3.1.66 \(\int x^3 (a+b \tanh ^{-1}(c x^2))^2 \, dx\) [66]

Optimal. Leaf size=91 \[ \frac {a b x^2}{2 c}+\frac {b^2 x^2 \tanh ^{-1}\left (c x^2\right )}{2 c}-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {b^2 \log \left (1-c^2 x^4\right )}{4 c^2} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6039, 6037, 6127, 6021, 266, 6095} \begin {gather*} -\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 c^2}+\frac {a b x^2}{2 c}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {b^2 \log \left (1-c^2 x^4\right )}{4 c^2}+\frac {b^2 x^2 \tanh ^{-1}\left (c x^2\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x^2])^2,x]

[Out]

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

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 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^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x^3 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {1}{2} b x^3 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{4} b^2 x^3 \log ^2\left (1+c x^2\right )\right ) \, dx\\ &=\frac {1}{4} \int x^3 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \, dx-\frac {1}{2} b \int x^3 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right ) \, dx+\frac {1}{4} b^2 \int x^3 \log ^2\left (1+c x^2\right ) \, dx\\ &=\frac {1}{8} \text {Subst}\left (\int x (2 a-b \log (1-c x))^2 \, dx,x,x^2\right )-\frac {1}{4} b \text {Subst}\left (\int x (-2 a+b \log (1-c x)) \log (1+c x) \, dx,x,x^2\right )+\frac {1}{8} b^2 \text {Subst}\left (\int x \log ^2(1+c x) \, dx,x,x^2\right )\\ &=\frac {1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{8} \text {Subst}\left (\int \left (\frac {(2 a-b \log (1-c x))^2}{c}-\frac {(1-c x) (2 a-b \log (1-c x))^2}{c}\right ) \, dx,x,x^2\right )+\frac {1}{8} b^2 \text {Subst}\left (\int \left (-\frac {\log ^2(1+c x)}{c}+\frac {(1+c x) \log ^2(1+c x)}{c}\right ) \, dx,x,x^2\right )+\frac {1}{8} (b c) \text {Subst}\left (\int \frac {x^2 (-2 a+b \log (1-c x))}{1+c x} \, dx,x,x^2\right )-\frac {1}{8} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^2 \log (1+c x)}{1-c x} \, dx,x,x^2\right )\\ &=\frac {1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {\text {Subst}\left (\int (2 a-b \log (1-c x))^2 \, dx,x,x^2\right )}{8 c}-\frac {\text {Subst}\left (\int (1-c x) (2 a-b \log (1-c x))^2 \, dx,x,x^2\right )}{8 c}-\frac {b^2 \text {Subst}\left (\int \log ^2(1+c x) \, dx,x,x^2\right )}{8 c}+\frac {b^2 \text {Subst}\left (\int (1+c x) \log ^2(1+c x) \, dx,x,x^2\right )}{8 c}+\frac {1}{8} (b c) \text {Subst}\left (\int \left (-\frac {-2 a+b \log (1-c x)}{c^2}+\frac {x (-2 a+b \log (1-c x))}{c}+\frac {-2 a+b \log (1-c x)}{c^2 (1+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{8} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {\log (1+c x)}{c^2}-\frac {x \log (1+c x)}{c}-\frac {\log (1+c x)}{c^2 (-1+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{8} b \text {Subst}\left (\int x (-2 a+b \log (1-c x)) \, dx,x,x^2\right )+\frac {1}{8} b^2 \text {Subst}\left (\int x \log (1+c x) \, dx,x,x^2\right )-\frac {\text {Subst}\left (\int (2 a-b \log (x))^2 \, dx,x,1-c x^2\right )}{8 c^2}+\frac {\text {Subst}\left (\int x (2 a-b \log (x))^2 \, dx,x,1-c x^2\right )}{8 c^2}-\frac {b^2 \text {Subst}\left (\int \log ^2(x) \, dx,x,1+c x^2\right )}{8 c^2}+\frac {b^2 \text {Subst}\left (\int x \log ^2(x) \, dx,x,1+c x^2\right )}{8 c^2}-\frac {b \text {Subst}\left (\int (-2 a+b \log (1-c x)) \, dx,x,x^2\right )}{8 c}+\frac {b \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^2\right )}{8 c}+\frac {b^2 \text {Subst}\left (\int \log (1+c x) \, dx,x,x^2\right )}{8 c}+\frac {b^2 \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^2\right )}{8 c}\\ &=\frac {a b x^2}{4 c}-\frac {1}{16} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c^2}+\frac {1}{16} b^2 x^4 \log \left (1+c x^2\right )+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 c^2}+\frac {1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c^2}+\frac {b^2 \left (1+c x^2\right )^2 \log ^2\left (1+c x^2\right )}{16 c^2}+\frac {b \text {Subst}\left (\int x (2 a-b \log (x)) \, dx,x,1-c x^2\right )}{8 c^2}-\frac {b \text {Subst}\left (\int (2 a-b \log (x)) \, dx,x,1-c x^2\right )}{4 c^2}+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{8 c^2}-\frac {b^2 \text {Subst}\left (\int x \log (x) \, dx,x,1+c x^2\right )}{8 c^2}+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{4 c^2}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )}{8 c}-\frac {b^2 \text {Subst}\left (\int \log (1-c x) \, dx,x,x^2\right )}{8 c}+\frac {b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )}{8 c}+\frac {1}{16} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^2}{1-c x} \, dx,x,x^2\right )-\frac {1}{16} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^2}{1+c x} \, dx,x,x^2\right )\\ &=\frac {3 a b x^2}{4 c}-\frac {3 b^2 x^2}{8 c}+\frac {b^2 \left (1-c x^2\right )^2}{32 c^2}+\frac {b^2 \left (1+c x^2\right )^2}{32 c^2}-\frac {1}{16} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac {b \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )}{16 c^2}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c^2}+\frac {1}{16} b^2 x^4 \log \left (1+c x^2\right )+\frac {3 b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{8 c^2}-\frac {b^2 \left (1+c x^2\right )^2 \log \left (1+c x^2\right )}{16 c^2}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 c^2}+\frac {1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c^2}+\frac {b^2 \left (1+c x^2\right )^2 \log ^2\left (1+c x^2\right )}{16 c^2}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{8 c^2}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{8 c^2}+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{8 c^2}+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{4 c^2}+\frac {1}{16} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{16} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {1}{c^2}+\frac {x}{c}+\frac {1}{c^2 (1+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {3 a b x^2}{4 c}-\frac {b^2 x^4}{16}+\frac {b^2 \left (1-c x^2\right )^2}{32 c^2}+\frac {b^2 \left (1+c x^2\right )^2}{32 c^2}-\frac {b^2 \log \left (1-c x^2\right )}{16 c^2}+\frac {3 b^2 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{8 c^2}-\frac {1}{16} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac {b \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )}{16 c^2}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c^2}-\frac {b^2 \log \left (1+c x^2\right )}{16 c^2}+\frac {1}{16} b^2 x^4 \log \left (1+c x^2\right )+\frac {3 b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{8 c^2}-\frac {b^2 \left (1+c x^2\right )^2 \log \left (1+c x^2\right )}{16 c^2}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 c^2}+\frac {1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c^2}+\frac {b^2 \left (1+c x^2\right )^2 \log ^2\left (1+c x^2\right )}{16 c^2}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{8 c^2}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c^2}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 106, normalized size = 1.16 \begin {gather*} \frac {2 a b c x^2+a^2 c^2 x^4+2 b c x^2 \left (b+a c x^2\right ) \tanh ^{-1}\left (c x^2\right )+b^2 \left (-1+c^2 x^4\right ) \tanh ^{-1}\left (c x^2\right )^2+b (a+b) \log \left (1-c x^2\right )-a b \log \left (1+c x^2\right )+b^2 \log \left (1+c x^2\right )}{4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(286\) vs. \(2(81)=162\).
time = 0.17, size = 287, normalized size = 3.15

