3.1.70 \(\int \frac {(a+b \tanh ^{-1}(c x^2))^2}{x^5} \, dx\) [70]

Optimal. Leaf size=88 \[ -\frac {b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{2 x^2}+\frac {1}{4} c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 x^4}+b^2 c^2 \log (x)-\frac {1}{4} b^2 c^2 \log \left (1-c^2 x^4\right ) \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6039, 6037, 6129, 272, 36, 29, 31, 6095} \begin {gather*} \frac {1}{4} c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{2 x^2}-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 x^4}-\frac {1}{4} b^2 c^2 \log \left (1-c^2 x^4\right )+b^2 c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])^2/x^5,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{x^5} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{4 x^5}-\frac {b \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{2 x^5}+\frac {b^2 \log ^2\left (1+c x^2\right )}{4 x^5}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{x^5} \, dx-\frac {1}{2} b \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{x^5} \, dx+\frac {1}{4} b^2 \int \frac {\log ^2\left (1+c x^2\right )}{x^5} \, dx\\ &=\frac {1}{8} \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x^3} \, dx,x,x^2\right )-\frac {1}{4} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log (1+c x)}{x^3} \, dx,x,x^2\right )+\frac {1}{8} b^2 \text {Subst}\left (\int \frac {\log ^2(1+c x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}-\frac {b^2 \log ^2\left (1+c x^2\right )}{16 x^4}+\frac {1}{8} (b c) \text {Subst}\left (\int \frac {2 a-b \log (1-c x)}{x^2 (1-c x)} \, dx,x,x^2\right )-\frac {1}{8} (b c) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x^2 (1+c x)} \, dx,x,x^2\right )+\frac {1}{8} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x^2 (1-c x)} \, dx,x,x^2\right )+\frac {1}{8} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x^2 (1+c x)} \, dx,x,x^2\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}-\frac {b^2 \log ^2\left (1+c x^2\right )}{16 x^4}-\frac {1}{8} b \text {Subst}\left (\int \frac {2 a-b \log (x)}{x \left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^2\right )-\frac {1}{8} (b c) \text {Subst}\left (\int \left (\frac {-2 a+b \log (1-c x)}{x^2}-\frac {c (-2 a+b \log (1-c x))}{x}+\frac {c^2 (-2 a+b \log (1-c x))}{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)}{x^2}+\frac {c \log (1+c x)}{x}-\frac {c^2 \log (1+c x)}{-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)}{x^2}-\frac {c \log (1+c x)}{x}+\frac {c^2 \log (1+c x)}{1+c x}\right ) \, dx,x,x^2\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}-\frac {b^2 \log ^2\left (1+c x^2\right )}{16 x^4}-\frac {1}{8} b \text {Subst}\left (\int \frac {2 a-b \log (x)}{\left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^2\right )-\frac {1}{8} (b c) \text {Subst}\left (\int \frac {2 a-b \log (x)}{x \left (\frac {1}{c}-\frac {x}{c}\right )} \, dx,x,1-c x^2\right )-\frac {1}{8} (b c) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x^2} \, dx,x,x^2\right )+2 \left (\frac {1}{8} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x^2} \, dx,x,x^2\right )\right )+\frac {1}{8} \left (b c^2\right ) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x} \, dx,x,x^2\right )-\frac {1}{8} \left (b c^3\right ) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^2\right )-\frac {1}{8} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^2\right )+\frac {1}{8} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{1+c x} \, dx,x,x^2\right )\\ &=-\frac {1}{2} a b c^2 \log (x)-\frac {b c \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac {b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}+\frac {1}{8} b c^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )-\frac {1}{8} b^2 c^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}-\frac {b^2 \log ^2\left (1+c x^2\right )}{16 x^4}-\frac {1}{8} (b c) \text {Subst}\left (\int \frac {2 a-b \log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^2\right )-\frac {1}{8} \left (b^2 c\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^2\right )-\frac {1}{8} \left (b c^2\right ) \text {Subst}\left (\int \frac {2 a-b \log (x)}{x} \, dx,x,1-c x^2\right )+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x (1-c x)} \, dx,x,x^2\right )+2 \left (-\frac {b^2 c \log \left (1+c x^2\right )}{8 x^2}+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x (1+c x)} \, dx,x,x^2\right )\right )+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+c x^2\right )+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,x^2\right )+\frac {1}{8} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )-\frac {1}{8} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )\\ &=\frac {1}{4} b^2 c^2 \log (x)-\frac {b c \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac {b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}+\frac {1}{16} c^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}+\frac {1}{8} b c^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )-\frac {1}{8} b^2 c^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}+\frac {1}{16} b^2 c^2 \log ^2\left (1+c x^2\right )-\frac {b^2 \log ^2\left (1+c x^2\right )}{16 x^4}-\frac {1}{8} b^2 c^2 \text {Li}_2\left (c x^2\right )+\frac {1}{8} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^2\right )+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )+\frac {1}{8} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {1}{1-c x} \, dx,x,x^2\right )+2 \left (-\frac {b^2 c \log \left (1+c x^2\right )}{8 x^2}+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{8} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {1}{1+c x} \, dx,x,x^2\right )\right )\\ &=\frac {1}{2} b^2 c^2 \log (x)-\frac {1}{8} b^2 c^2 \log \left (1-c x^2\right )-\frac {b c \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac {b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}+\frac {1}{16} c^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}+\frac {1}{8} b c^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )-\frac {1}{8} b^2 c^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}+\frac {1}{16} b^2 c^2 \log ^2\left (1+c x^2\right )-\frac {b^2 \log ^2\left (1+c x^2\right )}{16 x^4}+2 \left (\frac {1}{4} b^2 c^2 \log (x)-\frac {1}{8} b^2 c^2 \log \left (1+c x^2\right )-\frac {b^2 c \log \left (1+c x^2\right )}{8 x^2}\right )-\frac {1}{8} b^2 c^2 \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )-\frac {1}{8} b^2 c^2 \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 111, normalized size = 1.26 \begin {gather*} \frac {1}{4} \left (-\frac {a^2}{x^4}-\frac {2 a b c}{x^2}-\frac {2 b \left (a+b c x^2\right ) \tanh ^{-1}\left (c x^2\right )}{x^4}+\frac {b^2 \left (-1+c^2 x^4\right ) \tanh ^{-1}\left (c x^2\right )^2}{x^4}+4 b^2 c^2 \log (x)-b (a+b) c^2 \log \left (1-c x^2\right )+(a-b) b c^2 \log \left (1+c x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])^2/x^5,x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(256\) vs. \(2(80)=160\).
time = 0.15, size = 257, normalized size = 2.92

