3.27 \(\int \frac{(a+b \tanh ^{-1}(c+d x))^3}{(c e+d e x)^3} \, dx\)

Optimal. Leaf size=166 \[ -\frac{3 b^3 \text{PolyLog}\left (2,\frac{2}{c+d x+1}-1\right )}{2 d e^3}+\frac{3 b^2 \log \left (2-\frac{2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3}-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}+\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3} \]

[Out]

(3*b*(a + b*ArcTanh[c + d*x])^2)/(2*d*e^3) - (3*b*(a + b*ArcTanh[c + d*x])^2)/(2*d*e^3*(c + d*x)) + (a + b*Arc
Tanh[c + d*x])^3/(2*d*e^3) - (a + b*ArcTanh[c + d*x])^3/(2*d*e^3*(c + d*x)^2) + (3*b^2*(a + b*ArcTanh[c + d*x]
)*Log[2 - 2/(1 + c + d*x)])/(d*e^3) - (3*b^3*PolyLog[2, -1 + 2/(1 + c + d*x)])/(2*d*e^3)

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Rubi [A]  time = 0.33502, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {6107, 12, 5916, 5982, 5988, 5932, 2447, 5948} \[ -\frac{3 b^3 \text{PolyLog}\left (2,\frac{2}{c+d x+1}-1\right )}{2 d e^3}+\frac{3 b^2 \log \left (2-\frac{2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3}-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}+\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])^3/(c*e + d*e*x)^3,x]

[Out]

(3*b*(a + b*ArcTanh[c + d*x])^2)/(2*d*e^3) - (3*b*(a + b*ArcTanh[c + d*x])^2)/(2*d*e^3*(c + d*x)) + (a + b*Arc
Tanh[c + d*x])^3/(2*d*e^3) - (a + b*ArcTanh[c + d*x])^3/(2*d*e^3*(c + d*x)^2) + (3*b^2*(a + b*ArcTanh[c + d*x]
)*Log[2 - 2/(1 + c + d*x)])/(d*e^3) - (3*b^3*PolyLog[2, -1 + 2/(1 + c + d*x)])/(2*d*e^3)

Rule 6107

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5982

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTanh[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 5988

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

Rule 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{(c e+d e x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^3}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^3}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{x^2 \left (1-x^2\right )} \, dx,x,c+d x\right )}{2 d e^3}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{2 d e^3}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{2 d e^3}\\ &=-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^3}\\ &=\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{x (1+x)} \, dx,x,c+d x\right )}{d e^3}\\ &=\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac{3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (2-\frac{2}{1+c+d x}\right )}{d e^3}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\log \left (2-\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e^3}\\ &=\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac{3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (2-\frac{2}{1+c+d x}\right )}{d e^3}-\frac{3 b^3 \text{Li}_2\left (-1+\frac{2}{1+c+d x}\right )}{2 d e^3}\\ \end{align*}

Mathematica [C]  time = 1.12085, size = 335, normalized size = 2.02 \[ \frac{-12 b^3 (c+d x)^2 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c+d x)}\right )-12 a^2 b c-12 a^2 b d x-4 a^3+12 b \tanh ^{-1}(c+d x) \left (a \left (a \left (c^2+2 c d x+d^2 x^2-1\right )-2 b (c+d x)\right )+2 b^2 (c+d x)^2 \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+24 a b^2 c^2 \log \left (\frac{c+d x}{\sqrt{1-(c+d x)^2}}\right )+24 a b^2 d^2 x^2 \log \left (\frac{c+d x}{\sqrt{1-(c+d x)^2}}\right )+48 a b^2 c d x \log \left (\frac{c+d x}{\sqrt{1-(c+d x)^2}}\right )+12 b^2 (c+d x-1) \tanh ^{-1}(c+d x)^2 (a (c+d x+1)+b (c+d x))+4 b^3 \left (c^2+2 c d x+d^2 x^2-1\right ) \tanh ^{-1}(c+d x)^3+2 i \pi ^3 b^3 c^2 d x+i \pi ^3 b^3 c^3+i \pi ^3 b^3 c d^2 x^2}{8 d e^3 (c+d x)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c + d*x])^3/(c*e + d*e*x)^3,x]

