3.7 \(\int \frac {a+b \tanh ^{-1}(c x)}{(d+e x)^3} \, dx\)
Optimal. Leaf size=130 \[ -\frac {a+b \tanh ^{-1}(c x)}{2 e (d+e x)^2}+\frac {b c}{2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {b c^2 \log (1-c x)}{4 e (c d+e)^2}+\frac {b c^2 \log (c x+1)}{4 e (c d-e)^2}-\frac {b c^3 d \log (d+e x)}{\left (c^2 d^2-e^2\right )^2} \]
[Out]
1/2*b*c/(c^2*d^2-e^2)/(e*x+d)+1/2*(-a-b*arctanh(c*x))/e/(e*x+d)^2-1/4*b*c^2*ln(-c*x+1)/e/(c*d+e)^2+1/4*b*c^2*l
n(c*x+1)/(c*d-e)^2/e-b*c^3*d*ln(e*x+d)/(c^2*d^2-e^2)^2
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Rubi [A] time = 0.13, antiderivative size = 130, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used =
{5926, 710, 801} \[ -\frac {a+b \tanh ^{-1}(c x)}{2 e (d+e x)^2}+\frac {b c}{2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {b c^3 d \log (d+e x)}{\left (c^2 d^2-e^2\right )^2}-\frac {b c^2 \log (1-c x)}{4 e (c d+e)^2}+\frac {b c^2 \log (c x+1)}{4 e (c d-e)^2} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTanh[c*x])/(d + e*x)^3,x]
[Out]
(b*c)/(2*(c^2*d^2 - e^2)*(d + e*x)) - (a + b*ArcTanh[c*x])/(2*e*(d + e*x)^2) - (b*c^2*Log[1 - c*x])/(4*e*(c*d
+ e)^2) + (b*c^2*Log[1 + c*x])/(4*(c*d - e)^2*e) - (b*c^3*d*Log[d + e*x])/(c^2*d^2 - e^2)^2
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 5926
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[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 \tanh ^{-1}(c x)}{(d+e x)^3} \, dx &=-\frac {a+b \tanh ^{-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 \tanh ^{-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 \tanh ^{-1}(c x)}{2 e (d+e x)^2}+\frac {\left (b c^3\right ) \int \left (\frac {-c d+e}{2 (c d+e) (-1+c x)}+\frac {c d+e}{2 (c d-e) (1+c x)}+\frac {2 d e^2}{(-c d+e) (c d+e) (d+e x)}\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 \tanh ^{-1}(c x)}{2 e (d+e x)^2}-\frac {b c^2 \log (1-c x)}{4 e (c d+e)^2}+\frac {b c^2 \log (1+c x)}{4 (c d-e)^2 e}-\frac {b c^3 d \log (d+e x)}{\left (c^2 d^2-e^2\right )^2}\\ \end {align*}
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Mathematica [A] time = 0.16, size = 133, normalized size = 1.02 \[ \frac {1}{4} \left (-\frac {2 a}{e (d+e x)^2}+\frac {2 b c}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {b c^2 \log (1-c x)}{e (c d+e)^2}+\frac {b c^2 \log (c x+1)}{e (e-c d)^2}-\frac {4 b c^3 d \log (d+e x)}{\left (e^2-c^2 d^2\right )^2}-\frac {2 b \tanh ^{-1}(c x)}{e (d+e x)^2}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcTanh[c*x])/(d + e*x)^3,x]
[Out]
((-2*a)/(e*(d + e*x)^2) + (2*b*c)/((c^2*d^2 - e^2)*(d + e*x)) - (2*b*ArcTanh[c*x])/(e*(d + e*x)^2) - (b*c^2*Lo
g[1 - c*x])/(e*(c*d + e)^2) + (b*c^2*Log[1 + c*x])/(e*(-(c*d) + e)^2) - (4*b*c^3*d*Log[d + e*x])/(-(c^2*d^2) +
e^2)^2)/4