method	result	size
risch	\(\frac {b^{2} \left (c^{2} x^{4}-1\right ) \ln \left (c \,x^{2}+1\right )^{2}}{16 x^{4}}-\frac {b \left (b \,c^{2} \ln \left (-c \,x^{2}+1\right ) x^{4}+2 b c \,x^{2}-b \ln \left (-c \,x^{2}+1\right )+2 a \right ) \ln \left (c \,x^{2}+1\right )}{8 x^{4}}+\frac {b^{2} c^{2} x^{4} \ln \left (-c \,x^{2}+1\right )^{2}+16 b^{2} c^{2} \ln \left (x \right ) x^{4}-4 b \,c^{2} \ln \left (c \,x^{2}-1\right ) x^{4} a -4 b^{2} c^{2} \ln \left (c \,x^{2}-1\right ) x^{4}+4 b \,c^{2} \ln \left (c \,x^{2}+1\right ) x^{4} a -4 b^{2} c^{2} \ln \left (c \,x^{2}+1\right ) x^{4}+4 b^{2} c \,x^{2} \ln \left (-c \,x^{2}+1\right )-8 a b c \,x^{2}-b^{2} \ln \left (-c \,x^{2}+1\right )^{2}+4 b \ln \left (-c \,x^{2}+1\right ) a -4 a^{2}}{16 x^{4}}\)	\(257\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))^2/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (80) = 160\).
time = 0.26, size = 175, normalized size = 1.99 \begin {gather*} \frac {1}{4} \, {\left ({\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac {2}{x^{2}}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{4}}\right )} a b + \frac {1}{16} \, {\left ({\left (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 ) + 16 \, \log \left (x\right )\right )} c^{2} + 4 \, {\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac {2}{x^{2}}\right )} c \operatorname {artanh}\left (c x^{2}\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (c x^{2}\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))^2/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 151, normalized size = 1.72 \begin {gather*} \frac {16 \, b^{2} c^{2} x^{4} \log \left (x\right ) + 4 \, {\left (a b - b^{2}\right )} c^{2} x^{4} \log \left (c x^{2} + 1\right ) - 4 \, {\left (a b + b^{2}\right )} c^{2} x^{4} \log \left (c x^{2} - 1\right ) - 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 \, a^{2} - 4 \, {\left (b^{2} c x^{2} + a b\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{16 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))^2/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (80) = 160\).
time = 7.11, size = 175, normalized size = 1.99 \begin {gather*} \begin {cases} - \frac {a^{2}}{4 x^{4}} + \frac {a b c^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{2} - \frac {a b c}{2 x^{2}} - \frac {a b \operatorname {atanh}{\left (c x^{2} \right )}}{2 x^{4}} + b^{2} c^{2} \log {\left (x \right )} - \frac {b^{2} c^{2} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{2} - \frac {b^{2} c^{2} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{2} + \frac {b^{2} c^{2} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{4} + \frac {b^{2} c^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{2} - \frac {b^{2} c \operatorname {atanh}{\left (c x^{2} \right )}}{2 x^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**2))**2/x**5,x)

[Out]

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

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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^2))^2/x^5,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^2) + a)^2/x^5, x)

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Mupad [B]
time = 1.50, size = 278, normalized size = 3.16 \begin {gather*} \frac {b^2\,c^2\,{\ln \left (c\,x^2+1\right )}^2}{16}-\frac {b^2\,c^2\,\ln \left (c\,x^2-1\right )}{4}-\frac {b^2\,c^2\,\ln \left (c\,x^2+1\right )}{4}-\frac {a^2}{4\,x^4}+\frac {b^2\,c^2\,{\ln \left (1-c\,x^2\right )}^2}{16}-\frac {b^2\,{\ln \left (c\,x^2+1\right )}^2}{16\,x^4}-\frac {b^2\,{\ln \left (1-c\,x^2\right )}^2}{16\,x^4}+b^2\,c^2\,\ln \left (x\right )-\frac {a\,b\,c^2\,\ln \left (c\,x^2-1\right )}{4}+\frac {a\,b\,c^2\,\ln \left (c\,x^2+1\right )}{4}-\frac {a\,b\,c}{2\,x^2}-\frac {a\,b\,\ln \left (c\,x^2+1\right )}{4\,x^4}+\frac {a\,b\,\ln \left (1-c\,x^2\right )}{4\,x^4}-\frac {b^2\,c^2\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{8}-\frac {b^2\,c\,\ln \left (c\,x^2+1\right )}{4\,x^2}+\frac {b^2\,c\,\ln \left (1-c\,x^2\right )}{4\,x^2}+\frac {b^2\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{8\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))^2/x^5,x)

[Out]

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

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