3.7 \(\int \frac {a+b \tan ^{-1}(c x)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=146 \[ -\frac {a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}-\frac {b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}+\frac {b c^2 (c d-e) (c d+e) \tan ^{-1}(c x)}{2 e \left (c^2 d^2+e^2\right )^2}-\frac {b c^3 d \log \left (c^2 x^2+1\right )}{2 \left (c^2 d^2+e^2\right )^2}+\frac {b c^3 d \log (d+e x)}{\left (c^2 d^2+e^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4862, 710, 801, 635, 203, 260} \[ -\frac {a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}-\frac {b c^3 d \log \left (c^2 x^2+1\right )}{2 \left (c^2 d^2+e^2\right )^2}-\frac {b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}+\frac {b c^3 d \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac {b c^2 (c d-e) (c d+e) \tan ^{-1}(c x)}{2 e \left (c^2 d^2+e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 260

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 635

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 710

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 4862

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

Rubi steps

\begin {align*} \int \frac {a+b \tan ^{-1}(c x)}{(d+e x)^3} \, dx &=-\frac {a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}+\frac {(b c) \int \frac {1}{(d+e x)^2 \left (1+c^2 x^2\right )} \, dx}{2 e}\\ &=-\frac {b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}+\frac {\left (b c^3\right ) \int \frac {d-e x}{(d+e x) \left (1+c^2 x^2\right )} \, dx}{2 e \left (c^2 d^2+e^2\right )}\\ &=-\frac {b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}+\frac {\left (b c^3\right ) \int \left (\frac {2 d e^2}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac {c^2 d^2-e^2-2 c^2 d e x}{\left (c^2 d^2+e^2\right ) \left (1+c^2 x^2\right )}\right ) \, dx}{2 e \left (c^2 d^2+e^2\right )}\\ &=-\frac {b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}+\frac {b c^3 d \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac {\left (b c^3\right ) \int \frac {c^2 d^2-e^2-2 c^2 d e x}{1+c^2 x^2} \, dx}{2 e \left (c^2 d^2+e^2\right )^2}\\ &=-\frac {b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}+\frac {b c^3 d \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}-\frac {\left (b c^5 d\right ) \int \frac {x}{1+c^2 x^2} \, dx}{\left (c^2 d^2+e^2\right )^2}+\frac {\left (b c^3 (c d-e) (c d+e)\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 e \left (c^2 d^2+e^2\right )^2}\\ &=-\frac {b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}+\frac {b c^2 (c d-e) (c d+e) \tan ^{-1}(c x)}{2 e \left (c^2 d^2+e^2\right )^2}-\frac {a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}+\frac {b c^3 d \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}-\frac {b c^3 d \log \left (1+c^2 x^2\right )}{2 \left (c^2 d^2+e^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 192, normalized size = 1.32 \[ -\frac {2 \left (a+b \tan ^{-1}(c x)\right )+\frac {b c (d+e x) \left (2 e \left (c^2 d^2+e^2\right )-\left (c^2 d \left (\sqrt {-c^2} d-2 e\right )-\sqrt {-c^2} e^2\right ) \log \left (1-\sqrt {-c^2} x\right ) (d+e x)-\left (\sqrt {-c^2} e^2-c^2 d \left (\sqrt {-c^2} d+2 e\right )\right ) \log \left (\sqrt {-c^2} x+1\right ) (d+e x)-4 c^2 d e (d+e x) \log (d+e x)\right )}{\left (c^2 d^2+e^2\right )^2}}{4 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(2*(a + b*ArcTan[c*x]) + (b*c*(d + e*x)*(2*e*(c^2*d^2 + e^2) - (c^2*d*(Sqrt[-c^2]*d - 2*e) - Sqrt[-c^2]*e
^2)*(d + e*x)*Log[1 - Sqrt[-c^2]*x] - (Sqrt[-c^2]*e^2 - c^2*d*(Sqrt[-c^2]*d + 2*e))*(d + e*x)*Log[1 + Sqrt[-c^
2]*x] - 4*c^2*d*e*(d + e*x)*Log[d + e*x]))/(c^2*d^2 + e^2)^2)/(e*(d + e*x)^2)