[Out]

(-4*a^3 - 12*a^2*b*c + I*b^3*c^3*Pi^3 - 12*a^2*b*d*x + (2*I)*b^3*c^2*d*Pi^3*x + I*b^3*c*d^2*Pi^3*x^2 + 12*b^2*
(-1 + c + d*x)*(b*(c + d*x) + a*(1 + c + d*x))*ArcTanh[c + d*x]^2 + 4*b^3*(-1 + c^2 + 2*c*d*x + d^2*x^2)*ArcTa
nh[c + d*x]^3 + 12*b*ArcTanh[c + d*x]*(a*(-2*b*(c + d*x) + a*(-1 + c^2 + 2*c*d*x + d^2*x^2)) + 2*b^2*(c + d*x)
^2*Log[1 - E^(-2*ArcTanh[c + d*x])]) + 24*a*b^2*c^2*Log[(c + d*x)/Sqrt[1 - (c + d*x)^2]] + 48*a*b^2*c*d*x*Log[
(c + d*x)/Sqrt[1 - (c + d*x)^2]] + 24*a*b^2*d^2*x^2*Log[(c + d*x)/Sqrt[1 - (c + d*x)^2]] - 12*b^3*(c + d*x)^2*
PolyLog[2, E^(-2*ArcTanh[c + d*x])])/(8*d*e^3*(c + d*x)^2)

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Maple [C]  time = 0.437, size = 5796, normalized size = 34.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^3,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^3,x, algorithm="maxima")

[Out]

-3/4*(d*(2/(d^3*e^3*x + c*d^2*e^3) - log(d*x + c + 1)/(d^2*e^3) + log(d*x + c - 1)/(d^2*e^3)) + 2*arctanh(d*x
+ c)/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3))*a^2*b - 3/8*(d^2*((log(d*x + c + 1)^2 - 2*log(d*x + c + 1)*log
(d*x + c - 1) + log(d*x + c - 1)^2 + 4*log(d*x + c - 1))/(d^3*e^3) + 4*log(d*x + c + 1)/(d^3*e^3) - 8*log(d*x
+ c)/(d^3*e^3)) + 4*d*(2/(d^3*e^3*x + c*d^2*e^3) - log(d*x + c + 1)/(d^2*e^3) + log(d*x + c - 1)/(d^2*e^3))*ar
ctanh(d*x + c))*a*b^2 - 1/16*b^3*(((d^2*x^2 + 2*c*d*x + c^2 - 1)*log(-d*x - c + 1)^3 + 3*(2*d*x - (d^2*x^2 + 2
*c*d*x + c^2 - 1)*log(d*x + c + 1) + 2*c)*log(-d*x - c + 1)^2)/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3) + 2*i
ntegrate(-((d*x + c - 1)*log(d*x + c + 1)^3 + 3*(2*d^2*x^2 + 4*c*d*x - (d*x + c - 1)*log(d*x + c + 1)^2 + 2*c^
2 - (d^3*x^3 + 3*c*d^2*x^2 + c^3 + (3*c^2*d - d)*x - c)*log(d*x + c + 1))*log(-d*x - c + 1))/(d^4*e^3*x^4 + c^
4*e^3 - c^3*e^3 + (4*c*d^3*e^3 - d^3*e^3)*x^3 + 3*(2*c^2*d^2*e^3 - c*d^2*e^3)*x^2 + (4*c^3*d*e^3 - 3*c^2*d*e^3
)*x), x)) - 3/2*a*b^2*arctanh(d*x + c)^2/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3) - 1/2*a^3/(d^3*e^3*x^2 + 2*
c*d^2*e^3*x + c^2*d*e^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \operatorname{artanh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\left (d x + c\right ) + a^{3}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

integral((b^3*arctanh(d*x + c)^3 + 3*a*b^2*arctanh(d*x + c)^2 + 3*a^2*b*arctanh(d*x + c) + a^3)/(d^3*e^3*x^3 +
3*c*d^2*e^3*x^2 + 3*c^2*d*e^3*x + c^3*e^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))**3/(d*e*x+c*e)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(d*x + c) + a)^3/(d*e*x + c*e)^3, x)