3.2.58 \(\int x^5 (a+b \tanh ^{-1}(\frac {c}{x^2})) \, dx\) [158]

Optimal. Leaf size=45 \[ \frac {1}{12} b c x^4+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{12} b c^3 \log \left (c^2-x^4\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 45, 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, 45} \begin {gather*} \frac {1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{12} b c^3 \log \left (c^2-x^4\right )+\frac {1}{12} b c x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcTanh[c/x^2]),x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 x^5 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right ) \, dx &=\frac {1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{3} (b c) \int \frac {x^3}{1-\frac {c^2}{x^4}} \, dx\\ &=\frac {1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{3} (b c) \int \frac {x^7}{-c^2+x^4} \, dx\\ &=\frac {1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{12} (b c) \text {Subst}\left (\int \frac {x}{-c^2+x} \, dx,x,x^4\right )\\ &=\frac {1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{12} (b c) \text {Subst}\left (\int \left (1-\frac {c^2}{c^2-x}\right ) \, dx,x,x^4\right )\\ &=\frac {1}{12} b c x^4+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{12} b c^3 \log \left (c^2-x^4\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 50, normalized size = 1.11 \begin {gather*} \frac {1}{12} b c x^4+\frac {a x^6}{6}+\frac {1}{6} b x^6 \tanh ^{-1}\left (\frac {c}{x^2}\right )+\frac {1}{12} b c^3 \log \left (-c^2+x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.39, size = 65, normalized size = 1.44

method	result	size
derivativedivides	\(\frac {x^{6} a}{6}+\frac {b \,x^{6} \arctanh \left (\frac {c}{x^{2}}\right )}{6}+\frac {b c \,x^{4}}{12}-\frac {b \,c^{3} \ln \left (\frac {1}{x}\right )}{3}+\frac {b \,c^{3} \ln \left (1+\frac {c}{x^{2}}\right )}{12}+\frac {b \,c^{3} \ln \left (\frac {c}{x^{2}}-1\right )}{12}\)	\(65\)
default	\(\frac {x^{6} a}{6}+\frac {b \,x^{6} \arctanh \left (\frac {c}{x^{2}}\right )}{6}+\frac {b c \,x^{4}}{12}-\frac {b \,c^{3} \ln \left (\frac {1}{x}\right )}{3}+\frac {b \,c^{3} \ln \left (1+\frac {c}{x^{2}}\right )}{12}+\frac {b \,c^{3} \ln \left (\frac {c}{x^{2}}-1\right )}{12}\)	\(65\)
risch	\(\frac {x^{6} b \ln \left (x^{2}+c \right )}{12}-\frac {x^{6} b \ln \left (-x^{2}+c \right )}{12}-\frac {i \pi b \,x^{6} \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{3}}{24}+\frac {i \pi b \,x^{6} \mathrm {csgn}\left (i \left (x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{2}}{24}-\frac {i \pi b \,x^{6} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}{24}+\frac {i \pi b \,x^{6} \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}}{12}-\frac {i \pi b \,x^{6} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}}{24}-\frac {i \pi b \,x^{6} \mathrm {csgn}\left (i \left (-x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}}{24}+\frac {i \pi b \,x^{6} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{2}}{24}+\frac {i \pi b \,x^{6} \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (-x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}{24}-\frac {i \pi b \,x^{6}}{12}-\frac {i \pi b \,x^{6} \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{3}}{24}+\frac {x^{6} a}{6}+\frac {b c \,x^{4}}{12}+\frac {b \,c^{3} \ln \left (x^{4}-c^{2}\right )}{12}\)	\(337\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 42, normalized size = 0.93 \begin {gather*} \frac {1}{6} \, a x^{6} + \frac {1}{12} \, {\left (2 \, x^{6} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + {\left (x^{4} + c^{2} \log \left (x^{4} - c^{2}\right )\right )} c\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 52, normalized size = 1.16 \begin {gather*} \frac {1}{12} \, b x^{6} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{6} \, a x^{6} + \frac {1}{12} \, b c x^{4} + \frac {1}{12} \, b c^{3} \log \left (x^{4} - c^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (37) = 74\).
time = 2.26, size = 75, normalized size = 1.67 \begin {gather*} \frac {a x^{6}}{6} + \frac {b c^{3} \log {\left (x - \sqrt {- c} \right )}}{6} + \frac {b c^{3} \log {\left (x + \sqrt {- c} \right )}}{6} - \frac {b c^{3} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{6} + \frac {b c x^{4}}{12} + \frac {b x^{6} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*atanh(c/x**2)),x)

[Out]

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

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Giac [A]
time = 0.42, size = 52, normalized size = 1.16 \begin {gather*} \frac {1}{12} \, b x^{6} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{6} \, a x^{6} + \frac {1}{12} \, b c x^{4} + \frac {1}{12} \, b c^{3} \log \left (x^{4} - c^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.81, size = 56, normalized size = 1.24 \begin {gather*} \frac {a\,x^6}{6}+\frac {b\,c^3\,\ln \left (x^4-c^2\right )}{12}+\frac {b\,x^6\,\ln \left (x^2+c\right )}{12}+\frac {b\,c\,x^4}{12}-\frac {b\,x^6\,\ln \left (x^2-c\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*atanh(c/x^2)),x)

[Out]

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

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