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

Optimal. Leaf size=221 \[ -\frac{b^2 (d e-c f) \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right )}{d^2}-\frac{\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d^2}-\frac{2 b (d e-c f) \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d^2}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}+\frac{a b f x}{d}+\frac{b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}+\frac{b^2 f (c+d x) \coth ^{-1}(c+d x)}{d^2} \]

[Out]

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

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Rubi [A]  time = 0.442664, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {6112, 5929, 5911, 260, 6049, 5949, 5985, 5919, 2402, 2315} \[ -\frac{b^2 (d e-c f) \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right )}{d^2}+\frac{\left (-\frac{\left (c^2+1\right ) f}{d}+2 c e-\frac{d e^2}{f}\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d^2}-\frac{2 b (d e-c f) \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d^2}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}+\frac{a b f x}{d}+\frac{b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}+\frac{b^2 f (c+d x) \coth ^{-1}(c+d x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5929

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(
a + b*ArcCoth[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcCoth[c*x])^(p
- 1), (d + e*x)^(q + 1)/(1 - c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] &
& NeQ[q, -1]

Rule 5911

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcCoth[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

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

Rule 6049

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcCoth[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x]
&& IGtQ[p, 0] && EqQ[c^2*d + e, 0] && IGtQ[m, 0]

Rule 5949

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 5985

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcCoth[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 5919

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcCoth[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

\begin{align*} \int (e+f x) \left (a+b \coth ^{-1}(c+d x)\right )^2 \, 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 )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{f^2 \left (a+b \coth ^{-1}(x)\right )}{d^2}+\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x\right ) \left (a+b \coth ^{-1}(x)\right )}{d^2 \left (1-x^2\right )}\right ) \, dx,x,c+d x\right )}{f}\\ &=\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}-\frac{b \operatorname{Subst}\left (\int \frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x\right ) \left (a+b \coth ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2 f}+\frac{(b f) \operatorname{Subst}\left (\int \left (a+b \coth ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d^2}\\ &=\frac{a b f x}{d}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}-\frac{b \operatorname{Subst}\left (\int \left (\frac{d^2 e^2 \left (1+\frac{f \left (-2 c d e+f+c^2 f\right )}{d^2 e^2}\right ) \left (a+b \coth ^{-1}(x)\right )}{1-x^2}-\frac{2 f (-d e+c f) x \left (a+b \coth ^{-1}(x)\right )}{1-x^2}\right ) \, dx,x,c+d x\right )}{d^2 f}+\frac{\left (b^2 f\right ) \operatorname{Subst}\left (\int \coth ^{-1}(x) \, dx,x,c+d x\right )}{d^2}\\ &=\frac{a b f x}{d}+\frac{b^2 f (c+d x) \coth ^{-1}(c+d x)}{d^2}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}-\frac{\left (b^2 f\right ) \operatorname{Subst}\left (\int \frac{x}{1-x^2} \, dx,x,c+d x\right )}{d^2}-\frac{(2 b (d e-c f)) \operatorname{Subst}\left (\int \frac{x \left (a+b \coth ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}-\frac{\left (b \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )\right ) \operatorname{Subst}\left (\int \frac{a+b \coth ^{-1}(x)}{1-x^2} \, dx,x,c+d x\right )}{d^2 f}\\ &=\frac{a b f x}{d}+\frac{b^2 f (c+d x) \coth ^{-1}(c+d x)}{d^2}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d^2}-\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}+\frac{b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}-\frac{(2 b (d e-c f)) \operatorname{Subst}\left (\int \frac{a+b \coth ^{-1}(x)}{1-x} \, dx,x,c+d x\right )}{d^2}\\ &=\frac{a b f x}{d}+\frac{b^2 f (c+d x) \coth ^{-1}(c+d x)}{d^2}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d^2}-\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}-\frac{2 b (d e-c f) \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac{2}{1-c-d x}\right )}{d^2}+\frac{b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}+\frac{\left (2 b^2 (d e-c f)\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}\\ &=\frac{a b f x}{d}+\frac{b^2 f (c+d x) \coth ^{-1}(c+d x)}{d^2}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d^2}-\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}-\frac{2 b (d e-c f) \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac{2}{1-c-d x}\right )}{d^2}+\frac{b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}-\frac{\left (2 b^2 (d e-c f)\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c-d x}\right )}{d^2}\\ &=\frac{a b f x}{d}+\frac{b^2 f (c+d x) \coth ^{-1}(c+d x)}{d^2}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d^2}-\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}-\frac{2 b (d e-c f) \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac{2}{1-c-d x}\right )}{d^2}+\frac{b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}-\frac{b^2 (d e-c f) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.593743, size = 295, normalized size = 1.33 \[ \frac{2 b^2 (d e-c f) \text{PolyLog}\left (2,e^{-2 \coth ^{-1}(c+d x)}\right )-a^2 c^2 f+2 a^2 c d e+2 a^2 d^2 e x+a^2 d^2 f x^2+2 b \coth ^{-1}(c+d x) \left (-(c+d x) (a c f-a d (2 e+f x)-b f)-2 b (d e-c f) \log \left (1-e^{-2 \coth ^{-1}(c+d x)}\right )\right )-4 a b d e \log \left (\frac{1}{(c+d x) \sqrt{1-\frac{1}{(c+d x)^2}}}\right )+a b f \log (-c-d x+1)-a b f \log (c+d x+1)+4 a b c f \log \left (\frac{1}{(c+d x) \sqrt{1-\frac{1}{(c+d x)^2}}}\right )+2 a b c f+2 a b d f x+b^2 (c+d x-1) \coth ^{-1}(c+d x)^2 (-c f+2 d e+d f x+f)-2 b^2 f \log \left (\frac{1}{(c+d x) \sqrt{1-\frac{1}{(c+d x)^2}}}\right )}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.062, size = 857, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d^2*b^2*ln(1/2+1/2*d*x+1/2*c)*ln(d*x+c-1)*f+1/d*b^2*arccoth(d*x+c)*f*x+1/d^2*b^2*arccoth(d*x+c)*f*c-1/2/d
^2*b^2*arccoth(d*x+c)^2*f*c^2+1/2/d^2*b^2*arccoth(d*x+c)*ln(d*x+c-1)*f+1/d^2*a*b*c*f+1/2/d^2*b^2*ln(-1/2*d*x-1
/2*c+1/2)*ln(1/2+1/2*d*x+1/2*c)*c*f-1/d^2*b^2*arccoth(d*x+c)*ln(d*x+c+1)*c*f+1/2*a^2*x^2*f+a^2*x*e-1/d*b^2*dil
og(1/2+1/2*d*x+1/2*c)*e-1/4/d*b^2*ln(d*x+c+1)^2*e+1/2/d^2*b^2*ln(d*x+c+1)*f+1/2/d^2*b^2*ln(d*x+c-1)*f+1/8/d^2*
b^2*ln(d*x+c+1)^2*f+1/8/d^2*b^2*ln(d*x+c-1)^2*f+1/4/d*b^2*ln(d*x+c-1)^2*e-1/2/d^2*b^2*ln(-1/2*d*x-1/2*c+1/2)*l
n(d*x+c+1)*c*f-1/d^2*a*b*arccoth(d*x+c)*c^2*f+2/d*arccoth(d*x+c)*a*b*c*e-1/d^2*a*b*ln(d*x+c-1)*c*f-1/d^2*a*b*l
n(d*x+c+1)*c*f-1/d^2*b^2*arccoth(d*x+c)*ln(d*x+c-1)*c*f+1/2/d^2*b^2*ln(1/2+1/2*d*x+1/2*c)*ln(d*x+c-1)*c*f-1/2/
d^2*a^2*c^2*f+1/d*a^2*c*e+1/2*b^2*arccoth(d*x+c)^2*f*x^2+arccoth(d*x+c)^2*x*b^2*e+1/4/d^2*b^2*ln(-1/2*d*x-1/2*
c+1/2)*ln(1/2+1/2*d*x+1/2*c)*f-1/4/d^2*b^2*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)*f+a*b*arccoth(d*x+c)*f*x^2+2*arc
coth(d*x+c)*x*a*b*e-1/2/d*b^2*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*x+1/2*c)*e+1/d*b^2*arccoth(d*x+c)*ln(d*x+c-1
)*e+1/2/d*b^2*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)*e-1/2/d*b^2*ln(1/2+1/2*d*x+1/2*c)*ln(d*x+c-1)*e+1/d*b^2*arcco
th(d*x+c)*ln(d*x+c+1)*e+1/d*arccoth(d*x+c)^2*b^2*c*e+1/d*a*b*ln(d*x+c-1)*e+1/d*a*b*ln(d*x+c+1)*e-1/2/d^2*b^2*a
rccoth(d*x+c)*ln(d*x+c+1)*f-1/4/d^2*b^2*ln(d*x+c-1)^2*c*f+1/d^2*b^2*dilog(1/2+1/2*d*x+1/2*c)*c*f-1/2/d^2*a*b*l
n(d*x+c+1)*f+1/4/d^2*b^2*ln(d*x+c+1)^2*c*f+1/2/d^2*a*b*ln(d*x+c-1)*f+a*b*f*x/d

