\(\int \frac {(d+e x^2)^3 (a+b \arctan (c x))}{x^3} \, dx\) [1143]

Optimal result

Integrand size = 21, antiderivative size = 200 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^3} \, dx=-\frac {b c d^3}{2 x}-\frac {3 b d e^2 x}{2 c}+\frac {b e^3 x}{4 c^3}-\frac {b e^3 x^3}{12 c}-\frac {1}{2} b c^2 d^3 \arctan (c x)+\frac {3 b d e^2 \arctan (c x)}{2 c^2}-\frac {b e^3 \arctan (c x)}{4 c^4}-\frac {d^3 (a+b \arctan (c x))}{2 x^2}+\frac {3}{2} d e^2 x^2 (a+b \arctan (c x))+\frac {1}{4} e^3 x^4 (a+b \arctan (c x))+3 a d^2 e \log (x)+\frac {3}{2} i b d^2 e \operatorname {PolyLog}(2,-i c x)-\frac {3}{2} i b d^2 e \operatorname {PolyLog}(2,i c x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {5100, 4946, 331, 209, 4940, 2438, 327, 308} \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^3} \, dx=-\frac {d^3 (a+b \arctan (c x))}{2 x^2}+\frac {3}{2} d e^2 x^2 (a+b \arctan (c x))+\frac {1}{4} e^3 x^4 (a+b \arctan (c x))+3 a d^2 e \log (x)-\frac {b e^3 \arctan (c x)}{4 c^4}-\frac {1}{2} b c^2 d^3 \arctan (c x)+\frac {3 b d e^2 \arctan (c x)}{2 c^2}+\frac {b e^3 x}{4 c^3}-\frac {b c d^3}{2 x}+\frac {3}{2} i b d^2 e \operatorname {PolyLog}(2,-i c x)-\frac {3}{2} i b d^2 e \operatorname {PolyLog}(2,i c x)-\frac {3 b d e^2 x}{2 c}-\frac {b e^3 x^3}{12 c} \]

[In]

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

[Out]

-1/2*(b*c*d^3)/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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 308

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 327

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 331

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 2438

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 4940

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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 5100

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*} \text {integral}& = \int \left (\frac {d^3 (a+b \arctan (c x))}{x^3}+\frac {3 d^2 e (a+b \arctan (c x))}{x}+3 d e^2 x (a+b \arctan (c x))+e^3 x^3 (a+b \arctan (c x))\right ) \, dx \\ & = d^3 \int \frac {a+b \arctan (c x)}{x^3} \, dx+\left (3 d^2 e\right ) \int \frac {a+b \arctan (c x)}{x} \, dx+\left (3 d e^2\right ) \int x (a+b \arctan (c x)) \, dx+e^3 \int x^3 (a+b \arctan (c x)) \, dx \\ & = -\frac {d^3 (a+b \arctan (c x))}{2 x^2}+\frac {3}{2} d e^2 x^2 (a+b \arctan (c x))+\frac {1}{4} e^3 x^4 (a+b \arctan (c x))+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 (a+b \arctan (c x))}{2 x^2}+\frac {3}{2} d e^2 x^2 (a+b \arctan (c x))+\frac {1}{4} e^3 x^4 (a+b \arctan (c x))+3 a d^2 e \log (x)+\frac {3}{2} i b d^2 e \operatorname {PolyLog}(2,-i c x)-\frac {3}{2} i b d^2 e \operatorname {PolyLog}(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 \arctan (c x)+\frac {3 b d e^2 \arctan (c x)}{2 c^2}-\frac {d^3 (a+b \arctan (c x))}{2 x^2}+\frac {3}{2} d e^2 x^2 (a+b \arctan (c x))+\frac {1}{4} e^3 x^4 (a+b \arctan (c x))+3 a d^2 e \log (x)+\frac {3}{2} i b d^2 e \operatorname {PolyLog}(2,-i c x)-\frac {3}{2} i b d^2 e \operatorname {PolyLog}(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 \arctan (c x)+\frac {3 b d e^2 \arctan (c x)}{2 c^2}-\frac {b e^3 \arctan (c x)}{4 c^4}-\frac {d^3 (a+b \arctan (c x))}{2 x^2}+\frac {3}{2} d e^2 x^2 (a+b \arctan (c x))+\frac {1}{4} e^3 x^4 (a+b \arctan (c x))+3 a d^2 e \log (x)+\frac {3}{2} i b d^2 e \operatorname {PolyLog}(2,-i c x)-\frac {3}{2} i b d^2 e \operatorname {PolyLog}(2,i c x) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.85 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^3} \, dx=\frac {1}{12} \left (-\frac {18 b d e^2 (c x-\arctan (c x))}{c^2}-\frac {b e^3 \left (-3 c x+c^3 x^3+3 \arctan (c x)\right )}{c^4}-\frac {6 d^3 (a+b \arctan (c x))}{x^2}+18 d e^2 x^2 (a+b \arctan (c x))+3 e^3 x^4 (a+b \arctan (c x))-\frac {6 b c d^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )}{x}+36 a d^2 e \log (x)+18 i b d^2 e \operatorname {PolyLog}(2,-i c x)-18 i b d^2 e \operatorname {PolyLog}(2,i c x)\right ) \]

