3.273 \(\int \frac {(c+a^2 c x^2)^2 \tan ^{-1}(a x)^2}{x^4} \, dx\)

Optimal. Leaf size=216 \[ a^4 c^2 x \tan ^{-1}(a x)^2-\frac {5}{3} i a^3 c^2 \text {Li}_2\left (\frac {2}{1-i a x}-1\right )+i a^3 c^2 \text {Li}_2\left (1-\frac {2}{i a x+1}\right )-\frac {2}{3} i a^3 c^2 \tan ^{-1}(a x)^2-\frac {1}{3} a^3 c^2 \tan ^{-1}(a x)+2 a^3 c^2 \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)+\frac {10}{3} a^3 c^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)-\frac {a^2 c^2}{3 x}-\frac {2 a^2 c^2 \tan ^{-1}(a x)^2}{x}-\frac {c^2 \tan ^{-1}(a x)^2}{3 x^3}-\frac {a c^2 \tan ^{-1}(a x)}{3 x^2} \]

[Out]

-1/3*a^2*c^2/x-1/3*a^3*c^2*arctan(a*x)-1/3*a*c^2*arctan(a*x)/x^2-2/3*I*a^3*c^2*arctan(a*x)^2-1/3*c^2*arctan(a*
x)^2/x^3-2*a^2*c^2*arctan(a*x)^2/x+a^4*c^2*x*arctan(a*x)^2+2*a^3*c^2*arctan(a*x)*ln(2/(1+I*a*x))+10/3*a^3*c^2*
arctan(a*x)*ln(2-2/(1-I*a*x))-5/3*I*a^3*c^2*polylog(2,-1+2/(1-I*a*x))+I*a^3*c^2*polylog(2,1-2/(1+I*a*x))

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Rubi [A]  time = 0.44, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {4948, 4846, 4920, 4854, 2402, 2315, 4852, 4918, 325, 203, 4924, 4868, 2447} \[ -\frac {5}{3} i a^3 c^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )+i a^3 c^2 \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )-\frac {a^2 c^2}{3 x}+a^4 c^2 x \tan ^{-1}(a x)^2-\frac {2}{3} i a^3 c^2 \tan ^{-1}(a x)^2-\frac {1}{3} a^3 c^2 \tan ^{-1}(a x)-\frac {2 a^2 c^2 \tan ^{-1}(a x)^2}{x}+2 a^3 c^2 \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)+\frac {10}{3} a^3 c^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)-\frac {a c^2 \tan ^{-1}(a x)}{3 x^2}-\frac {c^2 \tan ^{-1}(a x)^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((c + a^2*c*x^2)^2*ArcTan[a*x]^2)/x^4,x]

[Out]

-(a^2*c^2)/(3*x) - (a^3*c^2*ArcTan[a*x])/3 - (a*c^2*ArcTan[a*x])/(3*x^2) - ((2*I)/3)*a^3*c^2*ArcTan[a*x]^2 - (
c^2*ArcTan[a*x]^2)/(3*x^3) - (2*a^2*c^2*ArcTan[a*x]^2)/x + a^4*c^2*x*ArcTan[a*x]^2 + 2*a^3*c^2*ArcTan[a*x]*Log
[2/(1 + I*a*x)] + (10*a^3*c^2*ArcTan[a*x]*Log[2 - 2/(1 - I*a*x)])/3 - ((5*I)/3)*a^3*c^2*PolyLog[2, -1 + 2/(1 -
I*a*x)] + I*a^3*c^2*PolyLog[2, 1 - 2/(1 + I*a*x)]

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 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 4846

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

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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[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 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 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[c*x])^(p + 1))/(b*e*(p + 1)), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b,
c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 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 4948

