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]

(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)

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Rubi [A]  time = 0.234421, antiderivative size = 167, normalized size of antiderivative = 1., 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 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 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]

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*}

Mathematica [A]  time = 0.346272, 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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Maple [A]  time = 0.052, size = 236, normalized size = 1.4 \begin{align*} -{\frac{a{d}^{2}}{2\, \left ( dfx+de \right ) ^{2}f}}-{\frac{b{d}^{2}{\rm arccoth} \left (dx+c\right )}{2\, \left ( dfx+de \right ) ^{2}f}}-{\frac{b{d}^{2}\ln \left ( dx+c-1 \right ) }{4\,f \left ( cf-de-f \right ) ^{2}}}+{\frac{b{d}^{2}\ln \left ( dx+c+1 \right ) }{4\,f \left ( cf-de+f \right ) ^{2}}}+{\frac{b{d}^{2}}{ \left ( 2\,cf-2\,de-2\,f \right ) \left ( cf-de+f \right ) \left ( dfx+de \right ) }}+{\frac{b{d}^{2}f\ln \left ( \left ( dx+c \right ) f-cf+de \right ) c}{ \left ( cf-de-f \right ) ^{2} \left ( cf-de+f \right ) ^{2}}}-{\frac{b{d}^{3}\ln \left ( \left ( dx+c \right ) f-cf+de \right ) e}{ \left ( cf-de-f \right ) ^{2} \left ( cf-de+f \right ) ^{2}}} \end{align*}

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)-1/4*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)+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

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Maxima [A]  time = 1.04528, size = 393, normalized size = 2.35 \begin{align*} \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 )}} \end{align*}

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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Fricas [B]  time = 9.18629, size = 1766, normalized size = 10.57 \begin{align*} \text{result too large to display} \end{align*}

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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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)**3,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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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