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

Optimal. Leaf size=140 \[ \frac{1}{3} i b^2 c^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )+\frac{1}{3} i c^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{2}{3} b c^3 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac{b^2 c^2}{3 x}-\frac{1}{3} b^2 c^3 \tan ^{-1}(c x) \]

[Out]

-(b^2*c^2)/(3*x) - (b^2*c^3*ArcTan[c*x])/3 - (b*c*(a + b*ArcTan[c*x]))/(3*x^2) + (I/3)*c^3*(a + b*ArcTan[c*x])
^2 - (a + b*ArcTan[c*x])^2/(3*x^3) - (2*b*c^3*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/3 + (I/3)*b^2*c^3*Po
lyLog[2, -1 + 2/(1 - I*c*x)]

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Rubi [A]  time = 0.229991, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4852, 4918, 325, 203, 4924, 4868, 2447} \[ \frac{1}{3} i b^2 c^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )+\frac{1}{3} i c^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{2}{3} b c^3 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac{b^2 c^2}{3 x}-\frac{1}{3} b^2 c^3 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^2/x^4,x]

[Out]

-(b^2*c^2)/(3*x) - (b^2*c^3*ArcTan[c*x])/3 - (b*c*(a + b*ArcTan[c*x]))/(3*x^2) + (I/3)*c^3*(a + b*ArcTan[c*x])
^2 - (a + b*ArcTan[c*x])^2/(3*x^3) - (2*b*c^3*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/3 + (I/3)*b^2*c^3*Po
lyLog[2, -1 + 2/(1 - I*c*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 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 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 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 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 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 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]]

Rubi steps

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

Mathematica [A]  time = 0.364181, size = 153, normalized size = 1.09 \[ -\frac{-i b^2 c^3 x^3 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+a^2+2 a b c^3 x^3 \log (c x)-a b c^3 x^3 \log \left (c^2 x^2+1\right )+b \tan ^{-1}(c x) \left (2 a+b c^3 x^3+2 b c^3 x^3 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+b c x\right )+a b c x+b^2 c^2 x^2+b^2 \left (1-i c^3 x^3\right ) \tan ^{-1}(c x)^2}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.014, size = 399, normalized size = 2.9 \begin{align*} -{\frac{{a}^{2}}{3\,{x}^{3}}}-{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{3\,{x}^{3}}}+{\frac{{c}^{3}{b}^{2}\arctan \left ( cx \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) }{3}}-{\frac{c{b}^{2}\arctan \left ( cx \right ) }{3\,{x}^{2}}}-{\frac{2\,{c}^{3}{b}^{2}\ln \left ( cx \right ) \arctan \left ( cx \right ) }{3}}-{\frac{i}{6}}{c}^{3}{b}^{2}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) -{\frac{i}{12}}{c}^{3}{b}^{2} \left ( \ln \left ( cx-i \right ) \right ) ^{2}+{\frac{i}{3}}{c}^{3}{b}^{2}{\it dilog} \left ( 1-icx \right ) +{\frac{i}{6}}{c}^{3}{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx-i \right ) -{\frac{i}{3}}{c}^{3}{b}^{2}\ln \left ( cx \right ) \ln \left ( 1+icx \right ) +{\frac{i}{12}}{c}^{3}{b}^{2} \left ( \ln \left ( cx+i \right ) \right ) ^{2}-{\frac{i}{3}}{c}^{3}{b}^{2}{\it dilog} \left ( 1+icx \right ) -{\frac{i}{6}}{c}^{3}{b}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) -{\frac{{b}^{2}{c}^{3}\arctan \left ( cx \right ) }{3}}-{\frac{{b}^{2}{c}^{2}}{3\,x}}+{\frac{i}{6}}{c}^{3}{b}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) +{\frac{i}{6}}{c}^{3}{b}^{2}\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) -{\frac{i}{6}}{c}^{3}{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx+i \right ) +{\frac{i}{3}}{c}^{3}{b}^{2}\ln \left ( cx \right ) \ln \left ( 1-icx \right ) -{\frac{2\,ab\arctan \left ( cx \right ) }{3\,{x}^{3}}}+{\frac{{c}^{3}ab\ln \left ({c}^{2}{x}^{2}+1 \right ) }{3}}-{\frac{cab}{3\,{x}^{2}}}-{\frac{2\,{c}^{3}ab\ln \left ( cx \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2/x^3-1/3*b^2/x^3*arctan(c*x)^2+1/3*c^3*b^2*arctan(c*x)*ln(c^2*x^2+1)-1/3*c*b^2*arctan(c*x)/x^2-2/3*c^3
*b^2*ln(c*x)*arctan(c*x)-1/6*I*c^3*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-1/12*I*c^3*b^2*ln(c*x-I)^2+1/3*I*c^3*b^2*d
ilog(1-I*c*x)+1/6*I*c^3*b^2*ln(c^2*x^2+1)*ln(c*x-I)-1/3*I*c^3*b^2*ln(c*x)*ln(1+I*c*x)+1/12*I*c^3*b^2*ln(c*x+I)
^2-1/3*I*c^3*b^2*dilog(1+I*c*x)-1/6*I*c^3*b^2*dilog(-1/2*I*(c*x+I))-1/3*b^2*c^3*arctan(c*x)-1/3*b^2*c^2/x+1/6*
I*c^3*b^2*dilog(1/2*I*(c*x-I))+1/6*I*c^3*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))-1/6*I*c^3*b^2*ln(c^2*x^2+1)*ln(c*x+I)
+1/3*I*c^3*b^2*ln(c*x)*ln(1-I*c*x)-2/3*a*b/x^3*arctan(c*x)+1/3*c^3*a*b*ln(c^2*x^2+1)-1/3*c*a*b/x^2-2/3*c^3*a*b
*ln(c*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac{1}{x^{2}}\right )} c - \frac{2 \, \arctan \left (c x\right )}{x^{3}}\right )} a b + \frac{\frac{1}{4} \,{\left (4 \, x^{3} \int -\frac{12 \, c^{2} x^{2} \log \left (c^{2} x^{2} + 1\right ) - 56 \, c x \arctan \left (c x\right ) - 108 \,{\left (c^{2} x^{2} + 1\right )} \arctan \left (c x\right )^{2} - 9 \,{\left (c^{2} x^{2} + 1\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{4 \,{\left (c^{2} x^{6} + x^{4}\right )}}\,{d x} - 28 \, \arctan \left (c x\right )^{2} + 3 \, \log \left (c^{2} x^{2} + 1\right )^{2}\right )} b^{2}}{48 \, x^{3}} - \frac{a^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*a*b + 1/48*(48*x^3*integrate(-1/48*(
4*c^2*x^2*log(c^2*x^2 + 1) - 8*c*x*arctan(c*x) - 36*(c^2*x^2 + 1)*arctan(c*x)^2 - 3*(c^2*x^2 + 1)*log(c^2*x^2
+ 1)^2)/(c^2*x^6 + x^4), x) - 4*arctan(c*x)^2 + log(c^2*x^2 + 1)^2)*b^2/x^3 - 1/3*a^2/x^3

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + a^{2}}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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