\(\int \frac {(d+e x^2)^2 (a+b \arctan (c x))}{x} \, dx\) [1129]

Optimal result

Integrand size = 21, antiderivative size = 137 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=-\frac {b d e x}{c}+\frac {b e^2 x}{4 c^3}-\frac {b e^2 x^3}{12 c}+\frac {b d e \arctan (c x)}{c^2}-\frac {b e^2 \arctan (c x)}{4 c^4}+d e x^2 (a+b \arctan (c x))+\frac {1}{4} e^2 x^4 (a+b \arctan (c x))+a d^2 \log (x)+\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,i c x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5100, 4940, 2438, 4946, 327, 209, 308} \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=d e x^2 (a+b \arctan (c x))+\frac {1}{4} e^2 x^4 (a+b \arctan (c x))+a d^2 \log (x)-\frac {b e^2 \arctan (c x)}{4 c^4}+\frac {b d e \arctan (c x)}{c^2}+\frac {b e^2 x}{4 c^3}+\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,i c x)-\frac {b d e x}{c}-\frac {b e^2 x^3}{12 c} \]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.90 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=-\frac {b d e (c x-\arctan (c x))}{c^2}-\frac {b e^2 \left (-3 c x+c^3 x^3+3 \arctan (c x)\right )}{12 c^4}+d e x^2 (a+b \arctan (c x))+\frac {1}{4} e^2 x^4 (a+b \arctan (c x))+a d^2 \log (x)+\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,i c x) \]

[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]

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.28

method	result	size
derivativedivides	\(a d e \,x^{2}+\frac {a \,e^{2} x^{4}}{4}+a \,d^{2} \ln \left (c x \right )+\frac {b \left (\arctan \left (c x \right ) d \,c^{4} e \,x^{2}+\frac {\arctan \left (c x \right ) e^{2} c^{4} x^{4}}{4}+\arctan \left (c x \right ) c^{4} d^{2} \ln \left (c x \right )-\frac {e \left (4 c^{3} x d +\frac {e \,c^{3} x^{3}}{3}-e c x +\left (-4 c^{2} d +e \right ) \arctan \left (c x \right )\right )}{4}-c^{4} d^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{4}}\)	\(175\)
default	\(a d e \,x^{2}+\frac {a \,e^{2} x^{4}}{4}+a \,d^{2} \ln \left (c x \right )+\frac {b \left (\arctan \left (c x \right ) d \,c^{4} e \,x^{2}+\frac {\arctan \left (c x \right ) e^{2} c^{4} x^{4}}{4}+\arctan \left (c x \right ) c^{4} d^{2} \ln \left (c x \right )-\frac {e \left (4 c^{3} x d +\frac {e \,c^{3} x^{3}}{3}-e c x +\left (-4 c^{2} d +e \right ) \arctan \left (c x \right )\right )}{4}-c^{4} d^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{4}}\)	\(175\)
parts	\(a \left (\frac {x^{4} e^{2}}{4}+x^{2} e d +d^{2} \ln \left (x \right )\right )+b \left (\frac {\arctan \left (c x \right ) e^{2} x^{4}}{4}+\arctan \left (c x \right ) d e \,x^{2}+\arctan \left (c x \right ) d^{2} \ln \left (c x \right )-\frac {e \left (4 c^{3} x d +\frac {e \,c^{3} x^{3}}{3}-e c x +\left (-4 c^{2} d +e \right ) \arctan \left (c x \right )\right )-2 i c^{4} d^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )+2 i c^{4} d^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )+2 i c^{4} d^{2} \operatorname {dilog}\left (-i c x +1\right )-2 i c^{4} d^{2} \operatorname {dilog}\left (i c x +1\right )}{4 c^{4}}\right )\)	\(181\)
risch	\(\frac {b \,e^{2} x}{4 c^{3}}-\frac {b \,e^{2} x^{3}}{12 c}-\frac {b d e x}{c}+a \,d^{2} \ln \left (-i c x \right )+\frac {a \,e^{2} x^{4}}{4}-\frac {i b \,d^{2} \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i b \,d^{2} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {a d e}{c^{2}}-\frac {a \,e^{2}}{4 c^{4}}+a d e \,x^{2}-\frac {i b e d \ln \left (i c x +1\right ) x^{2}}{2}-\frac {b \,e^{2} \arctan \left (c x \right )}{4 c^{4}}+\frac {i b \,e^{2} \ln \left (-i c x +1\right ) x^{4}}{8}-\frac {i b \,e^{2} \ln \left (i c x +1\right ) x^{4}}{8}+\frac {i b d e \ln \left (-i c x +1\right ) x^{2}}{2}+\frac {b d e \arctan \left (c x \right )}{c^{2}}\)	\(200\)

[In]

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

[Out]

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

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

[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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.26 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=\frac {1}{4} \, a e^{2} x^{4} + a d e x^{2} + a d^{2} \log \left (x\right ) - \frac {b c^{3} e^{2} x^{3} + 3 \, \pi b c^{4} d^{2} \log \left (c^{2} x^{2} + 1\right ) - 12 \, b c^{4} d^{2} \arctan \left (c x\right ) \log \left (c x\right ) + 6 i \, b c^{4} d^{2} {\rm Li}_2\left (i \, c x + 1\right ) - 6 i \, b c^{4} d^{2} {\rm Li}_2\left (-i \, c x + 1\right ) + 3 \, {\left (4 \, b c^{3} d e - b c e^{2}\right )} x - 3 \, {\left (b c^{4} e^{2} x^{4} + 4 \, b c^{4} d e x^{2} + 4 \, b c^{2} d e - b e^{2}\right )} \arctan \left (c x\right )}{12 \, c^{4}} \]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=\left \{\begin {array}{cl} \frac {a\,\left (4\,d^2\,\ln \left (x\right )+e^2\,x^4+4\,d\,e\,x^2\right )}{4} & \text {\ if\ \ }c=0\\ \frac {a\,\left (4\,d^2\,\ln \left (x\right )+e^2\,x^4+4\,d\,e\,x^2\right )}{4}-2\,b\,d\,e\,\left (\frac {x}{2\,c}-\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\right )-\frac {b\,e^2\,\left (3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3\right )}{12\,c^4}+\frac {b\,e^2\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}-\frac {b\,d^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

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

[Out]

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