3.2.40 \(\int \frac {a+b \text {ArcTan}(c x)}{x^3 (d+e x)} \, dx\) [140]

Optimal. Leaf size=293 \[ -\frac {b c}{2 d x}-\frac {b c^2 \text {ArcTan}(c x)}{2 d}-\frac {a+b \text {ArcTan}(c x)}{2 d x^2}+\frac {e (a+b \text {ArcTan}(c x))}{d^2 x}-\frac {b c e \log (x)}{d^2}+\frac {a e^2 \log (x)}{d^3}+\frac {e^2 (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {e^2 (a+b \text {ArcTan}(c x)) \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {b c e \log \left (1+c^2 x^2\right )}{2 d^2}+\frac {i b e^2 \text {PolyLog}(2,-i c x)}{2 d^3}-\frac {i b e^2 \text {PolyLog}(2,i c x)}{2 d^3}-\frac {i b e^2 \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 d^3}+\frac {i b e^2 \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.737, Rules used = {4996, 4946, 331, 209, 272, 36, 29, 31, 4940, 2438, 4966, 2449, 2352, 2497} \begin {gather*} \frac {e^2 \log \left (\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))}{d^3}-\frac {e^2 (a+b \text {ArcTan}(c x)) \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac {e (a+b \text {ArcTan}(c x))}{d^2 x}-\frac {a+b \text {ArcTan}(c x)}{2 d x^2}+\frac {a e^2 \log (x)}{d^3}-\frac {b c^2 \text {ArcTan}(c x)}{2 d}+\frac {b c e \log \left (c^2 x^2+1\right )}{2 d^2}+\frac {i b e^2 \text {Li}_2(-i c x)}{2 d^3}-\frac {i b e^2 \text {Li}_2(i c x)}{2 d^3}-\frac {i b e^2 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 d^3}+\frac {i b e^2 \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}-\frac {b c e \log (x)}{d^2}-\frac {b c}{2 d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

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 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 4966

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x]))*(Log[2/(1
- I*c*x)]/e), x] + (Dist[b*(c/e), Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[b*(c/e), Int[Log[2*c*((
d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(1 + c^2*x^2), x], x] + Simp[(a + b*ArcTan[c*x])*(Log[2*c*((d + e*x)/((c*
d + I*e)*(1 - I*c*x)))]/e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 4996

