^3 \left (a^2 x^2+1\right )}+\frac {a^3 x}{16 c^3 \left (a^2 x^2+1\right )^2}-\frac {\tan ^{-1}(a x)}{2 c^3 x^2}-\frac {a}{2 c^3 x} \]

[Out]

-1/2*a/c^3/x+1/16*a^3*x/c^3/(a^2*x^2+1)^2+19/32*a^3*x/c^3/(a^2*x^2+1)+3/32*a^2*arctan(a*x)/c^3-1/2*arctan(a*x)
/c^3/x^2-1/4*a^2*arctan(a*x)/c^3/(a^2*x^2+1)^2-a^2*arctan(a*x)/c^3/(a^2*x^2+1)+3/2*I*a^2*arctan(a*x)^2/c^3-3*a
^2*arctan(a*x)*ln(2-2/(1-I*a*x))/c^3+3/2*I*a^2*polylog(2,-1+2/(1-I*a*x))/c^3

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Rubi [A]  time = 0.76, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {4966, 4918, 4852, 325, 203, 4924, 4868, 2447, 4930, 199, 205} \[ \frac {3 i a^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c^3}+\frac {19 a^3 x}{32 c^3 \left (a^2 x^2+1\right )}+\frac {a^3 x}{16 c^3 \left (a^2 x^2+1\right )^2}-\frac {a^2 \tan ^{-1}(a x)}{c^3 \left (a^2 x^2+1\right )}-\frac {a^2 \tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 i a^2 \tan ^{-1}(a x)^2}{2 c^3}+\frac {3 a^2 \tan ^{-1}(a x)}{32 c^3}-\frac {3 a^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^3}-\frac {\tan ^{-1}(a x)}{2 c^3 x^2}-\frac {a}{2 c^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(2*c^3*x) + (a^3*x)/(16*c^3*(1 + a^2*x^2)^2) + (19*a^3*x)/(32*c^3*(1 + a^2*x^2)) + (3*a^2*ArcTan[a*x])/(32*
c^3) - ArcTan[a*x]/(2*c^3*x^2) - (a^2*ArcTan[a*x])/(4*c^3*(1 + a^2*x^2)^2) - (a^2*ArcTan[a*x])/(c^3*(1 + a^2*x
^2)) + (((3*I)/2)*a^2*ArcTan[a*x]^2)/c^3 - (3*a^2*ArcTan[a*x]*Log[2 - 2/(1 - I*a*x)])/c^3 + (((3*I)/2)*a^2*Pol
yLog[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 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 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 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 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]

