3.1129 \(\int \frac{(d+e x^2)^2 (a+b \tan ^{-1}(c x))}{x} \, dx\)

Optimal. Leaf size=137 \[ \frac{1}{2} i b d^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^2 \text{PolyLog}(2,i c x)+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)+\frac{b d e \tan ^{-1}(c x)}{c^2}+\frac{b e^2 x}{4 c^3}-\frac{b e^2 \tan ^{-1}(c x)}{4 c^4}-\frac{b d e x}{c}-\frac{b e^2 x^3}{12 c} \]

[Out]

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

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Rubi [A]  time = 0.179517, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4980, 4848, 2391, 4852, 321, 203, 302} \[ \frac{1}{2} i b d^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^2 \text{PolyLog}(2,i c x)+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)+\frac{b d e \tan ^{-1}(c x)}{c^2}+\frac{b e^2 x}{4 c^3}-\frac{b e^2 \tan ^{-1}(c x)}{4 c^4}-\frac{b d e x}{c}-\frac{b e^2 x^3}{12 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.101077, size = 123, normalized size = 0.9 \[ \frac{1}{2} i b d^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^2 \text{PolyLog}(2,i c x)+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)-\frac{b d e \left (c x-\tan ^{-1}(c x)\right )}{c^2}-\frac{b e^2 \left (c^3 x^3-3 c x+3 \tan ^{-1}(c x)\right )}{12 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.051, size = 187, normalized size = 1.4 \begin{align*}{\frac{a{x}^{4}{e}^{2}}{4}}+a{x}^{2}de+a{d}^{2}\ln \left ( cx \right ) +{\frac{b\arctan \left ( cx \right ){x}^{4}{e}^{2}}{4}}+b\arctan \left ( cx \right ){x}^{2}de+b\arctan \left ( cx \right ){d}^{2}\ln \left ( cx \right ) -{\frac{b{e}^{2}{x}^{3}}{12\,c}}-{\frac{bdex}{c}}+{\frac{b{e}^{2}x}{4\,{c}^{3}}}+{\frac{bde\arctan \left ( cx \right ) }{{c}^{2}}}-{\frac{b{e}^{2}\arctan \left ( cx \right ) }{4\,{c}^{4}}}+{\frac{i}{2}}b{d}^{2}\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -{\frac{i}{2}}b{d}^{2}\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +{\frac{i}{2}}b{d}^{2}{\it dilog} \left ( 1+icx \right ) -{\frac{i}{2}}b{d}^{2}{\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,x)

[Out]

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

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Maxima [A]  time = 2.14796, size = 251, normalized size = 1.83 \begin{align*} \frac{1}{4} \, a e^{2} x^{4} + a d e x^{2} + a d^{2} \log \left (x\right ) - \frac{b c^{3} e^{2} x^{3} + 3 \, \pi b c^{4} d^{2} \log \left (c^{2} x^{2} + 1\right ) - 12 \, b c^{4} d^{2} \arctan \left (c x\right ) \log \left (x{\left | c \right |}\right ) + 6 i \, b c^{4} d^{2}{\rm Li}_2\left (i \, c x + 1\right ) - 6 i \, b c^{4} d^{2}{\rm Li}_2\left (-i \, c x + 1\right ) + 3 \,{\left (4 \, b c^{3} d e - b c e^{2}\right )} x -{\left (3 \, b c^{4} e^{2} x^{4} + 12 \, b c^{4} d e x^{2} + 12 i \, b c^{4} d^{2} \arctan \left (0, c\right ) + 12 \, b c^{2} d e - 3 \, b e^{2}\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)^2*(a+b*arctan(c*x))/x,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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