3.36 \(\int \frac {a+b \tanh ^{-1}(c+d x)}{(e+f x)^2} \, dx\)

Optimal. Leaf size=115 \[ -\frac {a+b \tanh ^{-1}(c+d x)}{f (e+f x)}-\frac {b d \log (-c-d x+1)}{2 f (-c f+d e+f)}+\frac {b d \log (c+d x+1)}{2 f (-c f+d e-f)}-\frac {b d \log (e+f x)}{(-c f+d e+f) (d e-(c+1) f)} \]

[Out]

(-a-b*arctanh(d*x+c))/f/(f*x+e)-1/2*b*d*ln(-d*x-c+1)/f/(-c*f+d*e+f)+1/2*b*d*ln(d*x+c+1)/f/(-c*f+d*e-f)-b*d*ln(
f*x+e)/(-c*f+d*e-f)/(-c*f+d*e+f)

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Rubi [A]  time = 0.17, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6109, 1982, 705, 31, 632} \[ -\frac {a+b \tanh ^{-1}(c+d x)}{f (e+f x)}-\frac {b d \log (-c-d x+1)}{2 f (-c f+d e+f)}+\frac {b d \log (c+d x+1)}{2 f (-c f+d e-f)}-\frac {b d \log (e+f x)}{(-c f+d e+f) (d e-(c+1) f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcTanh[c + d*x])/(f*(e + f*x))) - (b*d*Log[1 - c - d*x])/(2*f*(d*e + f - c*f)) + (b*d*Log[1 + c + d*
x])/(2*f*(d*e - f - c*f)) - (b*d*Log[e + f*x])/((d*e + f - c*f)*(d*e - (1 + c)*f))

Rule 31

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

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 1982

Int[(u_)^(m_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^p, x] /; FreeQ[{m, p}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 6109

