\(\int \frac {(a+b \arctan (c x)) (d+e \log (1+c^2 x^2))}{x^5} \, dx\) [1295]

Optimal result

Integrand size = 26, antiderivative size = 225 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=-\frac {a c^2 e}{4 x^2}-\frac {5 b c^3 e}{12 x}-\frac {11}{12} b c^4 e \arctan (c x)-\frac {b c^2 e \arctan (c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{4} a c^4 e \log \left (1+c^2 x^2\right )-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,-i c x)+\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,i c x) \]

[Out]

-1/4*a*c^2*e/x^2-5/12*b*c^3*e/x-11/12*b*c^4*e*arctan(c*x)-1/4*b*c^2*e*arctan(c*x)/x^2-1/2*a*c^4*e*ln(x)+1/4*a*
c^4*e*ln(c^2*x^2+1)-1/12*b*c*(d+e*ln(c^2*x^2+1))/x^3+1/4*b*c^3*(d+e*ln(c^2*x^2+1))/x+1/4*b*c^4*arctan(c*x)*(d+
e*ln(c^2*x^2+1))-1/4*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^4-1/4*I*b*c^4*e*polylog(2,-I*c*x)+1/4*I*b*c^4*e*p
olylog(2,I*c*x)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4946, 331, 209, 5141, 1816, 649, 266, 5100, 4940, 2438} \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x^4}-\frac {1}{2} a c^4 e \log (x)-\frac {a c^2 e}{4 x^2}+\frac {1}{4} a c^4 e \log \left (c^2 x^2+1\right )-\frac {11}{12} b c^4 e \arctan (c x)-\frac {b c^2 e \arctan (c x)}{4 x^2}+\frac {1}{4} b c^4 \arctan (c x) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,-i c x)+\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,i c x)-\frac {5 b c^3 e}{12 x}-\frac {b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{12 x^3}+\frac {b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x} \]

[In]

Int[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^5,x]

[Out]

-1/4*(a*c^2*e)/x^2 - (5*b*c^3*e)/(12*x) - (11*b*c^4*e*ArcTan[c*x])/12 - (b*c^2*e*ArcTan[c*x])/(4*x^2) - (a*c^4
*e*Log[x])/2 + (a*c^4*e*Log[1 + c^2*x^2])/4 - (b*c*(d + e*Log[1 + c^2*x^2]))/(12*x^3) + (b*c^3*(d + e*Log[1 +
c^2*x^2]))/(4*x) + (b*c^4*ArcTan[c*x]*(d + e*Log[1 + c^2*x^2]))/4 - ((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^
2]))/(4*x^4) - (I/4)*b*c^4*e*PolyLog[2, (-I)*c*x] + (I/4)*b*c^4*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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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])

Rule 5141

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> With
[{u = IntHide[x^m*(a + b*ArcTan[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegrand
[x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\left (2 c^2 e\right ) \int \left (\frac {-3 a-b c x+3 b c^3 x^3}{12 x^3 \left (1+c^2 x^2\right )}+\frac {b \left (-1+c^2 x^2\right ) \arctan (c x)}{4 x^3}\right ) \, dx \\ & = -\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac {1}{6} \left (c^2 e\right ) \int \frac {-3 a-b c x+3 b c^3 x^3}{x^3 \left (1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (b c^2 e\right ) \int \frac {\left (-1+c^2 x^2\right ) \arctan (c x)}{x^3} \, dx \\ & = -\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac {1}{6} \left (c^2 e\right ) \int \left (-\frac {3 a}{x^3}-\frac {b c}{x^2}+\frac {3 a c^2}{x}-\frac {c^3 (-4 b+3 a c x)}{1+c^2 x^2}\right ) \, dx-\frac {1}{2} \left (b c^2 e\right ) \int \left (-\frac {\arctan (c x)}{x^3}+\frac {c^2 \arctan (c x)}{x}\right ) \, dx \\ & = -\frac {a c^2 e}{4 x^2}-\frac {b c^3 e}{6 x}-\frac {1}{2} a c^4 e \log (x)-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}+\frac {1}{2} \left (b c^2 e\right ) \int \frac {\arctan (c x)}{x^3} \, dx-\frac {1}{2} \left (b c^4 e\right ) \int \frac {\arctan (c x)}{x} \, dx+\frac {1}{6} \left (c^5 e\right ) \int \frac {-4 b+3 a c x}{1+c^2 x^2} \, dx \\ & = -\frac {a c^2 e}{4 x^2}-\frac {b c^3 e}{6 x}-\frac {b c^2 e \arctan (c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} \left (b c^3 e\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac {1}{4} \left (i b c^4 e\right ) \int \frac {\log (1-i c x)}{x} \, dx+\frac {1}{4} \left (i b c^4 e\right ) \int \frac {\log (1+i c x)}{x} \, dx-\frac {1}{3} \left (2 b c^5 e\right ) \int \frac {1}{1+c^2 x^2} \, dx+\frac {1}{2} \left (a c^6 e\right ) \int \frac {x}{1+c^2 x^2} \, dx \\ & = -\frac {a c^2 e}{4 x^2}-\frac {5 b c^3 e}{12 x}-\frac {2}{3} b c^4 e \arctan (c x)-\frac {b c^2 e \arctan (c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{4} a c^4 e \log \left (1+c^2 x^2\right )-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,-i c x)+\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,i c x)-\frac {1}{4} \left (b c^5 e\right ) \int \frac {1}{1+c^2 x^2} \, dx \\ & = -\frac {a c^2 e}{4 x^2}-\frac {5 b c^3 e}{12 x}-\frac {11}{12} b c^4 e \arctan (c x)-\frac {b c^2 e \arctan (c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{4} a c^4 e \log \left (1+c^2 x^2\right )-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac {b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,-i c x)+\frac {1}{4} i b c^4 e \operatorname {PolyLog}(2,i c x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=-\frac {3 a d+b c d x+3 a c^2 e x^2-3 b c^3 d x^3+5 b c^3 e x^3+3 b d \arctan (c x)+3 b c^2 e x^2 \arctan (c x)-3 b c^4 d x^4 \arctan (c x)+11 b c^4 e x^4 \arctan (c x)+6 a c^4 e x^4 \log (x)+3 a e \log \left (1+c^2 x^2\right )+b c e x \log \left (1+c^2 x^2\right )-3 b c^3 e x^3 \log \left (1+c^2 x^2\right )-3 a c^4 e x^4 \log \left (1+c^2 x^2\right )+3 b e \arctan (c x) \log \left (1+c^2 x^2\right )-3 b c^4 e x^4 \arctan (c x) \log \left (1+c^2 x^2\right )+3 i b c^4 e x^4 \operatorname {PolyLog}(2,-i c x)-3 i b c^4 e x^4 \operatorname {PolyLog}(2,i c x)}{12 x^4} \]

