3.2.64 \(\int \frac {a+b \tanh ^{-1}(\frac {c}{x^2})}{x^7} \, dx\) [164]

Optimal. Leaf size=57 \[ -\frac {b}{12 c x^4}-\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{6 x^6}+\frac {b \log (x)}{3 c^3}-\frac {b \log \left (c^2-x^4\right )}{12 c^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6037, 269, 272, 46} \begin {gather*} -\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{6 x^6}+\frac {b \log (x)}{3 c^3}-\frac {b \log \left (c^2-x^4\right )}{12 c^3}-\frac {b}{12 c 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*Log[x])/(3*c^3) - (b*Log[c^2 - x^4])/(12*c^3)

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 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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 (\frac {c}{x^2}\right )}{x^7} \, dx &=-\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{6 x^6}-\frac {1}{3} (b c) \int \frac {1}{\left (1-\frac {c^2}{x^4}\right ) x^9} \, dx\\ &=-\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{6 x^6}-\frac {1}{3} (b c) \int \frac {1}{x^5 \left (-c^2+x^4\right )} \, dx\\ &=-\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{6 x^6}-\frac {1}{12} (b c) \text {Subst}\left (\int \frac {1}{x^2 \left (-c^2+x\right )} \, dx,x,x^4\right )\\ &=-\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{6 x^6}-\frac {1}{12} (b c) \text {Subst}\left (\int \left (-\frac {1}{c^4 \left (c^2-x\right )}-\frac {1}{c^2 x^2}-\frac {1}{c^4 x}\right ) \, dx,x,x^4\right )\\ &=-\frac {b}{12 c x^4}-\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{6 x^6}+\frac {b \log (x)}{3 c^3}-\frac {b \log \left (c^2-x^4\right )}{12 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 62, normalized size = 1.09 \begin {gather*} -\frac {a}{6 x^6}-\frac {b}{12 c x^4}-\frac {b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{6 x^6}+\frac {b \log (x)}{3 c^3}-\frac {b \log \left (-c^2+x^4\right )}{12 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 45, normalized size = 0.79

method	result	size
derivativedivides	\(-\frac {a}{6 x^{6}}-\frac {b \arctanh \left (\frac {c}{x^{2}}\right )}{6 x^{6}}-\frac {b}{12 c \,x^{4}}-\frac {b \ln \left (\frac {c^{2}}{x^{4}}-1\right )}{12 c^{3}}\)	\(45\)
default	\(-\frac {a}{6 x^{6}}-\frac {b \arctanh \left (\frac {c}{x^{2}}\right )}{6 x^{6}}-\frac {b}{12 c \,x^{4}}-\frac {b \ln \left (\frac {c^{2}}{x^{4}}-1\right )}{12 c^{3}}\)	\(45\)
risch	\(-\frac {b \ln \left (x^{2}+c \right )}{12 x^{6}}-\frac {i \pi b \,c^{3} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (-x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )-i \pi b \,c^{3} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}-2 i \pi b \,c^{3}-i \pi b \,c^{3} \mathrm {csgn}\left (i \left (-x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}+i \pi b \,c^{3} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{2}-i \pi b \,c^{3} \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{3}-i \pi b \,c^{3} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )+i \pi b \,c^{3} \mathrm {csgn}\left (i \left (x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{2}+2 i \pi b \,c^{3} \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}-i \pi b \,c^{3} \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{3}-8 b \ln \left (x \right ) x^{6}+2 b \ln \left (-x^{4}+c^{2}\right ) x^{6}-2 b \ln \left (-x^{2}+c \right ) c^{3}+2 b \,c^{2} x^{2}+4 a \,c^{3}}{24 c^{3} x^{6}}\)	\(356\)

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/x^4-1/12*b/c^3*ln(c^2/x^4-1)

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Maxima [A]
time = 0.26, size = 55, normalized size = 0.96 \begin {gather*} -\frac {1}{12} \, {\left (c {\left (\frac {\log \left (x^{4} - c^{2}\right )}{c^{4}} - \frac {\log \left (x^{4}\right )}{c^{4}} + \frac {1}{c^{2} x^{4}}\right )} + \frac {2 \, \operatorname {artanh}\left (\frac {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*(log(x^4 - c^2)/c^4 - log(x^4)/c^4 + 1/(c^2*x^4)) + 2*arctanh(c/x^2)/x^6)*b - 1/6*a/x^6

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Fricas [A]
time = 0.41, size = 67, normalized size = 1.18 \begin {gather*} -\frac {b x^{6} \log \left (x^{4} - c^{2}\right ) - 4 \, b x^{6} \log \left (x\right ) + b c^{2} x^{2} + b c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + 2 \, a c^{3}}{12 \, c^{3} 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*x^6*log(x^4 - c^2) - 4*b*x^6*log(x) + b*c^2*x^2 + b*c^3*log((x^2 + c)/(x^2 - c)) + 2*a*c^3)/(c^3*x^6)

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Sympy [A]
time = 8.24, size = 94, normalized size = 1.65 \begin {gather*} \begin {cases} - \frac {a}{6 x^{6}} - \frac {b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{6 x^{6}} - \frac {b}{12 c x^{4}} + \frac {b \log {\left (x \right )}}{3 c^{3}} - \frac {b \log {\left (x - \sqrt {- c} \right )}}{6 c^{3}} - \frac {b \log {\left (x + \sqrt {- c} \right )}}{6 c^{3}} + \frac {b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{6 c^{3}} & \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*atanh(c/x**2)/(6*x**6) - b/(12*c*x**4) + b*log(x)/(3*c**3) - b*log(x - sqrt(-c))/(6
*c**3) - b*log(x + sqrt(-c))/(6*c**3) + b*atanh(c/x**2)/(6*c**3), Ne(c, 0)), (-a/(6*x**6), True))

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Giac [A]
time = 0.41, size = 65, normalized size = 1.14 \begin {gather*} -\frac {b \log \left (x^{4} - c^{2}\right )}{12 \, c^{3}} + \frac {b \log \left (x\right )}{3 \, c^{3}} - \frac {b \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{12 \, x^{6}} - \frac {b x^{2} + 2 \, a c}{12 \, c 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*log(x^4 - c^2)/c^3 + 1/3*b*log(x)/c^3 - 1/12*b*log((x^2 + c)/(x^2 - c))/x^6 - 1/12*(b*x^2 + 2*a*c)/(c*
x^6)

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Mupad [B]
time = 0.89, size = 66, normalized size = 1.16 \begin {gather*} \frac {b\,\ln \left (x\right )}{3\,c^3}-\frac {b\,\ln \left (x^4-c^2\right )}{12\,c^3}-\frac {b}{12\,c\,x^4}-\frac {a}{6\,x^6}-\frac {b\,\ln \left (x^2+c\right )}{12\,x^6}+\frac {b\,\ln \left (x^2-c\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*log(x))/(3*c^3) - (b*log(x^4 - c^2))/(12*c^3) - b/(12*c*x^4) - a/(6*x^6) - (b*log(c + x^2))/(12*x^6) + (b*l
og(x^2 - c))/(12*x^6)

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