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

Optimal. Leaf size=138 \[ \frac{3}{2} i a^2 c^3 \text{PolyLog}(2,-i a x)-\frac{3}{2} i a^2 c^3 \text{PolyLog}(2,i a x)-\frac{1}{12} a^5 c^3 x^3+\frac{1}{4} a^6 c^3 x^4 \tan ^{-1}(a x)+\frac{3}{2} a^4 c^3 x^2 \tan ^{-1}(a x)-\frac{5}{4} a^3 c^3 x+\frac{3}{4} a^2 c^3 \tan ^{-1}(a x)-\frac{c^3 \tan ^{-1}(a x)}{2 x^2}-\frac{a c^3}{2 x} \]

[Out]

-(a*c^3)/(2*x) - (5*a^3*c^3*x)/4 - (a^5*c^3*x^3)/12 + (3*a^2*c^3*ArcTan[a*x])/4 - (c^3*ArcTan[a*x])/(2*x^2) +
(3*a^4*c^3*x^2*ArcTan[a*x])/2 + (a^6*c^3*x^4*ArcTan[a*x])/4 + ((3*I)/2)*a^2*c^3*PolyLog[2, (-I)*a*x] - ((3*I)/
2)*a^2*c^3*PolyLog[2, I*a*x]

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Rubi [A]  time = 0.152538, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4948, 4852, 325, 203, 4848, 2391, 321, 302} \[ \frac{3}{2} i a^2 c^3 \text{PolyLog}(2,-i a x)-\frac{3}{2} i a^2 c^3 \text{PolyLog}(2,i a x)-\frac{1}{12} a^5 c^3 x^3+\frac{1}{4} a^6 c^3 x^4 \tan ^{-1}(a x)+\frac{3}{2} a^4 c^3 x^2 \tan ^{-1}(a x)-\frac{5}{4} a^3 c^3 x+\frac{3}{4} a^2 c^3 \tan ^{-1}(a x)-\frac{c^3 \tan ^{-1}(a x)}{2 x^2}-\frac{a c^3}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c^3)/(2*x) - (5*a^3*c^3*x)/4 - (a^5*c^3*x^3)/12 + (3*a^2*c^3*ArcTan[a*x])/4 - (c^3*ArcTan[a*x])/(2*x^2) +
(3*a^4*c^3*x^2*ArcTan[a*x])/2 + (a^6*c^3*x^4*ArcTan[a*x])/4 + ((3*I)/2)*a^2*c^3*PolyLog[2, (-I)*a*x] - ((3*I)/
2)*a^2*c^3*PolyLog[2, I*a*x]

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])

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 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 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

