3.1133 \(\int \frac{(d+e x^2)^2 (a+b \tan ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=139 \[ \frac{1}{2} i b e^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b e^2 \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{d e \left (a+b \tan ^{-1}(c x)\right )}{x^2}+a e^2 \log (x)+\frac{b c^3 d^2}{4 x}+\frac{1}{4} b c^4 d^2 \tan ^{-1}(c x)-b c^2 d e \tan ^{-1}(c x)-\frac{b c d^2}{12 x^3}-\frac{b c d e}{x} \]

[Out]

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

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Rubi [A]  time = 0.168279, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4980, 4852, 325, 203, 4848, 2391} \[ \frac{1}{2} i b e^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b e^2 \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{d e \left (a+b \tan ^{-1}(c x)\right )}{x^2}+a e^2 \log (x)+\frac{b c^3 d^2}{4 x}+\frac{1}{4} b c^4 d^2 \tan ^{-1}(c x)-b c^2 d e \tan ^{-1}(c x)-\frac{b c d^2}{12 x^3}-\frac{b c d e}{x} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

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

Rubi steps

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

Mathematica [C]  time = 0.0945496, size = 130, normalized size = 0.94 \[ -\frac{b c d^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )}{12 x^3}-\frac{b c d e \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )}{x}+\frac{1}{2} i b e^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b e^2 \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{d e \left (a+b \tan ^{-1}(c x)\right )}{x^2}+a e^2 \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

-(d^2*(a + b*ArcTan[c*x]))/(4*x^4) - (d*e*(a + b*ArcTan[c*x]))/x^2 - (b*c*d^2*Hypergeometric2F1[-3/2, 1, -1/2,
-(c^2*x^2)])/(12*x^3) - (b*c*d*e*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/x + a*e^2*Log[x] + (I/2)*b*e^2*
PolyLog[2, (-I)*c*x] - (I/2)*b*e^2*PolyLog[2, I*c*x]

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Maple [A]  time = 0.059, size = 190, normalized size = 1.4 \begin{align*} -{\frac{aed}{{x}^{2}}}-{\frac{a{d}^{2}}{4\,{x}^{4}}}+a{e}^{2}\ln \left ( cx \right ) -{\frac{b\arctan \left ( cx \right ) ed}{{x}^{2}}}-{\frac{b{d}^{2}\arctan \left ( cx \right ) }{4\,{x}^{4}}}+b\arctan \left ( cx \right ){e}^{2}\ln \left ( cx \right ) +{\frac{i}{2}}b{e}^{2}\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -{\frac{i}{2}}b{e}^{2}\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +{\frac{i}{2}}b{e}^{2}{\it dilog} \left ( 1+icx \right ) -{\frac{i}{2}}b{e}^{2}{\it dilog} \left ( 1-icx \right ) +{\frac{b{c}^{4}{d}^{2}\arctan \left ( cx \right ) }{4}}-b{c}^{2}de\arctan \left ( cx \right ) +{\frac{b{c}^{3}{d}^{2}}{4\,x}}-{\frac{bcde}{x}}-{\frac{bc{d}^{2}}{12\,{x}^{3}}} \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^5,x)

[Out]

-a*e*d/x^2-1/4*a*d^2/x^4+a*e^2*ln(c*x)-b*arctan(c*x)*e*d/x^2-1/4*b*arctan(c*x)*d^2/x^4+b*arctan(c*x)*e^2*ln(c*
x)+1/2*I*b*e^2*ln(c*x)*ln(1+I*c*x)-1/2*I*b*e^2*ln(c*x)*ln(1-I*c*x)+1/2*I*b*e^2*dilog(1+I*c*x)-1/2*I*b*e^2*dilo
g(1-I*c*x)+1/4*b*c^4*d^2*arctan(c*x)-b*c^2*d*e*arctan(c*x)+1/4*b*c^3*d^2/x-b*c*d*e/x-1/12*b*c*d^2/x^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{12} \,{\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac{3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac{3 \, \arctan \left (c x\right )}{x^{4}}\right )} b d^{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 e + b e^{2} \int \frac{\arctan \left (c x\right )}{x}\,{d x} + a e^{2} \log \left (x\right ) - \frac{a d e}{x^{2}} - \frac{a d^{2}}{4 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

1/12*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d^2 - ((c*arctan(c*x) + 1/x)*c + arct
an(c*x)/x^2)*b*d*e + b*e^2*integrate(arctan(c*x)/x, x) + a*e^2*log(x) - a*d*e/x^2 - 1/4*a*d^2/x^4

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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^{5}}, 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^5,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^5, 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^{5}}\, 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**5,x)

[Out]

Integral((a + b*atan(c*x))*(d + e*x**2)**2/x**5, 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^{5}}\,{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^5,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^2*(b*arctan(c*x) + a)/x^5, x)