3.3.81 \(\int \frac {(c+a^2 c x^2)^3 \text {ArcTan}(a x)^2}{x^4} \, dx\) [281]

Optimal. Leaf size=250 \[ -\frac {a^2 c^3}{3 x}+\frac {1}{3} a^4 c^3 x-\frac {2}{3} a^3 c^3 \text {ArcTan}(a x)-\frac {a c^3 \text {ArcTan}(a x)}{3 x^2}-\frac {1}{3} a^5 c^3 x^2 \text {ArcTan}(a x)-\frac {c^3 \text {ArcTan}(a x)^2}{3 x^3}-\frac {3 a^2 c^3 \text {ArcTan}(a x)^2}{x}+3 a^4 c^3 x \text {ArcTan}(a x)^2+\frac {1}{3} a^6 c^3 x^3 \text {ArcTan}(a x)^2+\frac {16}{3} a^3 c^3 \text {ArcTan}(a x) \log \left (\frac {2}{1+i a x}\right )+\frac {16}{3} a^3 c^3 \text {ArcTan}(a x) \log \left (2-\frac {2}{1-i a x}\right )-\frac {8}{3} i a^3 c^3 \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )+\frac {8}{3} i a^3 c^3 \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right ) \]

[Out]

-1/3*a^2*c^3/x+1/3*a^4*c^3*x-2/3*a^3*c^3*arctan(a*x)-1/3*a*c^3*arctan(a*x)/x^2-1/3*a^5*c^3*x^2*arctan(a*x)-1/3
*c^3*arctan(a*x)^2/x^3-3*a^2*c^3*arctan(a*x)^2/x+3*a^4*c^3*x*arctan(a*x)^2+1/3*a^6*c^3*x^3*arctan(a*x)^2+16/3*
a^3*c^3*arctan(a*x)*ln(2/(1+I*a*x))+16/3*a^3*c^3*arctan(a*x)*ln(2-2/(1-I*a*x))-8/3*I*a^3*c^3*polylog(2,-1+2/(1
-I*a*x))+8/3*I*a^3*c^3*polylog(2,1-2/(1+I*a*x))

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Rubi [A]
time = 0.45, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 15, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.682, Rules used = {5068, 4930, 5040, 4964, 2449, 2352, 4946, 5038, 331, 209, 5044, 4988, 2497, 5036, 327} \begin {gather*} \frac {1}{3} a^6 c^3 x^3 \text {ArcTan}(a x)^2-\frac {1}{3} a^5 c^3 x^2 \text {ArcTan}(a x)+3 a^4 c^3 x \text {ArcTan}(a x)^2+\frac {1}{3} a^4 c^3 x-\frac {2}{3} a^3 c^3 \text {ArcTan}(a x)+\frac {16}{3} a^3 c^3 \text {ArcTan}(a x) \log \left (\frac {2}{1+i a x}\right )+\frac {16}{3} a^3 c^3 \text {ArcTan}(a x) \log \left (2-\frac {2}{1-i a x}\right )-\frac {8}{3} i a^3 c^3 \text {Li}_2\left (\frac {2}{1-i a x}-1\right )+\frac {8}{3} i a^3 c^3 \text {Li}_2\left (1-\frac {2}{i a x+1}\right )-\frac {3 a^2 c^3 \text {ArcTan}(a x)^2}{x}-\frac {a^2 c^3}{3 x}-\frac {c^3 \text {ArcTan}(a x)^2}{3 x^3}-\frac {a c^3 \text {ArcTan}(a x)}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((c + a^2*c*x^2)^3*ArcTan[a*x]^2)/x^4,x]

[Out]

