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

Optimal. Leaf size=120 \[ \frac{(e+f x)^3 \left (a+b \coth ^{-1}(c+d x)\right )}{3 f}+\frac{b f x (d e-c f)}{d^2}+\frac{b (-c f+d e+f)^3 \log (-c-d x+1)}{6 d^3 f}-\frac{b (d e-(c+1) f)^3 \log (c+d x+1)}{6 d^3 f}+\frac{b f^2 (c+d x)^2}{6 d^3} \]

[Out]

(b*f*(d*e - c*f)*x)/d^2 + (b*f^2*(c + d*x)^2)/(6*d^3) + ((e + f*x)^3*(a + b*ArcCoth[c + d*x]))/(3*f) + (b*(d*e
+ f - c*f)^3*Log[1 - c - d*x])/(6*d^3*f) - (b*(d*e - (1 + c)*f)^3*Log[1 + c + d*x])/(6*d^3*f)

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Rubi [A]  time = 0.203979, antiderivative size = 120, 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 = {6112, 5927, 702, 633, 31} \[ \frac{(e+f x)^3 \left (a+b \coth ^{-1}(c+d x)\right )}{3 f}+\frac{b f x (d e-c f)}{d^2}+\frac{b (-c f+d e+f)^3 \log (-c-d x+1)}{6 d^3 f}-\frac{b (d e-(c+1) f)^3 \log (c+d x+1)}{6 d^3 f}+\frac{b f^2 (c+d x)^2}{6 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*f*(d*e - c*f)*x)/d^2 + (b*f^2*(c + d*x)^2)/(6*d^3) + ((e + f*x)^3*(a + b*ArcCoth[c + d*x]))/(3*f) + (b*(d*e
+ f - c*f)^3*Log[1 - c - d*x])/(6*d^3*f) - (b*(d*e - (1 + c)*f)^3*Log[1 + c + d*x])/(6*d^3*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)^2 \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 )^2 \left (a+b \coth ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{(e+f x)^3 \left (a+b \coth ^{-1}(c+d x)\right )}{3 f}-\frac{b \operatorname{Subst}\left (\int \frac{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^3}{1-x^2} \, dx,x,c+d x\right )}{3 f}\\ &=\frac{(e+f x)^3 \left (a+b \coth ^{-1}(c+d x)\right )}{3 f}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{3 f^2 (d e-c f)}{d^3}-\frac{f^3 x}{d^3}+\frac{(d e-c f) \left (d^2 e^2-2 c d e f+3 f^2+c^2 f^2\right )+f \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) x}{d^3 \left (1-x^2\right )}\right ) \, dx,x,c+d x\right )}{3 f}\\ &=\frac{b f (d e-c f) x}{d^2}+\frac{b f^2 (c+d x)^2}{6 d^3}+\frac{(e+f x)^3 \left (a+b \coth ^{-1}(c+d x)\right )}{3 f}-\frac{b \operatorname{Subst}\left (\int \frac{(d e-c f) \left (d^2 e^2-2 c d e f+3 f^2+c^2 f^2\right )+f \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) x}{1-x^2} \, dx,x,c+d x\right )}{3 d^3 f}\\ &=\frac{b f (d e-c f) x}{d^2}+\frac{b f^2 (c+d x)^2}{6 d^3}+\frac{(e+f x)^3 \left (a+b \coth ^{-1}(c+d x)\right )}{3 f}-\frac{\left (b (d e+f-c f)^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,c+d x\right )}{6 d^3 f}+\frac{\left (b (d e-(1+c) f)^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,c+d x\right )}{6 d^3 f}\\ &=\frac{b f (d e-c f) x}{d^2}+\frac{b f^2 (c+d x)^2}{6 d^3}+\frac{(e+f x)^3 \left (a+b \coth ^{-1}(c+d x)\right )}{3 f}+\frac{b (d e+f-c f)^3 \log (1-c-d x)}{6 d^3 f}-\frac{b (d e-(1+c) f)^3 \log (1+c+d x)}{6 d^3 f}\\ \end{align*}