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^3 \left (c+a^2 c x^2\right )^3} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )^3} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)}{x^3 \left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=a^4 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^3} \, dx+\frac {\int \frac {\tan ^{-1}(a x)}{x^3 \left (c+a^2 c x^2\right )} \, dx}{c^2}-2 \frac {a^2 \int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=-\frac {a^2 \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {1}{4} a^3 \int \frac {1}{\left (c+a^2 c x^2\right )^3} \, dx+\frac {\int \frac {\tan ^{-1}(a x)}{x^3} \, dx}{c^3}-\frac {a^2 \int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )} \, dx}{c^2}-2 \left (\frac {a^2 \int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )} \, dx}{c^2}-\frac {a^4 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{c}\right )\\ &=\frac {a^3 x}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac {\tan ^{-1}(a x)}{2 c^3 x^2}-\frac {a^2 \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {i a^2 \tan ^{-1}(a x)^2}{2 c^3}+\frac {a \int \frac {1}{x^2 \left (1+a^2 x^2\right )} \, dx}{2 c^3}-\frac {\left (i a^2\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c^3}+\frac {\left (3 a^3\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{16 c}-2 \left (\frac {a^2 \tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac {i a^2 \tan ^{-1}(a x)^2}{2 c^3}+\frac {\left (i a^2\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c^3}-\frac {a^3 \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{2 c}\right )\\ &=-\frac {a}{2 c^3 x}+\frac {a^3 x}{16 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 a^3 x}{32 c^3 \left (1+a^2 x^2\right )}-\frac {\tan ^{-1}(a x)}{2 c^3 x^2}-\frac {a^2 \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {i a^2 \tan ^{-1}(a x)^2}{2 c^3}-\frac {a^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c^3}-\frac {a^3 \int \frac {1}{1+a^2 x^2} \, dx}{2 c^3}+\frac {a^3 \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c^3}+\frac {\left (3 a^3\right ) \int \frac {1}{c+a^2 c x^2} \, dx}{32 c^2}-2 \left (-\frac {a^3 x}{4 c^3 \left (1+a^2 x^2\right )}+\frac {a^2 \tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac {i a^2 \tan ^{-1}(a x)^2}{2 c^3}+\frac {a^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c^3}-\frac {a^3 \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c^3}-\frac {a^3 \int \frac {1}{c+a^2 c x^2} \, dx}{4 c^2}\right )\\ &=-\frac {a}{2 c^3 x}+\frac {a^3 x}{16 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 a^3 x}{32 c^3 \left (1+a^2 x^2\right )}-\frac {13 a^2 \tan ^{-1}(a x)}{32 c^3}-\frac {\tan ^{-1}(a x)}{2 c^3 x^2}-\frac {a^2 \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {i a^2 \tan ^{-1}(a x)^2}{2 c^3}-\frac {a^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c^3}+\frac {i a^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c^3}-2 \left (-\frac {a^3 x}{4 c^3 \left (1+a^2 x^2\right )}-\frac {a^2 \tan ^{-1}(a x)}{4 c^3}+\frac {a^2 \tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac {i a^2 \tan ^{-1}(a x)^2}{2 c^3}+\frac {a^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c^3}-\frac {i a^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c^3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 111, normalized size = 0.54 \[ \frac {a^2 \left (\tan ^{-1}(a x) \left (-\frac {64}{a^2 x^2}-384 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )-80 \cos \left (2 \tan ^{-1}(a x)\right )-4 \cos \left (4 \tan ^{-1}(a x)\right )-64\right )+192 i \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )-\frac {64}{a x}+192 i \tan ^{-1}(a x)^2+40 \sin \left (2 \tan ^{-1}(a x)\right )+\sin \left (4 \tan ^{-1}(a x)\right )\right )}{128 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*(-64/(a*x) + (192*I)*ArcTan[a*x]^2 + ArcTan[a*x]*(-64 - 64/(a^2*x^2) - 80*Cos[2*ArcTan[a*x]] - 4*Cos[4*Ar
cTan[a*x]] - 384*Log[1 - E^((2*I)*ArcTan[a*x])]) + (192*I)*PolyLog[2, E^((2*I)*ArcTan[a*x])] + 40*Sin[2*ArcTan
[a*x]] + Sin[4*ArcTan[a*x]]))/(128*c^3)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )}{a^{6} c^{3} x^{9} + 3 \, a^{4} c^{3} x^{7} + 3 \, a^{2} c^{3} x^{5} + c^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(a*x)/(a^6*c^3*x^9 + 3*a^4*c^3*x^7 + 3*a^2*c^3*x^5 + c^3*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(arctan(a*x)/x^3/(a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.13, size = 415, normalized size = 2.02 \[ -\frac {\arctan \left (a x \right )}{2 c^{3} x^{2}}-\frac {3 a^{2} \arctan \left (a x \right ) \ln \left (a x \right )}{c^{3}}+\frac {3 a^{2} \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 c^{3}}-\frac {a^{2} \arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {a^{2} \arctan \left (a x \right )}{c^{3} \left (a^{2} x^{2}+1\right )}-\frac {3 i a^{2} \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c^{3}}-\frac {3 i a^{2} \ln \left (a x \right ) \ln \left (i a x +1\right )}{2 c^{3}}+\frac {3 i a^{2} \dilog \left (-i a x +1\right )}{2 c^{3}}-\frac {3 i a^{2} \dilog \left (i a x +1\right )}{2 c^{3}}+\frac {3 i a^{2} \ln \left (a x +i\right )^{2}}{8 c^{3}}+\frac {3 i a^{2} \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2 c^{3}}+\frac {3 i a^{2} \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c^{3}}+\frac {3 i a^{2} \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c^{3}}-\frac {3 i a^{2} \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c^{3}}-\frac {3 i a^{2} \ln \left (a x -i\right )^{2}}{8 c^{3}}+\frac {3 i a^{2} \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c^{3}}-\frac {3 i a^{2} \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c^{3}}-\frac {a}{2 c^{3} x}+\frac {19 a^{5} x^{3}}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {21 a^{3} x}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a^{2} \arctan \left (a x \right )}{32 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctan(a*x)/c^3/x^2-3*a^2/c^3*arctan(a*x)*ln(a*x)+3/2*a^2/c^3*arctan(a*x)*ln(a^2*x^2+1)-1/4*a^2*arctan(a*
x)/c^3/(a^2*x^2+1)^2-a^2*arctan(a*x)/c^3/(a^2*x^2+1)+3/2*I*a^2/c^3*dilog(1-I*a*x)+3/4*I*a^2/c^3*dilog(1/2*I*(a
*x-I))-3/2*I*a^2/c^3*ln(a*x)*ln(1+I*a*x)-3/2*I*a^2/c^3*dilog(1+I*a*x)+3/8*I*a^2/c^3*ln(I+a*x)^2+3/2*I*a^2/c^3*
ln(a*x)*ln(1-I*a*x)+3/4*I*a^2/c^3*ln(a*x-I)*ln(a^2*x^2+1)+3/4*I*a^2/c^3*ln(I+a*x)*ln(1/2*I*(a*x-I))-3/4*I*a^2/
c^3*ln(I+a*x)*ln(a^2*x^2+1)-3/8*I*a^2/c^3*ln(a*x-I)^2-3/4*I*a^2/c^3*ln(a*x-I)*ln(-1/2*I*(I+a*x))-3/4*I*a^2/c^3
*dilog(-1/2*I*(I+a*x))-1/2*a/c^3/x+19/32*a^5/c^3/(a^2*x^2+1)^2*x^3+21/32*a^3*x/c^3/(a^2*x^2+1)^2+3/32*a^2*arct
an(a*x)/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}{\left (a x \right )}}{a^{6} x^{9} + 3 a^{4} x^{7} + 3 a^{2} x^{5} + x^{3}}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)/(a**6*x**9 + 3*a**4*x**7 + 3*a**2*x**5 + x**3), x)/c**3

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