-1/3*(a^2*c^3)/x + (a^4*c^3*x)/3 - (2*a^3*c^3*ArcTan[a*x])/3 - (a*c^3*ArcTan[a*x])/(3*x^2) - (a^5*c^3*x^2*ArcT
an[a*x])/3 - (c^3*ArcTan[a*x]^2)/(3*x^3) - (3*a^2*c^3*ArcTan[a*x]^2)/x + 3*a^4*c^3*x*ArcTan[a*x]^2 + (a^6*c^3*
x^3*ArcTan[a*x]^2)/3 + (16*a^3*c^3*ArcTan[a*x]*Log[2/(1 + I*a*x)])/3 + (16*a^3*c^3*ArcTan[a*x]*Log[2 - 2/(1 -
I*a*x)])/3 - ((8*I)/3)*a^3*c^3*PolyLog[2, -1 + 2/(1 - I*a*x)] + ((8*I)/3)*a^3*c^3*PolyLog[2, 1 - 2/(1 + I*a*x)
]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 327

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

Rule 331

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 2352

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

Rule 2449

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 2497

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 4930

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

Rule 4946

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

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 4988

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 5036

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d*(f^2/e), 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] && GtQ[m, 1]

Rule 5038

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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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 5044

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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 5068

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 )^3 \tan ^{-1}(a x)^2}{x^4} \, dx &=\int \left (3 a^4 c^3 \tan ^{-1}(a x)^2+\frac {c^3 \tan ^{-1}(a x)^2}{x^4}+\frac {3 a^2 c^3 \tan ^{-1}(a x)^2}{x^2}+a^6 c^3 x^2 \tan ^{-1}(a x)^2\right ) \, dx\\ &=c^3 \int \frac {\tan ^{-1}(a x)^2}{x^4} \, dx+\left (3 a^2 c^3\right ) \int \frac {\tan ^{-1}(a x)^2}{x^2} \, dx+\left (3 a^4 c^3\right ) \int \tan ^{-1}(a x)^2 \, dx+\left (a^6 c^3\right ) \int x^2 \tan ^{-1}(a x)^2 \, dx\\ &=-\frac {c^3 \tan ^{-1}(a x)^2}{3 x^3}-\frac {3 a^2 c^3 \tan ^{-1}(a x)^2}{x}+3 a^4 c^3 x \tan ^{-1}(a x)^2+\frac {1}{3} a^6 c^3 x^3 \tan ^{-1}(a x)^2+\frac {1}{3} \left (2 a c^3\right ) \int \frac {\tan ^{-1}(a x)}{x^3 \left (1+a^2 x^2\right )} \, dx+\left (6 a^3 c^3\right ) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx-\left (6 a^5 c^3\right ) \int \frac {x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx-\frac {1}{3} \left (2 a^7 c^3\right ) \int \frac {x^3 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=-\frac {c^3 \tan ^{-1}(a x)^2}{3 x^3}-\frac {3 a^2 c^3 \tan ^{-1}(a x)^2}{x}+3 a^4 c^3 x \tan ^{-1}(a x)^2+\frac {1}{3} a^6 c^3 x^3 \tan ^{-1}(a x)^2+\frac {1}{3} \left (2 a c^3\right ) \int \frac {\tan ^{-1}(a x)}{x^3} \, dx+\left (6 i a^3 c^3\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx-\frac {1}{3} \left (2 a^3 c^3\right ) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx+\left (6 a^4 c^3\right ) \int \frac {\tan ^{-1}(a x)}{i-a x} \, dx-\frac {1}{3} \left (2 a^5 c^3\right ) \int x \tan ^{-1}(a x) \, dx+\frac {1}{3} \left (2 a^5 c^3\right ) \int \frac {x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=-\frac {a c^3 \tan ^{-1}(a x)}{3 x^2}-\frac {1}{3} a^5 c^3 x^2 \tan ^{-1}(a x)-\frac {c^3 \tan ^{-1}(a x)^2}{3 x^3}-\frac {3 a^2 c^3 \tan ^{-1}(a x)^2}{x}+3 a^4 c^3 x \tan ^{-1}(a x)^2+\frac {1}{3} a^6 c^3 x^3 \tan ^{-1}(a x)^2+6 a^3 c^3 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )+6 a^3 c^3 