Mathematica [A]  time = 0.161644, size = 174, normalized size = 1.45 \[ \frac{2 d x \left (3 a d^2 e^2+b f (3 d e-2 c f)\right )+d^2 f x^2 (6 a d e+b f)+2 a d^3 f^2 x^3+2 b d^3 x \left (3 e^2+3 e f x+f^2 x^2\right ) \coth ^{-1}(c+d x)-b (c-1) \left (-3 (c-1) d e f+(c-1)^2 f^2+3 d^2 e^2\right ) \log (-c-d x+1)+b (c+1) \left (-3 (c+1) d e f+(c+1)^2 f^2+3 d^2 e^2\right ) \log (c+d x+1)}{6 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(3*a*d^2*e^2 + b*f*(3*d*e - 2*c*f))*x + d^2*f*(6*a*d*e + b*f)*x^2 + 2*a*d^3*f^2*x^3 + 2*b*d^3*x*(3*e^2 +
3*e*f*x + f^2*x^2)*ArcCoth[c + d*x] - b*(-1 + c)*(3*d^2*e^2 - 3*(-1 + c)*d*e*f + (-1 + c)^2*f^2)*Log[1 - c - d
*x] + b*(1 + c)*(3*d^2*e^2 - 3*(1 + c)*d*e*f + (1 + c)^2*f^2)*Log[1 + c + d*x])/(6*d^3)

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Maple [B]  time = 0.04, size = 477, normalized size = 4. \begin{align*}{\frac{bf\ln \left ( dx+c-1 \right ){c}^{2}e}{2\,{d}^{2}}}+{\frac{a{f}^{2}{x}^{3}}{3}}+ax{e}^{2}-{\frac{2\,b{f}^{2}cx}{3\,{d}^{2}}}+{\frac{b{f}^{2}\ln \left ( dx+c+1 \right ) }{6\,{d}^{3}}}+{\frac{b\ln \left ( dx+c-1 \right ){e}^{2}}{2\,d}}+af{x}^{2}e-{\frac{b\ln \left ( dx+c+1 \right ){e}^{3}}{6\,f}}+{\frac{b\ln \left ( dx+c-1 \right ){e}^{3}}{6\,f}}+{\frac{b{\rm arccoth} \left (dx+c\right ){e}^{3}}{3\,f}}+{\rm arccoth} \left (dx+c\right )xb{e}^{2}+{\frac{b{f}^{2}{\rm arccoth} \left (dx+c\right ){x}^{3}}{3}}+{\frac{b\ln \left ( dx+c+1 \right ){e}^{2}}{2\,d}}+{\frac{b{f}^{2}\ln \left ( dx+c-1 \right ) }{6\,{d}^{3}}}+{\frac{b{f}^{2}{x}^{2}}{6\,d}}-{\frac{bf\ln \left ( dx+c+1 \right ){c}^{2}e}{2\,{d}^{2}}}-{\frac{bf\ln \left ( dx+c+1 \right ) ce}{{d}^{2}}}-{\frac{bf\ln \left ( dx+c-1 \right ) ce}{{d}^{2}}}+{\frac{a{e}^{3}}{3\,f}}-{\frac{5\,b{f}^{2}{c}^{2}}{6\,{d}^{3}}}+{\frac{bfce}{{d}^{2}}}+{\frac{bfex}{d}}+{\frac{b{f}^{2}\ln \left ( dx+c+1 \right ){c}^{3}}{6\,{d}^{3}}}+{\frac{b{f}^{2}\ln \left ( dx+c+1 \right ){c}^{2}}{2\,{d}^{3}}}-{\frac{b{f}^{2}\ln \left ( dx+c-1 \right ) c}{2\,{d}^{3}}}-{\frac{b{f}^{2}\ln \left ( dx+c-1 \right ){c}^{3}}{6\,{d}^{3}}}+{\frac{b{f}^{2}\ln \left ( dx+c+1 \right ) c}{2\,{d}^{3}}}-{\frac{bf\ln \left ( dx+c+1 \right ) e}{2\,{d}^{2}}}+{\frac{bf\ln \left ( dx+c-1 \right ) e}{2\,{d}^{2}}}+{\frac{b{f}^{2}\ln \left ( dx+c-1 \right ){c}^{2}}{2\,{d}^{3}}}+bf{\rm arccoth} \left (dx+c\right )e{x}^{2}-{\frac{b\ln \left ( dx+c-1 \right ) c{e}^{2}}{2\,d}}+{\frac{b\ln \left ( dx+c+1 \right ) c{e}^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d^2*b*f*ln(d*x+c-1)*c^2*e+1/3*a*f^2*x^3+a*x*e^2-2/3*b/d^2*f^2*c*x+1/6/d^3*b*f^2*ln(d*x+c+1)+1/2/d*b*ln(d*x
+c-1)*e^2+a*f*x^2*e-1/6*b/f*ln(d*x+c+1)*e^3+1/6*b/f*ln(d*x+c-1)*e^3+1/3*b/f*arccoth(d*x+c)*e^3+arccoth(d*x+c)*
x*b*e^2+1/3*b*f^2*arccoth(d*x+c)*x^3+1/2/d*b*ln(d*x+c+1)*e^2+1/6/d^3*b*f^2*ln(d*x+c-1)+1/6/d*b*f^2*x^2-1/2/d^2
*b*f*ln(d*x+c+1)*c^2*e-1/d^2*b*f*ln(d*x+c+1)*c*e-1/d^2*b*f*ln(d*x+c-1)*c*e+1/3*a/f*e^3-5/6/d^3*b*f^2*c^2+1/d^2
*b*f*c*e+b/d*f*e*x+1/6/d^3*b*f^2*ln(d*x+c+1)*c^3+1/2/d^3*b*f^2*ln(d*x+c+1)*c^2-1/2/d^3*b*f^2*ln(d*x+c-1)*c-1/6
/d^3*b*f^2*ln(d*x+c-1)*c^3+1/2/d^3*b*f^2*ln(d*x+c+1)*c-1/2/d^2*b*f*ln(d*x+c+1)*e+1/2/d^2*b*f*ln(d*x+c-1)*e+1/2
/d^3*b*f^2*ln(d*x+c-1)*c^2+b*f*arccoth(d*x+c)*e*x^2-1/2/d*b*ln(d*x+c-1)*c*e^2+1/2/d*b*ln(d*x+c+1)*c*e^2

