3.2.96 \(\int x^2 (a+b \tanh ^{-1}(c \sqrt {x}))^2 \, dx\) [196]

Optimal. Leaf size=173 \[ \frac {2 a b \sqrt {x}}{3 c^5}+\frac {8 b^2 x}{45 c^4}+\frac {b^2 x^2}{30 c^2}+\frac {2 b^2 \sqrt {x} \tanh ^{-1}\left (c \sqrt {x}\right )}{3 c^5}+\frac {2 b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{9 c^3}+\frac {2 b x^{5/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{15 c}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{3 c^6}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+\frac {23 b^2 \log \left (1-c^2 x\right )}{45 c^6} \]

[Out]

8/45*b^2*x/c^4+1/30*b^2*x^2/c^2+2/9*b*x^(3/2)*(a+b*arctanh(c*x^(1/2)))/c^3+2/15*b*x^(5/2)*(a+b*arctanh(c*x^(1/
2)))/c-1/3*(a+b*arctanh(c*x^(1/2)))^2/c^6+1/3*x^3*(a+b*arctanh(c*x^(1/2)))^2+23/45*b^2*ln(-c^2*x+1)/c^6+2/3*a*
b*x^(1/2)/c^5+2/3*b^2*arctanh(c*x^(1/2))*x^(1/2)/c^5

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Rubi [A]
time = 0.27, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6039, 6037, 6127, 272, 45, 6021, 266, 6095} \begin {gather*} -\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{3 c^6}+\frac {2 a b \sqrt {x}}{3 c^5}+\frac {2 b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{9 c^3}+\frac {2 b x^{5/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{15 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+\frac {2 b^2 \sqrt {x} \tanh ^{-1}\left (c \sqrt {x}\right )}{3 c^5}+\frac {8 b^2 x}{45 c^4}+\frac {b^2 x^2}{30 c^2}+\frac {23 b^2 \log \left (1-c^2 x\right )}{45 c^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c*Sqrt[x]])^2,x]

[Out]

(2*a*b*Sqrt[x])/(3*c^5) + (8*b^2*x)/(45*c^4) + (b^2*x^2)/(30*c^2) + (2*b^2*Sqrt[x]*ArcTanh[c*Sqrt[x]])/(3*c^5)
+ (2*b*x^(3/2)*(a + b*ArcTanh[c*Sqrt[x]]))/(9*c^3) + (2*b*x^(5/2)*(a + b*ArcTanh[c*Sqrt[x]]))/(15*c) - (a + b
*ArcTanh[c*Sqrt[x]])^2/(3*c^6) + (x^3*(a + b*ArcTanh[c*Sqrt[x]])^2)/3 + (23*b^2*Log[1 - c^2*x])/(45*c^6)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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]

Rubi steps

\begin {align*} \int x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx &=\int x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 194, normalized size = 1.12 \begin {gather*} \frac {60 a b c \sqrt {x}+16 b^2 c^2 x+20 a b c^3 x^{3/2}+3 b^2 c^4 x^2+12 a b c^5 x^{5/2}+30 a^2 c^6 x^3+4 b c \sqrt {x} \left (15 a c^5 x^{5/2}+b \left (15+5 c^2 x+3 c^4 x^2\right )\right ) \tanh ^{-1}\left (c \sqrt {x}\right )+30 b^2 \left (-1+c^6 x^3\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^2+2 b (15 a+23 b) \log \left (1-c \sqrt {x}\right )-30 a b \log \left (1+c \sqrt {x}\right )+46 b^2 \log \left (1+c \sqrt {x}\right )}{90 c^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTanh[c*Sqrt[x]])^2,x]

[Out]

