\(\int \frac {(d+e x^2)^2 (a+b \arctan (c x))}{x^5} \, dx\) [1133]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^5} \, dx=-\frac {b c d^2}{12 x^3}+\frac {b c^3 d^2}{4 x}-\frac {b c d e}{x}+\frac {1}{4} b c^4 d^2 \arctan (c x)-b c^2 d e \arctan (c x)-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {d e (a+b \arctan (c x))}{x^2}+a e^2 \log (x)+\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,i c x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5100, 4946, 331, 209, 4940, 2438} \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^5} \, dx=-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {d e (a+b \arctan (c x))}{x^2}+a e^2 \log (x)+\frac {1}{4} b c^4 d^2 \arctan (c x)-b c^2 d e \arctan (c x)+\frac {b c^3 d^2}{4 x}-\frac {b c d^2}{12 x^3}-\frac {b c d e}{x}+\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,i c x) \]

[In]

Int[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

-1/12*(b*c*d^2)/x^3 + (b*c^3*d^2)/(4*x) - (b*c*d*e)/x + (b*c^4*d^2*ArcTan[c*x])/4 - b*c^2*d*e*ArcTan[c*x] - (d
^2*(a + b*ArcTan[c*x]))/(4*x^4) - (d*e*(a + b*ArcTan[c*x]))/x^2 + a*e^2*Log[x] + (I/2)*b*e^2*PolyLog[2, (-I)*c
*x] - (I/2)*b*e^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 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^2 (a+b \arctan (c x))}{x^5}+\frac {2 d e (a+b \arctan (c x))}{x^3}+\frac {e^2 (a+b \arctan (c x))}{x}\right ) \, dx \\ & = d^2 \int \frac {a+b \arctan (c x)}{x^5} \, dx+(2 d e) \int \frac {a+b \arctan (c x)}{x^3} \, dx+e^2 \int \frac {a+b \arctan (c x)}{x} \, dx \\ & = -\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {d e (a+b \arctan (c x))}{x^2}+a e^2 \log (x)+\frac {1}{4} \left (b c d^2\right ) \int \frac {1}{x^4 \left (1+c^2 x^2\right )} \, dx+(b c d e) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (i b e^2\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (i b e^2\right ) \int \frac {\log (1+i c x)}{x} \, dx \\ & = -\frac {b c d^2}{12 x^3}-\frac {b c d e}{x}-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {d e (a+b \arctan (c x))}{x^2}+a e^2 \log (x)+\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,i c x)-\frac {1}{4} \left (b c^3 d^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\left (b c^3 d e\right ) \int \frac {1}{1+c^2 x^2} \, dx \\ & = -\frac {b c d^2}{12 x^3}+\frac {b c^3 d^2}{4 x}-\frac {b c d e}{x}-b c^2 d e \arctan (c x)-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {d e (a+b \arctan (c x))}{x^2}+a e^2 \log (x)+\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,i c x)+\frac {1}{4} \left (b c^5 d^2\right ) \int \frac {1}{1+c^2 x^2} \, dx \\ & = -\frac {b c d^2}{12 x^3}+\frac {b c^3 d^2}{4 x}-\frac {b c d e}{x}+\frac {1}{4} b c^4 d^2 \arctan (c x)-b c^2 d e \arctan (c x)-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {d e (a+b \arctan (c x))}{x^2}+a e^2 \log (x)+\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e^2 \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 = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.94 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^5} \, dx=-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {d e (a+b \arctan (c x))}{x^2}-\frac {b c d^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-c^2 x^2\right )}{12 x^3}-\frac {b c d e \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )}{x}+a e^2 \log (x)+\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e^2 \operatorname {PolyLog}(2,i c x) \]

