3.3.7 \(\int \frac {(a+b \tanh ^{-1}(c \sqrt {x}))^3}{x^2} \, dx\) [207]
Optimal. Leaf size=142 \[ 3 b c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2-\frac {3 b c \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{\sqrt {x}}+c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{x}+6 b^2 c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (2-\frac {2}{1+c \sqrt {x}}\right )-3 b^3 c^2 \text {PolyLog}\left (2,-1+\frac {2}{1+c \sqrt {x}}\right ) \]
[Out]
3*b*c^2*(a+b*arctanh(c*x^(1/2)))^2+c^2*(a+b*arctanh(c*x^(1/2)))^3-(a+b*arctanh(c*x^(1/2)))^3/x+6*b^2*c^2*(a+b*
arctanh(c*x^(1/2)))*ln(2-2/(1+c*x^(1/2)))-3*b^3*c^2*polylog(2,-1+2/(1+c*x^(1/2)))-3*b*c*(a+b*arctanh(c*x^(1/2)
))^2/x^(1/2)
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Rubi [A]
time = 0.25, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6039, 6037,
6129, 6135, 6079, 2497, 6095} \begin {gather*} 6 b^2 c^2 \log \left (2-\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )+3 b c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-\frac {3 b c \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{\sqrt {x}}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{x}-3 b^3 c^2 \text {Li}_2\left (\frac {2}{\sqrt {x} c+1}-1\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTanh[c*Sqrt[x]])^3/x^2,x]
[Out]
3*b*c^2*(a + b*ArcTanh[c*Sqrt[x]])^2 - (3*b*c*(a + b*ArcTanh[c*Sqrt[x]])^2)/Sqrt[x] + c^2*(a + b*ArcTanh[c*Sqr
t[x]])^3 - (a + b*ArcTanh[c*Sqrt[x]])^3/x + 6*b^2*c^2*(a + b*ArcTanh[c*Sqrt[x]])*Log[2 - 2/(1 + c*Sqrt[x])] -
3*b^3*c^2*PolyLog[2, -1 + 2/(1 + c*Sqrt[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 6039
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[S
implify[(m + 1)/n]]
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]
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx &=\int \frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx\\ \end {align*}
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Mathematica [A]
time = 0.21, size = 230, normalized size = 1.62 \begin {gather*} \frac {6 b^2 \left (-1+c \sqrt {x}\right ) \left (a+a c \sqrt {x}+b c \sqrt {x}\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^2+2 b^3 \left (-1+c^2 x\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^3-6 b \tanh ^{-1}\left (c \sqrt {x}\right ) \left (a^2+2 a b c \sqrt {x}-2 b^2 c^2 x \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right )+a \left (-2 a^2-6 a b c \sqrt {x}-3 a b c^2 x \log \left (1-c \sqrt {x}\right )+3 a b c^2 x \log \left (1+c \sqrt {x}\right )+12 b^2 c^2 x \log \left (\frac {c \sqrt {x}}{\sqrt {1-c^2 x}}\right )\right )-6 b^3 c^2 x \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )}{2 x} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcTanh[c*Sqrt[x]])^3/x^2,x]
[Out]
(6*b^2*(-1 + c*Sqrt[x])*(a + a*c*Sqrt[x] + b*c*Sqrt[x])*ArcTanh[c*Sqrt[x]]^2 + 2*b^3*(-1 + c^2*x)*ArcTanh[c*Sq
rt[x]]^3 - 6*b*ArcTanh[c*Sqrt[x]]*(a^2 + 2*a*b*c*Sqrt[x] - 2*b^2*c^2*x*Log[1 - E^(-2*ArcTanh[c*Sqrt[x]])]) + a
*(-2*a^2 - 6*a*b*c*Sqrt[x] - 3*a*b*c^2*x*Log[1 - c*Sqrt[x]] + 3*a*b*c^2*x*Log[1 + c*Sqrt[x]] + 12*b^2*c^2*x*Lo
g[(c*Sqrt[x])/Sqrt[1 - c^2*x]]) - 6*b^3*c^2*x*PolyLog[2, E^(-2*ArcTanh[c*Sqrt[x]])])/(2*x)
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 7.18, size = 4974, normalized size = 35.03
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| method |
result |
size |
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| derivativedivides |
\(\text {Expression too large to display}\) |
\(4974\) |
| default |
\(\text {Expression too large to display}\) |
\(4974\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctanh(c*x^(1/2)))^3/x^2,x,method=_RETURNVERBOSE)
[Out]
2*c^2*(-3/8*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))*csgn(I*(1+c*x^(1
/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*arctanh(c*x^(1/2))*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/4*I
*b^3*Pi*csgn(I*(1+c*x^(1/2))/(-c^2*x+1)^(1/2))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))^2*dilog((1+c*x^(1/2))/(-c^2*x
+1)^(1/2))+3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*dilog((1+c*x^(1/2))
/(-c^2*x+1)^(1/2))-3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*dilog(1+(1+
c*x^(1/2))/(-c^2*x+1)^(1/2))-3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))^3*dilog(1+(1+c*x^(1/2))/(-c^2*x+1)
^(1/2))+3/4*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*polylog(2,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/4*I*