[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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.20

method	result	size
parts	\(a \left (\frac {e^{3} x^{4}}{4}+\frac {3 d \,e^{2} x^{2}}{2}+3 e \,d^{2} \ln \left (x \right )-\frac {d^{3}}{2 x^{2}}\right )+b \,c^{2} \left (\frac {\arctan \left (c x \right ) e^{3} x^{4}}{4 c^{2}}+\frac {3 \arctan \left (c x \right ) e^{2} d \,x^{2}}{2 c^{2}}+\frac {3 \arctan \left (c x \right ) d^{2} e \ln \left (c x \right )}{c^{2}}-\frac {\arctan \left (c x \right ) d^{3}}{2 c^{2} x^{2}}-\frac {\frac {e^{3} c^{3} x^{3}}{3}+6 c^{3} x d \,e^{2}-c x \,e^{3}+\frac {2 c^{5} d^{3}}{x}+\left (2 c^{6} d^{3}-6 e^{2} d \,c^{2}+e^{3}\right ) \arctan \left (c x \right )+12 c^{4} d^{2} e \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )}{4 c^{6}}\right )\)	\(240\)
derivativedivides	\(c^{2} \left (\frac {a \left (\frac {3 d \,c^{4} e^{2} x^{2}}{2}+\frac {e^{3} c^{4} x^{4}}{4}+3 c^{4} d^{2} e \ln \left (c x \right )-\frac {c^{4} d^{3}}{2 x^{2}}\right )}{c^{6}}+\frac {b \left (\frac {3 \arctan \left (c x \right ) d \,c^{4} e^{2} x^{2}}{2}+\frac {\arctan \left (c x \right ) e^{3} c^{4} x^{4}}{4}+3 \arctan \left (c x \right ) c^{4} d^{2} e \ln \left (c x \right )-\frac {\arctan \left (c x \right ) c^{4} d^{3}}{2 x^{2}}-\frac {3 c^{3} x d \,e^{2}}{2}-\frac {e^{3} c^{3} x^{3}}{12}+\frac {c x \,e^{3}}{4}-\frac {c^{5} d^{3}}{2 x}+\frac {\left (-2 c^{6} d^{3}+6 e^{2} d \,c^{2}-e^{3}\right ) \arctan \left (c x \right )}{4}-3 c^{4} d^{2} e \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{6}}\right )\)	\(258\)
default	\(c^{2} \left (\frac {a \left (\frac {3 d \,c^{4} e^{2} x^{2}}{2}+\frac {e^{3} c^{4} x^{4}}{4}+3 c^{4} d^{2} e \ln \left (c x \right )-\frac {c^{4} d^{3}}{2 x^{2}}\right )}{c^{6}}+\frac {b \left (\frac {3 \arctan \left (c x \right ) d \,c^{4} e^{2} x^{2}}{2}+\frac {\arctan \left (c x \right ) e^{3} c^{4} x^{4}}{4}+3 \arctan \left (c x \right ) c^{4} d^{2} e \ln \left (c x \right )-\frac {\arctan \left (c x \right ) c^{4} d^{3}}{2 x^{2}}-\frac {3 c^{3} x d \,e^{2}}{2}-\frac {e^{3} c^{3} x^{3}}{12}+\frac {c x \,e^{3}}{4}-\frac {c^{5} d^{3}}{2 x}+\frac {\left (-2 c^{6} d^{3}+6 e^{2} d \,c^{2}-e^{3}\right ) \arctan \left (c x \right )}{4}-3 c^{4} d^{2} e \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{6}}\right )\)	\(258\)
risch	\(\frac {b \,e^{3} x}{4 c^{3}}-\frac {b \,e^{3} x^{3}}{12 c}-\frac {b \,e^{3} \arctan \left (c x \right )}{4 c^{4}}-\frac {3 b d \,e^{2} x}{2 c}+\frac {3 b d \,e^{2} \arctan \left (c x \right )}{2 c^{2}}-\frac {b c \,d^{3}}{2 x}-\frac {b \,c^{2} d^{3} \arctan \left (c x \right )}{2}+\frac {3 a \,e^{2} d}{2 c^{2}}+\frac {3 a \,e^{2} d \,x^{2}}{2}+3 a \,d^{2} e \ln \left (-i c x \right )-\frac {a \,e^{3}}{4 c^{4}}-\frac {a \,d^{3}}{2 x^{2}}+\frac {a \,e^{3} x^{4}}{4}-\frac {i b \,d^{3} \ln \left (-i c x +1\right )}{4 x^{2}}+\frac {i b \,d^{3} \ln \left (i c x +1\right )}{4 x^{2}}-\frac {i b \,c^{2} d^{3} \ln \left (i c x \right )}{4}+\frac {3 i b e \,d^{2} \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i b \,e^{3} \ln \left (i c x +1\right ) x^{4}}{8}+\frac {i c^{2} b \,d^{3} \ln \left (-i c x \right )}{4}+\frac {3 i b \,e^{2} d \ln \left (-i c x +1\right ) x^{2}}{4}-\frac {3 i b \,d^{2} e \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i b \,e^{3} \ln \left (-i c x +1\right ) x^{4}}{8}-\frac {3 i b \,e^{2} d \ln \left (i c x +1\right ) x^{2}}{4}\)	\(319\)

