3.3.3 \(\int x^2 (a+b \tanh ^{-1}(c \sqrt {x}))^3 \, dx\) [203]
Optimal. Leaf size=304 \[ \frac {19 b^3 \sqrt {x}}{30 c^5}+\frac {b^3 x^{3/2}}{30 c^3}-\frac {19 b^3 \tanh ^{-1}\left (c \sqrt {x}\right )}{30 c^6}+\frac {8 b^2 x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{15 c^4}+\frac {b^2 x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{10 c^2}+\frac {23 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{15 c^6}+\frac {b \sqrt {x} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{c^5}+\frac {b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{3 c^3}+\frac {b x^{5/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{5 c}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{3 c^6}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-\frac {46 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1-c \sqrt {x}}\right )}{15 c^6}-\frac {23 b^3 \text {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )}{15 c^6} \]
[Out]
1/30*b^3*x^(3/2)/c^3-19/30*b^3*arctanh(c*x^(1/2))/c^6+8/15*b^2*x*(a+b*arctanh(c*x^(1/2)))/c^4+1/10*b^2*x^2*(a+
b*arctanh(c*x^(1/2)))/c^2+23/15*b*(a+b*arctanh(c*x^(1/2)))^2/c^6+1/3*b*x^(3/2)*(a+b*arctanh(c*x^(1/2)))^2/c^3+
1/5*b*x^(5/2)*(a+b*arctanh(c*x^(1/2)))^2/c-1/3*(a+b*arctanh(c*x^(1/2)))^3/c^6+1/3*x^3*(a+b*arctanh(c*x^(1/2)))
^3-46/15*b^2*(a+b*arctanh(c*x^(1/2)))*ln(2/(1-c*x^(1/2)))/c^6-23/15*b^3*polylog(2,1-2/(1-c*x^(1/2)))/c^6+19/30
*b^3*x^(1/2)/c^5+b*(a+b*arctanh(c*x^(1/2)))^2*x^(1/2)/c^5
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Rubi [A]
time = 0.73, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps
used = 34, number of rules used = 12, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6039, 6037,
6127, 308, 212, 327, 6131, 6055, 2449, 2352, 6021, 6095} \begin {gather*} -\frac {46 b^2 \log \left (\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{15 c^6}+\frac {8 b^2 x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{15 c^4}+\frac {b^2 x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{10 c^2}+\frac {23 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{15 c^6}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{3 c^6}+\frac {b \sqrt {x} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{c^5}+\frac {b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{3 c^3}+\frac {b x^{5/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{5 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-\frac {23 b^3 \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )}{15 c^6}-\frac {19 b^3 \tanh ^{-1}\left (c \sqrt {x}\right )}{30 c^6}+\frac {19 b^3 \sqrt {x}}{30 c^5}+\frac {b^3 x^{3/2}}{30 c^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x^2*(a + b*ArcTanh[c*Sqrt[x]])^3,x]
[Out]
(19*b^3*Sqrt[x])/(30*c^5) + (b^3*x^(3/2))/(30*c^3) - (19*b^3*ArcTanh[c*Sqrt[x]])/(30*c^6) + (8*b^2*x*(a + b*Ar
cTanh[c*Sqrt[x]]))/(15*c^4) + (b^2*x^2*(a + b*ArcTanh[c*Sqrt[x]]))/(10*c^2) + (23*b*(a + b*ArcTanh[c*Sqrt[x]])
