3.1.15 \(\int x^3 (a+b \tanh ^{-1}(c x))^2 \, dx\) [15]

Optimal. Leaf size=113 \[ \frac {a b x}{2 c^3}+\frac {b^2 x^2}{12 c^2}+\frac {b^2 x \tanh ^{-1}(c x)}{2 c^3}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b^2 \log \left (1-c^2 x^2\right )}{3 c^4} \]

[Out]

1/2*a*b*x/c^3+1/12*b^2*x^2/c^2+1/2*b^2*x*arctanh(c*x)/c^3+1/6*b*x^3*(a+b*arctanh(c*x))/c-1/4*(a+b*arctanh(c*x)
)^2/c^4+1/4*x^4*(a+b*arctanh(c*x))^2+1/3*b^2*ln(-c^2*x^2+1)/c^4

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Rubi [A]
time = 0.17, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6037, 6127, 272, 45, 6021, 266, 6095} \begin {gather*} -\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4}+\frac {a b x}{2 c^3}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {b^2 x \tanh ^{-1}(c x)}{2 c^3}+\frac {b^2 x^2}{12 c^2}+\frac {b^2 \log \left (1-c^2 x^2\right )}{3 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x])^2,x]

[Out]

(a*b*x)/(2*c^3) + (b^2*x^2)/(12*c^2) + (b^2*x*ArcTanh[c*x])/(2*c^3) + (b*x^3*(a + b*ArcTanh[c*x]))/(6*c) - (a
+ b*ArcTanh[c*x])^2/(4*c^4) + (x^4*(a + b*ArcTanh[c*x])^2)/4 + (b^2*Log[1 - c^2*x^2])/(3*c^4)

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 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^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{2} (b c) \int \frac {x^4 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c}-\frac {b \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{2 c}\\ &=\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{6} b^2 \int \frac {x^3}{1-c^2 x^2} \, dx+\frac {b \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{2 c^3}\\ &=\frac {a b x}{2 c^3}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{12} b^2 \text {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )+\frac {b^2 \int \tanh ^{-1}(c x) \, dx}{2 c^3}\\ &=\frac {a b x}{2 c^3}+\frac {b^2 x \tanh ^{-1}(c x)}{2 c^3}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{12} b^2 \text {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )-\frac {b^2 \int \frac {x}{1-c^2 x^2} \, dx}{2 c^2}\\ &=\frac {a b x}{2 c^3}+\frac {b^2 x^2}{12 c^2}+\frac {b^2 x \tanh ^{-1}(c x)}{2 c^3}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b^2 \log \left (1-c^2 x^2\right )}{3 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 132, normalized size = 1.17 \begin {gather*} \frac {6 a b c x+b^2 c^2 x^2+2 a b c^3 x^3+3 a^2 c^4 x^4+2 b c x \left (3 a c^3 x^3+b \left (3+c^2 x^2\right )\right ) \tanh ^{-1}(c x)+3 b^2 \left (-1+c^4 x^4\right ) \tanh ^{-1}(c x)^2+b (3 a+4 b) \log (1-c x)-3 a b \log (1+c x)+4 b^2 \log (1+c x)}{12 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTanh[c*x])^2,x]

[Out]

(6*a*b*c*x + b^2*c^2*x^2 + 2*a*b*c^3*x^3 + 3*a^2*c^4*x^4 + 2*b*c*x*(3*a*c^3*x^3 + b*(3 + c^2*x^2))*ArcTanh[c*x
] + 3*b^2*(-1 + c^4*x^4)*ArcTanh[c*x]^2 + b*(3*a + 4*b)*Log[1 - c*x] - 3*a*b*Log[1 + c*x] + 4*b^2*Log[1 + c*x]
)/(12*c^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(253\) vs. \(2(99)=198\).
time = 0.03, size = 254, normalized size = 2.25