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )+\frac {1}{3} \left (a^2 c^3\right ) \int \frac {1}{x^2 \left (1+a^2 x^2\right )} \, dx-\frac {1}{3} \left (2 i a^3 c^3\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx-\frac {1}{3} \left (2 a^4 c^3\right ) \int \frac {\tan ^{-1}(a x)}{i-a x} \, dx-\left (6 a^4 c^3\right ) \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx-\left (6 a^4 c^3\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx+\frac {1}{3} \left (a^6 c^3\right ) \int \frac {x^2}{1+a^2 x^2} \, dx\\ &=-\frac {a^2 c^3}{3 x}+\frac {1}{3} a^4 c^3 x-\frac {a c^3 \tan ^{-1}(a x)}{3 x^2}-\frac {1}{3} a^5 c^3 x^2 \tan ^{-1}(a x)-\frac {c^3 \tan ^{-1}(a x)^2}{3 x^3}-\frac {3 a^2 c^3 \tan ^{-1}(a x)^2}{x}+3 a^4 c^3 x \tan ^{-1}(a x)^2+\frac {1}{3} a^6 c^3 x^3 \tan ^{-1}(a x)^2+\frac {16}{3} a^3 c^3 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )+\frac {16}{3} a^3 c^3 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )-3 i a^3 c^3 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+\left (6 i a^3 c^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )-2 \left (\frac {1}{3} \left (a^4 c^3\right ) \int \frac {1}{1+a^2 x^2} \, dx\right )+\frac {1}{3} \left (2 a^4 c^3\right ) \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx+\frac {1}{3} \left (2 a^4 c^3\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {a^2 c^3}{3 x}+\frac {1}{3} a^4 c^3 x-\frac {2}{3} a^3 c^3 \tan ^{-1}(a x)-\frac {a c^3 \tan ^{-1}(a x)}{3 x^2}-\frac {1}{3} a^5 c^3 x^2 \tan ^{-1}(a x)-\frac {c^3 \tan ^{-1}(a x)^2}{3 x^3}-\frac {3 a^2 c^3 \tan ^{-1}(a x)^2}{x}+3 a^4 c^3 x \tan ^{-1}(a x)^2+\frac {1}{3} a^6 c^3 x^3 \tan ^{-1}(a x)^2+\frac {16}{3} a^3 c^3 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )+\frac {16}{3} a^3 c^3 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )-\frac {8}{3} i a^3 c^3 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+3 i a^3 c^3 \text {Li}_2\left (1-\frac {2}{1+i a x}\right )-\frac {1}{3} \left (2 i a^3 c^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )\\ &=-\frac {a^2 c^3}{3 x}+\frac {1}{3} a^4 c^3 x-\frac {2}{3} a^3 c^3 \tan ^{-1}(a x)-\frac {a c^3 \tan ^{-1}(a x)}{3 x^2}-\frac {1}{3} a^5 c^3 x^2 \tan ^{-1}(a x)-\frac {c^3 \tan ^{-1}(a x)^2}{3 x^3}-\frac {3 a^2 c^3 \tan ^{-1}(a x)^2}{x}+3 a^4 c^3 x \tan ^{-1}(a x)^2+\frac {1}{3} a^6 c^3 x^3 \tan ^{-1}(a x)^2+\frac {16}{3} a^3 c^3 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )+\frac {16}{3} a^3 c^3 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )-\frac {8}{3} i a^3 c^3 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )+\frac {8}{3} i a^3 c^3 \text {Li}_2\left (1-\frac {2}{1+i a x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.42, size = 221, normalized size = 0.88 \begin {gather*} \frac {c^3 \left (-a^2 x^2+a^4 x^4-a x \text {ArcTan}(a x)-2 a^3 x^3 \text {ArcTan}(a x)-a^5 x^5 \text {ArcTan}(a x)-\text {ArcTan}(a x)^2-9 a^2 x^2 \text {ArcTan}(a x)^2-16 i a^3 x^3 \text {ArcTan}(a x)^2+9 a^4 x^4 \text {ArcTan}(a x)^2+a^6 x^6 \text {ArcTan}(a x)^2+16 a^3 x^3 \text {ArcTan}(a x) \log \left (1-e^{2 i \text {ArcTan}(a x)}\right )+16 a^3 x^3 \text {ArcTan}(a x) \log \left (1+e^{2 i \text {ArcTan}(a x)}\right )-8 i a^3 x^3 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(a x)}\right )-8 i a^3 x^3 \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(a x)}\right )\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((c + a^2*c*x^2)^3*ArcTan[a*x]^2)/x^4,x]