[In]

Integrate[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^5,x]

[Out]

-1/12*(3*a*d + b*c*d*x + 3*a*c^2*e*x^2 - 3*b*c^3*d*x^3 + 5*b*c^3*e*x^3 + 3*b*d*ArcTan[c*x] + 3*b*c^2*e*x^2*Arc
Tan[c*x] - 3*b*c^4*d*x^4*ArcTan[c*x] + 11*b*c^4*e*x^4*ArcTan[c*x] + 6*a*c^4*e*x^4*Log[x] + 3*a*e*Log[1 + c^2*x
^2] + b*c*e*x*Log[1 + c^2*x^2] - 3*b*c^3*e*x^3*Log[1 + c^2*x^2] - 3*a*c^4*e*x^4*Log[1 + c^2*x^2] + 3*b*e*ArcTa
n[c*x]*Log[1 + c^2*x^2] - 3*b*c^4*e*x^4*ArcTan[c*x]*Log[1 + c^2*x^2] + (3*I)*b*c^4*e*x^4*PolyLog[2, (-I)*c*x]
- (3*I)*b*c^4*e*x^4*PolyLog[2, I*c*x])/x^4

Maple [F]

\[\int \frac {\left (a +b \arctan \left (c x \right )\right ) \left (d +e \ln \left (c^{2} x^{2}+1\right )\right )}{x^{5}}d x\]

[In]

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^5,x)

[Out]

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^5,x)

Fricas [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \]

[In]

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^5,x, algorithm="fricas")

[Out]

integral((b*d*arctan(c*x) + a*d + (b*e*arctan(c*x) + a*e)*log(c^2*x^2 + 1))/x^5, x)

Sympy [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e \log {\left (c^{2} x^{2} + 1 \right )}\right )}{x^{5}}\, dx \]

[In]

integrate((a+b*atan(c*x))*(d+e*ln(c**2*x**2+1))/x**5,x)

[Out]

Integral((a + b*atan(c*x))*(d + e*log(c**2*x**2 + 1))/x**5, x)

Maxima [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \]

[In]

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/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 + 1/4*((c^2*log(c^2*x^2 + 1) - c^2*
log(x^2) - 1/x^2)*c^2 - log(c^2*x^2 + 1)/x^4)*a*e - 1/12*(72*c^6*x^4*integrate(1/12*x*arctan(c*x)/(c^2*x^2 + 1
), x) + 8*c^4*x^4*arctan(c*x) - 72*c^2*x^4*integrate(1/12*arctan(c*x)/(c^2*x^5 + x^3), x) + 2*c^3*x^3 - (3*c^3
*x^3 - c*x + 3*(c^4*x^4 - 1)*arctan(c*x))*log(c^2*x^2 + 1))*b*e/x^4 - 1/4*a*d/x^4

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\text {Timed out} \]

[In]

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^5,x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (c^2\,x^2+1\right )\right )}{x^5} \,d x \]

[In]

int(((a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)))/x^5,x)

[Out]

int(((a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)))/x^5, x)