3.1.57 \(\int \frac {a+b \tanh ^{-1}(c x^2)}{x^7} \, dx\) [57]

Optimal. Leaf size=56 \[ -\frac {b c}{12 x^4}-\frac {a+b \tanh ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{3} b c^3 \log (x)-\frac {1}{12} b c^3 \log \left (1-c^2 x^4\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6037, 272, 46} \begin {gather*} -\frac {a+b \tanh ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{3} b c^3 \log (x)-\frac {1}{12} b c^3 \log \left (1-c^2 x^4\right )-\frac {b c}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])/x^7,x]

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 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^2\right )}{x^7} \, dx &=-\frac {a+b \tanh ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{3} (b c) \int \frac {1}{x^5 \left (1-c^2 x^4\right )} \, dx\\ &=-\frac {a+b \tanh ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{12} (b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^4\right )\\ &=-\frac {a+b \tanh ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{12} (b c) \text {Subst}\left (\int \left (\frac {1}{x^2}+\frac {c^2}{x}-\frac {c^4}{-1+c^2 x}\right ) \, dx,x,x^4\right )\\ &=-\frac {b c}{12 x^4}-\frac {a+b \tanh ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{3} b c^3 \log (x)-\frac {1}{12} b c^3 \log \left (1-c^2 x^4\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 61, normalized size = 1.09 \begin {gather*} -\frac {a}{6 x^6}-\frac {b c}{12 x^4}-\frac {b \tanh ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{3} b c^3 \log (x)-\frac {1}{12} b c^3 \log \left (1-c^2 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])/x^7,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.04, size = 63, normalized size = 1.12

method	result	size
default	\(-\frac {a}{6 x^{6}}-\frac {b \arctanh \left (c \,x^{2}\right )}{6 x^{6}}-\frac {b \,c^{3} \ln \left (c \,x^{2}-1\right )}{12}-\frac {b \,c^{3} \ln \left (c \,x^{2}+1\right )}{12}-\frac {b c}{12 x^{4}}+\frac {b \,c^{3} \ln \left (x \right )}{3}\)	\(63\)
risch	\(-\frac {b \ln \left (c \,x^{2}+1\right )}{12 x^{6}}+\frac {4 b \,c^{3} \ln \left (x \right ) x^{6}-b \,c^{3} \ln \left (c^{2} x^{4}-1\right ) x^{6}-b c \,x^{2}+b \ln \left (-c \,x^{2}+1\right )-2 a}{12 x^{6}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))/x^7,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 51, normalized size = 0.91 \begin {gather*} -\frac {1}{12} \, {\left ({\left (c^{2} \log \left (c^{2} x^{4} - 1\right ) - c^{2} \log \left (x^{4}\right ) + \frac {1}{x^{4}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{6}}\right )} b - \frac {a}{6 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^7,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 65, normalized size = 1.16 \begin {gather*} -\frac {b c^{3} x^{6} \log \left (c^{2} x^{4} - 1\right ) - 4 \, b c^{3} x^{6} \log \left (x\right ) + b c x^{2} + b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^7,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 8.95, size = 97, normalized size = 1.73 \begin {gather*} \begin {cases} - \frac {a}{6 x^{6}} + \frac {b c^{3} \log {\left (x \right )}}{3} - \frac {b c^{3} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{6} - \frac {b c^{3} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{6} + \frac {b c^{3} \operatorname {atanh}{\left (c x^{2} \right )}}{6} - \frac {b c}{12 x^{4}} - \frac {b \operatorname {atanh}{\left (c x^{2} \right )}}{6 x^{6}} & \text {for}\: c \neq 0 \\- \frac {a}{6 x^{6}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**2))/x**7,x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.43, size = 65, normalized size = 1.16 \begin {gather*} -\frac {1}{12} \, b c^{3} \log \left (c^{2} x^{4} - 1\right ) + \frac {1}{3} \, b c^{3} \log \left (x\right ) - \frac {b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{12 \, x^{6}} - \frac {b c x^{2} + 2 \, a}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^7,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.88, size = 67, normalized size = 1.20 \begin {gather*} \frac {b\,c^3\,\ln \left (x\right )}{3}-\frac {b\,c^3\,\ln \left (c^2\,x^4-1\right )}{12}-\frac {a}{6\,x^6}-\frac {b\,c}{12\,x^4}-\frac {b\,\ln \left (c\,x^2+1\right )}{12\,x^6}+\frac {b\,\ln \left (1-c\,x^2\right )}{12\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))/x^7,x)

[Out]

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

________________________________________________________________________________________