3.165 \(\int x^4 (a+b \tanh ^{-1}(\frac{c}{x^2})) \, dx\)

Optimal. Leaf size=63 \[ \frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )+\frac{1}{5} b c^{5/2} \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )-\frac{1}{5} b c^{5/2} \tanh ^{-1}\left (\frac{x}{\sqrt{c}}\right )+\frac{2}{15} b c x^3 \]

[Out]

(2*b*c*x^3)/15 + (b*c^(5/2)*ArcTan[x/Sqrt[c]])/5 + (x^5*(a + b*ArcTanh[c/x^2]))/5 - (b*c^(5/2)*ArcTanh[x/Sqrt[
c]])/5

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Rubi [A]  time = 0.0347141, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6097, 263, 321, 298, 203, 206} \[ \frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )+\frac{1}{5} b c^{5/2} \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )-\frac{1}{5} b c^{5/2} \tanh ^{-1}\left (\frac{x}{\sqrt{c}}\right )+\frac{2}{15} b c x^3 \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTanh[c/x^2]),x]

[Out]

(2*b*c*x^3)/15 + (b*c^(5/2)*ArcTan[x/Sqrt[c]])/5 + (x^5*(a + b*ArcTanh[c/x^2]))/5 - (b*c^(5/2)*ArcTanh[x/Sqrt[
c]])/5

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 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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

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

Rule 206

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

Rubi steps

\begin{align*} \int x^4 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right ) \, dx &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )+\frac{1}{5} (2 b c) \int \frac{x^2}{1-\frac{c^2}{x^4}} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )+\frac{1}{5} (2 b c) \int \frac{x^6}{-c^2+x^4} \, dx\\ &=\frac{2}{15} b c x^3+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )+\frac{1}{5} \left (2 b c^3\right ) \int \frac{x^2}{-c^2+x^4} \, dx\\ &=\frac{2}{15} b c x^3+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )-\frac{1}{5} \left (b c^3\right ) \int \frac{1}{c-x^2} \, dx+\frac{1}{5} \left (b c^3\right ) \int \frac{1}{c+x^2} \, dx\\ &=\frac{2}{15} b c x^3+\frac{1}{5} b c^{5/2} \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )-\frac{1}{5} b c^{5/2} \tanh ^{-1}\left (\frac{x}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0205633, size = 88, normalized size = 1.4 \[ \frac{a x^5}{5}+\frac{1}{10} b c^{5/2} \log \left (\sqrt{c}-x\right )-\frac{1}{10} b c^{5/2} \log \left (\sqrt{c}+x\right )+\frac{1}{5} b c^{5/2} \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )+\frac{2}{15} b c x^3+\frac{1}{5} b x^5 \tanh ^{-1}\left (\frac{c}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*c*x^3)/15 + (a*x^5)/5 + (b*c^(5/2)*ArcTan[x/Sqrt[c]])/5 + (b*x^5*ArcTanh[c/x^2])/5 + (b*c^(5/2)*Log[Sqrt[
c] - x])/10 - (b*c^(5/2)*Log[Sqrt[c] + x])/10

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Maple [A]  time = 0.014, size = 53, normalized size = 0.8 \begin{align*}{\frac{a{x}^{5}}{5}}+{\frac{b{x}^{5}}{5}{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) }+{\frac{b}{5}{c}^{{\frac{5}{2}}}\arctan \left ({x{\frac{1}{\sqrt{c}}}} \right ) }-{\frac{b}{5}{c}^{{\frac{5}{2}}}{\it Artanh} \left ({\frac{1}{x}\sqrt{c}} \right ) }+{\frac{2\,bc{x}^{3}}{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c/x^2)),x)

[Out]

1/5*a*x^5+1/5*b*x^5*arctanh(c/x^2)+1/5*b*c^(5/2)*arctan(x/c^(1/2))-1/5*b*c^(5/2)*arctanh(1/x*c^(1/2))+2/15*b*c
*x^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c/x^2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83608, size = 440, normalized size = 6.98 \begin{align*} \left [\frac{1}{10} \, b x^{5} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + \frac{1}{5} \, a x^{5} + \frac{2}{15} \, b c x^{3} + \frac{1}{5} \, b c^{\frac{5}{2}} \arctan \left (\frac{x}{\sqrt{c}}\right ) + \frac{1}{10} \, b c^{\frac{5}{2}} \log \left (\frac{x^{2} - 2 \, \sqrt{c} x + c}{x^{2} - c}\right ), \frac{1}{10} \, b x^{5} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + \frac{1}{5} \, a x^{5} + \frac{2}{15} \, b c x^{3} + \frac{1}{5} \, b \sqrt{-c} c^{2} \arctan \left (\frac{\sqrt{-c} x}{c}\right ) + \frac{1}{10} \, b \sqrt{-c} c^{2} \log \left (\frac{x^{2} + 2 \, \sqrt{-c} x - c}{x^{2} + c}\right )\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c/x^2)),x, algorithm="fricas")

[Out]

[1/10*b*x^5*log((x^2 + c)/(x^2 - c)) + 1/5*a*x^5 + 2/15*b*c*x^3 + 1/5*b*c^(5/2)*arctan(x/sqrt(c)) + 1/10*b*c^(
5/2)*log((x^2 - 2*sqrt(c)*x + c)/(x^2 - c)), 1/10*b*x^5*log((x^2 + c)/(x^2 - c)) + 1/5*a*x^5 + 2/15*b*c*x^3 +
1/5*b*sqrt(-c)*c^2*arctan(sqrt(-c)*x/c) + 1/10*b*sqrt(-c)*c^2*log((x^2 + 2*sqrt(-c)*x - c)/(x^2 + c))]

