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

Optimal. Leaf size=62 \[ \frac{1}{2} i c \text{PolyLog}(2,-i a x)-\frac{1}{2} i c \text{PolyLog}(2,i a x)+\frac{1}{2} a^2 c x^2 \tan ^{-1}(a x)-\frac{a c x}{2}+\frac{1}{2} c \tan ^{-1}(a x) \]

[Out]

-(a*c*x)/2 + (c*ArcTan[a*x])/2 + (a^2*c*x^2*ArcTan[a*x])/2 + (I/2)*c*PolyLog[2, (-I)*a*x] - (I/2)*c*PolyLog[2,
I*a*x]

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Rubi [A]  time = 0.0661995, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4950, 4848, 2391, 4852, 321, 203} \[ \frac{1}{2} i c \text{PolyLog}(2,-i a x)-\frac{1}{2} i c \text{PolyLog}(2,i a x)+\frac{1}{2} a^2 c x^2 \tan ^{-1}(a x)-\frac{a c x}{2}+\frac{1}{2} c \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c*x)/2 + (c*ArcTan[a*x])/2 + (a^2*c*x^2*ArcTan[a*x])/2 + (I/2)*c*PolyLog[2, (-I)*a*x] - (I/2)*c*PolyLog[2,
I*a*x]

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]))

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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.0039463, size = 62, normalized size = 1. \[ \frac{1}{2} i c \text{PolyLog}(2,-i a x)-\frac{1}{2} i c \text{PolyLog}(2,i a x)+\frac{1}{2} a^2 c x^2 \tan ^{-1}(a x)-\frac{a c x}{2}+\frac{1}{2} c \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c*x)/2 + (c*ArcTan[a*x])/2 + (a^2*c*x^2*ArcTan[a*x])/2 + (I/2)*c*PolyLog[2, (-I)*a*x] - (I/2)*c*PolyLog[2,
I*a*x]

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Maple [A]  time = 0.037, size = 93, normalized size = 1.5 \begin{align*}{\frac{{a}^{2}c{x}^{2}\arctan \left ( ax \right ) }{2}}+c\arctan \left ( ax \right ) \ln \left ( ax \right ) +{\frac{i}{2}}\ln \left ( ax \right ) \ln \left ( 1+iax \right ) c-{\frac{i}{2}}\ln \left ( ax \right ) \ln \left ( 1-iax \right ) c+{\frac{i}{2}}{\it dilog} \left ( 1+iax \right ) c-{\frac{i}{2}}{\it dilog} \left ( 1-iax \right ) c-{\frac{acx}{2}}+{\frac{c\arctan \left ( ax \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.62166, size = 101, normalized size = 1.63 \begin{align*} -\frac{1}{2} \, a c x - \frac{1}{4} \, \pi c \log \left (a^{2} x^{2} + 1\right ) + c \arctan \left (a x\right ) \log \left (x{\left | a \right |}\right ) + \frac{1}{2} \,{\left (a^{2} c x^{2} + c{\left (2 i \, \arctan \left (0, a\right ) + 1\right )}\right )} \arctan \left (a x\right ) - \frac{1}{2} i \, c{\rm Li}_2\left (i \, a x + 1\right ) + \frac{1}{2} i \, c{\rm Li}_2\left (-i \, a x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*c*x - 1/4*pi*c*log(a^2*x^2 + 1) + c*arctan(a*x)*log(x*abs(a)) + 1/2*(a^2*c*x^2 + c*(2*I*arctan2(0, a) +
1))*arctan(a*x) - 1/2*I*c*dilog(I*a*x + 1) + 1/2*I*c*dilog(-I*a*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

c*(Integral(atan(a*x)/x, x) + Integral(a**2*x*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 )} \arctan \left (a x\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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