^2)/(15*c^6) + (b*Sqrt[x]*(a + b*ArcTanh[c*Sqrt[x]])^2)/c^5 + (b*x^(3/2)*(a + b*ArcTanh[c*Sqrt[x]])^2)/(3*c^3)
+ (b*x^(5/2)*(a + b*ArcTanh[c*Sqrt[x]])^2)/(5*c) - (a + b*ArcTanh[c*Sqrt[x]])^3/(3*c^6) + (x^3*(a + b*ArcTanh
[c*Sqrt[x]])^3)/3 - (46*b^2*(a + b*ArcTanh[c*Sqrt[x]])*Log[2/(1 - c*Sqrt[x])])/(15*c^6) - (23*b^3*PolyLog[2, 1
- 2/(1 - c*Sqrt[x])])/(15*c^6)
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 308
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]
Rule 327
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
Rule 2352
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]
Rule 2449
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 6021
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])
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 6055
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 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 6127
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 6131
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]
Rubi steps
\begin {align*} \int x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3 \, dx &=\int x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3 \, dx\\ \end {align*}
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Mathematica [A]
time = 0.58, size = 351, normalized size = 1.15 \begin {gather*} \frac {-19 a b^2+30 a^2 b c \sqrt {x}+19 b^3 c \sqrt {x}+16 a b^2 c^2 x+10 a^2 b c^3 x^{3/2}+b^3 c^3 x^{3/2}+3 a b^2 c^4 x^2+6 a^2 b c^5 x^{5/2}+10 a^3 c^6 x^3+2 b^2 \left (b \left (-23+15 c \sqrt {x}+5 c^3 x^{3/2}+3 c^5 x^{5/2}\right )+15 a \left (-1+c^6 x^3\right )\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^2+10 b^3 \left (-1+c^6 x^3\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^3+b \tanh ^{-1}\left (c \sqrt {x}\right ) \left (30 a^2 c^6 x^3+4 a b c \sqrt {x} \left (15+5 c^2 x+3 c^4 x^2\right )+b^2 \left (-19+16 c^2 x+3 c^4 x^2\right )-92 b^2 \log \left (1+e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right )+15 a^2 b \log \left (1-c \sqrt {x}\right )-15 a^2 b \log \left (1+c \sqrt {x}\right )+46 a b^2 \log \left (1-c^2 x\right )+46 b^3 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )}{30 c^6} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x^2*(a + b*ArcTanh[c*Sqrt[x]])^3,x]
[Out]
(-19*a*b^2 + 30*a^2*b*c*Sqrt[x] + 19*b^3*c*Sqrt[x] + 16*a*b^2*c^2*x + 10*a^2*b*c^3*x^(3/2) + b^3*c^3*x^(3/2) +
3*a*b^2*c^4*x^2 + 6*a^2*b*c^5*x^(5/2) + 10*a^3*c^6*x^3 + 2*b^2*(b*(-23 + 15*c*Sqrt[x] + 5*c^3*x^(3/2) + 3*c^5
*x^(5/2)) + 15*a*(-1 + c^6*x^3))*ArcTanh[c*Sqrt[x]]^2 + 10*b^3*(-1 + c^6*x^3)*ArcTanh[c*Sqrt[x]]^3 + b*ArcTanh
[c*Sqrt[x]]*(30*a^2*c^6*x^3 + 4*a*b*c*Sqrt[x]*(15 + 5*c^2*x + 3*c^4*x^2) + b^2*(-19 + 16*c^2*x + 3*c^4*x^2) -
92*b^2*Log[1 + E^(-2*ArcTanh[c*Sqrt[x]])]) + 15*a^2*b*Log[1 - c*Sqrt[x]] - 15*a^2*b*Log[1 + c*Sqrt[x]] + 46*a*
