3.1.26 \(\int \frac {(a+b \tanh ^{-1}(c+d x))^3}{(c e+d e x)^2} \, dx\) [26]
Optimal. Leaf size=143 \[ \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^3 \text {PolyLog}\left (3,-1+\frac {2}{1+c+d x}\right )}{2 d e^2} \]
[Out]
(a+b*arctanh(d*x+c))^3/d/e^2-(a+b*arctanh(d*x+c))^3/d/e^2/(d*x+c)+3*b*(a+b*arctanh(d*x+c))^2*ln(2-2/(d*x+c+1))
/d/e^2-3*b^2*(a+b*arctanh(d*x+c))*polylog(2,-1+2/(d*x+c+1))/d/e^2-3/2*b^3*polylog(3,-1+2/(d*x+c+1))/d/e^2
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Rubi [A]
time = 0.21, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {6242, 12, 6037,
6135, 6079, 6095, 6203, 6745} \begin {gather*} -\frac {3 b^2 \text {Li}_2\left (\frac {2}{c+d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}+\frac {3 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2}-\frac {3 b^3 \text {Li}_3\left (\frac {2}{c+d x+1}-1\right )}{2 d e^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTanh[c + d*x])^3/(c*e + d*e*x)^2,x]
[Out]
(a + b*ArcTanh[c + d*x])^3/(d*e^2) - (a + b*ArcTanh[c + d*x])^3/(d*e^2*(c + d*x)) + (3*b*(a + b*ArcTanh[c + d*
x])^2*Log[2 - 2/(1 + c + d*x)])/(d*e^2) - (3*b^2*(a + b*ArcTanh[c + d*x])*PolyLog[2, -1 + 2/(1 + c + d*x)])/(d
*e^2) - (3*b^3*PolyLog[3, -1 + 2/(1 + c + d*x)])/(2*d*e^2)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 6203
Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x] - Dist[b*(p/2), Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 - u]/
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2
/(1 + c*x))^2, 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]
Rule 6745
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
; !FalseQ[w]] /; FreeQ[n, x]
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{(c e+d e x)^2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x (1+x)} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {\left (6 b^2\right ) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \log \left (2-\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{d e^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^3 \text {Li}_3\left (-1+\frac {2}{1+c+d x}\right )}{2 d e^2}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 0.63, size = 248, normalized size = 1.73 \begin {gather*} \frac {-\frac {2 a^3}{c+d x}-\frac {6 a^2 b \tanh ^{-1}(c+d x)}{c+d x}+6 a^2 b \log (c+d x)-3 a^2 b \log \left (1-c^2-2 c d x-d^2 x^2\right )+6 a b^2 \left (\tanh ^{-1}(c+d x) \left (\left (1-\frac {1}{c+d x}\right ) \tanh ^{-1}(c+d x)+2 \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )-\text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+2 b^3 \left (\frac {i \pi ^3}{8}-\tanh ^{-1}(c+d x)^3-\frac {\tanh ^{-1}(c+d x)^3}{c+d x}+3 \tanh ^{-1}(c+d x)^2 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+3 \tanh ^{-1}(c+d x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(c+d x)}\right )-\frac {3}{2} \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(c+d x)}\right )\right )}{2 d e^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcTanh[c + d*x])^3/(c*e + d*e*x)^2,x]
[Out]
((-2*a^3)/(c + d*x) - (6*a^2*b*ArcTanh[c + d*x])/(c + d*x) + 6*a^2*b*Log[c + d*x] - 3*a^2*b*Log[1 - c^2 - 2*c*
d*x - d^2*x^2] + 6*a*b^2*(ArcTanh[c + d*x]*((1 - (c + d*x)^(-1))*ArcTanh[c + d*x] + 2*Log[1 - E^(-2*ArcTanh[c
+ d*x])]) - PolyLog[2, E^(-2*ArcTanh[c + d*x])]) + 2*b^3*((I/8)*Pi^3 - ArcTanh[c + d*x]^3 - ArcTanh[c + d*x]^3
/(c + d*x) + 3*ArcTanh[c + d*x]^2*Log[1 - E^(2*ArcTanh[c + d*x])] + 3*ArcTanh[c + d*x]*PolyLog[2, E^(2*ArcTanh
[c + d*x])] - (3*PolyLog[3, E^(2*ArcTanh[c + d*x])])/2))/(2*d*e^2)
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 2.88, size = 1867, normalized size = 13.06
| | |
| method |
result |
size |
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| derivativedivides |
\(\text {Expression too large to display}\) |
\(1867\) |
| default |
\(\text {Expression too large to display}\) |
\(1867\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^2,x,method=_RETURNVERBOSE)
[Out]
1/d*(-a^3/e^2/(d*x+c)+3/2*I*b^3/e^2*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^
2)))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*arctanh(d*x+c)^2-3/4*I*b^3/e^2*Pi*arc
tanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^2/(1-(d*x+c)
^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))+3/2*I*b^3/e^2*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(
1-(d*x+c)^2)))^3*arctanh(d*x+c)^2-3/2*a^2*b/e^2*ln(d*x+c+1)+3*a^2*b/e^2*ln(d*x+c)-3/2*a^2*b/e^2*ln(d*x+c-1)+3*
b^3/e^2*ln(2)*arctanh(d*x+c)^2+3*b^3/e^2*arctanh(d*x+c)^2*ln((d*x+c+1)/(1-(d*x+c)^2)^(1/2))-b^3/e^2/(d*x+c)*ar