method	result	size
risch	\(\frac {b^{2} \left (c^{2} x^{4}-1\right ) \ln \left (c \,x^{2}+1\right )^{2}}{16 c^{2}}+\frac {b \left (-2 x^{4} b \ln \left (-c \,x^{2}+1\right ) a \,c^{2}+4 a^{2} c^{2} x^{4}+4 a b c \,x^{2}+2 b \ln \left (-c \,x^{2}+1\right ) a +b^{2}\right ) \ln \left (c \,x^{2}+1\right )}{16 c^{2} a}+\frac {b^{2} x^{4} \ln \left (-c \,x^{2}+1\right )^{2}}{16}-\frac {a b \,x^{4} \ln \left (-c \,x^{2}+1\right )}{4}+\frac {a^{2} x^{4}}{4}-\frac {b^{2} x^{2} \ln \left (-c \,x^{2}+1\right )}{4 c}+\frac {a b \,x^{2}}{2 c}-\frac {b^{2} \ln \left (-c \,x^{2}+1\right )^{2}}{16 c^{2}}+\frac {a \ln \left (-c \,x^{2}+1\right ) b}{4 c^{2}}+\frac {b^{2} \ln \left (-c \,x^{2}+1\right )}{4 c^{2}}-\frac {a \ln \left (-c \,x^{2}-1\right ) b}{4 c^{2}}+\frac {\ln \left (-c \,x^{2}-1\right ) b^{2}}{4 c^{2}}-\frac {\ln \left (-c \,x^{2}-1\right ) b^{3}}{16 c^{2} a}+\frac {b^{2}}{4 c^{2}}\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (81) = 162\).
time = 0.26, size = 186, normalized size = 2.04 \begin {gather*} \frac {1}{4} \, b^{2} x^{4} \operatorname {artanh}\left (c x^{2}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{4} \, {\left (2 \, x^{4} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {\log \left (c x^{2} + 1\right )}{c^{3}} + \frac {\log \left (c x^{2} - 1\right )}{c^{3}}\right )}\right )} a b + \frac {1}{16} \, {\left (4 \, c {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {\log \left (c x^{2} + 1\right )}{c^{3}} + \frac {\log \left (c x^{2} - 1\right )}{c^{3}}\right )} \operatorname {artanh}\left (c x^{2}\right ) - \frac {2 \, {\left (\log \left (c x^{2} - 1\right ) - 2\right )} \log \left (c x^{2} + 1\right ) - \log \left (c x^{2} + 1\right )^{2} - \log \left (c x^{2} - 1\right )^{2} - 4 \, \log \left (c x^{2} - 1\right )}{c^{2}}\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))^2,x, algorithm="maxima")

[Out]

1/4*b^2*x^4*arctanh(c*x^2)^2 + 1/4*a^2*x^4 + 1/4*(2*x^4*arctanh(c*x^2) + c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + l
og(c*x^2 - 1)/c^3))*a*b + 1/16*(4*c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c^3)*arctanh(c*x^2) - (2*
(log(c*x^2 - 1) - 2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 - log(c*x^2 - 1)^2 - 4*log(c*x^2 - 1))/c^2)*b^2

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Fricas [A]
time = 0.34, size = 138, normalized size = 1.52 \begin {gather*} \frac {4 \, a^{2} c^{2} x^{4} + 8 \, a b c x^{2} + {\left (b^{2} c^{2} x^{4} - b^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2} - 4 \, {\left (a b - b^{2}\right )} \log \left (c x^{2} + 1\right ) + 4 \, {\left (a b + b^{2}\right )} \log \left (c x^{2} - 1\right ) + 4 \, {\left (a b c^{2} x^{4} + b^{2} c x^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{16 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*a^2*c^2*x^4 + 8*a*b*c*x^2 + (b^2*c^2*x^4 - b^2)*log(-(c*x^2 + 1)/(c*x^2 - 1))^2 - 4*(a*b - b^2)*log(c*
x^2 + 1) + 4*(a*b + b^2)*log(c*x^2 - 1) + 4*(a*b*c^2*x^4 + b^2*c*x^2)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (78) = 156\).
time = 5.29, size = 163, normalized size = 1.79 \begin {gather*} \begin {cases} \frac {a^{2} x^{4}}{4} + \frac {a b x^{4} \operatorname {atanh}{\left (c x^{2} \right )}}{2} + \frac {a b x^{2}}{2 c} - \frac {a b \operatorname {atanh}{\left (c x^{2} \right )}}{2 c^{2}} + \frac {b^{2} x^{4} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{4} + \frac {b^{2} x^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{2 c} + \frac {b^{2} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{2 c^{2}} + \frac {b^{2} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{2 c^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{4 c^{2}} - \frac {b^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{2 c^{2}} & \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))**2,x)