method	result	size
derivativedivides	\(\frac {\frac {c^{4} x^{4} a^{2}}{4}+\frac {b^{2} c^{4} x^{4} \arctanh \left (c x \right )^{2}}{4}+\frac {b^{2} \arctanh \left (c x \right ) c^{3} x^{3}}{6}+\frac {b^{2} \arctanh \left (c x \right ) c x}{2}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{4}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}+\frac {b^{2} \ln \left (c x -1\right )^{2}}{16}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{8}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{16}+\frac {b^{2} c^{2} x^{2}}{12}+\frac {b^{2} \ln \left (c x -1\right )}{3}+\frac {b^{2} \ln \left (c x +1\right )}{3}+\frac {c^{4} x^{4} a b \arctanh \left (c x \right )}{2}+\frac {a b \,c^{3} x^{3}}{6}+\frac {a b c x}{2}+\frac {a b \ln \left (c x -1\right )}{4}-\frac {a b \ln \left (c x +1\right )}{4}}{c^{4}}\)	\(254\)
default	\(\frac {\frac {c^{4} x^{4} a^{2}}{4}+\frac {b^{2} c^{4} x^{4} \arctanh \left (c x \right )^{2}}{4}+\frac {b^{2} \arctanh \left (c x \right ) c^{3} x^{3}}{6}+\frac {b^{2} \arctanh \left (c x \right ) c x}{2}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{4}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}+\frac {b^{2} \ln \left (c x -1\right )^{2}}{16}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{8}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{16}+\frac {b^{2} c^{2} x^{2}}{12}+\frac {b^{2} \ln \left (c x -1\right )}{3}+\frac {b^{2} \ln \left (c x +1\right )}{3}+\frac {c^{4} x^{4} a b \arctanh \left (c x \right )}{2}+\frac {a b \,c^{3} x^{3}}{6}+\frac {a b c x}{2}+\frac {a b \ln \left (c x -1\right )}{4}-\frac {a b \ln \left (c x +1\right )}{4}}{c^{4}}\)	\(254\)
risch	\(\frac {b^{2} \left (c^{4} x^{4}-1\right ) \ln \left (c x +1\right )^{2}}{16 c^{4}}+\frac {b \left (-3 x^{4} b \ln \left (-c x +1\right ) c^{4}+6 c^{4} x^{4} a +2 b \,c^{3} x^{3}+6 b c x +3 b \ln \left (-c x +1\right )\right ) \ln \left (c x +1\right )}{24 c^{4}}+\frac {\ln \left (-c x +1\right )^{2} b^{2} x^{4}}{16}-\frac {\ln \left (-c x +1\right ) a b \,x^{4}}{4}+\frac {a^{2} x^{4}}{4}-\frac {b^{2} x^{3} \ln \left (-c x +1\right )}{12 c}+\frac {a b \,x^{3}}{6 c}+\frac {b^{2} x^{2}}{12 c^{2}}-\frac {b^{2} x \ln \left (-c x +1\right )}{4 c^{3}}-\frac {b^{2} \ln \left (-c x +1\right )^{2}}{16 c^{4}}+\frac {a b x}{2 c^{3}}+\frac {b \ln \left (-c x +1\right ) a}{4 c^{4}}+\frac {b^{2} \ln \left (-c x +1\right )}{3 c^{4}}-\frac {b \ln \left (c x +1\right ) a}{4 c^{4}}+\frac {b^{2} \ln \left (c x +1\right )}{3 c^{4}}\)	\(264\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

1/c^4*(1/4*c^4*x^4*a^2+1/4*b^2*c^4*x^4*arctanh(c*x)^2+1/6*b^2*arctanh(c*x)*c^3*x^3+1/2*b^2*arctanh(c*x)*c*x+1/
4*b^2*arctanh(c*x)*ln(c*x-1)-1/4*b^2*arctanh(c*x)*ln(c*x+1)-1/8*b^2*ln(c*x-1)*ln(1/2*c*x+1/2)+1/16*b^2*ln(c*x-
1)^2+1/8*b^2*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)-1/8*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/16*b^2*ln(c*x+1)^2+1/12*b^2
*c^2*x^2+1/3*b^2*ln(c*x-1)+1/3*b^2*ln(c*x+1)+1/2*c^4*x^4*a*b*arctanh(c*x)+1/6*a*b*c^3*x^3+1/2*a*b*c*x+1/4*a*b*
ln(c*x-1)-1/4*a*b*ln(c*x+1))

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Maxima [A]
time = 0.26, size = 189, normalized size = 1.67 \begin {gather*} \frac {1}{4} \, b^{2} x^{4} \operatorname {artanh}\left (c x\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{12} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} a b + \frac {1}{48} \, {\left (4 \, c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )} \operatorname {artanh}\left (c x\right ) + \frac {4 \, c^{2} x^{2} - 2 \, {\left (3 \, \log \left (c x - 1\right ) - 8\right )} \log \left (c x + 1\right ) + 3 \, \log \left (c x + 1\right )^{2} + 3 \, \log \left (c x - 1\right )^{2} + 16 \, \log \left (c x - 1\right )}{c^{4}}\right )} b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))^2,x, algorithm="maxima")