[In]

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

[Out]

a*(1/4*e^3*x^4+3/2*d*e^2*x^2+3*e*d^2*ln(x)-1/2*d^3/x^2)+b*c^2*(1/4*arctan(c*x)/c^2*e^3*x^4+3/2*arctan(c*x)/c^2
*e^2*d*x^2+3*arctan(c*x)/c^2*d^2*e*ln(c*x)-1/2*arctan(c*x)*d^3/c^2/x^2-1/4/c^6*(1/3*e^3*c^3*x^3+6*c^3*x*d*e^2-
c*x*e^3+2*c^5*d^3/x+(2*c^6*d^3-6*c^2*d*e^2+e^3)*arctan(c*x)+12*c^4*d^2*e*(-1/2*I*ln(c*x)*ln(1+I*c*x)+1/2*I*ln(
c*x)*ln(1-I*c*x)-1/2*I*dilog(1+I*c*x)+1/2*I*dilog(1-I*c*x))))

Fricas [F]

\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^3} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x^{3}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.12 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^3} \, dx=\frac {1}{4} \, a e^{3} x^{4} + \frac {3}{2} \, a d e^{2} x^{2} - \frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b d^{3} + 3 \, a d^{2} e \log \left (x\right ) - \frac {a d^{3}}{2 \, x^{2}} - \frac {b c^{3} e^{3} x^{3} + 9 \, \pi b c^{4} d^{2} e \log \left (c^{2} x^{2} + 1\right ) - 36 \, b c^{4} d^{2} e \arctan \left (c x\right ) \log \left (c x\right ) + 18 i \, b c^{4} d^{2} e {\rm Li}_2\left (i \, c x + 1\right ) - 18 i \, b c^{4} d^{2} e {\rm Li}_2\left (-i \, c x + 1\right ) + 3 \, {\left (6 \, b c^{3} d e^{2} - b c e^{3}\right )} x - 3 \, {\left (b c^{4} e^{3} x^{4} + 6 \, b c^{4} d e^{2} x^{2} + 6 \, b c^{2} d e^{2} - b e^{3}\right )} \arctan \left (c x\right )}{12 \, c^{4}} \]

[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(c*x) +
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 + 6*b*c^4*d*e^2*x^2 + 6*b*c^2*d*e^2 - b*e^3)*arctan(c*x))/c^4

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.12 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^3} \, dx=\left \{\begin {array}{cl} \frac {a\,e^3\,x^4}{4}-\frac {a\,d^3}{2\,x^2}+\frac {3\,a\,d\,e^2\,x^2}{2}+3\,a\,d^2\,e\,\ln \left (x\right ) & \text {\ if\ \ }c=0\\ \frac {a\,e^3\,x^4}{4}-\frac {a\,d^3}{2\,x^2}-\frac {b\,d^3\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )}{2\,c}-3\,b\,d\,e^2\,\left (\frac {x}{2\,c}-\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\right )+\frac {3\,a\,d\,e^2\,x^2}{2}+3\,a\,d^2\,e\,\ln \left (x\right )-\frac {b\,e^3\,\left (3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3\right )}{12\,c^4}-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{2\,x^2}+\frac {b\,e^3\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}-\frac {b\,d^2\,e\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {b\,d^2\,e\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

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

[Out]

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