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fricas [B]  time = 0.69, size = 313, normalized size = 2.14 \[ -\frac {a c^{4} d^{4} + b c^{3} d^{3} e + 2 \, a c^{2} d^{2} e^{2} + b c d e^{3} + a e^{4} + {\left (b c^{3} d^{2} e^{2} + b c e^{4}\right )} x + {\left (3 \, b c^{2} d^{2} e^{2} + b e^{4} - {\left (b c^{4} d^{2} e^{2} - b c^{2} e^{4}\right )} x^{2} - 2 \, {\left (b c^{4} d^{3} e - b c^{2} d e^{3}\right )} x\right )} \arctan \left (c x\right ) + {\left (b c^{3} d e^{3} x^{2} + 2 \, b c^{3} d^{2} e^{2} x + b c^{3} d^{3} e\right )} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (b c^{3} d e^{3} x^{2} + 2 \, b c^{3} d^{2} e^{2} x + b c^{3} d^{3} e\right )} \log \left (e x + d\right )}{2 \, {\left (c^{4} d^{6} e + 2 \, c^{2} d^{4} e^{3} + d^{2} e^{5} + {\left (c^{4} d^{4} e^{3} + 2 \, c^{2} d^{2} e^{5} + e^{7}\right )} x^{2} + 2 \, {\left (c^{4} d^{5} e^{2} + 2 \, c^{2} d^{3} e^{4} + d e^{6}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*c^4*d^4 + b*c^3*d^3*e + 2*a*c^2*d^2*e^2 + b*c*d*e^3 + a*e^4 + (b*c^3*d^2*e^2 + b*c*e^4)*x + (3*b*c^2*d
^2*e^2 + b*e^4 - (b*c^4*d^2*e^2 - b*c^2*e^4)*x^2 - 2*(b*c^4*d^3*e - b*c^2*d*e^3)*x)*arctan(c*x) + (b*c^3*d*e^3
*x^2 + 2*b*c^3*d^2*e^2*x + b*c^3*d^3*e)*log(c^2*x^2 + 1) - 2*(b*c^3*d*e^3*x^2 + 2*b*c^3*d^2*e^2*x + b*c^3*d^3*
e)*log(e*x + d))/(c^4*d^6*e + 2*c^2*d^4*e^3 + d^2*e^5 + (c^4*d^4*e^3 + 2*c^2*d^2*e^5 + e^7)*x^2 + 2*(c^4*d^5*e
^2 + 2*c^2*d^3*e^4 + d*e^6)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.04, size = 184, normalized size = 1.26 \[ -\frac {c^{2} a}{2 \left (c e x +d c \right )^{2} e}-\frac {c^{2} b \arctan \left (c x \right )}{2 \left (c e x +d c \right )^{2} e}-\frac {c^{2} b}{2 \left (c^{2} d^{2}+e^{2}\right ) \left (c e x +d c \right )}+\frac {c^{3} b d \ln \left (c e x +d c \right )}{\left (c^{2} d^{2}+e^{2}\right )^{2}}+\frac {c^{4} b \arctan \left (c x \right ) d^{2}}{2 e \left (c^{2} d^{2}+e^{2}\right )^{2}}-\frac {b \,c^{3} d \ln \left (c^{2} x^{2}+1\right )}{2 \left (c^{2} d^{2}+e^{2}\right )^{2}}-\frac {c^{2} b e \arctan \left (c x \right )}{2 \left (c^{2} d^{2}+e^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 214, normalized size = 1.47 \[ -\frac {1}{2} \, {\left ({\left (\frac {c^{2} d \log \left (c^{2} x^{2} + 1\right )}{c^{4} d^{4} + 2 \, c^{2} d^{2} e^{2} + e^{4}} - \frac {2 \, c^{2} d \log \left (e x + d\right )}{c^{4} d^{4} + 2 \, c^{2} d^{2} e^{2} + e^{4}} - \frac {{\left (c^{4} d^{2} - c^{2} e^{2}\right )} \arctan \left (c x\right )}{{\left (c^{4} d^{4} e + 2 \, c^{2} d^{2} e^{3} + e^{5}\right )} c} + \frac {1}{c^{2} d^{3} + d e^{2} + {\left (c^{2} d^{2} e + e^{3}\right )} x}\right )} c + \frac {\arctan \left (c x\right )}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right )} b - \frac {a}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((c^2*d*log(c^2*x^2 + 1)/(c^4*d^4 + 2*c^2*d^2*e^2 + e^4) - 2*c^2*d*log(e*x + d)/(c^4*d^4 + 2*c^2*d^2*e^2
+ e^4) - (c^4*d^2 - c^2*e^2)*arctan(c*x)/((c^4*d^4*e + 2*c^2*d^2*e^3 + e^5)*c) + 1/(c^2*d^3 + d*e^2 + (c^2*d^2
*e + e^3)*x))*c + arctan(c*x)/(e^3*x^2 + 2*d*e^2*x + d^2*e))*b - 1/2*a/(e^3*x^2 + 2*d*e^2*x + d^2*e)