b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*dilog(1+(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*b^3*Pi*csgn(I*(1
+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/8*I*b^3
*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*arctanh(c*x^(1/2))^2+3/8*I*b^3*Pi*csgn(
I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*polylog(2,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/4*I*
b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*polylog(2,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/4*I*b^3*Pi*csgn(I/
(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))^2+3/4*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*a
rctanh(c*x^(1/2))^2+3/4*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*dilog(1+(1+c*x^(1/2))/(-c^2*x+1)^(1/
2))-3/4*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*dilog((1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/4*I*b^3*Pi*c
sgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*dilog((1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/4*I*b^3*Pi*csgn(I/(1+(1+c*x^(1
/2))^2/(-c^2*x+1)))^3*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/4*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*
x+1)))^2*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/2*a*b^2*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)+3/2*a*b^2*arc
tanh(c*x^(1/2))*ln(1+c*x^(1/2))+3/4*a*b^2*ln(c*x^(1/2)-1)*ln(1/2*c*x^(1/2)+1/2)+3/4*a*b^2*ln(-1/2*c*x^(1/2)+1/
2)*ln(1+c*x^(1/2))-3/4*a*b^2*ln(-1/2*c*x^(1/2)+1/2)*ln(1/2*c*x^(1/2)+1/2)+3/8*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))
^2/(-c^2*x+1)))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*polylog(2,(1+c*x^(1/2))/(-c
^2*x+1)^(1/2))-3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))/(-c^2*x+1)^(1/2))^2*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))*dilog(1
+(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))/(-c^2*x+1)^(1/2))^2*csgn(I*(1+c*x^(1/2))^2/
(c^2*x-1))*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/4*I*b^3*Pi*csgn(I*(1+c*x^(1/2))/(-c^2*x+1)^(1/2))*csgn
(I*(1+c*x^(1/2))^2/(c^2*x-1))^2*polylog(2,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))/(-
c^2*x+1)^(1/2))^2*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))*dilog((1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*b^3*Pi*csgn(I/
(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*dilog((1+c*
x^(1/2))/(-c^2*x+1)^(1/2))-3/8*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1
)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))^2+3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))*csgn(I
*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*dilog(1+(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/4*I*b^3
*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*b^3
*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*
polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/4*I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*arctanh(c*x^(
1/2))*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))*csgn(I*(1+c*x^(1/2))
^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*dilog((1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/8*a*b^2*ln(c*x^(1/2)-1)
^2-3/8*a*b^2*ln(1+c*x^(1/2))^2-3/2*a*b^2*ln(c*x^(1/2)-1)-3/2*a*b^2*ln(1+c*x^(1/2))-3/4*a^2*b*ln(c*x^(1/2)-1)+3
/4*a^2*b*ln(1+c*x^(1/2))-3/4*b^3*arctanh(c*x^(1/2))^2*ln(c*x^(1/2)-1)+3/4*b^3*arctanh(c*x^(1/2))^2*ln(1+c*x^(1
/2))-3/2*b^3*arctanh(c*x^(1/2))^2*ln((1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*
x-1))^3*arctanh(c*x^(1/2))*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))
*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))^2-3/8*I*b^3*Pi*csgn(I*(
1+c*x^(1/2))^2/(c^2*x-1))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*polylog(2,-(1+c*x
^(1/2))/(-c^2*x+1)^(1/2))-3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))/(-c^2*x+1)^(1/2))^2*csgn(I*(1+c*x^(1/2))^2/(c^2*x-
1))*arctanh(c*x^(1/2))^2+3/4*I*b^3*Pi*csgn(I*(1+c*x^(1/2))/(-c^2*x+1)^(1/2))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))
^2*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1+c*x^(1/2))^2
/(-c^2*x+1)))^3*arctanh(c*x^(1/2))*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/8*I*b^3*Pi*csgn(I*(1+c*x^(1/2))^2/(c
^2*x-1))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1+(1...
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 528 vs.
\(2 (125) = 250\).