b^2*Log[1 - c^2*x] + 46*b^3*PolyLog[2, -E^(-2*ArcTanh[c*Sqrt[x]])])/(30*c^6)
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 7.97, size = 1341, normalized size = 4.41
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(\text {Expression too large to display}\) |
\(1341\) |
| default |
\(\text {Expression too large to display}\) |
\(1341\) |
| | |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*(a+b*arctanh(c*x^(1/2)))^3,x,method=_RETURNVERBOSE)
[Out]
2/c^6*(-1/8*I*b^3*arctanh(c*x^(1/2))^2*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)))^2-1/3*b^3+1/5*a*b^2*arctanh(c*x^(1/2))*c^5*x^(5/2)+1/20*a*b^2*c^4*x^2+1/10*a^2*
b*c^5*x^(5/2)+1/6*a^2*b*c^3*x^(3/2)+1/2*a^2*b*c*x^(1/2)+1/2*a*b^2*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)-1/2*a*b^2
*arctanh(c*x^(1/2))*ln(1+c*x^(1/2))-1/4*a*b^2*ln(c*x^(1/2)-1)*ln(1/2*c*x^(1/2)+1/2)-1/4*a*b^2*ln(-1/2*c*x^(1/2
)+1/2)*ln(1+c*x^(1/2))+1/4*a*b^2*ln(-1/2*c*x^(1/2)+1/2)*ln(1/2*c*x^(1/2)+1/2)+1/20*b^3*arctanh(c*x^(1/2))*c^4*
x^2+4/15*b^3*arctanh(c*x^(1/2))*c^2*x+1/10*b^3*arctanh(c*x^(1/2))^2*c^5*x^(5/2)+1/6*b^3*arctanh(c*x^(1/2))^2*c
^3*x^(3/2)+1/2*b^3*arctanh(c*x^(1/2))^2*c*x^(1/2)-1/4*I*b^3*arctanh(c*x^(1/2))^2*Pi+1/6*b^3*c^6*x^3*arctanh(c*
x^(1/2))^3+1/60*b^3*c^3*x^(3/2)+19/60*b^3*c*x^(1/2)+1/8*a*b^2*ln(c*x^(1/2)-1)^2+1/8*a*b^2*ln(1+c*x^(1/2))^2+23
/30*a*b^2*ln(c*x^(1/2)-1)+23/30*a*b^2*ln(1+c*x^(1/2))+1/4*a^2*b*ln(c*x^(1/2)-1)-1/4*a^2*b*ln(1+c*x^(1/2))-23/1
5*b^3*arctanh(c*x^(1/2))*ln(1+I*(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-23/15*b^3*arctanh(c*x^(1/2))*ln(1-I*(1+c*x^(1/
2))/(-c^2*x+1)^(1/2))+1/4*b^3*arctanh(c*x^(1/2))^2*ln(c*x^(1/2)-1)-1/4*b^3*arctanh(c*x^(1/2))^2*ln(1+c*x^(1/2)
)+1/2*b^3*arctanh(c*x^(1/2))^2*ln((1+c*x^(1/2))/(-c^2*x+1)^(1/2))+4/15*a*b^2*x*c^2+1/2*a*b^2*c^6*x^3*arctanh(c
*x^(1/2))^2+1/8*I*b^3*arctanh(c*x^(1/2))^2*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+1/8*I*b^3*arctanh(c*x^(1/2))^2*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))^3+1/4*I*b^3*arctanh(c*x^(1/2))^2*Pi*csg
n(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2-1/4*I*b^3*arctanh(c*x^(1/2))^2*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))
^3+1/2*a^2*b*c^6*x^3*arctanh(c*x^(1/2))+1/3*a*b^2*arctanh(c*x^(1/2))*c^3*x^(3/2)+a*b^2*arctanh(c*x^(1/2))*c*x^
(1/2)+1/6*c^6*x^3*a^3-1/8*I*b^3*arctanh(c*x^(1/2))^2*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)))+1/8*I*b^3*arctanh(c*x^(1/2
))^2*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+1/4*I*b^3*arctanh(c*x^(1/2))^2*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+1/8*I*b^3*arctanh(c*x^(1/2))^2*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)
)+23/30*b^3*arctanh(c*x^(1/2))^2-1/6*b^3*arctanh(c*x^(1/2))^3-23/15*b^3*dilog(1-I*(1+c*x^(1/2))/(-c^2*x+1)^(1/
2))-23/15*b^3*dilog(1+I*(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-19/60*b^3*arctanh(c*x^(1/2)))
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1579 vs.