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(c^{4} \left (\frac {a \,e^{2} \ln \left (c x \right )}{c^{4}}-\frac {a \,d^{2}}{4 c^{4} x^{4}}-\frac {a d e}{c^{4} x^{2}}+\frac {b \left (\arctan \left (c x \right ) e^{2} \ln \left (c x \right )-\frac {\arctan \left (c x \right ) d^{2}}{4 x^{4}}-\frac {\arctan \left (c x \right ) d e}{x^{2}}+\frac {i e^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i e^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i e^{2} \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i e^{2} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {d \,c^{2} \left (\left (c^{2} d -4 e \right ) \arctan \left (c x \right )-\frac {-c^{2} d +4 e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{4}\right )}{c^{4}}\right )\)	\(194\)
default	\(c^{4} \left (\frac {a \,e^{2} \ln \left (c x \right )}{c^{4}}-\frac {a \,d^{2}}{4 c^{4} x^{4}}-\frac {a d e}{c^{4} x^{2}}+\frac {b \left (\arctan \left (c x \right ) e^{2} \ln \left (c x \right )-\frac {\arctan \left (c x \right ) d^{2}}{4 x^{4}}-\frac {\arctan \left (c x \right ) d e}{x^{2}}+\frac {i e^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i e^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i e^{2} \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i e^{2} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {d \,c^{2} \left (\left (c^{2} d -4 e \right ) \arctan \left (c x \right )-\frac {-c^{2} d +4 e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{4}\right )}{c^{4}}\right )\)	\(194\)
parts	\(a \left (e^{2} \ln \left (x \right )-\frac {d^{2}}{4 x^{4}}-\frac {e d}{x^{2}}\right )+b \,c^{4} \left (\frac {\arctan \left (c x \right ) \ln \left (c x \right ) e^{2}}{c^{4}}-\frac {\arctan \left (c x \right ) d^{2}}{4 c^{4} x^{4}}-\frac {\arctan \left (c x \right ) d e}{c^{4} x^{2}}-\frac {-2 i e^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )+2 i e^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )-2 i e^{2} \operatorname {dilog}\left (i c x +1\right )+2 i e^{2} \operatorname {dilog}\left (-i c x +1\right )-d \,c^{2} \left (\left (c^{2} d -4 e \right ) \arctan \left (c x \right )-\frac {-c^{2} d +4 e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{4 c^{4}}\right )\)	\(194\)
risch	\(\frac {b \,c^{4} d^{2} \arctan \left (c x \right )}{8}-\frac {b c \,d^{2}}{12 x^{3}}+\frac {b \,c^{3} d^{2}}{4 x}-\frac {b c d e}{x}-\frac {b \,c^{2} d e \arctan \left (c x \right )}{2}+\frac {i b \,e^{2} \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {a d e}{x^{2}}-\frac {a \,d^{2}}{4 x^{4}}+a \,e^{2} \ln \left (-i c x \right )+\frac {i b \,d^{2} \ln \left (i c x +1\right )}{8 x^{4}}-\frac {i b \,c^{4} d^{2} \ln \left (i c x +1\right )}{8}-\frac {i b d e \ln \left (-i c x +1\right )}{2 x^{2}}+\frac {i b e d \ln \left (i c x +1\right )}{2 x^{2}}-\frac {i c^{4} b \,d^{2} \ln \left (-i c x \right )}{8}-\frac {i b \,e^{2} \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {i b \,d^{2} \ln \left (-i c x +1\right )}{8 x^{4}}-\frac {i b \,c^{2} e d \ln \left (i c x \right )}{2}+\frac {i b \,c^{2} e d \ln \left (i c x +1\right )}{2}+\frac {i b \,c^{4} d^{2} \ln \left (i c x \right )}{8}-\frac {i c^{2} b d e \ln \left (c^{2} x^{2}+1\right )}{4}+\frac {i c^{4} b \,d^{2} \ln \left (c^{2} x^{2}+1\right )}{16}+\frac {i c^{2} b d e \ln \left (-i c x \right )}{2}\)	\(322\)

[In]

int((e*x^2+d)^2*(a+b*arctan(c*x))/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^5,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^5, x)

Sympy [F]

\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{5}}\, dx \]

[In]

integrate((e*x**2+d)**2*(a+b*atan(c*x))/x**5,x)

[Out]

Integral((a + b*atan(c*x))*(d + e*x**2)**2/x**5, x)

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^5,x, algorithm="maxima")

[Out]

1/12*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d^2 - ((c*arctan(c*x) + 1/x)*c + arct
an(c*x)/x^2)*b*d*e + b*e^2*integrate(arctan(c*x)/x, x) + a*e^2*log(x) - a*d*e/x^2 - 1/4*a*d^2/x^4

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^5,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.90 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.27 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^5} \, dx=\left \{\begin {array}{cl} a\,e^2\,\ln \left (x\right )-\frac {\frac {a\,d^2}{4}+a\,e\,d\,x^2}{x^4} & \text {\ if\ \ }c=0\\ a\,e^2\,\ln \left (x\right )-\frac {\frac {a\,d^2}{4}+a\,e\,d\,x^2}{x^4}-\frac {b\,d^2\,\left (\frac {\frac {c^2}{3}-c^4\,x^2}{x^3}-c^5\,\mathrm {atan}\left (c\,x\right )\right )}{4\,c}-2\,b\,d\,e\,\left (\frac {c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}}{2\,c}+\frac {\mathrm {atan}\left (c\,x\right )}{2\,x^2}\right )-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{4\,x^4}-\frac {b\,e^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,e^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^2)/x^5,x)

[Out]

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