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

Optimal. Leaf size=97 \[ \frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}+\frac{b (-c f+d e+f)^2 \log (-c-d x+1)}{4 d^2 f}-\frac{b (d e-(c+1) f)^2 \log (c+d x+1)}{4 d^2 f}+\frac{b f x}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.171195, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {6112, 5927, 702, 633, 31} \[ \frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}+\frac{b (-c f+d e+f)^2 \log (-c-d x+1)}{4 d^2 f}-\frac{b (d e-(c+1) f)^2 \log (c+d x+1)}{4 d^2 f}+\frac{b f x}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6112

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

Rule 5927

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcCoth[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]

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 633

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

Rule 31

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

Rubi steps

\begin{align*} \int (e+f x) \left (a+b \coth ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \left (\frac{d e-c f}{d}+\frac{f x}{d}\right ) \left (a+b \coth ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac{b \operatorname{Subst}\left (\int \frac{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^2}{1-x^2} \, dx,x,c+d x\right )}{2 f}\\ &=\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{f^2}{d^2}+\frac{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x}{d^2 \left (1-x^2\right )}\right ) \, dx,x,c+d x\right )}{2 f}\\ &=\frac{b f x}{2 d}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac{b \operatorname{Subst}\left (\int \frac{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x}{1-x^2} \, dx,x,c+d x\right )}{2 d^2 f}\\ &=\frac{b f x}{2 d}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac{\left (b (d e+f-c f)^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,c+d x\right )}{4 d^2 f}+\frac{\left (b (d e-(1+c) f)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,c+d x\right )}{4 d^2 f}\\ &=\frac{b f x}{2 d}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}+\frac{b (d e+f-c f)^2 \log (1-c-d x)}{4 d^2 f}-\frac{b (d e-(1+c) f)^2 \log (1+c+d x)}{4 d^2 f}\\ \end{align*}

Mathematica [A]  time = 0.0427247, size = 138, normalized size = 1.42 \[ a e x+\frac{1}{2} a f x^2+\frac{b \left (c^2-2 c+1\right ) f \log (-c-d x+1)}{4 d^2}+\frac{b \left (-c^2-2 c-1\right ) f \log (c+d x+1)}{4 d^2}+\frac{b e ((c+1) \log (c+d x+1)-(c-1) \log (-c-d x+1))}{2 d}+b e x \coth ^{-1}(c+d x)+\frac{1}{2} b f x^2 \coth ^{-1}(c+d x)+\frac{b f x}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.04, size = 184, normalized size = 1.9 \begin{align*}{\frac{a{x}^{2}f}{2}}-{\frac{a{c}^{2}f}{2\,{d}^{2}}}+axe+{\frac{ace}{d}}+{\frac{b{\rm arccoth} \left (dx+c\right )f{x}^{2}}{2}}-{\frac{{\rm arccoth} \left (dx+c\right )b{c}^{2}f}{2\,{d}^{2}}}+{\rm arccoth} \left (dx+c\right )xbe+{\frac{b{\rm arccoth} \left (dx+c\right )ce}{d}}+{\frac{bfx}{2\,d}}+{\frac{bcf}{2\,{d}^{2}}}-{\frac{b\ln \left ( dx+c-1 \right ) cf}{2\,{d}^{2}}}+{\frac{b\ln \left ( dx+c-1 \right ) e}{2\,d}}+{\frac{b\ln \left ( dx+c-1 \right ) f}{4\,{d}^{2}}}-{\frac{b\ln \left ( dx+c+1 \right ) cf}{2\,{d}^{2}}}+{\frac{b\ln \left ( dx+c+1 \right ) e}{2\,d}}-{\frac{b\ln \left ( dx+c+1 \right ) f}{4\,{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.960865, size = 147, normalized size = 1.52 \begin{align*} \frac{1}{2} \, a f x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \operatorname{arcoth}\left (d x + c\right ) + d{\left (\frac{2 \, x}{d^{2}} - \frac{{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac{{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} b f + a e x + \frac{{\left (2 \,{\left (d x + c\right )} \operatorname{arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b e}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.8918, size = 319, normalized size = 3.29 \begin{align*} \frac{2 \, a d^{2} f x^{2} + 2 \,{\left (2 \, a d^{2} e + b d f\right )} x +{\left (2 \,{\left (b c + b\right )} d e -{\left (b c^{2} + 2 \, b c + b\right )} f\right )} \log \left (d x + c + 1\right ) -{\left (2 \,{\left (b c - b\right )} d e -{\left (b c^{2} - 2 \, b c + b\right )} f\right )} \log \left (d x + c - 1\right ) +{\left (b d^{2} f x^{2} + 2 \, b d^{2} e x\right )} \log \left (\frac{d x + c + 1}{d x + c - 1}\right )}{4 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.6259, size = 173, normalized size = 1.78 \begin{align*} \begin{cases} a e x + \frac{a f x^{2}}{2} - \frac{b c^{2} f \operatorname{acoth}{\left (c + d x \right )}}{2 d^{2}} + \frac{b c e \operatorname{acoth}{\left (c + d x \right )}}{d} - \frac{b c f \log{\left (\frac{c}{d} + x + \frac{1}{d} \right )}}{d^{2}} + \frac{b c f \operatorname{acoth}{\left (c + d x \right )}}{d^{2}} + b e x \operatorname{acoth}{\left (c + d x \right )} + \frac{b f x^{2} \operatorname{acoth}{\left (c + d x \right )}}{2} + \frac{b e \log{\left (\frac{c}{d} + x + \frac{1}{d} \right )}}{d} - \frac{b e \operatorname{acoth}{\left (c + d x \right )}}{d} + \frac{b f x}{2 d} - \frac{b f \operatorname{acoth}{\left (c + d x \right )}}{2 d^{2}} & \text{for}\: d \neq 0 \\\left (a + b \operatorname{acoth}{\left (c \right )}\right ) \left (e x + \frac{f x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*e*x + a*f*x**2/2 - b*c**2*f*acoth(c + d*x)/(2*d**2) + b*c*e*acoth(c + d*x)/d - b*c*f*log(c/d + x
+ 1/d)/d**2 + b*c*f*acoth(c + d*x)/d**2 + b*e*x*acoth(c + d*x) + b*f*x**2*acoth(c + d*x)/2 + b*e*log(c/d + x +
1/d)/d - b*e*acoth(c + d*x)/d + b*f*x/(2*d) - b*f*acoth(c + d*x)/(2*d**2), Ne(d, 0)), ((a + b*acoth(c))*(e*x
+ f*x**2/2), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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