3.2.59 \(\int x^3 (a+b \tanh ^{-1}(\frac {c}{x^2})) \, dx\) [159]

Optimal. Leaf size=43 \[ \frac {1}{4} b c x^2+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )-\frac {1}{4} b c^2 \tanh ^{-1}\left (\frac {x^2}{c}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 43, 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, 327, 213} \begin {gather*} \frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )-\frac {1}{4} b c^2 \tanh ^{-1}\left (\frac {x^2}{c}\right )+\frac {1}{4} b c x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c/x^2]),x]

[Out]

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

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 x^3 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right ) \, dx &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{2} (b c) \int \frac {x}{1-\frac {c^2}{x^4}} \, dx\\ &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{2} (b c) \int \frac {x^5}{-c^2+x^4} \, dx\\ &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{4} (b c) \text {Subst}\left (\int \frac {x^2}{-c^2+x^2} \, dx,x,x^2\right )\\ &=\frac {1}{4} b c x^2+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{4} \left (b c^3\right ) \text {Subst}\left (\int \frac {1}{-c^2+x^2} \, dx,x,x^2\right )\\ &=\frac {1}{4} b c x^2+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )-\frac {1}{4} b c^2 \tanh ^{-1}\left (\frac {x^2}{c}\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 62, normalized size = 1.44 \begin {gather*} \frac {1}{4} b c x^2+\frac {a x^4}{4}+\frac {1}{4} b x^4 \tanh ^{-1}\left (\frac {c}{x^2}\right )+\frac {1}{8} b c^2 \log \left (-c+x^2\right )-\frac {1}{8} b c^2 \log \left (c+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTanh[c/x^2]),x]

[Out]

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

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Maple [A]
time = 0.21, size = 55, normalized size = 1.28

method	result	size
derivativedivides	\(\frac {x^{4} a}{4}+\frac {\arctanh \left (\frac {c}{x^{2}}\right ) b \,x^{4}}{4}-\frac {b \,c^{2} \ln \left (1+\frac {c}{x^{2}}\right )}{8}+\frac {b \,c^{2} \ln \left (\frac {c}{x^{2}}-1\right )}{8}+\frac {b c \,x^{2}}{4}\)	\(55\)
default	\(\frac {x^{4} a}{4}+\frac {\arctanh \left (\frac {c}{x^{2}}\right ) b \,x^{4}}{4}-\frac {b \,c^{2} \ln \left (1+\frac {c}{x^{2}}\right )}{8}+\frac {b \,c^{2} \ln \left (\frac {c}{x^{2}}-1\right )}{8}+\frac {b c \,x^{2}}{4}\)	\(55\)
risch	\(\text {Expression too large to display}\)	\(4322\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c/x^2)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 49, normalized size = 1.14 \begin {gather*} \frac {1}{4} \, a x^{4} + \frac {1}{8} \, {\left (2 \, x^{4} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + {\left (2 \, x^{2} - c \log \left (x^{2} + c\right ) + c \log \left (x^{2} - c\right )\right )} c\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c/x^2)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 44, normalized size = 1.02 \begin {gather*} \frac {1}{4} \, a x^{4} + \frac {1}{4} \, b c x^{2} + \frac {1}{8} \, {\left (b x^{4} - b c^{2}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c/x^2)),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 1.68, size = 41, normalized size = 0.95 \begin {gather*} \frac {a x^{4}}{4} - \frac {b c^{2} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{4} + \frac {b c x^{2}}{4} + \frac {b x^{4} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c/x**2)),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (37) = 74\).
time = 0.43, size = 162, normalized size = 3.77 \begin {gather*} \frac {\frac {{\left (x^{2} + c\right )} b c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{{\left (x^{2} - c\right )} {\left (\frac {{\left (x^{2} + c\right )}^{2}}{{\left (x^{2} - c\right )}^{2}} - \frac {2 \, {\left (x^{2} + c\right )}}{x^{2} - c} + 1\right )}} + \frac {\frac {2 \, {\left (x^{2} + c\right )} a c^{3}}{x^{2} - c} + \frac {{\left (x^{2} + c\right )} b c^{3}}{x^{2} - c} - b c^{3}}{\frac {{\left (x^{2} + c\right )}^{2}}{{\left (x^{2} - c\right )}^{2}} - \frac {2 \, {\left (x^{2} + c\right )}}{x^{2} - c} + 1}}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c/x^2)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.86, size = 57, normalized size = 1.33 \begin {gather*} \frac {a\,x^4}{4}+\frac {b\,x^4\,\ln \left (x^2+c\right )}{8}+\frac {b\,c\,x^2}{4}-\frac {b\,x^4\,\ln \left (x^2-c\right )}{8}+\frac {b\,c^2\,\mathrm {atan}\left (\frac {x^2\,1{}\mathrm {i}}{c}\right )\,1{}\mathrm {i}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c/x^2)),x)

[Out]

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

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