(60*a*b*c*Sqrt[x] + 16*b^2*c^2*x + 20*a*b*c^3*x^(3/2) + 3*b^2*c^4*x^2 + 12*a*b*c^5*x^(5/2) + 30*a^2*c^6*x^3 +
4*b*c*Sqrt[x]*(15*a*c^5*x^(5/2) + b*(15 + 5*c^2*x + 3*c^4*x^2))*ArcTanh[c*Sqrt[x]] + 30*b^2*(-1 + c^6*x^3)*Arc
Tanh[c*Sqrt[x]]^2 + 2*b*(15*a + 23*b)*Log[1 - c*Sqrt[x]] - 30*a*b*Log[1 + c*Sqrt[x]] + 46*b^2*Log[1 + c*Sqrt[x
]])/(90*c^6)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(334\) vs. \(2(137)=274\).
time = 0.18, size = 335, normalized size = 1.94

method	result	size
derivativedivides	\(\frac {\frac {c^{6} x^{3} a^{2}}{3}+\frac {b^{2} c^{6} x^{3} \arctanh \left (c \sqrt {x}\right )^{2}}{3}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{15}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{9}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) c \sqrt {x}}{3}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{3}-\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{3}-\frac {b^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} \ln \left (c \sqrt {x}-1\right )^{2}}{12}+\frac {b^{2} \ln \left (1+c \sqrt {x}\right )^{2}}{12}-\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{6}+\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} c^{4} x^{2}}{30}+\frac {8 b^{2} c^{2} x}{45}+\frac {23 b^{2} \ln \left (c \sqrt {x}-1\right )}{45}+\frac {23 b^{2} \ln \left (1+c \sqrt {x}\right )}{45}+\frac {2 a b \,c^{6} x^{3} \arctanh \left (c \sqrt {x}\right )}{3}+\frac {2 a b \,c^{5} x^{\frac {5}{2}}}{15}+\frac {2 a b \,c^{3} x^{\frac {3}{2}}}{9}+\frac {2 a b c \sqrt {x}}{3}+\frac {a b \ln \left (c \sqrt {x}-1\right )}{3}-\frac {a b \ln \left (1+c \sqrt {x}\right )}{3}}{c^{6}}\)	\(335\)
default	\(\frac {\frac {c^{6} x^{3} a^{2}}{3}+\frac {b^{2} c^{6} x^{3} \arctanh \left (c \sqrt {x}\right )^{2}}{3}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{15}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{9}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) c \sqrt {x}}{3}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{3}-\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{3}-\frac {b^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} \ln \left (c \sqrt {x}-1\right )^{2}}{12}+\frac {b^{2} \ln \left (1+c \sqrt {x}\right )^{2}}{12}-\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{6}+\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} c^{4} x^{2}}{30}+\frac {8 b^{2} c^{2} x}{45}+\frac {23 b^{2} \ln \left (c \sqrt {x}-1\right )}{45}+\frac {23 b^{2} \ln \left (1+c \sqrt {x}\right )}{45}+\frac {2 a b \,c^{6} x^{3} \arctanh \left (c \sqrt {x}\right )}{3}+\frac {2 a b \,c^{5} x^{\frac {5}{2}}}{15}+\frac {2 a b \,c^{3} x^{\frac {3}{2}}}{9}+\frac {2 a b c \sqrt {x}}{3}+\frac {a b \ln \left (c \sqrt {x}-1\right )}{3}-\frac {a b \ln \left (1+c \sqrt {x}\right )}{3}}{c^{6}}\)	\(335\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctanh(c*x^(1/2)))^2,x,method=_RETURNVERBOSE)

[Out]

