3.11 \(\int (d+e x) (a+b \tanh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=160 \[ -\frac{b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c}-\frac{\left (\frac{e^2}{c^2}+d^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{c}-\frac{2 b d \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{a b e x}{c}+\frac{b^2 e \log \left (1-c^2 x^2\right )}{2 c^2}+\frac{b^2 e x \tanh ^{-1}(c x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.329908, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.562, Rules used = {5928, 5910, 260, 6048, 5948, 5984, 5918, 2402, 2315} \[ -\frac{b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c}-\frac{\left (\frac{e^2}{c^2}+d^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{c}-\frac{2 b d \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{a b e x}{c}+\frac{b^2 e \log \left (1-c^2 x^2\right )}{2 c^2}+\frac{b^2 e x \tanh ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5928

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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[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 6048

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 5948

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 5984

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 5918

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

Mathematica [A]  time = 0.423255, size = 174, normalized size = 1.09 \[ \frac{2 b^2 c d \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+2 a^2 c^2 d x+a^2 c^2 e x^2+2 a b c d \log \left (1-c^2 x^2\right )+2 b c \tanh ^{-1}(c x) \left (a c x (2 d+e x)-2 b d \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+b e x\right )+2 a b c e x+a b e \log (1-c x)-a b e \log (c x+1)+b^2 e \log \left (1-c^2 x^2\right )+b^2 (c x-1) \tanh ^{-1}(c x)^2 (2 c d+c e x+e)}{2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a^2*c^2*d*x + 2*a*b*c*e*x + a^2*c^2*e*x^2 + b^2*(-1 + c*x)*(2*c*d + e + c*e*x)*ArcTanh[c*x]^2 + 2*b*c*ArcTa
nh[c*x]*(b*e*x + a*c*x*(2*d + e*x) - 2*b*d*Log[1 + E^(-2*ArcTanh[c*x])]) + a*b*e*Log[1 - c*x] - a*b*e*Log[1 +
c*x] + 2*a*b*c*d*Log[1 - c^2*x^2] + b^2*e*Log[1 - c^2*x^2] + 2*b^2*c*d*PolyLog[2, -E^(-2*ArcTanh[c*x])])/(2*c^
2)

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Maple [B]  time = 0.052, size = 462, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.76164, size = 423, normalized size = 2.64 \begin{align*} \frac{1}{2} \, a^{2} e x^{2} + \frac{1}{2} \,{\left (2 \, x^{2} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \, x}{c^{2}} - \frac{\log \left (c x + 1\right )}{c^{3}} + \frac{\log \left (c x - 1\right )}{c^{3}}\right )}\right )} a b e + a^{2} d x + \frac{{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a b d}{c} + \frac{{\left (\log \left (c x + 1\right ) \log \left (-\frac{1}{2} \, c x + \frac{1}{2}\right ) +{\rm Li}_2\left (\frac{1}{2} \, c x + \frac{1}{2}\right )\right )} b^{2} d}{c} + \frac{b^{2} e \log \left (c x + 1\right )}{2 \, c^{2}} + \frac{b^{2} e \log \left (c x - 1\right )}{2 \, c^{2}} + \frac{4 \, b^{2} c e x \log \left (c x + 1\right ) +{\left (b^{2} c^{2} e x^{2} + 2 \, b^{2} c^{2} d x +{\left (2 \, c d - e\right )} b^{2}\right )} \log \left (c x + 1\right )^{2} +{\left (b^{2} c^{2} e x^{2} + 2 \, b^{2} c^{2} d x -{\left (2 \, c d + e\right )} b^{2}\right )} \log \left (-c x + 1\right )^{2} - 2 \,{\left (2 \, b^{2} c e x +{\left (b^{2} c^{2} e x^{2} + 2 \, b^{2} c^{2} d x +{\left (2 \, c d - e\right )} b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*e*x^2 + 1/2*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*a*b*e + a^2*d*x +
(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a*b*d/c + (log(c*x + 1)*log(-1/2*c*x + 1/2) + dilog(1/2*c*x + 1/2))*
b^2*d/c + 1/2*b^2*e*log(c*x + 1)/c^2 + 1/2*b^2*e*log(c*x - 1)/c^2 + 1/8*(4*b^2*c*e*x*log(c*x + 1) + (b^2*c^2*e
*x^2 + 2*b^2*c^2*d*x + (2*c*d - e)*b^2)*log(c*x + 1)^2 + (b^2*c^2*e*x^2 + 2*b^2*c^2*d*x - (2*c*d + e)*b^2)*log
(-c*x + 1)^2 - 2*(2*b^2*c*e*x + (b^2*c^2*e*x^2 + 2*b^2*c^2*d*x + (2*c*d - e)*b^2)*log(c*x + 1))*log(-c*x + 1))
/c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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