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

Rule 321

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rubi steps

\begin{align*} \int \frac{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)}{x^3} \, dx &=\int \left (\frac{c^3 \tan ^{-1}(a x)}{x^3}+\frac{3 a^2 c^3 \tan ^{-1}(a x)}{x}+3 a^4 c^3 x \tan ^{-1}(a x)+a^6 c^3 x^3 \tan ^{-1}(a x)\right ) \, dx\\ &=c^3 \int \frac{\tan ^{-1}(a x)}{x^3} \, dx+\left (3 a^2 c^3\right ) \int \frac{\tan ^{-1}(a x)}{x} \, dx+\left (3 a^4 c^3\right ) \int x \tan ^{-1}(a x) \, dx+\left (a^6 c^3\right ) \int x^3 \tan ^{-1}(a x) \, dx\\ &=-\frac{c^3 \tan ^{-1}(a x)}{2 x^2}+\frac{3}{2} a^4 c^3 x^2 \tan ^{-1}(a x)+\frac{1}{4} a^6 c^3 x^4 \tan ^{-1}(a x)+\frac{1}{2} \left (a c^3\right ) \int \frac{1}{x^2 \left (1+a^2 x^2\right )} \, dx+\frac{1}{2} \left (3 i a^2 c^3\right ) \int \frac{\log (1-i a x)}{x} \, dx-\frac{1}{2} \left (3 i a^2 c^3\right ) \int \frac{\log (1+i a x)}{x} \, dx-\frac{1}{2} \left (3 a^5 c^3\right ) \int \frac{x^2}{1+a^2 x^2} \, dx-\frac{1}{4} \left (a^7 c^3\right ) \int \frac{x^4}{1+a^2 x^2} \, dx\\ &=-\frac{a c^3}{2 x}-\frac{3}{2} a^3 c^3 x-\frac{c^3 \tan ^{-1}(a x)}{2 x^2}+\frac{3}{2} a^4 c^3 x^2 \tan ^{-1}(a x)+\frac{1}{4} a^6 c^3 x^4 \tan ^{-1}(a x)+\frac{3}{2} i a^2 c^3 \text{Li}_2(-i a x)-\frac{3}{2} i a^2 c^3 \text{Li}_2(i a x)-\frac{1}{2} \left (a^3 c^3\right ) \int \frac{1}{1+a^2 x^2} \, dx+\frac{1}{2} \left (3 a^3 c^3\right ) \int \frac{1}{1+a^2 x^2} \, dx-\frac{1}{4} \left (a^7 c^3\right ) \int \left (-\frac{1}{a^4}+\frac{x^2}{a^2}+\frac{1}{a^4 \left (1+a^2 x^2\right )}\right ) \, dx\\ &=-\frac{a c^3}{2 x}-\frac{5}{4} a^3 c^3 x-\frac{1}{12} a^5 c^3 x^3+a^2 c^3 \tan ^{-1}(a x)-\frac{c^3 \tan ^{-1}(a x)}{2 x^2}+\frac{3}{2} a^4 c^3 x^2 \tan ^{-1}(a x)+\frac{1}{4} a^6 c^3 x^4 \tan ^{-1}(a x)+\frac{3}{2} i a^2 c^3 \text{Li}_2(-i a x)-\frac{3}{2} i a^2 c^3 \text{Li}_2(i a x)-\frac{1}{4} \left (a^3 c^3\right ) \int \frac{1}{1+a^2 x^2} \, dx\\ &=-\frac{a c^3}{2 x}-\frac{5}{4} a^3 c^3 x-\frac{1}{12} a^5 c^3 x^3+\frac{3}{4} a^2 c^3 \tan ^{-1}(a x)-\frac{c^3 \tan ^{-1}(a x)}{2 x^2}+\frac{3}{2} a^4 c^3 x^2 \tan ^{-1}(a x)+\frac{1}{4} a^6 c^3 x^4 \tan ^{-1}(a x)+\frac{3}{2} i a^2 c^3 \text{Li}_2(-i a x)-\frac{3}{2} i a^2 c^3 \text{Li}_2(i a x)\\ \end{align*}

Mathematica [C]  time = 0.0433007, size = 154, normalized size = 1.12 \[ -\frac{a c^3 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-a^2 x^2\right )}{2 x}+\frac{3}{2} i a^2 c^3 \text{PolyLog}(2,-i a x)-\frac{3}{2} i a^2 c^3 \text{PolyLog}(2,i a x)-\frac{1}{12} a^5 c^3 x^3+\frac{1}{4} a^6 c^3 x^4 \tan ^{-1}(a x)+\frac{3}{2} a^4 c^3 x^2 \tan ^{-1}(a x)-\frac{5}{4} a^3 c^3 x+\frac{5}{4} a^2 c^3 \tan ^{-1}(a x)-\frac{c^3 \tan ^{-1}(a x)}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-5*a^3*c^3*x)/4 - (a^5*c^3*x^3)/12 + (5*a^2*c^3*ArcTan[a*x])/4 - (c^3*ArcTan[a*x])/(2*x^2) + (3*a^4*c^3*x^2*A
rcTan[a*x])/2 + (a^6*c^3*x^4*ArcTan[a*x])/4 - (a*c^3*Hypergeometric2F1[-1/2, 1, 1/2, -(a^2*x^2)])/(2*x) + ((3*
I)/2)*a^2*c^3*PolyLog[2, (-I)*a*x] - ((3*I)/2)*a^2*c^3*PolyLog[2, I*a*x]

