∫ a+b tanh−1(cx2)___________

Optimal. Leaf size=40 \[ -\frac {a+b \tanh ^{-1}\left (c x^2\right )}{2 x^2}+b c \log (x)-\frac {1}{4} b c \log \left (1-c^2 x^4\right ) \]

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Rubi [A]
time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6037, 272, 36, 29, 31} \begin {gather*} -\frac {a+b \tanh ^{-1}\left (c x^2\right )}{2 x^2}-\frac {1}{4} b c \log \left (1-c^2 x^4\right )+b c \log (x) \end {gather*}

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

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]

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]]

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

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

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Mathematica [A]
time = 0.01, size = 45, normalized size = 1.12 \begin {gather*} -\frac {a}{2 x^2}-\frac {b \tanh ^{-1}\left (c x^2\right )}{2 x^2}+b c \log (x)-\frac {1}{4} b c \log \left (1-c^2 x^4\right ) \end {gather*}

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method	result	size

default	\(-\frac {a}{2 x^{2}}-\frac {b \arctanh \left (c \,x^{2}\right )}{2 x^{2}}-\frac {b c \ln \left (c \,x^{2}-1\right )}{4}-\frac {b c \ln \left (c \,x^{2}+1\right )}{4}+b c \ln \left (x \right )\)	\(49\)
risch	\(-\frac {b \ln \left (c \,x^{2}+1\right )}{4 x^{2}}+\frac {4 b c \ln \left (x \right ) x^{2}-b c \ln \left (c^{2} x^{4}-1\right ) x^{2}+b \ln \left (-c \,x^{2}+1\right )-2 a}{4 x^{2}}\)	\(62\)

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

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Maxima [A]
time = 0.26, size = 41, normalized size = 1.02 \begin {gather*} -\frac {1}{4} \, {\left (c {\left (\log \left (c^{2} x^{4} - 1\right ) - \log \left (x^{4}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{2}}\right )} b - \frac {a}{2 \, x^{2}} \end {gather*}

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Fricas [A]
time = 0.39, size = 55, normalized size = 1.38 \begin {gather*} -\frac {b c x^{2} \log \left (c^{2} x^{4} - 1\right ) - 4 \, b c x^{2} \log \left (x\right ) + b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a}{4 \, x^{2}} \end {gather*}

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (39) = 78\).
time = 5.46, size = 80, normalized size = 2.00 \begin {gather*} \begin {cases} - \frac {a}{2 x^{2}} + b c \log {\left (x \right )} - \frac {b c \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{2} - \frac {b c \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{2} + \frac {b c \operatorname {atanh}{\left (c x^{2} \right )}}{2} - \frac {b \operatorname {atanh}{\left (c x^{2} \right )}}{2 x^{2}} & \text {for}\: c \neq 0 \\- \frac {a}{2 x^{2}} & \text {otherwise} \end {cases} \end {gather*}

Piecewise((-a/(2*x**2) + b*c*log(x) - b*c*log(x - sqrt(-1/c))/2 - b*c*log(x + sqrt(-1/c))/2 + b*c*atanh(c*x**2

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Giac [A]
time = 0.43, size = 51, normalized size = 1.28 \begin {gather*} -\frac {1}{4} \, b c \log \left (c^{2} x^{4} - 1\right ) + b c \log \left (x\right ) - \frac {b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{4 \, x^{2}} - \frac {a}{2 \, x^{2}} \end {gather*}

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

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Mupad [B]
time = 0.85, size = 55, normalized size = 1.38 \begin {gather*} b\,c\,\ln \left (x\right )-\frac {a}{2\,x^2}-\frac {b\,c\,\ln \left (c^2\,x^4-1\right )}{4}-\frac {b\,\ln \left (c\,x^2+1\right )}{4\,x^2}+\frac {b\,\ln \left (1-c\,x^2\right )}{4\,x^2} \end {gather*}

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

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3.1.55 \(\int \frac {a+b \tanh ^{-1}(c x^2)}{x^3} \, dx\) [55]