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fricas [B] time = 0.85, size = 454, normalized size = 3.49 \[ -\frac {2 \, a c^{4} d^{4} - 2 \, b c^{3} d^{3} e - 4 \, a c^{2} d^{2} e^{2} + 2 \, b c d e^{3} + 2 \, a e^{4} - 2 \, {\left (b c^{3} d^{2} e^{2} - b c e^{4}\right )} x - {\left (b c^{4} d^{4} + 2 \, b c^{3} d^{3} e + b c^{2} d^{2} e^{2} + {\left (b c^{4} d^{2} e^{2} + 2 \, b c^{3} d e^{3} + b c^{2} e^{4}\right )} x^{2} + 2 \, {\left (b c^{4} d^{3} e + 2 \, b c^{3} d^{2} e^{2} + b c^{2} d e^{3}\right )} x\right )} \log \left (c x + 1\right ) + {\left (b c^{4} d^{4} - 2 \, b c^{3} d^{3} e + b c^{2} d^{2} e^{2} + {\left (b c^{4} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + b c^{2} e^{4}\right )} x^{2} + 2 \, {\left (b c^{4} d^{3} e - 2 \, b c^{3} d^{2} e^{2} + b c^{2} d e^{3}\right )} x\right )} \log \left (c x - 1\right ) + 4 \, {\left (b c^{3} d e^{3} x^{2} + 2 \, b c^{3} d^{2} e^{2} x + b c^{3} d^{3} e\right )} \log \left (e x + d\right ) + {\left (b c^{4} d^{4} - 2 \, b c^{2} d^{2} e^{2} + b e^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, {\left (c^{4} d^{6} e - 2 \, c^{2} d^{4} e^{3} + d^{2} e^{5} + {\left (c^{4} d^{4} e^{3} - 2 \, c^{2} d^{2} e^{5} + e^{7}\right )} x^{2} + 2 \, {\left (c^{4} d^{5} e^{2} - 2 \, c^{2} d^{3} e^{4} + d e^{6}\right )} x\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))/(e*x+d)^3,x, algorithm="fricas")
[Out]
-1/4*(2*a*c^4*d^4 - 2*b*c^3*d^3*e - 4*a*c^2*d^2*e^2 + 2*b*c*d*e^3 + 2*a*e^4 - 2*(b*c^3*d^2*e^2 - b*c*e^4)*x -
(b*c^4*d^4 + 2*b*c^3*d^3*e + b*c^2*d^2*e^2 + (b*c^4*d^2*e^2 + 2*b*c^3*d*e^3 + b*c^2*e^4)*x^2 + 2*(b*c^4*d^3*e
+ 2*b*c^3*d^2*e^2 + b*c^2*d*e^3)*x)*log(c*x + 1) + (b*c^4*d^4 - 2*b*c^3*d^3*e + b*c^2*d^2*e^2 + (b*c^4*d^2*e^2
- 2*b*c^3*d*e^3 + b*c^2*e^4)*x^2 + 2*(b*c^4*d^3*e - 2*b*c^3*d^2*e^2 + b*c^2*d*e^3)*x)*log(c*x - 1) + 4*(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) + (b*c^4*d^4 - 2*b*c^2*d^2*e^2 + b*e^4)*log(-(c*x +
1)/(c*x - 1)))/(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 [B] time = 0.19, size = 1197, normalized size = 9.21 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))/(e*x+d)^3,x, algorithm="giac")
[Out]
-((c*x + 1)^2*b*c^4*d^3*log(-(c*x + 1)*c*d/(c*x - 1) + c*d - (c*x + 1)*e/(c*x - 1) - e)/(c*x - 1)^2 - 2*(c*x +
1)*b*c^4*d^3*log(-(c*x + 1)*c*d/(c*x - 1) + c*d - (c*x + 1)*e/(c*x - 1) - e)/(c*x - 1) + b*c^4*d^3*log(-(c*x
+ 1)*c*d/(c*x - 1) + c*d - (c*x + 1)*e/(c*x - 1) - e) - (c*x + 1)^2*b*c^4*d^3*log(-(c*x + 1)/(c*x - 1))/(c*x -
1)^2 + (c*x + 1)*b*c^4*d^3*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) - 2*(c*x + 1)*a*c^4*d^3/(c*x - 1) + 2*a*c^4*d^
3 + 2*(c*x + 1)^2*b*c^3*d^2*e*log(-(c*x + 1)*c*d/(c*x - 1) + c*d - (c*x + 1)*e/(c*x - 1) - e)/(c*x - 1)^2 - 2*
b*c^3*d^2*e*log(-(c*x + 1)*c*d/(c*x - 1) + c*d - (c*x + 1)*e/(c*x - 1) - e) - 2*(c*x + 1)^2*b*c^3*d^2*e*log(-(
c*x + 1)/(c*x - 1))/(c*x - 1)^2 + (c*x + 1)*b*c^3*d^2*e*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) + 2*(c*x + 1)*a*c^
3*d^2*e/(c*x - 1) - 4*a*c^3*d^2*e + (c*x + 1)*b*c^3*d^2*e/(c*x - 1) - b*c^3*d^2*e + (c*x + 1)^2*b*c^2*d*e^2*lo
g(-(c*x + 1)*c*d/(c*x - 1) + c*d - (c*x + 1)*e/(c*x - 1) - e)/(c*x - 1)^2 + 2*(c*x + 1)*b*c^2*d*e^2*log(-(c*x
+ 1)*c*d/(c*x - 1) + c*d - (c*x + 1)*e/(c*x - 1) - e)/(c*x - 1) + b*c^2*d*e^2*log(-(c*x + 1)*c*d/(c*x - 1) + c
*d - (c*x + 1)*e/(c*x - 1) - e) - (c*x + 1)^2*b*c^2*d*e^2*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^2 - (c*x + 1)*b*
c^2*d*e^2*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) + 2*(c*x + 1)*a*c^2*d*e^2/(c*x - 1) + 2*a*c^2*d*e^2 + 2*b*c^2*d*
