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

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

[Out]

-((a + b*ArcCoth[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))

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Rubi [A]  time = 0.168724, antiderivative size = 115, normalized size of antiderivative = 1., 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 = {6110, 1982, 705, 31, 632} \[ -\frac{a+b \coth ^{-1}(c+d x)}{f (e+f x)}-\frac{b d \log (-c-d x+1)}{2 f (-c f+d e+f)}+\frac{b d \log (c+d x+1)}{2 f (-c f+d e-f)}-\frac{b d \log (e+f x)}{(-c f+d e+f) (d e-(c+1) f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcCoth[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 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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{a+b \coth ^{-1}(c+d x)}{(e+f x)^2} \, dx &=-\frac{a+b \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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*}

Mathematica [A]  time = 0.200933, size = 125, normalized size = 1.09 \[ \frac{1}{2} \left (-\frac{2 a}{f (e+f x)}-\frac{2 b d \log (e+f x)}{\left (c^2-1\right ) f^2-2 c d e f+d^2 e^2}+\frac{b d \log (-c-d x+1)}{f ((c-1) f-d e)}-\frac{b d \log (c+d x+1)}{f (c f-d e+f)}-\frac{2 b \coth ^{-1}(c+d x)}{f (e+f x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a)/(f*(e + f*x)) - (2*b*ArcCoth[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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Maple [A]  time = 0.043, size = 141, normalized size = 1.2 \begin{align*} -{\frac{ad}{ \left ( dfx+de \right ) f}}-{\frac{bd{\rm arccoth} \left (dx+c\right )}{ \left ( dfx+de \right ) f}}+{\frac{bd\ln \left ( dx+c-1 \right ) }{f \left ( 2\,cf-2\,de-2\,f \right ) }}-{\frac{bd\ln \left ( dx+c+1 \right ) }{f \left ( 2\,cf-2\,de+2\,f \right ) }}-{\frac{bd\ln \left ( \left ( dx+c \right ) f-cf+de \right ) }{ \left ( cf-de-f \right ) \left ( cf-de+f \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccoth(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*arccoth(d*x+c)+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)-d*b/(c*f-d*e-f)/(c*f-d*e+f)*ln((d*x+c)*f-c*f+d*e)

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Maxima [A]  time = 0.996587, size = 163, normalized size = 1.42 \begin{align*} \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{arcoth}\left (d x + c\right )}{f^{2} x + e f}\right )} b - \frac{a}{f^{2} x + e f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [B]  time = 3.11053, size = 583, normalized size = 5.07 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcoth}\left (d x + c\right ) + a}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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