3.2.63 \(\int \frac {a+b \tanh ^{-1}(\frac {c}{x^2})}{x^5} \, dx\) [163]

Optimal. Leaf size=45 \[ -\frac {b}{4 c x^2}-\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{4 x^4}+\frac {b \tanh ^{-1}\left (\frac {x^2}{c}\right )}{4 c^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 45, 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, 269, 281, 331, 213} \begin {gather*} -\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{4 x^4}+\frac {b \tanh ^{-1}\left (\frac {x^2}{c}\right )}{4 c^2}-\frac {b}{4 c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x^2])/x^5,x]

[Out]

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 64, normalized size = 1.42 \begin {gather*} -\frac {a}{4 x^4}-\frac {b}{4 c x^2}-\frac {b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{4 x^4}-\frac {b \log \left (-c+x^2\right )}{8 c^2}+\frac {b \log \left (c+x^2\right )}{8 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*a/x^4 - b/(4*c*x^2) - (b*ArcTanh[c/x^2])/(4*x^4) - (b*Log[-c + x^2])/(8*c^2) + (b*Log[c + x^2])/(8*c^2)

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 57, normalized size = 1.27

method	result	size
derivativedivides	\(-\frac {a}{4 x^{4}}-\frac {b \arctanh \left (\frac {c}{x^{2}}\right )}{4 x^{4}}-\frac {b}{4 c \,x^{2}}-\frac {b \ln \left (\frac {c}{x^{2}}-1\right )}{8 c^{2}}+\frac {b \ln \left (1+\frac {c}{x^{2}}\right )}{8 c^{2}}\)	\(57\)
default	\(-\frac {a}{4 x^{4}}-\frac {b \arctanh \left (\frac {c}{x^{2}}\right )}{4 x^{4}}-\frac {b}{4 c \,x^{2}}-\frac {b \ln \left (\frac {c}{x^{2}}-1\right )}{8 c^{2}}+\frac {b \ln \left (1+\frac {c}{x^{2}}\right )}{8 c^{2}}\)	\(57\)
risch	\(-\frac {b \ln \left (x^{2}+c \right )}{8 x^{4}}-\frac {2 i \pi b \,c^{2} \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}+i \pi b \,c^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (-x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )-2 i \pi b \,c^{2}-i \pi b \,c^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}-i \pi b \,c^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )-i \pi b \,c^{2} \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{3}+i \pi b \,c^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{2}+i \pi b \,c^{2} \mathrm {csgn}\left (i \left (x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{2}-i \pi b \,c^{2} \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{3}-i \pi b \,c^{2} \mathrm {csgn}\left (i \left (-x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}-2 b \ln \left (-x^{2}-c \right ) x^{4}+2 b \ln \left (-x^{2}+c \right ) x^{4}-2 b \ln \left (-x^{2}+c \right ) c^{2}+4 b c \,x^{2}+4 a \,c^{2}}{16 c^{2} x^{4}}\)	\(360\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 56, normalized size = 1.24 \begin {gather*} \frac {1}{8} \, {\left (c {\left (\frac {\log \left (x^{2} + c\right )}{c^{3}} - \frac {\log \left (x^{2} - c\right )}{c^{3}} - \frac {2}{c^{2} x^{2}}\right )} - \frac {2 \, \operatorname {artanh}\left (\frac {c}{x^{2}}\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(c*(log(x^2 + c)/c^3 - log(x^2 - c)/c^3 - 2/(c^2*x^2)) - 2*arctanh(c/x^2)/x^4)*b - 1/4*a/x^4

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 52, normalized size = 1.16 \begin {gather*} -\frac {2 \, b c x^{2} + 2 \, a c^{2} - {\left (b x^{4} - b c^{2}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{8 \, c^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 5.82, size = 49, normalized size = 1.09 \begin {gather*} \begin {cases} - \frac {a}{4 x^{4}} - \frac {b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{4 x^{4}} - \frac {b}{4 c x^{2}} + \frac {b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\- \frac {a}{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))/x**5,x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.42, size = 66, normalized size = 1.47 \begin {gather*} \frac {b \log \left (x^{2} + c\right )}{8 \, c^{2}} - \frac {b \log \left (-x^{2} + c\right )}{8 \, c^{2}} - \frac {b \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{8 \, x^{4}} - \frac {b x^{2} + a c}{4 \, c x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 1.00, size = 59, normalized size = 1.31 \begin {gather*} \frac {\frac {b\,x^4\,\mathrm {atanh}\left (\frac {x^2}{c}\right )}{4}-\frac {b\,c\,x^2}{4}}{c^2\,x^4}-\frac {\frac {a}{4}-\frac {b\,\ln \left (x^2-c\right )}{8}+\frac {b\,\ln \left (x^2+c\right )}{8}}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c/x^2))/x^5,x)

[Out]

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

________________________________________________________________________________________