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Int[Ex
pandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e,
c^2*d] && IGtQ[p, 0] && IGtQ[q, 1] && (EqQ[p, 1] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^2}{x^4} \, dx &=\int \left (a^4 c^2 \tan ^{-1}(a x)^2+\frac {c^2 \tan ^{-1}(a x)^2}{x^4}+\frac {2 a^2 c^2 \tan ^{-1}(a x)^2}{x^2}\right ) \, dx\\ &=c^2 \int \frac {\tan ^{-1}(a x)^2}{x^4} \, dx+\left (2 a^2 c^2\right ) \int \frac {\tan ^{-1}(a x)^2}{x^2} \, dx+\left (a^4 c^2\right ) \int \tan ^{-1}(a x)^2 \, dx\\ &=-\frac {c^2 \tan ^{-1}(a x)^2}{3 x^3}-\frac {2 a^2 c^2 \tan ^{-1}(a x)^2}{x}+a^4 c^2 x \tan ^{-1}(a x)^2+\frac {1}{3} \left (2 a c^2\right ) \int \frac {\tan ^{-1}(a x)}{x^3 \left (1+a^2 x^2\right )} \, dx+\left (4 a^3 c^2\right ) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx-\left (2 a^5 c^2\right ) \int \frac {x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=-i a^3 c^2 \tan ^{-1}(a x)^2-\frac {c^2 \tan ^{-1}(a x)^2}{3 x^3}-\frac {2 a^2 c^2 \tan ^{-1}(a x)^2}{x}+a^4 c^2 x \tan ^{-1}(a x)^2+\frac {1}{3} \left (2 a c^2\right ) \int \frac {\tan ^{-1}(a x)}{x^3} \, dx+\left (4 i a^3 c^2\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx-\frac {1}{3} \left (2 a^3 c^2\right ) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx+\left (2 a^4 c^2\right ) \int \frac {\tan ^{-1}(a x)}{i-a x} \, dx\\ &=-\frac {a c^2 \tan ^{-1}(a x)}{3 x^2}-\frac {2}{3} i a^3 c^2 \tan ^{-1}(a x)^2-\frac {c^2 \tan ^{-1}(a x)^2}{3 x^3}-\frac {2 a^2 c^2 \tan ^{-1}(a x)^2}{x}+a^4 c^2 x \tan ^{-1}(a x)^2+2 a^3 c^2 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )+4 a^3 c^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )+\frac {1}{3} \left (a^2 c^2\right ) \int \frac {1}{x^2 \left (1+a^2 x^2\right )} \, dx-\frac {1}{3} \left (2 i a^3 c^2\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx-\left (2 a^4 c^2\right ) \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx-\left (4 a^4 c^2\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {a^2 c^2}{3 x}-\frac {a c^2 \tan ^{-1}(a x)}{3 x^2}-\frac {2}{3} i a^3 c^2 \tan ^{-1}(a x)^2-\frac {c^2 \tan ^{-1}(a x)^2}{3 x^3}-\frac {2 a^2 c^2 \tan ^{-1}(a x)^2}{x}+a^4 c^2 x \tan ^{-1}(a x)^2+2 a^3 c^2 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )+\frac {10}{3} a^3 c^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )-2 i a^3 c^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+\left (2 i a^3 c^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )-\frac {1}{3} \left (a^4 c^2\right ) \int \frac {1}{1+a^2 x^2} \, dx+\frac {1}{3} \left (2 a^4 c^2\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {a^2 c^2}{3 x}-\frac {1}{3} a^3 c^2 \tan ^{-1}(a x)-\frac {a c^2 \tan ^{-1}(a x)}{3 x^2}-\frac {2}{3} i a^3 c^2 \tan ^{-1}(a x)^2-\frac {c^2 \tan ^{-1}(a x)^2}{3 x^3}-\frac {2 a^2 c^2 \tan ^{-1}(a x)^2}{x}+a^4 c^2 x \tan ^{-1}(a x)^2+2 a^3 c^2 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )+\frac {10}{3} a^3 c^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )-\frac {5}{3} i a^3 c^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+i a^3 c^2 \text {Li}_2\left (1-\frac {2}{1+i a x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 189, normalized size = 0.88 \[ \frac {c^2 \left (3 a^4 x^4 \tan ^{-1}(a x)^2-3 i a^3 x^3 \text {Li}_2\left (-e^{2 i \tan ^{-1}(a x)}\right )-5 i a^3 x^3 \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )-8 i a^3 x^3 \tan ^{-1}(a x)^2-a^3 x^3 \tan ^{-1}(a x)+10 a^3 x^3 \tan ^{-1}(a x) \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )+6 a^3 x^3 \tan ^{-1}(a x) \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )-a^2 x^2-6 a^2 x^2 \tan ^{-1}(a x)^2-a x \tan ^{-1}(a x)-\tan ^{-1}(a x)^2\right )}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((c + a^2*c*x^2)^2*ArcTan[a*x]^2)/x^4,x]

[Out]