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Maxima [A]  time = 0.970765, size = 279, normalized size = 2.32 \begin{align*} \frac{1}{3} \, a f^{2} x^{3} + a e f x^{2} + \frac{1}{2} \,{\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 e f + \frac{1}{6} \,{\left (2 \, x^{3} \operatorname{arcoth}\left (d x + c\right ) + d{\left (\frac{d x^{2} - 4 \, c x}{d^{3}} + \frac{{\left (c^{3} + 3 \, c^{2} + 3 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{4}} - \frac{{\left (c^{3} - 3 \, c^{2} + 3 \, c - 1\right )} \log \left (d x + c - 1\right )}{d^{4}}\right )}\right )} b f^{2} + a e^{2} 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}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.96481, size = 548, normalized size = 4.57 \begin{align*} \frac{2 \, a d^{3} f^{2} x^{3} +{\left (6 \, a d^{3} e f + b d^{2} f^{2}\right )} x^{2} + 2 \,{\left (3 \, a d^{3} e^{2} + 3 \, b d^{2} e f - 2 \, b c d f^{2}\right )} x +{\left (3 \,{\left (b c + b\right )} d^{2} e^{2} - 3 \,{\left (b c^{2} + 2 \, b c + b\right )} d e f +{\left (b c^{3} + 3 \, b c^{2} + 3 \, b c + b\right )} f^{2}\right )} \log \left (d x + c + 1\right ) -{\left (3 \,{\left (b c - b\right )} d^{2} e^{2} - 3 \,{\left (b c^{2} - 2 \, b c + b\right )} d e f +{\left (b c^{3} - 3 \, b c^{2} + 3 \, b c - b\right )} f^{2}\right )} \log \left (d x + c - 1\right ) +{\left (b d^{3} f^{2} x^{3} + 3 \, b d^{3} e f x^{2} + 3 \, b d^{3} e^{2} x\right )} \log \left (\frac{d x + c + 1}{d x + c - 1}\right )}{6 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 18.1172, size = 369, normalized size = 3.08 \begin{align*} \begin{cases} a e^{2} x + a e f x^{2} + \frac{a f^{2} x^{3}}{3} + \frac{b c^{3} f^{2} \operatorname{acoth}{\left (c + d x \right )}}{3 d^{3}} - \frac{b c^{2} e f \operatorname{acoth}{\left (c + d x \right )}}{d^{2}} + \frac{b c^{2} f^{2} \log{\left (\frac{c}{d} + x + \frac{1}{d} \right )}}{d^{3}} - \frac{b c^{2} f^{2} \operatorname{acoth}{\left (c + d x \right )}}{d^{3}} + \frac{b c e^{2} \operatorname{acoth}{\left (c + d x \right )}}{d} - \frac{2 b c e f \log{\left (\frac{c}{d} + x + \frac{1}{d} \right )}}{d^{2}} + \frac{2 b c e f \operatorname{acoth}{\left (c + d x \right )}}{d^{2}} - \frac{2 b c f^{2} x}{3 d^{2}} + \frac{b c f^{2} \operatorname{acoth}{\left (c + d x \right )}}{d^{3}} + b e^{2} x \operatorname{acoth}{\left (c + d x \right )} + b e f x^{2} \operatorname{acoth}{\left (c + d x \right )} + \frac{b f^{2} x^{3} \operatorname{acoth}{\left (c + d x \right )}}{3} + \frac{b e^{2} \log{\left (\frac{c}{d} + x + \frac{1}{d} \right )}}{d} - \frac{b e^{2} \operatorname{acoth}{\left (c + d x \right )}}{d} + \frac{b e f x}{d} + \frac{b f^{2} x^{2}}{6 d} - \frac{b e f \operatorname{acoth}{\left (c + d x \right )}}{d^{2}} + \frac{b f^{2} \log{\left (\frac{c}{d} + x + \frac{1}{d} \right )}}{3 d^{3}} - \frac{b f^{2} \operatorname{acoth}{\left (c + d x \right )}}{3 d^{3}} & \text{for}\: d \neq 0 \\\left (a + b \operatorname{acoth}{\left (c \right )}\right ) \left (e^{2} x + e f x^{2} + \frac{f^{2} x^{3}}{3}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*e**2*x + a*e*f*x**2 + a*f**2*x**3/3 + b*c**3*f**2*acoth(c + d*x)/(3*d**3) - b*c**2*e*f*acoth(c +
d*x)/d**2 + b*c**2*f**2*log(c/d + x + 1/d)/d**3 - b*c**2*f**2*acoth(c + d*x)/d**3 + b*c*e**2*acoth(c + d*x)/d
- 2*b*c*e*f*log(c/d + x + 1/d)/d**2 + 2*b*c*e*f*acoth(c + d*x)/d**2 - 2*b*c*f**2*x/(3*d**2) + b*c*f**2*acoth(c
+ d*x)/d**3 + b*e**2*x*acoth(c + d*x) + b*e*f*x**2*acoth(c + d*x) + b*f**2*x**3*acoth(c + d*x)/3 + b*e**2*log
(c/d + x + 1/d)/d - b*e**2*acoth(c + d*x)/d + b*e*f*x/d + b*f**2*x**2/(6*d) - b*e*f*acoth(c + d*x)/d**2 + b*f*
*2*log(c/d + x + 1/d)/(3*d**3) - b*f**2*acoth(c + d*x)/(3*d**3), Ne(d, 0)), ((a + b*acoth(c))*(e**2*x + e*f*x*
*2 + f**2*x**3/3), True))

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

[Out]

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