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

Optimal. Leaf size=167 \[ -\frac {a+b \coth ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac {b d^2 \log (-c-d x+1)}{4 f (-c f+d e+f)^2}+\frac {b d^2 \log (c+d x+1)}{4 f (-c f+d e-f)^2}-\frac {b d^2 (d e-c f) \log (e+f x)}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b d}{2 (e+f x) (-c f+d e+f) (d e-(c+1) f)} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6110, 1982, 709, 800} \[ -\frac {a+b \coth ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac {b d^2 \log (-c-d x+1)}{4 f (-c f+d e+f)^2}+\frac {b d^2 \log (c+d x+1)}{4 f (-c f+d e-f)^2}-\frac {b d^2 (d e-c f) \log (e+f x)}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b d}{2 (e+f x) (-c f+d e+f) (d e-(c+1) f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d)/(2*(d*e + f - c*f)*(d*e - (1 + c)*f)*(e + f*x)) - (a + b*ArcCoth[c + d*x])/(2*f*(e + f*x)^2) - (b*d^2*Lo
g[1 - c - d*x])/(4*f*(d*e + f - c*f)^2) + (b*d^2*Log[1 + c + d*x])/(4*f*(d*e - f - c*f)^2) - (b*d^2*(d*e - c*f
)*Log[e + f*x])/((d*e + f - c*f)^2*(d*e - (1 + c)*f)^2)