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Maxima [A]  time = 1.96123, size = 540, normalized size = 2.44 \begin{align*} \frac{1}{2} \, a^{2} 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 )} a b f + a^{2} 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 )} a b e}{d} - \frac{{\left (d e - c f\right )}{\left (\log \left (d x + c - 1\right ) \log \left (\frac{1}{2} \, d x + \frac{1}{2} \, c + \frac{1}{2}\right ) +{\rm Li}_2\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c + \frac{1}{2}\right )\right )} b^{2}}{d^{2}} + \frac{{\left (c f + f\right )} b^{2} \log \left (d x + c + 1\right )}{2 \, d^{2}} + \frac{{\left (b^{2} d^{2} f x^{2} + 2 \, b^{2} d^{2} e x -{\left (c^{2} f - 2 \,{\left (d e - f\right )} c - 2 \, d e + f\right )} b^{2}\right )} \log \left (d x + c + 1\right )^{2} +{\left (b^{2} d^{2} f x^{2} + 2 \, b^{2} d^{2} e x -{\left (c^{2} f - 2 \,{\left (d e + f\right )} c + 2 \, d e + f\right )} b^{2}\right )} \log \left (d x + c - 1\right )^{2} + 2 \,{\left (2 \, b^{2} d f x -{\left (b^{2} d^{2} f x^{2} + 2 \, b^{2} d^{2} e x -{\left (c^{2} f - 2 \,{\left (d e + f\right )} c + 2 \, d e + f\right )} b^{2}\right )} \log \left (d x + c - 1\right )\right )} \log \left (d x + c + 1\right ) - 4 \,{\left (b^{2} d f x +{\left (c f - f\right )} b^{2}\right )} \log \left (d x + c - 1\right )}{8 \, 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))^2,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{2}\,{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))^2,x, algorithm="giac")

[Out]

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