(c^2*(-(a^2*x^2) - a*x*ArcTan[a*x] - a^3*x^3*ArcTan[a*x] - ArcTan[a*x]^2 - 6*a^2*x^2*ArcTan[a*x]^2 - (8*I)*a^3
*x^3*ArcTan[a*x]^2 + 3*a^4*x^4*ArcTan[a*x]^2 + 10*a^3*x^3*ArcTan[a*x]*Log[1 - E^((2*I)*ArcTan[a*x])] + 6*a^3*x
^3*ArcTan[a*x]*Log[1 + E^((2*I)*ArcTan[a*x])] - (3*I)*a^3*x^3*PolyLog[2, -E^((2*I)*ArcTan[a*x])] - (5*I)*a^3*x
^3*PolyLog[2, E^((2*I)*ArcTan[a*x])]))/(3*x^3)

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )^{2}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^2*arctan(a*x)^2/x^4,x, algorithm="fricas")

[Out]

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*arctan(a*x)^2/x^4, x)

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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((a^2*c*x^2+c)^2*arctan(a*x)^2/x^4,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.08, size = 375, normalized size = 1.74 \[ a^{4} c^{2} x \arctan \left (a x \right )^{2}-\frac {c^{2} \arctan \left (a x \right )^{2}}{3 x^{3}}-\frac {2 a^{2} c^{2} \arctan \left (a x \right )^{2}}{x}-\frac {a \,c^{2} \arctan \left (a x \right )}{3 x^{2}}+\frac {10 a^{3} c^{2} \arctan \left (a x \right ) \ln \left (a x \right )}{3}-\frac {8 a^{3} c^{2} \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{3}-\frac {a^{2} c^{2}}{3 x}-\frac {a^{3} c^{2} \arctan \left (a x \right )}{3}-\frac {2 i a^{3} c^{2} \ln \left (a x +i\right )^{2}}{3}+\frac {4 i a^{3} c^{2} \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{3}-\frac {4 i a^{3} c^{2} \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{3}+\frac {2 i a^{3} c^{2} \ln \left (a x -i\right )^{2}}{3}+\frac {5 i a^{3} c^{2} \ln \left (a x \right ) \ln \left (i a x +1\right )}{3}-\frac {4 i a^{3} c^{2} \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{3}-\frac {5 i a^{3} c^{2} \dilog \left (-i a x +1\right )}{3}-\frac {4 i a^{3} c^{2} \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{3}+\frac {4 i a^{3} c^{2} \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{3}-\frac {5 i a^{3} c^{2} \ln \left (a x \right ) \ln \left (-i a x +1\right )}{3}+\frac {5 i a^{3} c^{2} \dilog \left (i a x +1\right )}{3}+\frac {4 i a^{3} c^{2} \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^2*arctan(a*x)^2/x^4,x)

[Out]

a^4*c^2*x*arctan(a*x)^2-1/3*c^2*arctan(a*x)^2/x^3-2*a^2*c^2*arctan(a*x)^2/x-1/3*a*c^2*arctan(a*x)/x^2+10/3*a^3
*c^2*arctan(a*x)*ln(a*x)-8/3*a^3*c^2*arctan(a*x)*ln(a^2*x^2+1)-1/3*a^2*c^2/x-1/3*a^3*c^2*arctan(a*x)+5/3*I*a^3
*c^2*dilog(1+I*a*x)-2/3*I*a^3*c^2*ln(I+a*x)^2+4/3*I*a^3*c^2*ln(a*x-I)*ln(-1/2*I*(I+a*x))+4/3*I*a^3*c^2*dilog(-
1/2*I*(I+a*x))+2/3*I*a^3*c^2*ln(a*x-I)^2+5/3*I*a^3*c^2*ln(a*x)*ln(1+I*a*x)-4/3*I*a^3*c^2*ln(a*x-I)*ln(a^2*x^2+
1)-4/3*I*a^3*c^2*ln(I+a*x)*ln(1/2*I*(a*x-I))+4/3*I*a^3*c^2*ln(I+a*x)*ln(a^2*x^2+1)-5/3*I*a^3*c^2*dilog(1-I*a*x
)-5/3*I*a^3*c^2*ln(a*x)*ln(1-I*a*x)-4/3*I*a^3*c^2*dilog(1/2*I*(a*x-I))

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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((a^2*c*x^2+c)^2*arctan(a*x)^2/x^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atan(a*x)^2*(c + a^2*c*x^2)^2)/x^4,x)

[Out]

int((atan(a*x)^2*(c + a^2*c*x^2)^2)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**2*atan(a*x)**2/x**4,x)

[Out]

c**2*(Integral(a**4*atan(a*x)**2, x) + Integral(atan(a*x)**2/x**4, x) + Integral(2*a**2*atan(a*x)**2/x**2, x))

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