3.1131 \(\int \frac{(d+e x^2)^2 (a+b \tan ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=128 \[ i b d e \text{PolyLog}(2,-i c x)-i b d e \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{1}{2} e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+2 a d e \log (x)-\frac{1}{2} b c^2 d^2 \tan ^{-1}(c x)+\frac{b e^2 \tan ^{-1}(c x)}{2 c^2}-\frac{b c d^2}{2 x}-\frac{b e^2 x}{2 c} \]

[Out]

-(b*c*d^2)/(2*x) - (b*e^2*x)/(2*c) - (b*c^2*d^2*ArcTan[c*x])/2 + (b*e^2*ArcTan[c*x])/(2*c^2) - (d^2*(a + b*Arc
Tan[c*x]))/(2*x^2) + (e^2*x^2*(a + b*ArcTan[c*x]))/2 + 2*a*d*e*Log[x] + I*b*d*e*PolyLog[2, (-I)*c*x] - I*b*d*e
*PolyLog[2, I*c*x]

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Rubi [A]  time = 0.163266, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4980, 4852, 325, 203, 4848, 2391, 321} \[ i b d e \text{PolyLog}(2,-i c x)-i b d e \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{1}{2} e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+2 a d e \log (x)-\frac{1}{2} b c^2 d^2 \tan ^{-1}(c x)+\frac{b e^2 \tan ^{-1}(c x)}{2 c^2}-\frac{b c d^2}{2 x}-\frac{b e^2 x}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d^2)/(2*x) - (b*e^2*x)/(2*c) - (b*c^2*d^2*ArcTan[c*x])/2 + (b*e^2*ArcTan[c*x])/(2*c^2) - (d^2*(a + b*Arc
Tan[c*x]))/(2*x^2) + (e^2*x^2*(a + b*ArcTan[c*x]))/2 + 2*a*d*e*Log[x] + I*b*d*e*PolyLog[2, (-I)*c*x] - I*b*d*e
*PolyLog[2, I*c*x]

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || 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]

Rubi steps

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

Mathematica [C]  time = 0.102198, size = 118, normalized size = 0.92 \[ \frac{1}{2} \left (-\frac{b c d^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )}{x}+2 i b d e \text{PolyLog}(2,-i c x)-2 i b d e \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+4 a d e \log (x)-\frac{b e^2 \left (c x-\tan ^{-1}(c x)\right )}{c^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-((b*e^2*(c*x - ArcTan[c*x]))/c^2) - (d^2*(a + b*ArcTan[c*x]))/x^2 + e^2*x^2*(a + b*ArcTan[c*x]) - (b*c*d^2*H
ypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/x + 4*a*d*e*Log[x] + (2*I)*b*d*e*PolyLog[2, (-I)*c*x] - (2*I)*b*d*
e*PolyLog[2, I*c*x])/2

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Maple [A]  time = 0.058, size = 178, normalized size = 1.4 \begin{align*}{\frac{a{x}^{2}{e}^{2}}{2}}-{\frac{a{d}^{2}}{2\,{x}^{2}}}+2\,aed\ln \left ( cx \right ) +{\frac{b\arctan \left ( cx \right ){x}^{2}{e}^{2}}{2}}-{\frac{b{d}^{2}\arctan \left ( cx \right ) }{2\,{x}^{2}}}+2\,b\arctan \left ( cx \right ) ed\ln \left ( cx \right ) -{\frac{b{e}^{2}x}{2\,c}}-{\frac{b{c}^{2}{d}^{2}\arctan \left ( cx \right ) }{2}}+{\frac{b{e}^{2}\arctan \left ( cx \right ) }{2\,{c}^{2}}}-{\frac{bc{d}^{2}}{2\,x}}+ibed\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -ibed\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +ibed{\it dilog} \left ( 1+icx \right ) -ibed{\it dilog} \left ( 1-icx \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.14647, size = 223, normalized size = 1.74 \begin{align*} \frac{1}{2} \, a e^{2} x^{2} - \frac{1}{2} \,{\left ({\left (c \arctan \left (c x\right ) + \frac{1}{x}\right )} c + \frac{\arctan \left (c x\right )}{x^{2}}\right )} b d^{2} + 2 \, a d e \log \left (x\right ) - \frac{a d^{2}}{2 \, x^{2}} - \frac{\pi b c^{2} d e \log \left (c^{2} x^{2} + 1\right ) - 4 \, b c^{2} d e \arctan \left (c x\right ) \log \left (x{\left | c \right |}\right ) + 2 i \, b c^{2} d e{\rm Li}_2\left (i \, c x + 1\right ) - 2 i \, b c^{2} d e{\rm Li}_2\left (-i \, c x + 1\right ) + b c e^{2} x -{\left (b c^{2} e^{2} x^{2} + 4 i \, b c^{2} d e \arctan \left (0, c\right ) + b e^{2}\right )} \arctan \left (c x\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arctan(c*x))/x^3, 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 ) \left (d + e x^{2}\right )^{2}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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