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

Optimal. Leaf size=99 \[ \frac {1}{4} a^4 c^2 x^4 \tan ^{-1}(a x)-\frac {1}{12} a^3 c^2 x^3+a^2 c^2 x^2 \tan ^{-1}(a x)+\frac {1}{2} i c^2 \text {Li}_2(-i a x)-\frac {1}{2} i c^2 \text {Li}_2(i a x)-\frac {3}{4} a c^2 x+\frac {3}{4} c^2 \tan ^{-1}(a x) \]

[Out]

-3/4*a*c^2*x-1/12*a^3*c^2*x^3+3/4*c^2*arctan(a*x)+a^2*c^2*x^2*arctan(a*x)+1/4*a^4*c^2*x^4*arctan(a*x)+1/2*I*c^
2*polylog(2,-I*a*x)-1/2*I*c^2*polylog(2,I*a*x)

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Rubi [A]  time = 0.12, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4948, 4848, 2391, 4852, 321, 203, 302} \[ \frac {1}{2} i c^2 \text {PolyLog}(2,-i a x)-\frac {1}{2} i c^2 \text {PolyLog}(2,i a x)-\frac {1}{12} a^3 c^2 x^3+\frac {1}{4} a^4 c^2 x^4 \tan ^{-1}(a x)+a^2 c^2 x^2 \tan ^{-1}(a x)-\frac {3}{4} a c^2 x+\frac {3}{4} c^2 \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 99, normalized size = 1.00 \[ \frac {1}{4} a^4 c^2 x^4 \tan ^{-1}(a x)-\frac {1}{12} a^3 c^2 x^3+a^2 c^2 x^2 \tan ^{-1}(a x)+\frac {1}{2} i c^2 \text {Li}_2(-i a x)-\frac {1}{2} i c^2 \text {Li}_2(i a x)-\frac {3}{4} a c^2 x+\frac {3}{4} c^2 \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 134, normalized size = 1.35 \[ \frac {a^{4} c^{2} x^{4} \arctan \left (a x \right )}{4}+a^{2} c^{2} x^{2} \arctan \left (a x \right )+c^{2} \arctan \left (a x \right ) \ln \left (a x \right )-\frac {a^{3} c^{2} x^{3}}{12}-\frac {3 a \,c^{2} x}{4}+\frac {3 c^{2} \arctan \left (a x \right )}{4}+\frac {i c^{2} \ln \left (a x \right ) \ln \left (i a x +1\right )}{2}-\frac {i c^{2} \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2}+\frac {i c^{2} \dilog \left (i a x +1\right )}{2}-\frac {i c^{2} \dilog \left (-i a x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 104, normalized size = 1.05 \[ -\frac {1}{12} \, a^{3} c^{2} x^{3} - \frac {3}{4} \, a c^{2} x - \frac {1}{4} \, \pi c^{2} \log \left (a^{2} x^{2} + 1\right ) + c^{2} \arctan \left (a x\right ) \log \left (a x\right ) - \frac {1}{2} i \, c^{2} {\rm Li}_2\left (i \, a x + 1\right ) + \frac {1}{2} i \, c^{2} {\rm Li}_2\left (-i \, a x + 1\right ) + \frac {1}{4} \, {\left (a^{4} c^{2} x^{4} + 4 \, a^{2} c^{2} x^{2} + 3 \, c^{2}\right )} \arctan \left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*a^3*c^2*x^3 - 3/4*a*c^2*x - 1/4*pi*c^2*log(a^2*x^2 + 1) + c^2*arctan(a*x)*log(a*x) - 1/2*I*c^2*dilog(I*a
*x + 1) + 1/2*I*c^2*dilog(-I*a*x + 1) + 1/4*(a^4*c^2*x^4 + 4*a^2*c^2*x^2 + 3*c^2)*arctan(a*x)

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mupad [B]  time = 0.61, size = 105, normalized size = 1.06 \[ \left \{\begin {array}{cl} 0 & \text {\ if\ \ }a=0\\ 2\,a^2\,c^2\,\mathrm {atan}\left (a\,x\right )\,\left (\frac {1}{2\,a^2}+\frac {x^2}{2}\right )-a\,c^2\,x-\frac {c^2\,\left (3\,\mathrm {atan}\left (a\,x\right )-3\,a\,x+a^3\,x^3\right )}{12}+\frac {a^4\,c^2\,x^4\,\mathrm {atan}\left (a\,x\right )}{4}-\frac {c^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {c^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }a\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(a == 0, 0, a ~= 0, - (c^2*dilog(- a*x*1i + 1)*1i)/2 + (c^2*dilog(a*x*1i + 1)*1i)/2 - (c^2*(3*atan(a*
x) - 3*a*x + a^3*x^3))/12 - a*c^2*x + 2*a^2*c^2*atan(a*x)*(1/(2*a^2) + x^2/2) + (a^4*c^2*x^4*atan(a*x))/4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \left (\int \frac {\operatorname {atan}{\left (a x \right )}}{x}\, dx + \int 2 a^{2} x \operatorname {atan}{\left (a x \right )}\, dx + \int a^{4} x^{3} \operatorname {atan}{\left (a x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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