[Out]

Piecewise((a**2*x**4/4 + a*b*x**4*atanh(c*x**2)/2 + a*b*x**2/(2*c) - a*b*atanh(c*x**2)/(2*c**2) + b**2*x**4*at
anh(c*x**2)**2/4 + b**2*x**2*atanh(c*x**2)/(2*c) + b**2*log(x - sqrt(-1/c))/(2*c**2) + b**2*log(x + sqrt(-1/c)
)/(2*c**2) - b**2*atanh(c*x**2)**2/(4*c**2) - b**2*atanh(c*x**2)/(2*c**2), 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. 361 vs. \(2 (81) = 162\).
time = 0.45, size = 361, normalized size = 3.97 \begin {gather*} \frac {1}{4} \, {\left (\frac {{\left (c x^{2} + 1\right )} b^{2} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2}}{{\left (\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}\right )} {\left (c x^{2} - 1\right )}} + \frac {2 \, {\left (\frac {2 \, {\left (c x^{2} + 1\right )} a b}{c x^{2} - 1} + \frac {{\left (c x^{2} + 1\right )} b^{2}}{c x^{2} - 1} - b^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}} + \frac {4 \, {\left (\frac {{\left (c x^{2} + 1\right )} a^{2}}{c x^{2} - 1} + \frac {{\left (c x^{2} + 1\right )} a b}{c x^{2} - 1} - a b\right )}}{\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}} - \frac {2 \, b^{2} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1} + 1\right )}{c^{3}} + \frac {2 \, b^{2} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{c^{3}}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.26, size = 275, normalized size = 3.02 \begin {gather*} \frac {a^2\,x^4}{4}+\frac {b^2\,\ln \left (c\,x^2-1\right )}{4\,c^2}+\frac {b^2\,\ln \left (c\,x^2+1\right )}{4\,c^2}-\frac {b^2\,{\ln \left (c\,x^2+1\right )}^2}{16\,c^2}-\frac {b^2\,{\ln \left (1-c\,x^2\right )}^2}{16\,c^2}+\frac {b^2\,x^4\,{\ln \left (c\,x^2+1\right )}^2}{16}+\frac {b^2\,x^4\,{\ln \left (1-c\,x^2\right )}^2}{16}+\frac {b^2\,x^2\,\ln \left (c\,x^2+1\right )}{4\,c}-\frac {b^2\,x^2\,\ln \left (1-c\,x^2\right )}{4\,c}+\frac {a\,b\,\ln \left (c\,x^2-1\right )}{4\,c^2}-\frac {a\,b\,\ln \left (c\,x^2+1\right )}{4\,c^2}+\frac {a\,b\,x^4\,\ln \left (c\,x^2+1\right )}{4}-\frac {a\,b\,x^4\,\ln \left (1-c\,x^2\right )}{4}+\frac {b^2\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{8\,c^2}+\frac {a\,b\,x^2}{2\,c}-\frac {b^2\,x^4\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x^2))^2,x)

[Out]

(a^2*x^4)/4 + (b^2*log(c*x^2 - 1))/(4*c^2) + (b^2*log(c*x^2 + 1))/(4*c^2) - (b^2*log(c*x^2 + 1)^2)/(16*c^2) -
(b^2*log(1 - c*x^2)^2)/(16*c^2) + (b^2*x^4*log(c*x^2 + 1)^2)/16 + (b^2*x^4*log(1 - c*x^2)^2)/16 + (b^2*x^2*log
(c*x^2 + 1))/(4*c) - (b^2*x^2*log(1 - c*x^2))/(4*c) + (a*b*log(c*x^2 - 1))/(4*c^2) - (a*b*log(c*x^2 + 1))/(4*c
^2) + (a*b*x^4*log(c*x^2 + 1))/4 - (a*b*x^4*log(1 - c*x^2))/4 + (b^2*log(c*x^2 + 1)*log(1 - c*x^2))/(8*c^2) +
(a*b*x^2)/(2*c) - (b^2*x^4*log(c*x^2 + 1)*log(1 - c*x^2))/8

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