2/c^6*(1/6*c^6*x^3*a^2+1/6*b^2*c^6*x^3*arctanh(c*x^(1/2))^2+1/15*b^2*arctanh(c*x^(1/2))*c^5*x^(5/2)+1/9*b^2*ar
ctanh(c*x^(1/2))*c^3*x^(3/2)+1/3*b^2*arctanh(c*x^(1/2))*c*x^(1/2)+1/6*b^2*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)-1
/6*b^2*arctanh(c*x^(1/2))*ln(1+c*x^(1/2))-1/12*b^2*ln(c*x^(1/2)-1)*ln(1/2*c*x^(1/2)+1/2)+1/24*b^2*ln(c*x^(1/2)
-1)^2+1/24*b^2*ln(1+c*x^(1/2))^2-1/12*b^2*ln(-1/2*c*x^(1/2)+1/2)*ln(1+c*x^(1/2))+1/12*b^2*ln(-1/2*c*x^(1/2)+1/
2)*ln(1/2*c*x^(1/2)+1/2)+1/60*b^2*c^4*x^2+4/45*b^2*c^2*x+23/90*b^2*ln(c*x^(1/2)-1)+23/90*b^2*ln(1+c*x^(1/2))+1
/3*a*b*c^6*x^3*arctanh(c*x^(1/2))+1/15*a*b*c^5*x^(5/2)+1/9*a*b*c^3*x^(3/2)+1/3*a*b*c*x^(1/2)+1/6*a*b*ln(c*x^(1
/2)-1)-1/6*a*b*ln(1+c*x^(1/2)))

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Maxima [A]
time = 0.26, size = 241, normalized size = 1.39 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} \operatorname {artanh}\left (c \sqrt {x}\right )^{2} + \frac {1}{3} \, a^{2} x^{3} + \frac {1}{45} \, {\left (30 \, x^{3} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{\frac {5}{2}} + 5 \, c^{2} x^{\frac {3}{2}} + 15 \, \sqrt {x}\right )}}{c^{6}} - \frac {15 \, \log \left (c \sqrt {x} + 1\right )}{c^{7}} + \frac {15 \, \log \left (c \sqrt {x} - 1\right )}{c^{7}}\right )}\right )} a b + \frac {1}{180} \, {\left (4 \, c {\left (\frac {2 \, {\left (3 \, c^{4} x^{\frac {5}{2}} + 5 \, c^{2} x^{\frac {3}{2}} + 15 \, \sqrt {x}\right )}}{c^{6}} - \frac {15 \, \log \left (c \sqrt {x} + 1\right )}{c^{7}} + \frac {15 \, \log \left (c \sqrt {x} - 1\right )}{c^{7}}\right )} \operatorname {artanh}\left (c \sqrt {x}\right ) + \frac {6 \, c^{4} x^{2} + 32 \, c^{2} x - 2 \, {\left (15 \, \log \left (c \sqrt {x} - 1\right ) - 46\right )} \log \left (c \sqrt {x} + 1\right ) + 15 \, \log \left (c \sqrt {x} + 1\right )^{2} + 15 \, \log \left (c \sqrt {x} - 1\right )^{2} + 92 \, \log \left (c \sqrt {x} - 1\right )}{c^{6}}\right )} b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x^(1/2)))^2,x, algorithm="maxima")

[Out]

1/3*b^2*x^3*arctanh(c*sqrt(x))^2 + 1/3*a^2*x^3 + 1/45*(30*x^3*arctanh(c*sqrt(x)) + c*(2*(3*c^4*x^(5/2) + 5*c^2
*x^(3/2) + 15*sqrt(x))/c^6 - 15*log(c*sqrt(x) + 1)/c^7 + 15*log(c*sqrt(x) - 1)/c^7))*a*b + 1/180*(4*c*(2*(3*c^
4*x^(5/2) + 5*c^2*x^(3/2) + 15*sqrt(x))/c^6 - 15*log(c*sqrt(x) + 1)/c^7 + 15*log(c*sqrt(x) - 1)/c^7)*arctanh(c
*sqrt(x)) + (6*c^4*x^2 + 32*c^2*x - 2*(15*log(c*sqrt(x) - 1) - 46)*log(c*sqrt(x) + 1) + 15*log(c*sqrt(x) + 1)^
2 + 15*log(c*sqrt(x) - 1)^2 + 92*log(c*sqrt(x) - 1))/c^6)*b^2

