3.1.27 \(\int \frac {(a+b \tanh ^{-1}(c+d x))^3}{(c e+d e x)^3} \, dx\) [27]

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

[Out]

3/2*b*(a+b*arctanh(d*x+c))^2/d/e^3-3/2*b*(a+b*arctanh(d*x+c))^2/d/e^3/(d*x+c)+1/2*(a+b*arctanh(d*x+c))^3/d/e^3
-1/2*(a+b*arctanh(d*x+c))^3/d/e^3/(d*x+c)^2+3*b^2*(a+b*arctanh(d*x+c))*ln(2-2/(d*x+c+1))/d/e^3-3/2*b^3*polylog
(2,-1+2/(d*x+c+1))/d/e^3

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Rubi [A]
time = 0.24, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 = {6242, 12, 6037, 6129, 6135, 6079, 2497, 6095} \begin {gather*} \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}-\frac {3 b^3 \text {Li}_2\left (\frac {2}{c+d x+1}-1\right )}{2 d e^3} \end {gather*}

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 12

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

Rule 2497

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 6037

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

Rule 6079

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 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 6129

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 6135

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 6242

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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.83, size = 335, normalized size = 2.02 \begin {gather*} \frac {-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)) \tanh ^{-1}(c+d x)^2+4 b^3 \left (-1+c^2+2 c d x+d^2 x^2\right ) \tanh ^{-1}(c+d x)^3+12 b \tanh ^{-1}(c+d x) \left (a \left (-2 b (c+d x)+a \left (-1+c^2+2 c d x+d^2 x^2\right )\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 )+48 a b^2 c d x \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 )-12 b^3 (c+d x)^2 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c+d x)}\right )}{8 d e^3 (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 3.13, size = 5530, normalized size = 33.31

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(5530\)
default	\(\text {Expression too large to display}\)	\(5530\)

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,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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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((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)*e^(-3)/(d^3*x^
3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{3}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b^{3} \operatorname {atanh}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a^{2} b \operatorname {atanh}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \end {gather*}

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]

(Integral(a**3/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(b**3*atanh(c + d*x)**3/(c**3 + 3
*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(3*a*b**2*atanh(c + d*x)**2/(c**3 + 3*c**2*d*x + 3*c*d**2
*x**2 + d**3*x**3), x) + Integral(3*a**2*b*atanh(c + d*x)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x))
/e**3

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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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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