[Out]

1/4*b^2*x^4*arctanh(c*x)^2 + 1/4*a^2*x^4 + 1/12*(6*x^4*arctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1
)/c^5 + 3*log(c*x - 1)/c^5))*a*b + 1/48*(4*c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5)
*arctanh(c*x) + (4*c^2*x^2 - 2*(3*log(c*x - 1) - 8)*log(c*x + 1) + 3*log(c*x + 1)^2 + 3*log(c*x - 1)^2 + 16*lo
g(c*x - 1))/c^4)*b^2

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Fricas [A]
time = 0.35, size = 160, normalized size = 1.42 \begin {gather*} \frac {12 \, a^{2} c^{4} x^{4} + 8 \, a b c^{3} x^{3} + 4 \, b^{2} c^{2} x^{2} + 24 \, a b c x + 3 \, {\left (b^{2} c^{4} x^{4} - b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} - 4 \, {\left (3 \, a b - 4 \, b^{2}\right )} \log \left (c x + 1\right ) + 4 \, {\left (3 \, a b + 4 \, b^{2}\right )} \log \left (c x - 1\right ) + 4 \, {\left (3 \, a b c^{4} x^{4} + b^{2} c^{3} x^{3} + 3 \, b^{2} c x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{48 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))^2,x, algorithm="fricas")

[Out]

1/48*(12*a^2*c^4*x^4 + 8*a*b*c^3*x^3 + 4*b^2*c^2*x^2 + 24*a*b*c*x + 3*(b^2*c^4*x^4 - b^2)*log(-(c*x + 1)/(c*x
- 1))^2 - 4*(3*a*b - 4*b^2)*log(c*x + 1) + 4*(3*a*b + 4*b^2)*log(c*x - 1) + 4*(3*a*b*c^4*x^4 + b^2*c^3*x^3 + 3
*b^2*c*x)*log(-(c*x + 1)/(c*x - 1)))/c^4

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Sympy [A]
time = 0.39, size = 168, normalized size = 1.49 \begin {gather*} \begin {cases} \frac {a^{2} x^{4}}{4} + \frac {a b x^{4} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {a b x^{3}}{6 c} + \frac {a b x}{2 c^{3}} - \frac {a b \operatorname {atanh}{\left (c x \right )}}{2 c^{4}} + \frac {b^{2} x^{4} \operatorname {atanh}^{2}{\left (c x \right )}}{4} + \frac {b^{2} x^{3} \operatorname {atanh}{\left (c x \right )}}{6 c} + \frac {b^{2} x^{2}}{12 c^{2}} + \frac {b^{2} x \operatorname {atanh}{\left (c x \right )}}{2 c^{3}} + \frac {2 b^{2} \log {\left (x - \frac {1}{c} \right )}}{3 c^{4}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{4 c^{4}} + \frac {2 b^{2} \operatorname {atanh}{\left (c x \right )}}{3 c^{4}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{4}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c*x))**2,x)

[Out]

Piecewise((a**2*x**4/4 + a*b*x**4*atanh(c*x)/2 + a*b*x**3/(6*c) + a*b*x/(2*c**3) - a*b*atanh(c*x)/(2*c**4) + b
**2*x**4*atanh(c*x)**2/4 + b**2*x**3*atanh(c*x)/(6*c) + b**2*x**2/(12*c**2) + b**2*x*atanh(c*x)/(2*c**3) + 2*b
**2*log(x - 1/c)/(3*c**4) - b**2*atanh(c*x)**2/(4*c**4) + 2*b**2*atanh(c*x)/(3*c**4), Ne(c, 0)), (a**2*x**4/4,
True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 603 vs. \(2 (99) = 198\).
time = 0.43, size = 603, normalized size = 5.34 \begin {gather*} \frac {1}{6} \, {\left (\frac {3 \, {\left (\frac {{\left (c x + 1\right )}^{3} b^{2}}{{\left (c x - 1\right )}^{3}} + \frac {{\left (c x + 1\right )} b^{2}}{c x - 1}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2}}{\frac {{\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}} + \frac {2 \, {\left (\frac {6 \, {\left (c x + 1\right )}^{3} a b}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )} a b}{c x - 1} + \frac {3 \, {\left (c x + 1\right )}^{3} b^{2}}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} b^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} b^{2}}{c x - 1} - 2 \, b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}} + \frac {2 \, {\left (\frac {6 \, {\left (c x + 1\right )}^{3} a^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )} a^{2}}{c x - 1} + \frac {6 \, {\left (c x + 1\right )}^{3} a b}{{\left (c x - 1\right )}^{3}} - \frac {12 \, {\left (c x + 1\right )}^{2} a b}{{\left (c x - 1\right )}^{2}} + \frac {10 \, {\left (c x + 1\right )} a b}{c x - 1} - 4 \, a b + \frac {{\left (c x + 1\right )}^{3} b^{2}}{{\left (c x - 1\right )}^{3}} - \frac {2 \, {\left (c x + 1\right )}^{2} b^{2}}{{\left (c x - 1\right )}^{2}} + \frac {{\left (c x + 1\right )} b^{2}}{c x - 1}\right )}}{\frac {{\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}} - \frac {4 \, b^{2} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{5}} + \frac {4 \, b^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{5}}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))^2,x, algorithm="giac")