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Fricas [A]
time = 0.42, size = 241, normalized size = 1.39 \begin {gather*} \frac {60 \, a^{2} c^{6} x^{3} + 6 \, b^{2} c^{4} x^{2} + 32 \, b^{2} c^{2} x + 15 \, {\left (b^{2} c^{6} x^{3} - b^{2}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right )^{2} + 4 \, {\left (15 \, a b c^{6} - 15 \, a b + 23 \, b^{2}\right )} \log \left (c \sqrt {x} + 1\right ) - 4 \, {\left (15 \, a b c^{6} - 15 \, a b - 23 \, b^{2}\right )} \log \left (c \sqrt {x} - 1\right ) + 4 \, {\left (15 \, a b c^{6} x^{3} - 15 \, a b c^{6} + {\left (3 \, b^{2} c^{5} x^{2} + 5 \, b^{2} c^{3} x + 15 \, b^{2} c\right )} \sqrt {x}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 8 \, {\left (3 \, a b c^{5} x^{2} + 5 \, a b c^{3} x + 15 \, a b c\right )} \sqrt {x}}{180 \, c^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x^(1/2)))^2,x, algorithm="fricas")

[Out]

1/180*(60*a^2*c^6*x^3 + 6*b^2*c^4*x^2 + 32*b^2*c^2*x + 15*(b^2*c^6*x^3 - b^2)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(
c^2*x - 1))^2 + 4*(15*a*b*c^6 - 15*a*b + 23*b^2)*log(c*sqrt(x) + 1) - 4*(15*a*b*c^6 - 15*a*b - 23*b^2)*log(c*s
qrt(x) - 1) + 4*(15*a*b*c^6*x^3 - 15*a*b*c^6 + (3*b^2*c^5*x^2 + 5*b^2*c^3*x + 15*b^2*c)*sqrt(x))*log(-(c^2*x +
2*c*sqrt(x) + 1)/(c^2*x - 1)) + 8*(3*a*b*c^5*x^2 + 5*a*b*c^3*x + 15*a*b*c)*sqrt(x))/c^6

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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 )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*atanh(c*x**(1/2)))**2,x)

[Out]

Integral(x**2*(a + b*atanh(c*sqrt(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(x^2*(a+b*arctanh(c*x^(1/2)))^2,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*sqrt(x)) + a)^2*x^2, x)

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Mupad [B]
time = 1.57, size = 185, normalized size = 1.07 \begin {gather*} \frac {46\,b^2\,\ln \left (c^2\,x-1\right )-30\,b^2\,{\mathrm {atanh}\left (c\,\sqrt {x}\right )}^2-60\,a\,b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+16\,b^2\,c^2\,x+30\,a^2\,c^6\,x^3+3\,b^2\,c^4\,x^2+30\,b^2\,c^6\,x^3\,{\mathrm {atanh}\left (c\,\sqrt {x}\right )}^2+60\,b^2\,c\,\sqrt {x}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+60\,a\,b\,c\,\sqrt {x}+20\,b^2\,c^3\,x^{3/2}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+12\,b^2\,c^5\,x^{5/2}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+20\,a\,b\,c^3\,x^{3/2}+12\,a\,b\,c^5\,x^{5/2}+60\,a\,b\,c^6\,x^3\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{90\,c^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atanh(c*x^(1/2)))^2,x)

[Out]

(46*b^2*log(c^2*x - 1) - 30*b^2*atanh(c*x^(1/2))^2 - 60*a*b*atanh(c*x^(1/2)) + 16*b^2*c^2*x + 30*a^2*c^6*x^3 +
3*b^2*c^4*x^2 + 30*b^2*c^6*x^3*atanh(c*x^(1/2))^2 + 60*b^2*c*x^(1/2)*atanh(c*x^(1/2)) + 60*a*b*c*x^(1/2) + 20
*b^2*c^3*x^(3/2)*atanh(c*x^(1/2)) + 12*b^2*c^5*x^(5/2)*atanh(c*x^(1/2)) + 20*a*b*c^3*x^(3/2) + 12*a*b*c^5*x^(5
/2) + 60*a*b*c^6*x^3*atanh(c*x^(1/2)))/(90*c^6)

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