Rule 709

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 6110

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*(a + b*ArcCoth[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*A
rcCoth[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 \coth ^{-1}(c+d x)}{(e+f x)^3} \, dx &=-\frac {a+b \coth ^{-1}(c+d x)}{2 f (e+f x)^2}+\frac {(b d) \int \frac {1}{(e+f x)^2 \left (1-(c+d x)^2\right )} \, dx}{2 f}\\ &=-\frac {a+b \coth ^{-1}(c+d x)}{2 f (e+f x)^2}+\frac {(b d) \int \frac {1}{(e+f x)^2 \left (1-c^2-2 c d x-d^2 x^2\right )} \, dx}{2 f}\\ &=\frac {b d}{2 (d e+f-c f) (d e-(1+c) f) (e+f x)}-\frac {a+b \coth ^{-1}(c+d x)}{2 f (e+f x)^2}+\frac {(b d) \int \frac {-d (d e-2 c f)+d^2 f x}{(e+f x) \left (1-c^2-2 c d x-d^2 x^2\right )} \, dx}{2 f \left (-d^2 e^2+2 c d e f+\left (1-c^2\right ) f^2\right )}\\ &=\frac {b d}{2 (d e+f-c f) (d e-(1+c) f) (e+f x)}-\frac {a+b \coth ^{-1}(c+d x)}{2 f (e+f x)^2}+\frac {(b d) \int \left (\frac {d^2 (-d e+(1+c) f)}{2 (d e+f-c f) (1-c-d x)}+\frac {d^2 (-d e-f+c f)}{2 (d e-(1+c) f) (1+c+d x)}+\frac {2 d f^2 (d e-c f)}{(d e+(1-c) f) (d e-f-c f) (e+f x)}\right ) \, dx}{2 f \left (-d^2 e^2+2 c d e f+\left (1-c^2\right ) f^2\right )}\\ &=\frac {b d}{2 (d e+f-c f) (d e-(1+c) f) (e+f x)}-\frac {a+b \coth ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac {b d^2 \log (1-c-d x)}{4 f (d e+f-c f)^2}+\frac {b d^2 \log (1+c+d x)}{4 f (d e-f-c f)^2}-\frac {b d^2 (d e-c f) \log (e+f x)}{(d e+f-c f)^2 (d e-(1+c) f)^2}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 174, normalized size = 1.04 \[ \frac {1}{4} \left (-\frac {2 a}{f (e+f x)^2}+\frac {2 b d}{(e+f x) \left (\left (c^2-1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac {4 b d^2 (d e-c f) \log (e+f x)}{\left (\left (c^2-1\right ) f^2-2 c d e f+d^2 e^2\right )^2}-\frac {b d^2 \log (-c-d x+1)}{f (-c f+d e+f)^2}+\frac {b d^2 \log (c+d x+1)}{f (c f-d e+f)^2}-\frac {2 b \coth ^{-1}(c+d x)}{f (e+f x)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.70, size = 833, normalized size = 4.99 \[ -\frac {2 \, a d^{4} e^{4} - 2 \, {\left (4 \, a c + b\right )} d^{3} e^{3} f + 4 \, {\left (3 \, a c^{2} + b c - a\right )} d^{2} e^{2} f^{2} - 2 \, {\left (4 \, a c^{3} + b c^{2} - 4 \, a c - b\right )} d e f^{3} + 2 \, {\left (a c^{4} - 2 \, a c^{2} + a\right )} f^{4} - 2 \, {\left (b d^{3} e^{2} f^{2} - 2 \, b c d^{2} e f^{3} + {\left (b c^{2} - b\right )} d f^{4}\right )} x - {\left (b d^{4} e^{4} - 2 \, {\left (b c - b\right )} d^{3} e^{3} f + {\left (b c^{2} - 2 \, b c + b\right )} d^{2} e^{2} f^{2} + {\left (b d^{4} e^{2} f^{2} - 2 \, {\left (b c - b\right )} d^{3} e f^{3} + {\left (b c^{2} - 2 \, b c + b\right )} d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{4} e^{3} f - 2 \, {\left (b c - b\right )} d^{3} e^{2} f^{2} + {\left (b c^{2} - 2 \, b c + b\right )} d^{2} e f^{3}\right )} x\right )} \log \left (d x + c + 1\right ) + {\left (b d^{4} e^{4} - 2 \, {\left (b c + b\right )} d^{3} e^{3} f + {\left (b c^{2} + 2 \, b c + b\right )} d^{2} e^{2} f^{2} + {\left (b d^{4} e^{2} f^{2} - 2 \, {\left (b c + b\right )} d^{3} e f^{3} + {\left (b c^{2} + 2 \, b c + b\right )} d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{4} e^{3} f - 2 \, {\left (b c + b\right )} d^{3} e^{2} f^{2} + {\left (b c^{2} + 2 \, b c + b\right )} d^{2} e f^{3}\right )} x\right )} \log \left (d x + c - 1\right ) + 4 \, {\left (b d^{3} e^{3} f - b c d^{2} e^{2} f^{2} + {\left (b d^{3} e f^{3} - b c d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{3} e^{2} f^{2} - b c d^{2} e f^{3}\right )} x\right )} \log \left (f x + e\right ) + {\left (b d^{4} e^{4} - 4 \, b c d^{3} e^{3} f + 2 \, {\left (3 \, b c^{2} - b\right )} d^{2} e^{2} f^{2} - 4 \, {\left (b c^{3} - b c\right )} d e f^{3} + {\left (b c^{4} - 2 \, b c^{2} + b\right )} f^{4}\right )} \log \left (\frac {d x + c + 1}{d x + c - 1}\right )}{4 \, {\left (d^{4} e^{6} f - 4 \, c d^{3} e^{5} f^{2} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} e^{4} f^{3} - 4 \, {\left (c^{3} - c\right )} d e^{3} f^{4} + {\left (c^{4} - 2 \, c^{2} + 1\right )} e^{2} f^{5} + {\left (d^{4} e^{4} f^{3} - 4 \, c d^{3} e^{3} f^{4} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} e^{2} f^{5} - 4 \, {\left (c^{3} - c\right )} d e f^{6} + {\left (c^{4} - 2 \, c^{2} + 1\right )} f^{7}\right )} x^{2} + 2 \, {\left (d^{4} e^{5} f^{2} - 4 \, c d^{3} e^{4} f^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} e^{3} f^{4} - 4 \, {\left (c^{3} - c\right )} d e^{2} f^{5} + {\left (c^{4} - 2 \, c^{2} + 1\right )} e f^{6}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcoth}\left (d x + c\right ) + a}{{\left (f x + e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccoth(d*x + c) + a)/(f*x + e)^3, x)

