3.1141 \(\int \frac {(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x} \, dx\)

Optimal. Leaf size=228 \[ \frac {3}{2} d^2 e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac {b e^3 \tan ^{-1}(c x)}{6 c^6}-\frac {b e^3 x}{6 c^5}-\frac {3 b d e^2 \tan ^{-1}(c x)}{4 c^4}+\frac {3 b d e^2 x}{4 c^3}+\frac {b e^3 x^3}{18 c^3}+\frac {3 b d^2 e \tan ^{-1}(c x)}{2 c^2}+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x)-\frac {3 b d^2 e x}{2 c}-\frac {b d e^2 x^3}{4 c}-\frac {b e^3 x^5}{30 c} \]

[Out]

-3/2*b*d^2*e*x/c+3/4*b*d*e^2*x/c^3-1/6*b*e^3*x/c^5-1/4*b*d*e^2*x^3/c+1/18*b*e^3*x^3/c^3-1/30*b*e^3*x^5/c+3/2*b
*d^2*e*arctan(c*x)/c^2-3/4*b*d*e^2*arctan(c*x)/c^4+1/6*b*e^3*arctan(c*x)/c^6+3/2*d^2*e*x^2*(a+b*arctan(c*x))+3
/4*d*e^2*x^4*(a+b*arctan(c*x))+1/6*e^3*x^6*(a+b*arctan(c*x))+a*d^3*ln(x)+1/2*I*b*d^3*polylog(2,-I*c*x)-1/2*I*b
*d^3*polylog(2,I*c*x)

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Rubi [A]  time = 0.22, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 16, 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} \[ \frac {1}{2} i b d^3 \text {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \text {PolyLog}(2,i c x)+\frac {3}{2} d^2 e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac {3 b d^2 e \tan ^{-1}(c x)}{2 c^2}+\frac {3 b d e^2 x}{4 c^3}-\frac {3 b d e^2 \tan ^{-1}(c x)}{4 c^4}+\frac {b e^3 x^3}{18 c^3}-\frac {b e^3 x}{6 c^5}+\frac {b e^3 \tan ^{-1}(c x)}{6 c^6}-\frac {3 b d^2 e x}{2 c}-\frac {b d e^2 x^3}{4 c}-\frac {b e^3 x^5}{30 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*d^2*e*x)/(2*c) + (3*b*d*e^2*x)/(4*c^3) - (b*e^3*x)/(6*c^5) - (b*d*e^2*x^3)/(4*c) + (b*e^3*x^3)/(18*c^3)
- (b*e^3*x^5)/(30*c) + (3*b*d^2*e*ArcTan[c*x])/(2*c^2) - (3*b*d*e^2*ArcTan[c*x])/(4*c^4) + (b*e^3*ArcTan[c*x])
/(6*c^6) + (3*d^2*e*x^2*(a + b*ArcTan[c*x]))/2 + (3*d*e^2*x^4*(a + b*ArcTan[c*x]))/4 + (e^3*x^6*(a + b*ArcTan[
c*x]))/6 + a*d^3*Log[x] + (I/2)*b*d^3*PolyLog[2, (-I)*c*x] - (I/2)*b*d^3*PolyLog[2, I*c*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 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])

