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

Optimal. Leaf size=160 \[ -\frac{3 b^3 e \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right )}{2 d}-\frac{3 b^2 e \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac{3 b e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac{e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.262531, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {6107, 12, 5916, 5980, 5910, 5984, 5918, 2402, 2315, 5948} \[ -\frac{3 b^3 e \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right )}{2 d}-\frac{3 b^2 e \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac{3 b e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac{e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5980

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

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

Mathematica [A]  time = 1.16106, size = 213, normalized size = 1.33 \[ \frac{e \left (6 b^3 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )+a \left (-3 a b \left (c^2-1\right ) \log (-c-d x+1)+3 a b \left (c^2-1\right ) \log (c+d x+1)+2 a d x (2 a c+a d x+3 b)-12 b^2 \log \left (\frac{1}{\sqrt{1-(c+d x)^2}}\right )\right )+6 b^2 (c+d x-1) \tanh ^{-1}(c+d x)^2 (a (c+d x+1)+b)+6 b \tanh ^{-1}(c+d x) \left (a (a d x (2 c+d x)+2 b (c+d x))-2 b^2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )+2 b^3 \left (c^2+2 c d x+d^2 x^2-1\right ) \tanh ^{-1}(c+d x)^3\right )}{4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(e*(6*b^2*(-1 + c + d*x)*(b + a*(1 + c + d*x))*ArcTanh[c + d*x]^2 + 2*b^3*(-1 + c^2 + 2*c*d*x + d^2*x^2)*ArcTa
nh[c + d*x]^3 + 6*b*ArcTanh[c + d*x]*(a*(2*b*(c + d*x) + a*d*x*(2*c + d*x)) - 2*b^2*Log[1 + E^(-2*ArcTanh[c +
d*x])]) + a*(2*a*d*x*(3*b + 2*a*c + a*d*x) - 3*a*b*(-1 + c^2)*Log[1 - c - d*x] + 3*a*b*(-1 + c^2)*Log[1 + c +
d*x] - 12*b^2*Log[1/Sqrt[1 - (c + d*x)^2]]) + 6*b^3*PolyLog[2, -E^(-2*ArcTanh[c + d*x])]))/(4*d)

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Maple [C]  time = 0.43, size = 6834, normalized size = 42.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [B]  time = 1.94895, size = 849, normalized size = 5.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3*d*e*x^2 + 3/4*(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^2*b*d*e + a^3*c*e*x + 3/2*(2*(d*x + c)*arctanh(d*x + c) + log(-(d*x + c)^2 + 1)
)*a^2*b*c*e/d + 3/2*(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^3*e/d + 3/
2*(c*e + e)*a*b^2*log(d*x + c + 1)/d - 3/2*(c*e - e)*a*b^2*log(d*x + c - 1)/d + 1/16*(24*a*b^2*d*e*x*log(d*x +
c + 1) + (b^3*d^2*e*x^2 + 2*b^3*c*d*e*x + (c^2*e - e)*b^3)*log(d*x + c + 1)^3 - (b^3*d^2*e*x^2 + 2*b^3*c*d*e*
x + (c^2*e - e)*b^3)*log(-d*x - c + 1)^3 + 6*(a*b^2*d^2*e*x^2 + (c^2*e - e)*a*b^2 + (c*e + e)*b^3 + (2*a*b^2*c
*d*e + b^3*d*e)*x)*log(d*x + c + 1)^2 + 3*(2*a*b^2*d^2*e*x^2 + 2*(c^2*e - e)*a*b^2 + 2*(c*e - e)*b^3 + 2*(2*a*
b^2*c*d*e + b^3*d*e)*x + (b^3*d^2*e*x^2 + 2*b^3*c*d*e*x + (c^2*e - e)*b^3)*log(d*x + c + 1))*log(-d*x - c + 1)
^2 - 3*(8*a*b^2*d*e*x + (b^3*d^2*e*x^2 + 2*b^3*c*d*e*x + (c^2*e - e)*b^3)*log(d*x + c + 1)^2 + 4*(a*b^2*d^2*e*
x^2 + (c^2*e - e)*a*b^2 + (c*e + e)*b^3 + (2*a*b^2*c*d*e + b^3*d*e)*x)*log(d*x + c + 1))*log(-d*x - c + 1))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e \left (\int a^{3} c\, dx + \int a^{3} d x\, dx + \int b^{3} c \operatorname{atanh}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} c \operatorname{atanh}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b c \operatorname{atanh}{\left (c + d x \right )}\, dx + \int b^{3} d x \operatorname{atanh}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} d x \operatorname{atanh}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b d x \operatorname{atanh}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*(Integral(a**3*c, x) + Integral(a**3*d*x, x) + Integral(b**3*c*atanh(c + d*x)**3, x) + Integral(3*a*b**2*c*a
tanh(c + d*x)**2, x) + Integral(3*a**2*b*c*atanh(c + d*x), x) + Integral(b**3*d*x*atanh(c + d*x)**3, x) + Inte
gral(3*a*b**2*d*x*atanh(c + d*x)**2, x) + Integral(3*a**2*b*d*x*atanh(c + d*x), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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