e^2 - (c*x + 1)*b*c*e^3*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) - 2*(c*x + 1)*a*c*e^3/(c*x - 1) - (c*x + 1)*b*c*e^
3/(c*x - 1) - b*c*e^3)*c/((c*x + 1)^2*c^6*d^6/(c*x - 1)^2 - 2*(c*x + 1)*c^6*d^6/(c*x - 1) + c^6*d^6 + 2*(c*x +
1)^2*c^5*d^5*e/(c*x - 1)^2 - 2*c^5*d^5*e - (c*x + 1)^2*c^4*d^4*e^2/(c*x - 1)^2 + 6*(c*x + 1)*c^4*d^4*e^2/(c*x
- 1) - c^4*d^4*e^2 - 4*(c*x + 1)^2*c^3*d^3*e^3/(c*x - 1)^2 + 4*c^3*d^3*e^3 - (c*x + 1)^2*c^2*d^2*e^4/(c*x - 1
)^2 - 6*(c*x + 1)*c^2*d^2*e^4/(c*x - 1) - c^2*d^2*e^4 + 2*(c*x + 1)^2*c*d*e^5/(c*x - 1)^2 - 2*c*d*e^5 + (c*x +
1)^2*e^6/(c*x - 1)^2 + 2*(c*x + 1)*e^6/(c*x - 1) + e^6)
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maple [A] time = 0.04, size = 154, normalized size = 1.18 \[ -\frac {c^{2} a}{2 \left (c x e +c d \right )^{2} e}-\frac {c^{2} b \arctanh \left (c x \right )}{2 \left (c x e +c d \right )^{2} e}+\frac {c^{2} b}{2 \left (c d +e \right ) \left (c d -e \right ) \left (c x e +c d \right )}-\frac {c^{3} b d \ln \left (c x e +c d \right )}{\left (c d +e \right )^{2} \left (c d -e \right )^{2}}-\frac {c^{2} b \ln \left (c x -1\right )}{4 e \left (c d +e \right )^{2}}+\frac {b \,c^{2} \ln \left (c x +1\right )}{4 \left (c d -e \right )^{2} e} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctanh(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*arctanh(c*x)+1/2*c^2*b/(c*d+e)/(c*d-e)/(c*e*x+c*d)-c^3*b*
d/(c*d+e)^2/(c*d-e)^2*ln(c*e*x+c*d)-1/4*c^2*b/e/(c*d+e)^2*ln(c*x-1)+1/4*b*c^2*ln(c*x+1)/(c*d-e)^2/e
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maxima [A] time = 0.33, size = 190, normalized size = 1.46 \[ -\frac {1}{4} \, {\left ({\left (\frac {4 \, c^{2} d \log \left (e x + d\right )}{c^{4} d^{4} - 2 \, c^{2} d^{2} e^{2} + e^{4}} - \frac {c \log \left (c x + 1\right )}{c^{2} d^{2} e - 2 \, c d e^{2} + e^{3}} + \frac {c \log \left (c x - 1\right )}{c^{2} d^{2} e + 2 \, c d e^{2} + e^{3}} - \frac {2}{c^{2} d^{3} - d e^{2} + {\left (c^{2} d^{2} e - e^{3}\right )} x}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right )} b - \frac {a}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))/(e*x+d)^3,x, algorithm="maxima")
[Out]
-1/4*((4*c^2*d*log(e*x + d)/(c^4*d^4 - 2*c^2*d^2*e^2 + e^4) - c*log(c*x + 1)/(c^2*d^2*e - 2*c*d*e^2 + e^3) + c
*log(c*x - 1)/(c^2*d^2*e + 2*c*d*e^2 + e^3) - 2/(c^2*d^3 - d*e^2 + (c^2*d^2*e - e^3)*x))*c + 2*arctanh(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 = 4.00, size = 427, normalized size = 3.28 \[ \frac {b\,c^3\,d\,\ln \left (c^2\,x^2-1\right )}{2\,\left (c^4\,d^4-2\,c^2\,d^2\,e^2+e^4\right )}-\frac {\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )\,\left (b\,c^5\,d^2+b\,c^3\,e^2\right )}{2\,e^5\,\sqrt {-c^2}-c^2\,\left (2\,d^4\,e\,{\left (-c^2\right )}^{3/2}+4\,d^2\,e^3\,\sqrt {-c^2}\right )}-\frac {b\,c^3\,d\,\ln \left (d+e\,x\right )}{c^4\,d^4-2\,c^2\,d^2\,e^2+e^4}-\frac {\frac {b\,\mathrm {atanh}\left (c\,x\right )}{2\,e}-\frac {x\,\left (-a\,c^2\,d^2+\frac {b\,c\,d\,e}{2}+a\,e^2\right )}{d\,\left (e^2-c^2\,d^2\right )}-\frac {x^2\,\left (-\frac {a\,c^2\,d^2\,e}{2}+\frac {b\,c\,d\,e^2}{2}+\frac {a\,e^3}{2}\right )}{d^2\,\left (e^2-c^2\,d^2\right )}+\frac {x^4\,\left (-\frac {a\,c^4\,d^2\,e}{2}+\frac {b\,c^3\,d\,e^2}{2}+\frac {a\,c^2\,e^3}{2}\right )}{d^2\,\left (e^2-c^2\,d^2\right )}+\frac {x^3\,\left (-a\,c^4\,d^2+\frac {b\,c^3\,d\,e}{2}+a\,c^2\,e^2\right )}{d\,\left (e^2-c^2\,d^2\right )}-\frac {b\,c^2\,x^2\,\mathrm {atanh}\left (c\,x\right )}{2\,e}}{-c^2\,d^2\,x^2-2\,c^2\,d\,e\,x^3-c^2\,e^2\,x^4+d^2+2\,d\,e\,x+e^2\,x^2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*atanh(c*x))/(d + e*x)^3,x)
[Out]
(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))*(b*c^5*d^2 + b*c^
3*e^2))/(2*e^5*(-c^2)^(1/2) - c^2*(2*d^4*e*(-c^2)^(3/2) + 4*d^2*e^3*(-c^2)^(1/2))) - (b*c^3*d*log(d + e*x))/(e
^4 + c^4*d^4 - 2*c^2*d^2*e^2) - ((b*atanh(c*x))/(2*e) - (x*(a*e^2 - a*c^2*d^2 + (b*c*d*e)/2))/(d*(e^2 - c^2*d^
2)) - (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^2*e^2 - a*c^4*d^2 + (b*c^3*d*e)/2))/(d*(e^2 -
c^2*d^2)) - (b*c^2*x^2*atanh(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
)
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sympy [A] time = 7.13, size = 3216, normalized size = 24.74 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atanh(c*x))/(e*x+d)**3,x)
[Out]
Piecewise((a*x/d**3, Eq(c, 0) & 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*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*b*d**2/(8*d**4*e + 16*d**3*e**2*x + 8*d**2*e
**3*x**2) - 2*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) + b*d*e*x/(8*d**4*e + 16*d**
3*e**2*x + 8*d**2*e**3*x**2) - 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,
-e/d)), (-4*a*d**2/(8*d**4*e + 16*d**3*e**2*x + 8*d**2*e**3*x**2) - 3*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*b*d**2/(8*d**4*e + 16*d**3*e**2*x + 8*d**2*e**3*x**2) + 2*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) - b*d*e*x/(8*d**4*e + 16*d**3*e**2*x + 8*d**2*e**3*x**2) + 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, e/d)), ((a*x + b*x*atanh(c*x) + b*l
og(x - 1/c)/c + b*atanh(c*x)/c)/d**3, Eq(e, 0)), (-a/(2*d**2*e + 4*d*e**2*x + 2*e**3*x**2), Eq(c, 0)), (-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*atanh(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*atanh(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**3*d**3*e*log(x - 1/c)/(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) + 2*b*c**3*d**3*e*atanh(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/(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(x - 1/c)/(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) + 4*b*c**3*d**2*e**2*x*atanh(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**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) + 2*b*c**3*d*e**3*x**2*log(x - 1/c)
/(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) + 2*b*c**3*d*e**3*x**2*atanh(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) + 3*b*c**2*d**2*e**2*atanh(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*atanh(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*atanh(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*atanh(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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