3.2.4 \(\int \frac {a+b \tanh ^{-1}(c x^3)}{x^4} \, dx\) [104]

Optimal. Leaf size=40 \[ -\frac {a+b \tanh ^{-1}\left (c x^3\right )}{3 x^3}+b c \log (x)-\frac {1}{6} b c \log \left (1-c^2 x^6\right ) \]

[Out]

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

________________________________________________________________________________________

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^3\right )}{3 x^3}-\frac {1}{6} b c \log \left (1-c^2 x^6\right )+b c \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^3])/x^4,x]

[Out]

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

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
]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 45, normalized size = 1.12 \begin {gather*} -\frac {a}{3 x^3}-\frac {b \tanh ^{-1}\left (c x^3\right )}{3 x^3}+b c \log (x)-\frac {1}{6} b c \log \left (1-c^2 x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^3])/x^4,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 49, normalized size = 1.22

method	result	size
default	\(-\frac {a}{3 x^{3}}-\frac {b \arctanh \left (c \,x^{3}\right )}{3 x^{3}}-\frac {b c \ln \left (c \,x^{3}+1\right )}{6}-\frac {b c \ln \left (c \,x^{3}-1\right )}{6}+b c \ln \left (x \right )\)	\(49\)
risch	\(-\frac {b \ln \left (c \,x^{3}+1\right )}{6 x^{3}}+\frac {6 b c \ln \left (x \right ) x^{3}-b c \ln \left (c^{2} x^{6}-1\right ) x^{3}+b \ln \left (-c \,x^{3}+1\right )-2 a}{6 x^{3}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^3))/x^4,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 41, normalized size = 1.02 \begin {gather*} -\frac {1}{6} \, {\left (c {\left (\log \left (c^{2} x^{6} - 1\right ) - \log \left (x^{6}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x^{3}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 55, normalized size = 1.38 \begin {gather*} -\frac {b c x^{3} \log \left (c^{2} x^{6} - 1\right ) - 6 \, b c x^{3} \log \left (x\right ) + b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, a}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**3))/x**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 0.42, size = 51, normalized size = 1.28 \begin {gather*} -\frac {1}{6} \, b c \log \left (c^{2} x^{6} - 1\right ) + b c \log \left (x\right ) - \frac {b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )}{6 \, x^{3}} - \frac {a}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^4,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.85, size = 55, normalized size = 1.38 \begin {gather*} b\,c\,\ln \left (x\right )-\frac {a}{3\,x^3}-\frac {b\,c\,\ln \left (c^2\,x^6-1\right )}{6}-\frac {b\,\ln \left (c\,x^3+1\right )}{6\,x^3}+\frac {b\,\ln \left (1-c\,x^3\right )}{6\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^3))/x^4,x)

[Out]

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

________________________________________________________________________________________