[Out]

1/6*(3*((c*x + 1)^3*b^2/(c*x - 1)^3 + (c*x + 1)*b^2/(c*x - 1))*log(-(c*x + 1)/(c*x - 1))^2/((c*x + 1)^4*c^5/(c
*x - 1)^4 - 4*(c*x + 1)^3*c^5/(c*x - 1)^3 + 6*(c*x + 1)^2*c^5/(c*x - 1)^2 - 4*(c*x + 1)*c^5/(c*x - 1) + c^5) +
2*(6*(c*x + 1)^3*a*b/(c*x - 1)^3 + 6*(c*x + 1)*a*b/(c*x - 1) + 3*(c*x + 1)^3*b^2/(c*x - 1)^3 - 6*(c*x + 1)^2*
b^2/(c*x - 1)^2 + 5*(c*x + 1)*b^2/(c*x - 1) - 2*b^2)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^4*c^5/(c*x - 1)^4 -
4*(c*x + 1)^3*c^5/(c*x - 1)^3 + 6*(c*x + 1)^2*c^5/(c*x - 1)^2 - 4*(c*x + 1)*c^5/(c*x - 1) + c^5) + 2*(6*(c*x +
1)^3*a^2/(c*x - 1)^3 + 6*(c*x + 1)*a^2/(c*x - 1) + 6*(c*x + 1)^3*a*b/(c*x - 1)^3 - 12*(c*x + 1)^2*a*b/(c*x -
1)^2 + 10*(c*x + 1)*a*b/(c*x - 1) - 4*a*b + (c*x + 1)^3*b^2/(c*x - 1)^3 - 2*(c*x + 1)^2*b^2/(c*x - 1)^2 + (c*x
+ 1)*b^2/(c*x - 1))/((c*x + 1)^4*c^5/(c*x - 1)^4 - 4*(c*x + 1)^3*c^5/(c*x - 1)^3 + 6*(c*x + 1)^2*c^5/(c*x - 1
)^2 - 4*(c*x + 1)*c^5/(c*x - 1) + c^5) - 4*b^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^5 + 4*b^2*log(-(c*x + 1)/(c*x -
1))/c^5)*c

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Mupad [B]
time = 0.92, size = 134, normalized size = 1.19 \begin {gather*} \frac {4\,b^2\,\ln \left (c^2\,x^2-1\right )-3\,b^2\,{\mathrm {atanh}\left (c\,x\right )}^2+3\,a^2\,c^4\,x^4+b^2\,c^2\,x^2-6\,a\,b\,\mathrm {atanh}\left (c\,x\right )+2\,b^2\,c^3\,x^3\,\mathrm {atanh}\left (c\,x\right )+6\,b^2\,c\,x\,\mathrm {atanh}\left (c\,x\right )+3\,b^2\,c^4\,x^4\,{\mathrm {atanh}\left (c\,x\right )}^2+2\,a\,b\,c^3\,x^3+6\,a\,b\,c\,x+6\,a\,b\,c^4\,x^4\,\mathrm {atanh}\left (c\,x\right )}{12\,c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x))^2,x)

[Out]

(4*b^2*log(c^2*x^2 - 1) - 3*b^2*atanh(c*x)^2 + 3*a^2*c^4*x^4 + b^2*c^2*x^2 - 6*a*b*atanh(c*x) + 2*b^2*c^3*x^3*
atanh(c*x) + 6*b^2*c*x*atanh(c*x) + 3*b^2*c^4*x^4*atanh(c*x)^2 + 2*a*b*c^3*x^3 + 6*a*b*c*x + 6*a*b*c^4*x^4*ata
nh(c*x))/(12*c^4)

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