\(2 (243) = 486\).
time = 0.77, size = 1579, normalized size = 5.19 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*(a+b*arctanh(c*x^(1/2)))^3,x, algorithm="maxima")
[Out]
1/3*a^3*x^3 - 1/720*a*b^2*c*((20*c^5*x^3 + 39*c^3*x^2 + 138*c*x - 6*(10*c^5*x^3 + 12*c^4*x^(5/2) + 15*c^3*x^2
+ 20*c^2*x^(3/2) + 30*c*x + 60*sqrt(x))*log(c*sqrt(x) + 1))/c^6 - 222*log(c*sqrt(x) + 1)/c^7 - 222*log(c*sqrt(
x) - 1)/c^7) - 1/120*(60*x^3*log(c*sqrt(x) + 1) - c*((10*c^5*x^3 - 12*c^4*x^(5/2) + 15*c^3*x^2 - 20*c^2*x^(3/2
) + 30*c*x - 60*sqrt(x))/c^6 + 60*log(c*sqrt(x) + 1)/c^7))*a*b^2*log(-c*sqrt(x) + 1) + 1/120*(60*x^3*log(c*sqr
t(x) + 1) - c*((10*c^5*x^3 - 12*c^4*x^(5/2) + 15*c^3*x^2 - 20*c^2*x^(3/2) + 30*c*x - 60*sqrt(x))/c^6 + 60*log(
c*sqrt(x) + 1)/c^7))*a^2*b - 1/120*(60*x^3*log(-c*sqrt(x) + 1) - c*((10*c^5*x^3 + 12*c^4*x^(5/2) + 15*c^3*x^2
+ 20*c^2*x^(3/2) + 30*c*x + 60*sqrt(x))/c^6 + 60*log(c*sqrt(x) - 1)/c^7))*a^2*b + 1/7200*(100*(18*log(-c*sqrt(
x) + 1)^2 - 6*log(-c*sqrt(x) + 1) + 1)*(c*sqrt(x) - 1)^6 + 432*(25*log(-c*sqrt(x) + 1)^2 - 10*log(-c*sqrt(x) +
1) + 2)*(c*sqrt(x) - 1)^5 + 3375*(8*log(-c*sqrt(x) + 1)^2 - 4*log(-c*sqrt(x) + 1) + 1)*(c*sqrt(x) - 1)^4 + 40
00*(9*log(-c*sqrt(x) + 1)^2 - 6*log(-c*sqrt(x) + 1) + 2)*(c*sqrt(x) - 1)^3 + 13500*(2*log(-c*sqrt(x) + 1)^2 -
2*log(-c*sqrt(x) + 1) + 1)*(c*sqrt(x) - 1)^2 + 10800*(log(-c*sqrt(x) + 1)^2 - 2*log(-c*sqrt(x) + 1) + 2)*(c*sq
rt(x) - 1))*a*b^2/c^6 - 1/864000*(1000*(36*log(-c*sqrt(x) + 1)^3 - 18*log(-c*sqrt(x) + 1)^2 + 6*log(-c*sqrt(x)
+ 1) - 1)*(c*sqrt(x) - 1)^6 + 1728*(125*log(-c*sqrt(x) + 1)^3 - 75*log(-c*sqrt(x) + 1)^2 + 30*log(-c*sqrt(x)
+ 1) - 6)*(c*sqrt(x) - 1)^5 + 16875*(32*log(-c*sqrt(x) + 1)^3 - 24*log(-c*sqrt(x) + 1)^2 + 12*log(-c*sqrt(x) +
1) - 3)*(c*sqrt(x) - 1)^4 + 80000*(9*log(-c*sqrt(x) + 1)^3 - 9*log(-c*sqrt(x) + 1)^2 + 6*log(-c*sqrt(x) + 1)
- 2)*(c*sqrt(x) - 1)^3 + 135000*(4*log(-c*sqrt(x) + 1)^3 - 6*log(-c*sqrt(x) + 1)^2 + 6*log(-c*sqrt(x) + 1) - 3
)*(c*sqrt(x) - 1)^2 + 216000*(log(-c*sqrt(x) + 1)^3 - 3*log(-c*sqrt(x) + 1)^2 + 6*log(-c*sqrt(x) + 1) - 6)*(c*
sqrt(x) - 1))*b^3/c^6 + 23/15*(log(c*sqrt(x) + 1)*log(-1/2*c*sqrt(x) + 1/2) + dilog(1/2*c*sqrt(x) + 1/2))*b^3/