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maple [A]  time = 0.05, size = 236, normalized size = 1.41 \[ -\frac {d^{2} a}{2 \left (d f x +d e \right )^{2} f}-\frac {d^{2} b \,\mathrm {arccoth}\left (d x +c \right )}{2 \left (d f x +d e \right )^{2} f}+\frac {d^{2} b}{2 \left (c f -d e -f \right ) \left (c f -d e +f \right ) \left (d f x +d e \right )}+\frac {d^{2} b f \ln \left (\left (d x +c \right ) f -c f +d e \right ) c}{\left (c f -d e -f \right )^{2} \left (c f -d e +f \right )^{2}}-\frac {d^{3} b \ln \left (\left (d x +c \right ) f -c f +d e \right ) e}{\left (c f -d e -f \right )^{2} \left (c f -d e +f \right )^{2}}-\frac {d^{2} b \ln \left (d x +c -1\right )}{4 f \left (c f -d e -f \right )^{2}}+\frac {d^{2} b \ln \left (d x +c +1\right )}{4 f \left (c f -d e +f \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 291, normalized size = 1.74 \[ \frac {1}{4} \, {\left (d {\left (\frac {d \log \left (d x + c + 1\right )}{d^{2} e^{2} f - 2 \, {\left (c + 1\right )} d e f^{2} + {\left (c^{2} + 2 \, c + 1\right )} f^{3}} - \frac {d \log \left (d x + c - 1\right )}{d^{2} e^{2} f - 2 \, {\left (c - 1\right )} d e f^{2} + {\left (c^{2} - 2 \, c + 1\right )} f^{3}} - \frac {4 \, {\left (d^{2} e - c d f\right )} \log \left (f x + e\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} e^{2} f^{2} - 4 \, {\left (c^{3} - c\right )} d e f^{3} + {\left (c^{4} - 2 \, c^{2} + 1\right )} f^{4}} + \frac {2}{d^{2} e^{3} - 2 \, c d e^{2} f + {\left (c^{2} - 1\right )} e f^{2} + {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + {\left (c^{2} - 1\right )} f^{3}\right )} x}\right )} - \frac {2 \, \operatorname {arcoth}\left (d x + c\right )}{f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f}\right )} b - \frac {a}{2 \, {\left (f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.30, size = 422, normalized size = 2.53 \[ \frac {b\,\ln \left (1-\frac {1}{c+d\,x}\right )}{2\,f\,\left (2\,e^2+4\,e\,f\,x+2\,f^2\,x^2\right )}-\frac {\ln \left (e+f\,x\right )\,\left (b\,d^3\,e-b\,c\,d^2\,f\right )}{c^4\,f^4-4\,c^3\,d\,e\,f^3+6\,c^2\,d^2\,e^2\,f^2-2\,c^2\,f^4-4\,c\,d^3\,e^3\,f+4\,c\,d\,e\,f^3+d^4\,e^4-2\,d^2\,e^2\,f^2+f^4}-\frac {\frac {-a\,c^2\,f^2+2\,a\,c\,d\,e\,f-a\,d^2\,e^2+b\,d\,e\,f+a\,f^2}{-c^2\,f^2+2\,c\,d\,e\,f-d^2\,e^2+f^2}+\frac {b\,d\,f^2\,x}{-c^2\,f^2+2\,c\,d\,e\,f-d^2\,e^2+f^2}}{2\,e^2\,f+4\,e\,f^2\,x+2\,f^3\,x^2}-\frac {b\,d^2\,\ln \left (c+d\,x-1\right )}{4\,c^2\,f^3-8\,c\,d\,e\,f^2-8\,c\,f^3+4\,d^2\,e^2\,f+8\,d\,e\,f^2+4\,f^3}+\frac {b\,d^2\,\ln \left (c+d\,x+1\right )}{4\,c^2\,f^3-8\,c\,d\,e\,f^2+8\,c\,f^3+4\,d^2\,e^2\,f-8\,d\,e\,f^2+4\,f^3}-\frac {b\,\ln \left (\frac {1}{c+d\,x}+1\right )}{4\,f\,\left (e^2+2\,e\,f\,x+f^2\,x^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*log(1 - 1/(c + d*x)))/(2*f*(2*e^2 + 2*f^2*x^2 + 4*e*f*x)) - (log(e + f*x)*(b*d^3*e - b*c*d^2*f))/(f^4 - 2*c
^2*f^4 + c^4*f^4 + d^4*e^4 - 2*d^2*e^2*f^2 + 4*c*d*e*f^3 + 6*c^2*d^2*e^2*f^2 - 4*c*d^3*e^3*f - 4*c^3*d*e*f^3)
- ((a*f^2 - a*c^2*f^2 - a*d^2*e^2 + b*d*e*f + 2*a*c*d*e*f)/(f^2 - c^2*f^2 - d^2*e^2 + 2*c*d*e*f) + (b*d*f^2*x)
/(f^2 - c^2*f^2 - d^2*e^2 + 2*c*d*e*f))/(2*e^2*f + 2*f^3*x^2 + 4*e*f^2*x) - (b*d^2*log(c + d*x - 1))/(4*f^3 -
8*c*f^3 + 4*c^2*f^3 + 4*d^2*e^2*f + 8*d*e*f^2 - 8*c*d*e*f^2) + (b*d^2*log(c + d*x + 1))/(8*c*f^3 + 4*f^3 + 4*c
^2*f^3 + 4*d^2*e^2*f - 8*d*e*f^2 - 8*c*d*e*f^2) - (b*log(1/(c + d*x) + 1))/(4*f*(e^2 + f^2*x^2 + 2*e*f*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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