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

Optimal. Leaf size=70 \[ \frac {1}{2} i a^2 c \text {Li}_2(-i a x)-\frac {1}{2} i a^2 c \text {Li}_2(i a x)-\frac {1}{2} a^2 c \tan ^{-1}(a x)-\frac {c \tan ^{-1}(a x)}{2 x^2}-\frac {a c}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 70, normalized size of antiderivative = 1.00, 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, 4852, 325, 203, 4848, 2391} \[ \frac {1}{2} i a^2 c \text {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \text {PolyLog}(2,i a x)-\frac {1}{2} a^2 c \tan ^{-1}(a x)-\frac {c \tan ^{-1}(a x)}{2 x^2}-\frac {a c}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.01, size = 74, normalized size = 1.06 \[ -\frac {a c \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-a^2 x^2\right )}{2 x}+\frac {1}{2} i a^2 c \text {Li}_2(-i a x)-\frac {1}{2} i a^2 c \text {Li}_2(i a x)-\frac {c \tan ^{-1}(a x)}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(c*ArcTan[a*x])/x^2 - (a*c*Hypergeometric2F1[-1/2, 1, 1/2, -(a^2*x^2)])/(2*x) + (I/2)*a^2*c*PolyLog[2, (-
I)*a*x] - (I/2)*a^2*c*PolyLog[2, I*a*x]

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 110, normalized size = 1.57 \[ a^{2} c \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c \arctan \left (a x \right )}{2 x^{2}}-\frac {a c}{2 x}-\frac {a^{2} c \arctan \left (a x \right )}{2}+\frac {i a^{2} c \ln \left (a x \right ) \ln \left (i a x +1\right )}{2}-\frac {i a^{2} c \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2}+\frac {i a^{2} c \dilog \left (i a x +1\right )}{2}-\frac {i a^{2} c \dilog \left (-i a x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 95, normalized size = 1.36 \[ -\frac {\pi a^{2} c x^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{2} c x^{2} \arctan \left (a x\right ) \log \left (a x\right ) + 2 i \, a^{2} c x^{2} {\rm Li}_2\left (i \, a x + 1\right ) - 2 i \, a^{2} c x^{2} {\rm Li}_2\left (-i \, a x + 1\right ) + 2 \, a c x + 2 \, {\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.56, size = 71, normalized size = 1.01 \[ \left \{\begin {array}{cl} 0 & \text {\ if\ \ }a=0\\ -\frac {c\,\mathrm {atan}\left (a\,x\right )}{2\,x^2}-\frac {c\,\left (a^3\,\mathrm {atan}\left (a\,x\right )+\frac {a^2}{x}\right )}{2\,a}-\frac {a^2\,c\,{\mathrm {Li}}_{\mathrm {2}}\left (1-a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a^2\,c\,{\mathrm {Li}}_{\mathrm {2}}\left (1+a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }a\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(a == 0, 0, a ~= 0, - (c*atan(a*x))/(2*x^2) - (a^2*c*dilog(- a*x*1i + 1)*1i)/2 + (a^2*c*dilog(a*x*1i
+ 1)*1i)/2 - (c*(a^3*atan(a*x) + a^2/x))/(2*a))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c \left (\int \frac {\operatorname {atan}{\left (a x \right )}}{x^{3}}\, dx + \int \frac {a^{2} \operatorname {atan}{\left (a x \right )}}{x}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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