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*(a + b*ArcTanh[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*A
rcTanh[c + d*x])^(p - 1))/(1 - (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -
1]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 125, normalized size = 1.09 \[ \frac {1}{2} \left (-\frac {2 a}{f (e+f x)}-\frac {2 b d \log (e+f x)}{\left (c^2-1\right ) f^2-2 c d e f+d^2 e^2}+\frac {b d \log (-c-d x+1)}{f ((c-1) f-d e)}-\frac {b d \log (c+d x+1)}{f (c f-d e+f)}-\frac {2 b \tanh ^{-1}(c+d x)}{f (e+f x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.61, size = 263, normalized size = 2.29 \[ -\frac {2 \, a d^{2} e^{2} - 4 \, a c d e f + 2 \, {\left (a c^{2} - a\right )} f^{2} - {\left (b d^{2} e^{2} - {\left (b c - b\right )} d e f + {\left (b d^{2} e f - {\left (b c - b\right )} d f^{2}\right )} x\right )} \log \left (d x + c + 1\right ) + {\left (b d^{2} e^{2} - {\left (b c + b\right )} d e f + {\left (b d^{2} e f - {\left (b c + b\right )} d f^{2}\right )} x\right )} \log \left (d x + c - 1\right ) + 2 \, {\left (b d f^{2} x + b d e f\right )} \log \left (f x + e\right ) + {\left (b d^{2} e^{2} - 2 \, b c d e f + {\left (b c^{2} - b\right )} f^{2}\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{2 \, {\left (d^{2} e^{3} f - 2 \, c d e^{2} f^{2} + {\left (c^{2} - 1\right )} e f^{3} + {\left (d^{2} e^{2} f^{2} - 2 \, c d e f^{3} + {\left (c^{2} - 1\right )} f^{4}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))/(f*x+e)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.27, size = 896, normalized size = 7.79 \[ -\frac {{\left (\frac {{\left (d x + c + 1\right )} b c f \log \left (\frac {{\left (d x + c + 1\right )} c f}{d x + c - 1} - c f - \frac {{\left (d x + c + 1\right )} d e}{d x + c - 1} + d e - \frac {{\left (d x + c + 1\right )} f}{d x + c - 1} - f\right )}{d x + c - 1} - b c f \log \left (\frac {{\left (d x + c + 1\right )} c f}{d x + c - 1} - c f - \frac {{\left (d x + c + 1\right )} d e}{d x + c - 1} + d e - \frac {{\left (d x + c + 1\right )} f}{d x + c - 1} - f\right ) - \frac {{\left (d x + c + 1\right )} b d e \log \left (\frac {{\left (d x + c + 1\right )} c f}{d x + c - 1} - c f - \frac {{\left (d x + c + 1\right )} d e}{d x + c - 1} + d e - \frac {{\left (d x + c + 1\right )} f}{d x + c - 1} - f\right )}{d x + c - 1} + b d e \log \left (\frac {{\left (d x + c + 1\right )} c f}{d x + c - 1} - c f - \frac {{\left (d x + c + 1\right )} d e}{d x + c - 1} + d e - \frac {{\left (d x + c + 1\right )} f}{d x + c - 1} - f\right ) - \frac {{\left (d x + c + 1\right )} b c f \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} + \frac {{\left (d x + c + 1\right )} b d e \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} - 2 \, a c f + 2 \, a d e - \frac {{\left (d x + c + 1\right )} b f \log \left (\frac {{\left (d x + c + 1\right )} c f}{d x + c - 1} - c f - \frac {{\left (d x + c + 1\right )} d e}{d x + c - 1} + d e - \frac {{\left (d x + c + 1\right )} f}{d x + c - 1} - f\right )}{d x + c - 1} - b f \log \left (\frac {{\left (d x + c + 1\right )} c f}{d x + c - 1} - c f - \frac {{\left (d x + c + 1\right )} d e}{d x + c - 1} + d e - \frac {{\left (d x + c + 1\right )} f}{d x + c - 1} - f\right ) + \frac {{\left (d x + c + 1\right )} b f \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} - 2 \, a f\right )} {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )}}{2 \, {\left (\frac {{\left (d x + c + 1\right )} c^{3} f^{3}}{d x + c - 1} - c^{3} f^{3} - \frac {3 \, {\left (d x + c + 1\right )} c^{2} d f^{2} e}{d x + c - 1} + 3 \, c^{2} d f^{2} e - \frac {{\left (d x + c + 1\right )} c^{2} f^{3}}{d x + c - 1} - c^{2} f^{3} + \frac {3 \, {\left (d x + c + 1\right )} c d^{2} f e^{2}}{d x + c - 1} - 3 \, c d^{2} f e^{2} + \frac {2 \, {\left (d x + c + 1\right )} c d f^{2} e}{d x + c - 1} + 2 \, c d f^{2} e - \frac {{\left (d x + c + 1\right )} c f^{3}}{d x + c - 1} + c f^{3} - \frac {{\left (d x + c + 1\right )} d^{3} e^{3}}{d x + c - 1} + d^{3} e^{3} - \frac {{\left (d x + c + 1\right )} d^{2} f e^{2}}{d x + c - 1} - d^{2} f e^{2} + \frac {{\left (d x + c + 1\right )} d f^{2} e}{d x + c - 1} - d f^{2} e + \frac {{\left (d x + c + 1\right )} f^{3}}{d x + c - 1} + f^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 141, normalized size = 1.23 \[ -\frac {d a}{\left (d f x +d e \right ) f}-\frac {d b \arctanh \left (d x +c \right )}{\left (d f x +d e \right ) f}-\frac {d b \ln \left (\left (d x +c \right ) f -c f +d e \right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {d b \ln \left (d x +c -1\right )}{f \left (2 c f -2 d e -2 f \right )}-\frac {d b \ln \left (d x +c +1\right )}{f \left (2 c f -2 d e +2 f \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))/(f*x+e)^2,x)

[Out]

-d*a/(d*f*x+d*e)/f-d*b/(d*f*x+d*e)/f*arctanh(d*x+c)-d*b/(c*f-d*e-f)/(c*f-d*e+f)*ln((d*x+c)*f-c*f+d*e)+d*b/f/(2
*c*f-2*d*e-2*f)*ln(d*x+c-1)-d*b/f/(2*c*f-2*d*e+2*f)*ln(d*x+c+1)

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maxima [A]  time = 0.32, size = 121, normalized size = 1.05 \[ \frac {1}{2} \, {\left (d {\left (\frac {\log \left (d x + c + 1\right )}{d e f - {\left (c + 1\right )} f^{2}} - \frac {\log \left (d x + c - 1\right )}{d e f - {\left (c - 1\right )} f^{2}} - \frac {2 \, \log \left (f x + e\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} - 1\right )} f^{2}}\right )} - \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{f^{2} x + e f}\right )} b - \frac {a}{f^{2} x + e f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d*(log(d*x + c + 1)/(d*e*f - (c + 1)*f^2) - log(d*x + c - 1)/(d*e*f - (c - 1)*f^2) - 2*log(f*x + e)/(d^2*
e^2 - 2*c*d*e*f + (c^2 - 1)*f^2)) - 2*arctanh(d*x + c)/(f^2*x + e*f))*b - a/(f^2*x + e*f)

