3.1.40 \(\int (e+f x) (a+b \tanh ^{-1}(c+d x))^2 \, dx\) [40]

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

[Out]

a*b*f*x/d+b^2*f*(d*x+c)*arctanh(d*x+c)/d^2+(-c*f+d*e)*(a+b*arctanh(d*x+c))^2/d^2-1/2*(d^2*e^2-2*c*d*e*f+(c^2+1
)*f^2)*(a+b*arctanh(d*x+c))^2/d^2/f+1/2*(f*x+e)^2*(a+b*arctanh(d*x+c))^2/f-2*b*(-c*f+d*e)*(a+b*arctanh(d*x+c))
*ln(2/(-d*x-c+1))/d^2+1/2*b^2*f*ln(1-(d*x+c)^2)/d^2-b^2*(-c*f+d*e)*polylog(2,(-d*x-c-1)/(-d*x-c+1))/d^2

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Rubi [A]
time = 0.32, antiderivative size = 220, normalized size of antiderivative = 1.00, 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 = {6246, 6065, 6021, 266, 6195, 6095, 6131, 6055, 2449, 2352} \begin {gather*} \frac {\left (-\frac {\left (c^2+1\right ) f}{d}+2 c e-\frac {d e^2}{f}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {(d e-c f) \left (a+b \tanh ^{-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 \tanh ^{-1}(c+d x)\right )}{d^2}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}+\frac {a b f x}{d}-\frac {b^2 (d e-c f) \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{d^2}+\frac {b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}+\frac {b^2 f (c+d x) \tanh ^{-1}(c+d x)}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*f*x)/d + (b^2*f*(c + d*x)*ArcTanh[c + d*x])/d^2 + ((d*e - c*f)*(a + b*ArcTanh[c + d*x])^2)/d^2 + ((2*c*e
- (d*e^2)/f - ((1 + c^2)*f)/d)*(a + b*ArcTanh[c + d*x])^2)/(2*d) + ((e + f*x)^2*(a + b*ArcTanh[c + d*x])^2)/(2
*f) - (2*b*(d*e - c*f)*(a + b*ArcTanh[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 266

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 2352

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

Rule 2449

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 6021

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

Rule 6055

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)
*(Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTanh[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 6065

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((
a + b*ArcTanh[c*x])^p/(e*(q + 1))), x] - Dist[b*c*(p/(e*(q + 1))), Int[ExpandIntegrand[(a + b*ArcTanh[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 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[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 6131

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[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 6195

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcTanh[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 6246

Int[((a_.) + ArcTanh[(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*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &
& IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]
time = 0.32, size = 271, normalized size = 1.23 \begin {gather*} \frac {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) \tanh ^{-1}(c+d x)^2+2 b \tanh ^{-1}(c+d x) \left (-((c+d x) (-b f+a c f-a d (2 e+f x)))-2 b (d e-c f) \log \left (1+e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+a b f \log (1-c-d x)-a b f \log (1+c+d x)-4 a b d e \log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )-2 b^2 f \log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )+4 a b c f \log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )+2 b^2 (d e-c f) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )}{2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(811\) vs. \(2(217)=434\).
time = 0.32, size = 812, normalized size = 3.67 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*(a+b*arctanh(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 435 vs. \(2 (213) = 426\).
time = 0.45, size = 435, normalized size = 1.97 \begin {gather*} \frac {1}{2} \, a^{2} f x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\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} x e + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a b e}{d} + \frac {{\left (c f + f\right )} b^{2} \log \left (d x + c + 1\right )}{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} c f - b^{2} d e\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 )}}{d^{2}} + \frac {4 \, b^{2} d f x \log \left (d x + c + 1\right ) + {\left (b^{2} d^{2} f x^{2} + 2 \, b^{2} d^{2} x e + 2 \, {\left (c d + d\right )} b^{2} e - {\left (c^{2} f + 2 \, c f + 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} x e + 2 \, {\left (c d - d\right )} b^{2} e - {\left (c^{2} f - 2 \, c f + 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} x e + 2 \, {\left (c d + d\right )} b^{2} e - {\left (c^{2} f + 2 \, c f + f\right )} b^{2}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{8 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{2} \left (e + f x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (e+f\,x\right )\,{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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