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mupad [B]  time = 4.43, size = 591, normalized size = 4.05 \[ \frac {\frac {x\,\left (a\,c^2\,d^2+\frac {b\,c\,d\,e}{2}+a\,e^2\right )}{d\,\left (c^2\,d^2+e^2\right )}-\frac {b\,\mathrm {atan}\left (c\,x\right )}{2\,e}+\frac {x^2\,\left (\frac {a\,c^2\,d^2\,e}{2}+\frac {b\,c\,d\,e^2}{2}+\frac {a\,e^3}{2}\right )}{d^2\,\left (c^2\,d^2+e^2\right )}+\frac {x^4\,\left (\frac {a\,c^4\,d^2\,e}{2}+\frac {b\,c^3\,d\,e^2}{2}+\frac {a\,c^2\,e^3}{2}\right )}{d^2\,\left (c^2\,d^2+e^2\right )}+\frac {x^3\,\left (a\,c^4\,d^2+\frac {b\,c^3\,d\,e}{2}+a\,c^2\,e^2\right )}{d\,\left (c^2\,d^2+e^2\right )}-\frac {b\,c^2\,x^2\,\mathrm {atan}\left (c\,x\right )}{2\,e}}{c^2\,d^2\,x^2+2\,c^2\,d\,e\,x^3+c^2\,e^2\,x^4+d^2+2\,d\,e\,x+e^2\,x^2}+\frac {b\,c^3\,d\,\ln \left (d+e\,x\right )}{c^4\,d^4+2\,c^2\,d^2\,e^2+e^4}-\frac {b\,c^3\,d\,\ln \left (c^2\,x^2+1\right )}{2\,\left (c^4\,d^4+2\,c^2\,d^2\,e^2+e^4\right )}+\frac {\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {c^2}}\right )\,{\left (c^2\right )}^{7/2}\,\left (c^4\,d^4+8\,c^2\,d^2\,e^2+2\,e^4\right )\,\left (3\,c^6\,d^4+26\,c^4\,d^2\,e^2+4\,c^2\,e^4\right )\,\left (27\,b\,c^{10}\,d^{10}+23\,b\,c^8\,d^8\,e^2-34\,b\,c^6\,d^6\,e^4-26\,b\,c^4\,d^4\,e^6+7\,b\,c^2\,d^2\,e^8+3\,b\,e^{10}\right )}{2\,c\,\left (81\,c^{26}\,d^{20}\,e+1662\,c^{24}\,d^{18}\,e^3+11515\,c^{22}\,d^{16}\,e^5+32306\,c^{20}\,d^{14}\,e^7+43705\,c^{18}\,d^{12}\,e^9+28142\,c^{16}\,d^{10}\,e^{11}+4857\,c^{14}\,d^8\,e^{13}-3650\,c^{12}\,d^6\,e^{15}-2054\,c^{10}\,d^4\,e^{17}-380\,c^8\,d^2\,e^{19}-24\,c^6\,e^{21}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/(d + e*x)^3,x)

[Out]

((x*(a*e^2 + a*c^2*d^2 + (b*c*d*e)/2))/(d*(e^2 + c^2*d^2)) - (b*atan(c*x))/(2*e) + (x^2*((a*e^3)/2 + (b*c*d*e^
2)/2 + (a*c^2*d^2*e)/2))/(d^2*(e^2 + c^2*d^2)) + (x^4*((a*c^2*e^3)/2 + (a*c^4*d^2*e)/2 + (b*c^3*d*e^2)/2))/(d^
2*(e^2 + c^2*d^2)) + (x^3*(a*c^4*d^2 + a*c^2*e^2 + (b*c^3*d*e)/2))/(d*(e^2 + c^2*d^2)) - (b*c^2*x^2*atan(c*x))
/(2*e))/(d^2 + e^2*x^2 + 2*d*e*x + c^2*d^2*x^2 + c^2*e^2*x^4 + 2*c^2*d*e*x^3) + (b*c^3*d*log(d + e*x))/(e^4 +
c^4*d^4 + 2*c^2*d^2*e^2) - (b*c^3*d*log(c^2*x^2 + 1))/(2*(e^4 + c^4*d^4 + 2*c^2*d^2*e^2)) + (atan((c^2*x)/(c^2
)^(1/2))*(c^2)^(7/2)*(2*e^4 + c^4*d^4 + 8*c^2*d^2*e^2)*(3*c^6*d^4 + 4*c^2*e^4 + 26*c^4*d^2*e^2)*(3*b*e^10 + 27
*b*c^10*d^10 + 7*b*c^2*d^2*e^8 - 26*b*c^4*d^4*e^6 - 34*b*c^6*d^6*e^4 + 23*b*c^8*d^8*e^2))/(2*c*(81*c^26*d^20*e
- 24*c^6*e^21 - 380*c^8*d^2*e^19 - 2054*c^10*d^4*e^17 - 3650*c^12*d^6*e^15 + 4857*c^14*d^8*e^13 + 28142*c^16*
d^10*e^11 + 43705*c^18*d^12*e^9 + 32306*c^20*d^14*e^7 + 11515*c^22*d^16*e^5 + 1662*c^24*d^18*e^3))