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Sympy [A]  time = 42.2058, size = 588, normalized size = 9.33 \begin{align*} \begin{cases} \frac{a x^{5}}{5} & \text{for}\: c = 0 \\\frac{x^{5} \left (a - \infty b\right )}{5} & \text{for}\: c = - x^{2} \\\frac{x^{5} \left (a + \infty b\right )}{5} & \text{for}\: c = x^{2} \\- \frac{6 a c^{69} x^{5}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{6 a c^{67} x^{9}}{- 30 c^{69} + 30 c^{67} x^{4}} - \frac{6 b c^{\frac{143}{2}} \log{\left (- \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{3 b c^{\frac{143}{2}} \log{\left (- i \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{3 i b c^{\frac{143}{2}} \log{\left (- i \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{3 b c^{\frac{143}{2}} \log{\left (i \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} - \frac{3 i b c^{\frac{143}{2}} \log{\left (i \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} - \frac{6 b c^{\frac{143}{2}} \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{6 b c^{\frac{139}{2}} x^{4} \log{\left (- \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} - \frac{3 b c^{\frac{139}{2}} x^{4} \log{\left (- i \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} - \frac{3 i b c^{\frac{139}{2}} x^{4} \log{\left (- i \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} - \frac{3 b c^{\frac{139}{2}} x^{4} \log{\left (i \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{3 i b c^{\frac{139}{2}} x^{4} \log{\left (i \sqrt{c} + x \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{6 b c^{\frac{139}{2}} x^{4} \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} - \frac{4 b c^{70} x^{3}}{- 30 c^{69} + 30 c^{67} x^{4}} - \frac{6 b c^{69} x^{5} \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{4 b c^{68} x^{7}}{- 30 c^{69} + 30 c^{67} x^{4}} + \frac{6 b c^{67} x^{9} \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}}{- 30 c^{69} + 30 c^{67} x^{4}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**5/5, Eq(c, 0)), (x**5*(a - oo*b)/5, Eq(c, -x**2)), (x**5*(a + oo*b)/5, Eq(c, x**2)), (-6*a*c**
69*x**5/(-30*c**69 + 30*c**67*x**4) + 6*a*c**67*x**9/(-30*c**69 + 30*c**67*x**4) - 6*b*c**(143/2)*log(-sqrt(c)
+ x)/(-30*c**69 + 30*c**67*x**4) + 3*b*c**(143/2)*log(-I*sqrt(c) + x)/(-30*c**69 + 30*c**67*x**4) + 3*I*b*c**
(143/2)*log(-I*sqrt(c) + x)/(-30*c**69 + 30*c**67*x**4) + 3*b*c**(143/2)*log(I*sqrt(c) + x)/(-30*c**69 + 30*c*
*67*x**4) - 3*I*b*c**(143/2)*log(I*sqrt(c) + x)/(-30*c**69 + 30*c**67*x**4) - 6*b*c**(143/2)*atanh(c/x**2)/(-3
0*c**69 + 30*c**67*x**4) + 6*b*c**(139/2)*x**4*log(-sqrt(c) + x)/(-30*c**69 + 30*c**67*x**4) - 3*b*c**(139/2)*
x**4*log(-I*sqrt(c) + x)/(-30*c**69 + 30*c**67*x**4) - 3*I*b*c**(139/2)*x**4*log(-I*sqrt(c) + x)/(-30*c**69 +
30*c**67*x**4) - 3*b*c**(139/2)*x**4*log(I*sqrt(c) + x)/(-30*c**69 + 30*c**67*x**4) + 3*I*b*c**(139/2)*x**4*lo
g(I*sqrt(c) + x)/(-30*c**69 + 30*c**67*x**4) + 6*b*c**(139/2)*x**4*atanh(c/x**2)/(-30*c**69 + 30*c**67*x**4) -
4*b*c**70*x**3/(-30*c**69 + 30*c**67*x**4) - 6*b*c**69*x**5*atanh(c/x**2)/(-30*c**69 + 30*c**67*x**4) + 4*b*c
**68*x**7/(-30*c**69 + 30*c**67*x**4) + 6*b*c**67*x**9*atanh(c/x**2)/(-30*c**69 + 30*c**67*x**4), True))

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Giac [A]  time = 1.30346, size = 90, normalized size = 1.43 \begin{align*} \frac{1}{10} \, b x^{5} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + \frac{1}{5} \, a x^{5} + \frac{2}{15} \, b c x^{3} + \frac{1}{5} \, b c^{3}{\left (\frac{\arctan \left (\frac{x}{\sqrt{-c}}\right )}{\sqrt{-c}} + \frac{\arctan \left (\frac{x}{\sqrt{c}}\right )}{\sqrt{c}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c/x^2)),x, algorithm="giac")

[Out]

1/10*b*x^5*log((x^2 + c)/(x^2 - c)) + 1/5*a*x^5 + 2/15*b*c*x^3 + 1/5*b*c^3*(arctan(x/sqrt(-c))/sqrt(-c) + arct
an(x/sqrt(c))/sqrt(c))