c^6 - 8929/14400*b^3*log(c*sqrt(x) - 1)/c^6 + 1/120*(147*a*b^2 - 38*b^3)*log(c*sqrt(x) + 1)/c^6 + 1/864000*(10
00*(12*a*b^2*c^6 - b^3*c^6)*x^3 + 36000*(b^3*c^6*x^3 - b^3)*log(c*sqrt(x) + 1)^3 - 48*(660*a*b^2*c^5 + 91*b^3*
c^5)*x^(5/2) + 15*(4440*a*b^2*c^4 - 919*b^3*c^4)*x^2 + 14400*(15*a*b^2*c^6*x^3 + 3*b^3*c^5*x^(5/2) + 5*b^3*c^3
*x^(3/2) + 15*b^3*c*sqrt(x) - 15*a*b^2 + 23*b^3)*log(c*sqrt(x) + 1)^2 - 1800*(10*b^3*c^6*x^3 - 12*b^3*c^5*x^(5
/2) + 15*b^3*c^4*x^2 - 20*b^3*c^3*x^(3/2) + 30*b^3*c^2*x - 60*b^3*c*sqrt(x) + 37*b^3 - 60*(b^3*c^6*x^3 - b^3)*
log(c*sqrt(x) + 1))*log(-c*sqrt(x) + 1)^2 - 20*(6840*a*b^2*c^3 + 619*b^3*c^3)*x^(3/2) + 870*(360*a*b^2*c^2 - 1
61*b^3*c^2)*x - 7200*(10*a*b^2*c^6*x^3 - 12*a*b^2*c^5*x^(5/2) - 20*a*b^2*c^3*x^(3/2) - 60*a*b^2*c*sqrt(x) + 3*
(5*a*b^2*c^4 - 2*b^3*c^4)*x^2 + 2*(15*a*b^2*c^2 - 16*b^3*c^2)*x)*log(c*sqrt(x) + 1) + 60*(100*b^3*c^6*x^3 + 26
4*b^3*c^5*x^(5/2) - 165*b^3*c^4*x^2 + 1140*b^3*c^3*x^(3/2) - 1230*b^3*c^2*x + 8820*b^3*c*sqrt(x) - 1800*(b^3*c
^6*x^3 - b^3)*log(c*sqrt(x) + 1)^2 - 480*(3*b^3*c^5*x^(5/2) + 5*b^3*c^3*x^(3/2) + 15*b^3*c*sqrt(x) + 23*b^3)*l
og(c*sqrt(x) + 1))*log(-c*sqrt(x) + 1) - 60*(17640*a*b^2*c + 4369*b^3*c)*sqrt(x))/c^6
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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(x^2*(a+b*arctanh(c*x^(1/2)))^3,x, algorithm="fricas")
[Out]
integral(b^3*x^2*arctanh(c*sqrt(x))^3 + 3*a*b^2*x^2*arctanh(c*sqrt(x))^2 + 3*a^2*b*x^2*arctanh(c*sqrt(x)) + a^
3*x^2, x)
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )^{3}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*(a+b*atanh(c*x**(1/2)))**3,x)
[Out]
Integral(x**2*(a + b*atanh(c*sqrt(x)))**3, x)
________________________________________________________________________________________
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(x^2*(a+b*arctanh(c*x^(1/2)))^3,x, algorithm="giac")
[Out]
integrate((b*arctanh(c*sqrt(x)) + a)^3*x^2, x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,{\left (a+b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*(a + b*atanh(c*x^(1/2)))^3,x)
[Out]
int(x^2*(a + b*atanh(c*x^(1/2)))^3, x)
________________________________________________________________________________________