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sympy [A]  time = 9.65, size = 2887, normalized size = 19.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x/d**3, Eq(c, 0) & Eq(e, 0)), (-a/(2*d**2*e + 4*d*e**2*x + 2*e**3*x**2), Eq(c, 0)), ((a*x + b*x*a
tan(c*x) - b*log(x**2 + c**(-2))/(2*c))/d**3, Eq(e, 0)), (4*a*d**2/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2*e**3*x
**2) - 3*I*b*d**2*atanh(e*x/d)/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2*e**3*x**2) - 2*I*b*d**2/(-8*d**4*e - 16*d*
*3*e**2*x - 8*d**2*e**3*x**2) + 2*I*b*d*e*x*atanh(e*x/d)/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2*e**3*x**2) - I*b
*d*e*x/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2*e**3*x**2) + I*b*e**2*x**2*atanh(e*x/d)/(-8*d**4*e - 16*d**3*e**2*
x - 8*d**2*e**3*x**2), Eq(c, -I*e/d)), (4*a*d**2/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2*e**3*x**2) + 3*I*b*d**2*
atanh(e*x/d)/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2*e**3*x**2) + 2*I*b*d**2/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2
*e**3*x**2) - 2*I*b*d*e*x*atanh(e*x/d)/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2*e**3*x**2) + I*b*d*e*x/(-8*d**4*e
- 16*d**3*e**2*x - 8*d**2*e**3*x**2) - I*b*e**2*x**2*atanh(e*x/d)/(-8*d**4*e - 16*d**3*e**2*x - 8*d**2*e**3*x*
*2), Eq(c, I*e/d)), (-a*c**4*d**4/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e*
*3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - 2*a*c**2*d**2*e**2
/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*
d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - a*e**4/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4
*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2
*e**7*x**2) + 2*b*c**4*d**3*e*x*atan(c*x)/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2
*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) + b*c**4*d**
2*e**2*x**2*atan(c*x)/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*
d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - b*c**3*d**3*e*log(x**2 + c**(-
2))/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c*
*2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) + 2*b*c**3*d**3*e*log(d/e + x)/(2*c**4*d**6*e + 4*
c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d
**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - b*c**3*d**3*e/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x*
*2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) -
2*b*c**3*d**2*e**2*x*log(x**2 + c**(-2))/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2
*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) + 4*b*c**3*d
**2*e**2*x*log(d/e + x)/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**
2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - b*c**3*d**2*e**2*x/(2*c**4*d
**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*
x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - b*c**3*d*e**3*x**2*log(x**2 + c**(-2))/(2*c**4*d**6*e + 4*c**
4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2
*e**5 + 4*d*e**6*x + 2*e**7*x**2) + 2*b*c**3*d*e**3*x**2*log(d/e + x)/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*
c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x
+ 2*e**7*x**2) - 3*b*c**2*d**2*e**2*atan(c*x)/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4
*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - 2*b*c
**2*d*e**3*x*atan(c*x)/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2
*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - b*c**2*e**4*x**2*atan(c*x)/(2
*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**
2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - b*c*d*e**3/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**
4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x +
2*e**7*x**2) - b*c*e**4*x/(2*c**4*d**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c
**2*d**3*e**4*x + 4*c**2*d**2*e**5*x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) - b*e**4*atan(c*x)/(2*c**4*d
**6*e + 4*c**4*d**5*e**2*x + 2*c**4*d**4*e**3*x**2 + 4*c**2*d**4*e**3 + 8*c**2*d**3*e**4*x + 4*c**2*d**2*e**5*
x**2 + 2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2), True))

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