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

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

[Out]

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

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Rubi [A]  time = 0.0297014, antiderivative size = 43, normalized size of antiderivative = 1., 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 = {6097, 263, 275, 321, 207} \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )-\frac{1}{4} b c^2 \tanh ^{-1}\left (\frac{x^2}{c}\right )+\frac{1}{4} b c x^2 \]

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 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 263

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 275

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 321

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 207

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

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) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.0111031, size = 62, normalized size = 1.44 \[ \frac{a x^4}{4}+\frac{1}{8} b c^2 \log \left (x^2-c\right )-\frac{1}{8} b c^2 \log \left (c+x^2\right )+\frac{1}{4} b c x^2+\frac{1}{4} b x^4 \tanh ^{-1}\left (\frac{c}{x^2}\right ) \]

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 = 0.006, size = 55, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}}{4}{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) }-{\frac{b{c}^{2}}{8}\ln \left ( 1+{\frac{c}{{x}^{2}}} \right ) }+{\frac{b{c}^{2}}{8}\ln \left ({\frac{c}{{x}^{2}}}-1 \right ) }+{\frac{bc{x}^{2}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.960415, size = 66, normalized size = 1.53 \begin{align*} \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{align*}

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 = 1.585, size = 97, normalized size = 2.26 \begin{align*} \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{align*}

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 = 15.3665, size = 41, normalized size = 0.95 \begin{align*} \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{align*}

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 [A]  time = 1.25419, size = 84, normalized size = 1.95 \begin{align*} \frac{1}{8} \, b x^{4} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + \frac{1}{4} \, a x^{4} + \frac{1}{4} \, b c x^{2} - \frac{1}{8} \, b c^{2} \log \left (x^{2} + c\right ) + \frac{1}{8} \, b c^{2} \log \left (-x^{2} + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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