[Out]

(c^3*(-(a^2*x^2) + a^4*x^4 - a*x*ArcTan[a*x] - 2*a^3*x^3*ArcTan[a*x] - a^5*x^5*ArcTan[a*x] - ArcTan[a*x]^2 - 9
*a^2*x^2*ArcTan[a*x]^2 - (16*I)*a^3*x^3*ArcTan[a*x]^2 + 9*a^4*x^4*ArcTan[a*x]^2 + a^6*x^6*ArcTan[a*x]^2 + 16*a
^3*x^3*ArcTan[a*x]*Log[1 - E^((2*I)*ArcTan[a*x])] + 16*a^3*x^3*ArcTan[a*x]*Log[1 + E^((2*I)*ArcTan[a*x])] - (8
*I)*a^3*x^3*PolyLog[2, -E^((2*I)*ArcTan[a*x])] - (8*I)*a^3*x^3*PolyLog[2, E^((2*I)*ArcTan[a*x])]))/(3*x^3)

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Maple [A]
time = 0.33, size = 324, normalized size = 1.30

method	result	size
derivativedivides	\(a^{3} \left (\frac {a^{3} c^{3} x^{3} \arctan \left (a x \right )^{2}}{3}+3 a \,c^{3} x \arctan \left (a x \right )^{2}-\frac {3 c^{3} \arctan \left (a x \right )^{2}}{a x}-\frac {c^{3} \arctan \left (a x \right )^{2}}{3 a^{3} x^{3}}-\frac {2 c^{3} \left (\frac {\arctan \left (a x \right ) a^{2} x^{2}}{2}+8 \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )+\frac {\arctan \left (a x \right )}{2 a^{2} x^{2}}-8 \arctan \left (a x \right ) \ln \left (a x \right )-\frac {a x}{2}+\frac {1}{2 a x}+\arctan \left (a x \right )+4 i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )-4 i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )-4 i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )+4 i \dilog \left (-i a x +1\right )-4 i \ln \left (a x \right ) \ln \left (i a x +1\right )+4 i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )-4 i \dilog \left (i a x +1\right )+2 i \ln \left (a x +i\right )^{2}-2 i \ln \left (a x -i\right )^{2}-4 i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )+4 i \ln \left (a x \right ) \ln \left (-i a x +1\right )+4 i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )\right )}{3}\right )\)	\(324\)
default	\(a^{3} \left (\frac {a^{3} c^{3} x^{3} \arctan \left (a x \right )^{2}}{3}+3 a \,c^{3} x \arctan \left (a x \right )^{2}-\frac {3 c^{3} \arctan \left (a x \right )^{2}}{a x}-\frac {c^{3} \arctan \left (a x \right )^{2}}{3 a^{3} x^{3}}-\frac {2 c^{3} \left (\frac {\arctan \left (a x \right ) a^{2} x^{2}}{2}+8 \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )+\frac {\arctan \left (a x \right )}{2 a^{2} x^{2}}-8 \arctan \left (a x \right ) \ln \left (a x \right )-\frac {a x}{2}+\frac {1}{2 a x}+\arctan \left (a x \right )+4 i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )-4 i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )-4 i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )+4 i \dilog \left (-i a x +1\right )-4 i \ln \left (a x \right ) \ln \left (i a x +1\right )+4 i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )-4 i \dilog \left (i a x +1\right )+2 i \ln \left (a x +i\right )^{2}-2 i \ln \left (a x -i\right )^{2}-4 i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )+4 i \ln \left (a x \right ) \ln \left (-i a x +1\right )+4 i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )\right )}{3}\right )\)	\(324\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^3*arctan(a*x)^2/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^3*arctan(a*x)^2/x^4,x, algorithm="maxima")