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Maple [A]  time = 0.044, size = 177, normalized size = 1.3 \begin{align*}{\frac{{a}^{6}{c}^{3}{x}^{4}\arctan \left ( ax \right ) }{4}}+{\frac{3\,{a}^{4}{c}^{3}{x}^{2}\arctan \left ( ax \right ) }{2}}-{\frac{{c}^{3}\arctan \left ( ax \right ) }{2\,{x}^{2}}}+3\,{a}^{2}{c}^{3}\arctan \left ( ax \right ) \ln \left ( ax \right ) -{\frac{{a}^{5}{c}^{3}{x}^{3}}{12}}-{\frac{5\,{a}^{3}{c}^{3}x}{4}}+{\frac{3\,{a}^{2}{c}^{3}\arctan \left ( ax \right ) }{4}}-{\frac{a{c}^{3}}{2\,x}}+{\frac{3\,i}{2}}{a}^{2}{c}^{3}\ln \left ( ax \right ) \ln \left ( 1+iax \right ) -{\frac{3\,i}{2}}{a}^{2}{c}^{3}\ln \left ( ax \right ) \ln \left ( 1-iax \right ) +{\frac{3\,i}{2}}{a}^{2}{c}^{3}{\it dilog} \left ( 1+iax \right ) -{\frac{3\,i}{2}}{a}^{2}{c}^{3}{\it dilog} \left ( 1-iax \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^6*c^3*x^4*arctan(a*x)+3/2*a^4*c^3*x^2*arctan(a*x)-1/2*c^3*arctan(a*x)/x^2+3*a^2*c^3*arctan(a*x)*ln(a*x)-
1/12*a^5*c^3*x^3-5/4*a^3*c^3*x+3/4*a^2*c^3*arctan(a*x)-1/2*a*c^3/x+3/2*I*a^2*c^3*ln(a*x)*ln(1+I*a*x)-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)-3/2*I*a^2*c^3*dilog(1-I*a*x)

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Maxima [A]  time = 1.69919, size = 220, normalized size = 1.59 \begin{align*} -\frac{a^{5} c^{3} x^{5} + 15 \, a^{3} c^{3} x^{3} + 9 \, \pi a^{2} c^{3} x^{2} \log \left (a^{2} x^{2} + 1\right ) - 36 \, a^{2} c^{3} x^{2} \arctan \left (a x\right ) \log \left (x{\left | a \right |}\right ) + 18 i \, a^{2} c^{3} x^{2}{\rm Li}_2\left (i \, a x + 1\right ) - 18 i \, a^{2} c^{3} x^{2}{\rm Li}_2\left (-i \, a x + 1\right ) + 6 \, a c^{3} x - 3 \,{\left (a^{6} c^{3} x^{6} + 6 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2}{\left (4 i \, \arctan \left (0, a\right ) + 1\right )} - 2 \, c^{3}\right )} \arctan \left (a x\right )}{12 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(a^5*c^3*x^5 + 15*a^3*c^3*x^3 + 9*pi*a^2*c^3*x^2*log(a^2*x^2 + 1) - 36*a^2*c^3*x^2*arctan(a*x)*log(x*abs
(a)) + 18*I*a^2*c^3*x^2*dilog(I*a*x + 1) - 18*I*a^2*c^3*x^2*dilog(-I*a*x + 1) + 6*a*c^3*x - 3*(a^6*c^3*x^6 + 6
*a^4*c^3*x^4 + 3*a^2*c^3*x^2*(4*I*arctan2(0, a) + 1) - 2*c^3)*arctan(a*x))/x^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}\right )} \arctan \left (a x\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a^{2} c x^{2} + c\right )}^{3} \arctan \left (a x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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