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

Optimal. Leaf size=135 \[ -\frac {2}{3} i a^3 c \text {Li}_2\left (\frac {2}{1-i a x}-1\right )-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^2-\frac {1}{3} a^3 c \tan ^{-1}(a x)+\frac {4}{3} a^3 c \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)-\frac {a^2 c}{3 x}-\frac {a^2 c \tan ^{-1}(a x)^2}{x}-\frac {c \tan ^{-1}(a x)^2}{3 x^3}-\frac {a c \tan ^{-1}(a x)}{3 x^2} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4950, 4852, 4918, 325, 203, 4924, 4868, 2447} \[ -\frac {2}{3} i a^3 c \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )-\frac {a^2 c}{3 x}-\frac {2}{3} i a^3 c \tan ^{-1}(a x)^2-\frac {1}{3} a^3 c \tan ^{-1}(a x)-\frac {a^2 c \tan ^{-1}(a x)^2}{x}+\frac {4}{3} a^3 c \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)-\frac {a c \tan ^{-1}(a x)}{3 x^2}-\frac {c \tan ^{-1}(a x)^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*c)/(3*x) - (a^3*c*ArcTan[a*x])/3 - (a*c*ArcTan[a*x])/(3*x^2) - ((2*I)/3)*a^3*c*ArcTan[a*x]^2 - (c*ArcTan
[a*x]^2)/(3*x^3) - (a^2*c*ArcTan[a*x]^2)/x + (4*a^3*c*ArcTan[a*x]*Log[2 - 2/(1 - I*a*x)])/3 - ((2*I)/3)*a^3*c*
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 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 4950

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

Rubi steps

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

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Mathematica [A]  time = 0.63, size = 103, normalized size = 0.76 \[ \frac {c \left (-2 i a^3 x^3 \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )-a^2 x^2+a x \tan ^{-1}(a x) \left (-a^2 x^2+4 a^2 x^2 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )-1\right )+(1-2 i a x) (a x-i)^2 \tan ^{-1}(a x)^2\right )}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )^{2}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [B]  time = 0.11, size = 323, normalized size = 2.39 \[ -\frac {c \arctan \left (a x \right )^{2}}{3 x^{3}}-\frac {a^{2} c \arctan \left (a x \right )^{2}}{x}-\frac {a c \arctan \left (a x \right )}{3 x^{2}}+\frac {4 a^{3} c \arctan \left (a x \right ) \ln \left (a x \right )}{3}-\frac {2 a^{3} c \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{3}-\frac {a^{2} c}{3 x}-\frac {a^{3} c \arctan \left (a x \right )}{3}+\frac {2 i a^{3} c \ln \left (a x \right ) \ln \left (i a x +1\right )}{3}+\frac {2 i a^{3} c \dilog \left (i a x +1\right )}{3}+\frac {i a^{3} c \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{3}+\frac {i a^{3} c \ln \left (a x -i\right )^{2}}{6}-\frac {i a^{3} c \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{3}-\frac {i a^{3} c \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{3}-\frac {2 i a^{3} c \dilog \left (-i a x +1\right )}{3}-\frac {i a^{3} c \ln \left (a x +i\right )^{2}}{6}+\frac {i a^{3} c \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{3}-\frac {i a^{3} c \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{3}+\frac {i a^{3} c \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{3}-\frac {2 i a^{3} c \ln \left (a x \right ) \ln \left (-i a x +1\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*c*arctan(a*x)^2/x^3-a^2*c*arctan(a*x)^2/x-1/3*a*c*arctan(a*x)/x^2+4/3*a^3*c*arctan(a*x)*ln(a*x)-2/3*a^3*c
*arctan(a*x)*ln(a^2*x^2+1)-1/3*a^2*c/x-1/3*a^3*c*arctan(a*x)+2/3*I*a^3*c*ln(a*x)*ln(1+I*a*x)+1/3*I*a^3*c*ln(a*
x-I)*ln(-1/2*I*(I+a*x))+1/6*I*a^3*c*ln(a*x-I)^2-1/3*I*a^3*c*ln(a*x-I)*ln(a^2*x^2+1)-1/3*I*a^3*c*ln(I+a*x)*ln(1
/2*I*(a*x-I))-1/6*I*a^3*c*ln(I+a*x)^2+2/3*I*a^3*c*dilog(1+I*a*x)-1/3*I*a^3*c*dilog(1/2*I*(a*x-I))+1/3*I*a^3*c*
dilog(-1/2*I*(I+a*x))+1/3*I*a^3*c*ln(I+a*x)*ln(a^2*x^2+1)-2/3*I*a^3*c*dilog(1-I*a*x)-2/3*I*a^3*c*ln(a*x)*ln(1-
I*a*x)

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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)*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 = -0.01 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2\,\left (c\,a^2\,x^2+c\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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