time = 1.10, size = 528, normalized size = 3.72 \begin {gather*} -3 \, {\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right )\right )} b^{3} c^{2} - 3 \, {\left (\log \left (c \sqrt {x}\right ) \log \left (-c \sqrt {x} + 1\right ) + {\rm Li}_2\left (-c \sqrt {x} + 1\right )\right )} b^{3} c^{2} + 3 \, {\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x}\right ) + {\rm Li}_2\left (c \sqrt {x} + 1\right )\right )} b^{3} c^{2} - 3 \, a b^{2} c^{2} \log \left (c \sqrt {x} - 1\right ) - \frac {3}{4} \, {\left ({\left (2 \, c \log \left (c \sqrt {x} - 1\right ) - c \log \left (x\right ) + \frac {2}{\sqrt {x}}\right )} c - \frac {2 \, \log \left (-c \sqrt {x} + 1\right )}{x}\right )} a^{2} b - \frac {a^{3}}{x} + \frac {3}{2} \, {\left (a^{2} b c^{2} - 2 \, a b^{2} c^{2}\right )} \log \left (c \sqrt {x} + 1\right ) - \frac {3}{4} \, {\left (a^{2} b c^{2} - 4 \, a b^{2} c^{2}\right )} \log \left (x\right ) - \frac {12 \, a^{2} b c \sqrt {x} - {\left (b^{3} c^{2} x - b^{3}\right )} \log \left (c \sqrt {x} + 1\right )^{3} + {\left (b^{3} c^{2} x - b^{3}\right )} \log \left (-c \sqrt {x} + 1\right )^{3} + 6 \, {\left (b^{3} c \sqrt {x} + a b^{2} - {\left (a b^{2} c^{2} - b^{3} c^{2}\right )} x\right )} \log \left (c \sqrt {x} + 1\right )^{2} + 3 \, {\left (2 \, b^{3} c \sqrt {x} + 2 \, a b^{2} - 2 \, {\left (a b^{2} c^{2} + b^{3} c^{2}\right )} x - {\left (b^{3} c^{2} x - b^{3}\right )} \log \left (c \sqrt {x} + 1\right )\right )} \log \left (-c \sqrt {x} + 1\right )^{2} + 12 \, {\left (2 \, a b^{2} c \sqrt {x} + a^{2} b\right )} \log \left (c \sqrt {x} + 1\right ) - 3 \, {\left (8 \, a b^{2} c \sqrt {x} - {\left (b^{3} c^{2} x - b^{3}\right )} \log \left (c \sqrt {x} + 1\right )^{2} + 4 \, {\left (b^{3} c \sqrt {x} + a b^{2} - {\left (a b^{2} c^{2} - b^{3} c^{2}\right )} x\right )} \log \left (c \sqrt {x} + 1\right )\right )} \log \left (-c \sqrt {x} + 1\right )}{8 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x^(1/2)))^3/x^2,x, algorithm="maxima")
[Out]
-3*(log(c*sqrt(x) + 1)*log(-1/2*c*sqrt(x) + 1/2) + dilog(1/2*c*sqrt(x) + 1/2))*b^3*c^2 - 3*(log(c*sqrt(x))*log
(-c*sqrt(x) + 1) + dilog(-c*sqrt(x) + 1))*b^3*c^2 + 3*(log(c*sqrt(x) + 1)*log(-c*sqrt(x)) + dilog(c*sqrt(x) +
1))*b^3*c^2 - 3*a*b^2*c^2*log(c*sqrt(x) - 1) - 3/4*((2*c*log(c*sqrt(x) - 1) - c*log(x) + 2/sqrt(x))*c - 2*log(
-c*sqrt(x) + 1)/x)*a^2*b - a^3/x + 3/2*(a^2*b*c^2 - 2*a*b^2*c^2)*log(c*sqrt(x) + 1) - 3/4*(a^2*b*c^2 - 4*a*b^2
*c^2)*log(x) - 1/8*(12*a^2*b*c*sqrt(x) - (b^3*c^2*x - b^3)*log(c*sqrt(x) + 1)^3 + (b^3*c^2*x - b^3)*log(-c*sqr
t(x) + 1)^3 + 6*(b^3*c*sqrt(x) + a*b^2 - (a*b^2*c^2 - b^3*c^2)*x)*log(c*sqrt(x) + 1)^2 + 3*(2*b^3*c*sqrt(x) +
2*a*b^2 - 2*(a*b^2*c^2 + b^3*c^2)*x - (b^3*c^2*x - b^3)*log(c*sqrt(x) + 1))*log(-c*sqrt(x) + 1)^2 + 12*(2*a*b^
2*c*sqrt(x) + a^2*b)*log(c*sqrt(x) + 1) - 3*(8*a*b^2*c*sqrt(x) - (b^3*c^2*x - b^3)*log(c*sqrt(x) + 1)^2 + 4*(b
^3*c*sqrt(x) + a*b^2 - (a*b^2*c^2 - b^3*c^2)*x)*log(c*sqrt(x) + 1))*log(-c*sqrt(x) + 1))/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(c*x^(1/2)))^3/x^2,x, algorithm="fricas")
[Out]
integral((b^3*arctanh(c*sqrt(x))^3 + 3*a*b^2*arctanh(c*sqrt(x))^2 + 3*a^2*b*arctanh(c*sqrt(x)) + a^3)/x^2, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )^{3}}{x^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atanh(c*x**(1/2)))**3/x**2,x)
[Out]
Integral((a + b*atanh(c*sqrt(x)))**3/x**2, x)
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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(c*x^(1/2)))^3/x^2,x, algorithm="giac")
[Out]
integrate((b*arctanh(c*sqrt(x)) + a)^3/x^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\,\sqrt {x}\right )\right )}^3}{x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*atanh(c*x^(1/2)))^3/x^2,x)
[Out]
int((a + b*atanh(c*x^(1/2)))^3/x^2, x)
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