3.1143 \(\int \frac{(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=200 \[ \frac{3}{2} i b d^2 e \text{PolyLog}(2,-i c x)-\frac{3}{2} i b d^2 e \text{PolyLog}(2,i c x)-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{3}{2} d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+3 a d^2 e \log (x)-\frac{1}{2} b c^2 d^3 \tan ^{-1}(c x)+\frac{3 b d e^2 \tan ^{-1}(c x)}{2 c^2}+\frac{b e^3 x}{4 c^3}-\frac{b e^3 \tan ^{-1}(c x)}{4 c^4}-\frac{b c d^3}{2 x}-\frac{3 b d e^2 x}{2 c}-\frac{b e^3 x^3}{12 c} \]

[Out]

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

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Rubi [A]  time = 0.210537, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {4980, 4852, 325, 203, 4848, 2391, 321, 302} \[ \frac{3}{2} i b d^2 e \text{PolyLog}(2,-i c x)-\frac{3}{2} i b d^2 e \text{PolyLog}(2,i c x)-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{3}{2} d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+3 a d^2 e \log (x)-\frac{1}{2} b c^2 d^3 \tan ^{-1}(c x)+\frac{3 b d e^2 \tan ^{-1}(c x)}{2 c^2}+\frac{b e^3 x}{4 c^3}-\frac{b e^3 \tan ^{-1}(c x)}{4 c^4}-\frac{b c d^3}{2 x}-\frac{3 b d e^2 x}{2 c}-\frac{b e^3 x^3}{12 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

Mathematica [C]  time = 0.160364, size = 170, normalized size = 0.85 \[ \frac{1}{12} \left (-\frac{6 b c d^3 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )}{x}+18 i b d^2 e \text{PolyLog}(2,-i c x)-18 i b d^2 e \text{PolyLog}(2,i c x)-\frac{6 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+18 d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+3 e^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+36 a d^2 e \log (x)-\frac{18 b d e^2 \left (c x-\tan ^{-1}(c x)\right )}{c^2}-\frac{b e^3 \left (c^3 x^3-3 c x+3 \tan ^{-1}(c x)\right )}{c^4}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-18*b*d*e^2*(c*x - ArcTan[c*x]))/c^2 - (b*e^3*(-3*c*x + c^3*x^3 + 3*ArcTan[c*x]))/c^4 - (6*d^3*(a + b*ArcTan
[c*x]))/x^2 + 18*d*e^2*x^2*(a + b*ArcTan[c*x]) + 3*e^3*x^4*(a + b*ArcTan[c*x]) - (6*b*c*d^3*Hypergeometric2F1[
-1/2, 1, 1/2, -(c^2*x^2)])/x + 36*a*d^2*e*Log[x] + (18*I)*b*d^2*e*PolyLog[2, (-I)*c*x] - (18*I)*b*d^2*e*PolyLo
g[2, I*c*x])/12

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Maple [A]  time = 0.056, size = 251, normalized size = 1.3 \begin{align*}{\frac{a{e}^{3}{x}^{4}}{4}}+{\frac{3\,a{x}^{2}d{e}^{2}}{2}}-{\frac{{d}^{3}a}{2\,{x}^{2}}}+3\,a{d}^{2}e\ln \left ( cx \right ) +{\frac{b\arctan \left ( cx \right ){e}^{3}{x}^{4}}{4}}+{\frac{3\,b\arctan \left ( cx \right ){x}^{2}d{e}^{2}}{2}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{2\,{x}^{2}}}+3\,b\arctan \left ( cx \right ){d}^{2}e\ln \left ( cx \right ) -{\frac{b{e}^{3}{x}^{3}}{12\,c}}-{\frac{3\,bd{e}^{2}x}{2\,c}}+{\frac{b{e}^{3}x}{4\,{c}^{3}}}-{\frac{b{c}^{2}{d}^{3}\arctan \left ( cx \right ) }{2}}+{\frac{3\,bd{e}^{2}\arctan \left ( cx \right ) }{2\,{c}^{2}}}-{\frac{b\arctan \left ( cx \right ){e}^{3}}{4\,{c}^{4}}}-{\frac{bc{d}^{3}}{2\,x}}-{\frac{3\,i}{2}}b{d}^{2}e{\it dilog} \left ( 1-icx \right ) +{\frac{3\,i}{2}}b{d}^{2}e{\it dilog} \left ( 1+icx \right ) -{\frac{3\,i}{2}}b{d}^{2}e\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +{\frac{3\,i}{2}}b{d}^{2}e\ln \left ( cx \right ) \ln \left ( 1+icx \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.20896, size = 321, normalized size = 1.6 \begin{align*} \frac{1}{4} \, a e^{3} x^{4} + \frac{3}{2} \, a d 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^{3} + 3 \, a d^{2} e \log \left (x\right ) - \frac{a d^{3}}{2 \, x^{2}} - \frac{b c^{3} e^{3} x^{3} + 9 \, \pi b c^{4} d^{2} e \log \left (c^{2} x^{2} + 1\right ) - 36 \, b c^{4} d^{2} e \arctan \left (c x\right ) \log \left (x{\left | c \right |}\right ) + 18 i \, b c^{4} d^{2} e{\rm Li}_2\left (i \, c x + 1\right ) - 18 i \, b c^{4} d^{2} e{\rm Li}_2\left (-i \, c x + 1\right ) + 3 \,{\left (6 \, b c^{3} d e^{2} - b c e^{3}\right )} x -{\left (3 \, b c^{4} e^{3} x^{4} + 18 \, b c^{4} d e^{2} x^{2} + 36 i \, b c^{4} d^{2} e \arctan \left (0, c\right ) + 18 \, b c^{2} d e^{2} - 3 \, b e^{3}\right )} \arctan \left (c x\right )}{12 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*e^3*x^4 + 3/2*a*d*e^2*x^2 - 1/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*d^3 + 3*a*d^2*e*log(x) - 1
/2*a*d^3/x^2 - 1/12*(b*c^3*e^3*x^3 + 9*pi*b*c^4*d^2*e*log(c^2*x^2 + 1) - 36*b*c^4*d^2*e*arctan(c*x)*log(x*abs(
c)) + 18*I*b*c^4*d^2*e*dilog(I*c*x + 1) - 18*I*b*c^4*d^2*e*dilog(-I*c*x + 1) + 3*(6*b*c^3*d*e^2 - b*c*e^3)*x -
(3*b*c^4*e^3*x^4 + 18*b*c^4*d*e^2*x^2 + 36*I*b*c^4*d^2*e*arctan2(0, c) + 18*b*c^2*d*e^2 - 3*b*e^3)*arctan(c*x
))/c^4

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a e^{3} x^{6} + 3 \, a d e^{2} x^{4} + 3 \, a d^{2} e x^{2} + a d^{3} +{\left (b e^{3} x^{6} + 3 \, b d e^{2} x^{4} + 3 \, b d^{2} e x^{2} + b d^{3}\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)^3*(a+b*arctan(c*x))/x^3,x, algorithm="fricas")

[Out]

integral((a*e^3*x^6 + 3*a*d*e^2*x^4 + 3*a*d^2*e*x^2 + a*d^3 + (b*e^3*x^6 + 3*b*d*e^2*x^4 + 3*b*d^2*e*x^2 + b*d
^3)*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 )^{3}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c*x))*(d + e*x**2)**3/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 )}^{3}{\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)^3*(a+b*arctan(c*x))/x^3,x, algorithm="giac")

[Out]

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