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^3 \left (a+b \tan ^{-1}(c x)\right )}{x} \, dx &=\int \left (\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )+3 d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )\right ) \, dx\\ &=d^3 \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx+\left (3 d^2 e\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx+\left (3 d e^2\right ) \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx+e^3 \int x^5 \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=\frac {3}{2} d^2 e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac {1}{2} \left (i b d^3\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (i b d^3\right ) \int \frac {\log (1+i c x)}{x} \, dx-\frac {1}{2} \left (3 b c d^2 e\right ) \int \frac {x^2}{1+c^2 x^2} \, dx-\frac {1}{4} \left (3 b c d e^2\right ) \int \frac {x^4}{1+c^2 x^2} \, dx-\frac {1}{6} \left (b c e^3\right ) \int \frac {x^6}{1+c^2 x^2} \, dx\\ &=-\frac {3 b d^2 e x}{2 c}+\frac {3}{2} d^2 e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x)+\frac {\left (3 b d^2 e\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 c}-\frac {1}{4} \left (3 b c d 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 {1}{6} \left (b c e^3\right ) \int \left (\frac {1}{c^6}-\frac {x^2}{c^4}+\frac {x^4}{c^2}-\frac {1}{c^6 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {3 b d^2 e x}{2 c}+\frac {3 b d e^2 x}{4 c^3}-\frac {b e^3 x}{6 c^5}-\frac {b d e^2 x^3}{4 c}+\frac {b e^3 x^3}{18 c^3}-\frac {b e^3 x^5}{30 c}+\frac {3 b d^2 e \tan ^{-1}(c x)}{2 c^2}+\frac {3}{2} d^2 e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x)-\frac {\left (3 b d e^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 c^3}+\frac {\left (b e^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{6 c^5}\\ &=-\frac {3 b d^2 e x}{2 c}+\frac {3 b d e^2 x}{4 c^3}-\frac {b e^3 x}{6 c^5}-\frac {b d e^2 x^3}{4 c}+\frac {b e^3 x^3}{18 c^3}-\frac {b e^3 x^5}{30 c}+\frac {3 b d^2 e \tan ^{-1}(c x)}{2 c^2}-\frac {3 b d e^2 \tan ^{-1}(c x)}{4 c^4}+\frac {b e^3 \tan ^{-1}(c x)}{6 c^6}+\frac {3}{2} d^2 e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x)\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 190, normalized size = 0.83 \[ \frac {3}{2} d^2 e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)-\frac {3 b d^2 e \left (c x-\tan ^{-1}(c x)\right )}{2 c^2}-\frac {b d e^2 \left (c^3 x^3-3 c x+3 \tan ^{-1}(c x)\right )}{4 c^4}-\frac {b e^3 \left (3 c^5 x^5-5 c^3 x^3+15 c x-15 \tan ^{-1}(c x)\right )}{90 c^6}+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/90*(b*e^3*(15*c*x - 5*c^3*x^3 + 3*c^5*x^5 - 15*ArcTan[c*x]))/c^6 - (3*b*d^2*e*(c*x - ArcTan[c*x]))/(2*c^2)
- (b*d*e^2*(-3*c*x + c^3*x^3 + 3*ArcTan[c*x]))/(4*c^4) + (3*d^2*e*x^2*(a + b*ArcTan[c*x]))/2 + (3*d*e^2*x^4*(a
+ b*ArcTan[c*x]))/4 + (e^3*x^6*(a + b*ArcTan[c*x]))/6 + a*d^3*Log[x] + (I/2)*b*d^3*PolyLog[2, (-I)*c*x] - (I/
2)*b*d^3*PolyLog[2, I*c*x]

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a e^{3} x^{6} + 3 \, a d e^{2} x^{4} + 3 \, a d^{2} e x^{2} + a d^{3} + {\left (b e^{3} x^{6} + 3 \, b d e^{2} x^{4} + 3 \, b d^{2} e x^{2} + b d^{3}\right )} \arctan \left (c x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.06, size = 272, normalized size = 1.19 \[ \frac {a \,x^{6} e^{3}}{6}+\frac {3 a \,x^{4} d \,e^{2}}{4}+\frac {3 a \,x^{2} d^{2} e}{2}+d^{3} a \ln \left (c x \right )+\frac {b \arctan \left (c x \right ) x^{6} e^{3}}{6}+\frac {3 b \arctan \left (c x \right ) x^{4} d \,e^{2}}{4}+\frac {3 b \arctan \left (c x \right ) x^{2} d^{2} e}{2}+b \arctan \left (c x \right ) d^{3} \ln \left (c x \right )-\frac {b \,e^{3} x^{5}}{30 c}-\frac {b d \,e^{2} x^{3}}{4 c}-\frac {3 b \,d^{2} e x}{2 c}+\frac {b \,e^{3} x^{3}}{18 c^{3}}+\frac {3 b d \,e^{2} x}{4 c^{3}}-\frac {b \,e^{3} x}{6 c^{5}}+\frac {3 b \,d^{2} e \arctan \left (c x \right )}{2 c^{2}}-\frac {3 b d \,e^{2} \arctan \left (c x \right )}{4 c^{4}}+\frac {b \,e^{3} \arctan \left (c x \right )}{6 c^{6}}+\frac {i b \,d^{3} \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i b \,d^{3} \dilog \left (-i c x +1\right )}{2}+\frac {i b \,d^{3} \dilog \left (i c x +1\right )}{2}-\frac {i b \,d^{3} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*x^6*e^3+3/4*a*x^4*d*e^2+3/2*a*x^2*d^2*e+d^3*a*ln(c*x)+1/6*b*arctan(c*x)*x^6*e^3+3/4*b*arctan(c*x)*x^4*d*
e^2+3/2*b*arctan(c*x)*x^2*d^2*e+b*arctan(c*x)*d^3*ln(c*x)-1/30*b*e^3*x^5/c-1/4*b*d*e^2*x^3/c-3/2*b*d^2*e*x/c+1
/18*b*e^3*x^3/c^3+3/4*b*d*e^2*x/c^3-1/6*b*e^3*x/c^5+3/2*b*d^2*e*arctan(c*x)/c^2-3/4*b*d*e^2*arctan(c*x)/c^4+1/
6*b*e^3*arctan(c*x)/c^6-1/2*I*b*d^3*dilog(1-I*c*x)+1/2*I*b*d^3*ln(c*x)*ln(1+I*c*x)-1/2*I*b*d^3*ln(c*x)*ln(1-I*
c*x)+1/2*I*b*d^3*dilog(1+I*c*x)