ctanh(d*x+c)^3+3*a*b^2/e^2*dilog(1/2*d*x+1/2*c+1/2)-3/4*a*b^2/e^2*ln(d*x+c-1)^2+3/4*a*b^2/e^2*ln(d*x+c+1)^2-3*
a*b^2/e^2*dilog(d*x+c+1)-3*a*b^2/e^2*dilog(d*x+c)+3*b^3/e^2*ln(d*x+c)*arctanh(d*x+c)^2+6*b^3/e^2*arctanh(d*x+c
)*polylog(2,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3*b^3/e^2*arctanh(d*x+c)^2*ln(1-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3*b^
3/e^2*arctanh(d*x+c)^2*ln(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+6*b^3/e^2*arctanh(d*x+c)*polylog(2,-(d*x+c+1)/(1-(d
*x+c)^2)^(1/2))-3*b^3/e^2*arctanh(d*x+c)^2*ln((d*x+c+1)^2/(1-(d*x+c)^2)-1)-3/2*b^3/e^2*arctanh(d*x+c)^2*ln(d*x
+c-1)-3/2*b^3/e^2*arctanh(d*x+c)^2*ln(d*x+c+1)+3/2*a*b^2/e^2*ln(d*x+c-1)*ln(1/2*d*x+1/2*c+1/2)-3/2*a*b^2/e^2*l
n(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)+3/2*a*b^2/e^2*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2*d*x+1/2*c+1/2)-3*a*b^2/e^2*ln(d*
x+c)*ln(d*x+c+1)-3*a^2*b/e^2/(d*x+c)*arctanh(d*x+c)-3*a*b^2/e^2/(d*x+c)*arctanh(d*x+c)^2-3*a*b^2/e^2*arctanh(d
*x+c)*ln(d*x+c+1)+6*a*b^2/e^2*ln(d*x+c)*arctanh(d*x+c)-3*a*b^2/e^2*arctanh(d*x+c)*ln(d*x+c-1)+3/2*I*b^3/e^2*Pi
*arctanh(d*x+c)^2-6*b^3/e^2*polylog(3,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-6*b^3/e^2*polylog(3,-(d*x+c+1)/(1-(d*x+c)
^2)^(1/2))-b^3/e^2*arctanh(d*x+c)^3+3/2*I*b^3/e^2*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3+
3/4*I*b^3/e^2*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3+3/4*I*b^3/
e^2*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))^3-3/2*I*b^3/e^2*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c
+1)^2/(1-(d*x+c)^2)))^2-3/4*I*b^3/e^2*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^2/(1-(
d*x+c)^2)))^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))+3/2*I*b^3/e^2*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^
2-1))^2*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/4*I*b^3/e^2*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-
1))*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))^2-3/2*I*b^3/e^2*Pi*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*((d*
x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*arctanh(d*x+c)^2-3/2*I*b^3/e^2*Pi*csgn(I*((d*x+c+1)
^2/(1-(d*x+c)^2)-1))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*arctanh(d*x+c)^2+3/
4*I*b^3/e^2*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x
+c+1)^2/(1-(d*x+c)^2)))^2)
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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)^2,x, algorithm="maxima")
[Out]
-3/2*(d*(e^(-2)*log(d*x + c + 1)/d^2 - 2*e^(-2)*log(d*x + c)/d^2 + e^(-2)*log(d*x + c - 1)/d^2) + 2*arctanh(d*
x + c)/(d^2*x*e^2 + c*d*e^2))*a^2*b - a^3/(d^2*x*e^2 + c*d*e^2) - 1/8*((b^3*d*x + b^3*(c - 1))*log(-d*x - c +
1)^3 + 3*(2*a*b^2 + (b^3*d*x + b^3*(c + 1))*log(d*x + c + 1))*log(-d*x - c + 1)^2)/(d^2*x*e^2 + c*d*e^2) - int
egrate(-1/8*((b^3*d*x + b^3*(c - 1))*log(d*x + c + 1)^3 + 6*(a*b^2*d*x + a*b^2*(c - 1))*log(d*x + c + 1)^2 + 3
*(4*a*b^2*d*x + 4*a*b^2*c - (b^3*d*x + b^3*(c - 1))*log(d*x + c + 1)^2 + 2*(b^3*d^2*x^2 + (c^2 + c)*b^3 - 2*a*
b^2*(c - 1) + ((2*c*d + d)*b^3 - 2*a*b^2*d)*x)*log(d*x + c + 1))*log(-d*x - c + 1))/(d^3*x^3*e^2 + (3*c*d^2 -
d^2)*x^2*e^2 + (3*c^2*d - 2*c*d)*x*e^2 + (c^3 - c^2)*e^2), x)
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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)^2,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^(-2)/(d^2*x^
2 + 2*c*d*x + c^2), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{3}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{3} \operatorname {atanh}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {atanh}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \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)**2,x)
[Out]
(Integral(a**3/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(b**3*atanh(c + d*x)**3/(c**2 + 2*c*d*x + d**2*x**2)
, x) + Integral(3*a*b**2*atanh(c + d*x)**2/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(3*a**2*b*atanh(c + d*x)
/(c**2 + 2*c*d*x + d**2*x**2), x))/e**2
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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)^2,x, algorithm="giac")
[Out]
integrate((b*arctanh(d*x + c) + a)^3/(d*e*x + c*e)^2, 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 )}^2} \,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)^2,x)
[Out]
int((a + b*atanh(c + d*x))^3/(c*e + d*e*x)^2, x)
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