[Out]

1/48*(24*(72*a^8*c^3*integrate(1/48*x^8*arctan(a*x)^2/(a^2*x^6 + x^4), x) + 6*a^8*c^3*integrate(1/48*x^8*log(a
^2*x^2 + 1)^2/(a^2*x^6 + x^4), x) + 8*a^8*c^3*integrate(1/48*x^8*log(a^2*x^2 + 1)/(a^2*x^6 + x^4), x) - 16*a^7
*c^3*integrate(1/48*x^7*arctan(a*x)/(a^2*x^6 + x^4), x) + 288*a^6*c^3*integrate(1/48*x^6*arctan(a*x)^2/(a^2*x^
6 + x^4), x) + 24*a^6*c^3*integrate(1/48*x^6*log(a^2*x^2 + 1)^2/(a^2*x^6 + x^4), x) + 72*a^6*c^3*integrate(1/4
8*x^6*log(a^2*x^2 + 1)/(a^2*x^6 + x^4), x) + 3*a^3*c^3*arctan(a*x)^3 - 144*a^5*c^3*integrate(1/48*x^5*arctan(a
*x)/(a^2*x^6 + x^4), x) + 36*a^4*c^3*integrate(1/48*x^4*log(a^2*x^2 + 1)^2/(a^2*x^6 + x^4), x) - 72*a^4*c^3*in
tegrate(1/48*x^4*log(a^2*x^2 + 1)/(a^2*x^6 + x^4), x) + 144*a^3*c^3*integrate(1/48*x^3*arctan(a*x)/(a^2*x^6 +
x^4), x) + 288*a^2*c^3*integrate(1/48*x^2*arctan(a*x)^2/(a^2*x^6 + x^4), x) + 24*a^2*c^3*integrate(1/48*x^2*lo
g(a^2*x^2 + 1)^2/(a^2*x^6 + x^4), x) - 8*a^2*c^3*integrate(1/48*x^2*log(a^2*x^2 + 1)/(a^2*x^6 + x^4), x) + 16*
a*c^3*integrate(1/48*x*arctan(a*x)/(a^2*x^6 + x^4), x) + 72*c^3*integrate(1/48*arctan(a*x)^2/(a^2*x^6 + x^4),
x) + 6*c^3*integrate(1/48*log(a^2*x^2 + 1)^2/(a^2*x^6 + x^4), x))*x^3 + 4*(a^6*c^3*x^6 + 9*a^4*c^3*x^4 - 9*a^2
*c^3*x^2 - c^3)*arctan(a*x)^2 - (a^6*c^3*x^6 + 9*a^4*c^3*x^4 - 9*a^2*c^3*x^2 - c^3)*log(a^2*x^2 + 1)^2)/x^3

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^3*arctan(a*x)^2/x^4,x, algorithm="fricas")

[Out]

integral((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3)*arctan(a*x)^2/x^4, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} c^{3} \left (\int 3 a^{4} \operatorname {atan}^{2}{\left (a x \right )}\, dx + \int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{x^{4}}\, dx + \int \frac {3 a^{2} \operatorname {atan}^{2}{\left (a x \right )}}{x^{2}}\, dx + \int a^{6} x^{2} \operatorname {atan}^{2}{\left (a x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**3*atan(a*x)**2/x**4,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^3}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atan(a*x)^2*(c + a^2*c*x^2)^3)/x^4,x)

[Out]

int((atan(a*x)^2*(c + a^2*c*x^2)^3)/x^4, x)

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