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maxima [A]  time = 0.66, size = 251, normalized size = 1.10 \[ \frac {1}{6} \, a e^{3} x^{6} + \frac {3}{4} \, a d e^{2} x^{4} + \frac {3}{2} \, a d^{2} e x^{2} + a d^{3} \log \relax (x) - \frac {6 \, b c^{5} e^{3} x^{5} + 45 \, \pi b c^{6} d^{3} \log \left (c^{2} x^{2} + 1\right ) - 180 \, b c^{6} d^{3} \arctan \left (c x\right ) \log \left (c x\right ) + 90 i \, b c^{6} d^{3} {\rm Li}_2\left (i \, c x + 1\right ) - 90 i \, b c^{6} d^{3} {\rm Li}_2\left (-i \, c x + 1\right ) + 5 \, {\left (9 \, b c^{5} d e^{2} - 2 \, b c^{3} e^{3}\right )} x^{3} + 15 \, {\left (18 \, b c^{5} d^{2} e - 9 \, b c^{3} d e^{2} + 2 \, b c e^{3}\right )} x - 15 \, {\left (2 \, b c^{6} e^{3} x^{6} + 9 \, b c^{6} d e^{2} x^{4} + 18 \, b c^{6} d^{2} e x^{2} + 18 \, b c^{4} d^{2} e - 9 \, b c^{2} d e^{2} + 2 \, b e^{3}\right )} \arctan \left (c x\right )}{180 \, c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*e^3*x^6 + 3/4*a*d*e^2*x^4 + 3/2*a*d^2*e*x^2 + a*d^3*log(x) - 1/180*(6*b*c^5*e^3*x^5 + 45*pi*b*c^6*d^3*lo
g(c^2*x^2 + 1) - 180*b*c^6*d^3*arctan(c*x)*log(c*x) + 90*I*b*c^6*d^3*dilog(I*c*x + 1) - 90*I*b*c^6*d^3*dilog(-
I*c*x + 1) + 5*(9*b*c^5*d*e^2 - 2*b*c^3*e^3)*x^3 + 15*(18*b*c^5*d^2*e - 9*b*c^3*d*e^2 + 2*b*c*e^3)*x - 15*(2*b
*c^6*e^3*x^6 + 9*b*c^6*d*e^2*x^4 + 18*b*c^6*d^2*e*x^2 + 18*b*c^4*d^2*e - 9*b*c^2*d*e^2 + 2*b*e^3)*arctan(c*x))
/c^6

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mupad [B]  time = 0.78, size = 232, normalized size = 1.02 \[ \left \{\begin {array}{cl} \frac {a\,e^3\,x^6}{6}+a\,d^3\,\ln \relax (x)+\frac {3\,a\,d^2\,e\,x^2}{2}+\frac {3\,a\,d\,e^2\,x^4}{4} & \text {\ if\ \ }c=0\\ \frac {a\,e^3\,x^6}{6}+a\,d^3\,\ln \relax (x)-\frac {b\,e^3\,\left (\frac {x}{c^4}-\frac {\mathrm {atan}\left (c\,x\right )}{c^5}+\frac {x^5}{5}-\frac {x^3}{3\,c^2}\right )}{6\,c}-3\,b\,d^2\,e\,\left (\frac {x}{2\,c}-\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\right )+\frac {3\,a\,d^2\,e\,x^2}{2}+\frac {3\,a\,d\,e^2\,x^4}{4}-3\,b\,d\,e^2\,\left (\frac {3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3}{12\,c^4}-\frac {x^4\,\mathrm {atan}\left (c\,x\right )}{4}\right )+\frac {b\,e^3\,x^6\,\mathrm {atan}\left (c\,x\right )}{6}-\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(c == 0, (a*e^3*x^6)/6 + a*d^3*log(x) + (3*a*d^2*e*x^2)/2 + (3*a*d*e^2*x^4)/4, c ~= 0, (a*e^3*x^6)/6
+ a*d^3*log(x) - (b*d^3*dilog(- c*x*1i + 1)*1i)/2 + (b*d^3*dilog(c*x*1i + 1)*1i)/2 - (b*e^3*(x/c^4 - atan(c*x)
/c^5 + x^5/5 - x^3/(3*c^2)))/(6*c) - 3*b*d^2*e*(x/(2*c) - atan(c*x)*(1/(2*c^2) + x^2/2)) + (3*a*d^2*e*x^2)/2 +
(3*a*d*e^2*x^4)/4 - 3*b*d*e^2*((3*atan(c*x) - 3*c*x + c^3*x^3)/(12*c^4) - (x^4*atan(c*x))/4) + (b*e^3*x^6*ata
n(c*x))/6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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