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

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

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Mathematica [A]
time = 2.17, size = 441, normalized size = 1.51 \begin {gather*} -\frac {\frac {a d^3}{x^2}-\frac {2 a d^2 e}{x}-2 a d e^2 \log (x)+2 a d e^2 \log (d+e x)+\frac {b \left (\frac {c^2 d^3}{x}+c^3 d^3 \text {ArcTan}(c x)+i c d e^2 \pi \text {ArcTan}(c x)+\frac {c d^3 \text {ArcTan}(c x)}{x^2}-\frac {2 c d^2 e \text {ArcTan}(c x)}{x}-2 i c d e^2 \text {ArcTan}\left (\frac {c d}{e}\right ) \text {ArcTan}(c x)+i c d e^2 \text {ArcTan}(c x)^2+e^3 \text {ArcTan}(c x)^2-\sqrt {1+\frac {c^2 d^2}{e^2}} e^3 e^{i \text {ArcTan}\left (\frac {c d}{e}\right )} \text {ArcTan}(c x)^2+c d e^2 \pi \log \left (1+e^{-2 i \text {ArcTan}(c x)}\right )-2 c d e^2 \text {ArcTan}(c x) \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )+2 c d e^2 \text {ArcTan}\left (\frac {c d}{e}\right ) \log \left (1-e^{2 i \left (\text {ArcTan}\left (\frac {c d}{e}\right )+\text {ArcTan}(c x)\right )}\right )+2 c d e^2 \text {ArcTan}(c x) \log \left (1-e^{2 i \left (\text {ArcTan}\left (\frac {c d}{e}\right )+\text {ArcTan}(c x)\right )}\right )+2 c^2 d^2 e \log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )+\frac {1}{2} c d e^2 \pi \log \left (1+c^2 x^2\right )-2 c d e^2 \text {ArcTan}\left (\frac {c d}{e}\right ) \log \left (\sin \left (\text {ArcTan}\left (\frac {c d}{e}\right )+\text {ArcTan}(c x)\right )\right )+i c d e^2 \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )-i c d e^2 \text {PolyLog}\left (2,e^{2 i \left (\text {ArcTan}\left (\frac {c d}{e}\right )+\text {ArcTan}(c x)\right )}\right )\right )}{c}}{2 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a*d^3)/x^2 - (2*a*d^2*e)/x - 2*a*d*e^2*Log[x] + 2*a*d*e^2*Log[d + e*x] + (b*((c^2*d^3)/x + c^3*d^3*ArcT
an[c*x] + I*c*d*e^2*Pi*ArcTan[c*x] + (c*d^3*ArcTan[c*x])/x^2 - (2*c*d^2*e*ArcTan[c*x])/x - (2*I)*c*d*e^2*ArcTa
n[(c*d)/e]*ArcTan[c*x] + I*c*d*e^2*ArcTan[c*x]^2 + e^3*ArcTan[c*x]^2 - Sqrt[1 + (c^2*d^2)/e^2]*e^3*E^(I*ArcTan
[(c*d)/e])*ArcTan[c*x]^2 + c*d*e^2*Pi*Log[1 + E^((-2*I)*ArcTan[c*x])] - 2*c*d*e^2*ArcTan[c*x]*Log[1 - E^((2*I)
*ArcTan[c*x])] + 2*c*d*e^2*ArcTan[(c*d)/e]*Log[1 - E^((2*I)*(ArcTan[(c*d)/e] + ArcTan[c*x]))] + 2*c*d*e^2*ArcT
an[c*x]*Log[1 - E^((2*I)*(ArcTan[(c*d)/e] + ArcTan[c*x]))] + 2*c^2*d^2*e*Log[(c*x)/Sqrt[1 + c^2*x^2]] + (c*d*e
^2*Pi*Log[1 + c^2*x^2])/2 - 2*c*d*e^2*ArcTan[(c*d)/e]*Log[Sin[ArcTan[(c*d)/e] + ArcTan[c*x]]] + I*c*d*e^2*Poly
Log[2, E^((2*I)*ArcTan[c*x])] - I*c*d*e^2*PolyLog[2, E^((2*I)*(ArcTan[(c*d)/e] + ArcTan[c*x]))]))/c)/d^4

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Maple [A]
time = 0.15, size = 448, normalized size = 1.53