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mupad [B]  time = 1.64, size = 170, normalized size = 1.48 \[ \ln \left (e+f\,x\right )\,\left (\frac {b\,\left (c-1\right )}{2\,e\,\left (d\,e-f\,\left (c-1\right )\right )}-\frac {b\,\left (c+1\right )}{2\,e\,\left (d\,e-f\,\left (c+1\right )\right )}\right )-\frac {a}{x\,f^2+e\,f}+\frac {b\,\ln \left (1-d\,x-c\right )}{f\,\left (2\,e+2\,f\,x\right )}-\frac {b\,\ln \left (c+d\,x+1\right )}{2\,f\,\left (e+f\,x\right )}-\frac {b\,d\,\ln \left (c+d\,x-1\right )}{2\,f^2-2\,c\,f^2+2\,d\,e\,f}-\frac {b\,d\,\ln \left (c+d\,x+1\right )}{2\,c\,f^2+2\,f^2-2\,d\,e\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c + d*x))/(e + f*x)^2,x)

[Out]

log(e + f*x)*((b*(c - 1))/(2*e*(d*e - f*(c - 1))) - (b*(c + 1))/(2*e*(d*e - f*(c + 1)))) - a/(e*f + f^2*x) + (
b*log(1 - d*x - c))/(f*(2*e + 2*f*x)) - (b*log(c + d*x + 1))/(2*f*(e + f*x)) - (b*d*log(c + d*x - 1))/(2*f^2 -
2*c*f^2 + 2*d*e*f) - (b*d*log(c + d*x + 1))/(2*c*f^2 + 2*f^2 - 2*d*e*f)

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sympy [A]  time = 10.01, size = 1658, normalized size = 14.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))/(f*x+e)**2,x)

[Out]

Piecewise(((a*x + b*c*atanh(c + d*x)/d + b*x*atanh(c + d*x) + b*log(c/d + x + 1/d)/d - b*atanh(c + d*x)/d)/e**
2, Eq(f, 0)), (-2*a*f/(2*e*f**2 + 2*f**3*x) + b*d*e*atanh(d*e/f + d*x - 1)/(2*e*f**2 + 2*f**3*x) + b*d*f*x*ata
nh(d*e/f + d*x - 1)/(2*e*f**2 + 2*f**3*x) - 2*b*f*atanh(d*e/f + d*x - 1)/(2*e*f**2 + 2*f**3*x) - b*f/(2*e*f**2
+ 2*f**3*x), Eq(c, (d*e - f)/f)), (-2*a*f/(2*e*f**2 + 2*f**3*x) - b*d*e*atanh(d*e/f + d*x + 1)/(2*e*f**2 + 2*
f**3*x) - b*d*f*x*atanh(d*e/f + d*x + 1)/(2*e*f**2 + 2*f**3*x) - 2*b*f*atanh(d*e/f + d*x + 1)/(2*e*f**2 + 2*f*
*3*x) + b*f/(2*e*f**2 + 2*f**3*x), Eq(c, (d*e + f)/f)), (zoo*(a*x + b*c*atanh(c + d*x)/d + b*x*atanh(c + d*x)
+ b*log(c/d + x + 1/d)/d - b*atanh(c + d*x)/d), Eq(e, -f*x)), (-(a + b*atanh(c))/(e*f + f**2*x), Eq(d, 0)), (-
a*c**2*f**2/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e
*f**3 - f**4*x) + 2*a*c*d*e*f/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d*
*2*e**2*f**2*x - e*f**3 - f**4*x) - a*d**2*e**2/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x
+ d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + a*f**2/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*
c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) - b*c**2*f**2*atanh(c + d*x)/(c**2*e*f**3 + c
**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + b*c*d*e*f*
atanh(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x
- e*f**3 - f**4*x) - b*c*d*f**2*x*atanh(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x
+ d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + b*d**2*e*f*x*atanh(c + d*x)/(c**2*e*f**3 + c**2*f**4*x
- 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) - b*d*e*f*log(e/f + x)/
(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4
*x) + b*d*e*f*log(c/d + x + 1/d)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f +
d**2*e**2*f**2*x - e*f**3 - f**4*x) - b*d*e*f*atanh(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2
*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) - b*d*f**2*x*log(e/f + x)/(c**2*e*f**3 + c**
2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + b*d*f**2*x*l
og(c/d + x + 1/d)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2
*x - e*f**3 - f**4*x) - b*d*f**2*x*atanh(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*
x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + b*f**2*atanh(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c
*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x), True))

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