method	result	size
derivativedivides	\(c^{2} \left (-\frac {a \,e^{2} \ln \left (c e x +c d \right )}{c^{2} d^{3}}-\frac {a}{2 d \,c^{2} x^{2}}+\frac {a \,e^{2} \ln \left (c x \right )}{c^{2} d^{3}}+\frac {a e}{c^{2} d^{2} x}-\frac {b \arctan \left (c x \right ) e^{2} \ln \left (c e x +c d \right )}{c^{2} d^{3}}-\frac {b \arctan \left (c x \right )}{2 d \,c^{2} x^{2}}+\frac {b \arctan \left (c x \right ) e^{2} \ln \left (c x \right )}{c^{2} d^{3}}+\frac {b \arctan \left (c x \right ) e}{c^{2} d^{2} x}+\frac {i b \,e^{2} \dilog \left (\frac {c e x +i e}{-c d +i e}\right )}{2 c^{2} d^{3}}+\frac {i b \,e^{2} \ln \left (c e x +c d \right ) \ln \left (\frac {c e x +i e}{-c d +i e}\right )}{2 c^{2} d^{3}}-\frac {i b \,e^{2} \dilog \left (\frac {-c e x +i e}{c d +i e}\right )}{2 c^{2} d^{3}}+\frac {i b \,e^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 c^{2} d^{3}}-\frac {i b \,e^{2} \dilog \left (-i c x +1\right )}{2 c^{2} d^{3}}-\frac {i b \,e^{2} \ln \left (c e x +c d \right ) \ln \left (\frac {-c e x +i e}{c d +i e}\right )}{2 c^{2} d^{3}}-\frac {i b \,e^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 c^{2} d^{3}}+\frac {i b \,e^{2} \dilog \left (i c x +1\right )}{2 c^{2} d^{3}}+\frac {b e \ln \left (c^{2} x^{2}+1\right )}{2 c \,d^{2}}-\frac {b \arctan \left (c x \right )}{2 d}-\frac {b}{2 d c x}-\frac {b e \ln \left (c x \right )}{c \,d^{2}}\right )\)	\(448\)
default	\(c^{2} \left (-\frac {a \,e^{2} \ln \left (c e x +c d \right )}{c^{2} d^{3}}-\frac {a}{2 d \,c^{2} x^{2}}+\frac {a \,e^{2} \ln \left (c x \right )}{c^{2} d^{3}}+\frac {a e}{c^{2} d^{2} x}-\frac {b \arctan \left (c x \right ) e^{2} \ln \left (c e x +c d \right )}{c^{2} d^{3}}-\frac {b \arctan \left (c x \right )}{2 d \,c^{2} x^{2}}+\frac {b \arctan \left (c x \right ) e^{2} \ln \left (c x \right )}{c^{2} d^{3}}+\frac {b \arctan \left (c x \right ) e}{c^{2} d^{2} x}+\frac {i b \,e^{2} \dilog \left (\frac {c e x +i e}{-c d +i e}\right )}{2 c^{2} d^{3}}+\frac {i b \,e^{2} \ln \left (c e x +c d \right ) \ln \left (\frac {c e x +i e}{-c d +i e}\right )}{2 c^{2} d^{3}}-\frac {i b \,e^{2} \dilog \left (\frac {-c e x +i e}{c d +i e}\right )}{2 c^{2} d^{3}}+\frac {i b \,e^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 c^{2} d^{3}}-\frac {i b \,e^{2} \dilog \left (-i c x +1\right )}{2 c^{2} d^{3}}-\frac {i b \,e^{2} \ln \left (c e x +c d \right ) \ln \left (\frac {-c e x +i e}{c d +i e}\right )}{2 c^{2} d^{3}}-\frac {i b \,e^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 c^{2} d^{3}}+\frac {i b \,e^{2} \dilog \left (i c x +1\right )}{2 c^{2} d^{3}}+\frac {b e \ln \left (c^{2} x^{2}+1\right )}{2 c \,d^{2}}-\frac {b \arctan \left (c x \right )}{2 d}-\frac {b}{2 d c x}-\frac {b e \ln \left (c x \right )}{c \,d^{2}}\right )\)	\(448\)
risch	\(-\frac {i b \,e^{2} \ln \left (-i c x +1\right ) \ln \left (\frac {-i c d +\left (-i c x +1\right ) e -e}{-i c d -e}\right )}{2 d^{3}}-\frac {i b \,e^{2} \dilog \left (\frac {-i c d +\left (-i c x +1\right ) e -e}{-i c d -e}\right )}{2 d^{3}}-\frac {i b e \ln \left (i c x +1\right )}{2 d^{2} x}-\frac {i b \dilog \left (-i c x +1\right ) e^{2}}{2 d^{3}}-\frac {b c}{2 d x}-\frac {b \,c^{2} \arctan \left (c x \right )}{2 d}+\frac {i b \,e^{2} \ln \left (i c x +1\right ) \ln \left (\frac {i c d +\left (i c x +1\right ) e -e}{i c d -e}\right )}{2 d^{3}}+\frac {i b \ln \left (i c x +1\right )}{4 d \,x^{2}}+\frac {b c e \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}-\frac {c b e \ln \left (-i c x \right )}{2 d^{2}}+\frac {a \,e^{2} \ln \left (-i c x \right )}{d^{3}}-\frac {b c e \ln \left (i c x \right )}{2 d^{2}}-\frac {a \,e^{2} \ln \left (i c d -\left (-i c x +1\right ) e +e \right )}{d^{3}}+\frac {i b e \ln \left (-i c x +1\right )}{2 d^{2} x}-\frac {i b \ln \left (-i c x +1\right )}{4 d \,x^{2}}-\frac {a}{2 d \,x^{2}}+\frac {i b \dilog \left (i c x +1\right ) e^{2}}{2 d^{3}}+\frac {i c^{2} b \ln \left (-i c x \right )}{4 d}+\frac {i b \,e^{2} \dilog \left (\frac {i c d +\left (i c x +1\right ) e -e}{i c d -e}\right )}{2 d^{3}}+\frac {a e}{x \,d^{2}}-\frac {i b \,c^{2} \ln \left (i c x \right )}{4 d}\)	\(452\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctan(c*x) + a)/(x^4*e + d*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^3\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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