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

Optimal. Leaf size=65 \[ \frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac{b \tan ^{-1}\left (\sqrt{c} x\right )}{5 c^{5/2}}-\frac{b \tanh ^{-1}\left (\sqrt{c} x\right )}{5 c^{5/2}}+\frac{2 b x^3}{15 c} \]

[Out]

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

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Rubi [A]  time = 0.0352452, antiderivative size = 65, 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, 321, 298, 203, 206} \[ \frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac{b \tan ^{-1}\left (\sqrt{c} x\right )}{5 c^{5/2}}-\frac{b \tanh ^{-1}\left (\sqrt{c} x\right )}{5 c^{5/2}}+\frac{2 b x^3}{15 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTanh[c*x^2]),x]

[Out]

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

Mathematica [A]  time = 0.0239303, size = 93, normalized size = 1.43 \[ \frac{a x^5}{5}+\frac{b \log \left (1-\sqrt{c} x\right )}{10 c^{5/2}}-\frac{b \log \left (\sqrt{c} x+1\right )}{10 c^{5/2}}+\frac{b \tan ^{-1}\left (\sqrt{c} x\right )}{5 c^{5/2}}+\frac{2 b x^3}{15 c}+\frac{1}{5} b x^5 \tanh ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*x^3)/(15*c) + (a*x^5)/5 + (b*ArcTan[Sqrt[c]*x])/(5*c^(5/2)) + (b*x^5*ArcTanh[c*x^2])/5 + (b*Log[1 - Sqrt[
c]*x])/(10*c^(5/2)) - (b*Log[1 + Sqrt[c]*x])/(10*c^(5/2))

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Maple [A]  time = 0.011, size = 53, normalized size = 0.8 \begin{align*}{\frac{a{x}^{5}}{5}}+{\frac{{x}^{5}b{\it Artanh} \left ( c{x}^{2} \right ) }{5}}+{\frac{2\,b{x}^{3}}{15\,c}}+{\frac{b}{5}\arctan \left ( x\sqrt{c} \right ){c}^{-{\frac{5}{2}}}}-{\frac{b}{5}{\it Artanh} \left ( x\sqrt{c} \right ){c}^{-{\frac{5}{2}}}} \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*x^5*b*arctanh(c*x^2)+2/15*b*x^3/c+1/5*b*arctan(x*c^(1/2))/c^(5/2)-1/5*b*arctanh(x*c^(1/2))/c^(5/
2)

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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 = 2.00913, size = 473, normalized size = 7.28 \begin{align*} \left [\frac{3 \, b c^{3} x^{5} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a c^{3} x^{5} + 4 \, b c^{2} x^{3} + 6 \, b \sqrt{c} \arctan \left (\sqrt{c} x\right ) + 3 \, b \sqrt{c} \log \left (\frac{c x^{2} - 2 \, \sqrt{c} x + 1}{c x^{2} - 1}\right )}{30 \, c^{3}}, \frac{3 \, b c^{3} x^{5} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a c^{3} x^{5} + 4 \, b c^{2} x^{3} + 6 \, b \sqrt{-c} \arctan \left (\sqrt{-c} x\right ) - 3 \, b \sqrt{-c} \log \left (\frac{c x^{2} - 2 \, \sqrt{-c} x - 1}{c x^{2} + 1}\right )}{30 \, c^{3}}\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]

[1/30*(3*b*c^3*x^5*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 6*a*c^3*x^5 + 4*b*c^2*x^3 + 6*b*sqrt(c)*arctan(sqrt(c)*x) +
3*b*sqrt(c)*log((c*x^2 - 2*sqrt(c)*x + 1)/(c*x^2 - 1)))/c^3, 1/30*(3*b*c^3*x^5*log(-(c*x^2 + 1)/(c*x^2 - 1))
+ 6*a*c^3*x^5 + 4*b*c^2*x^3 + 6*b*sqrt(-c)*arctan(sqrt(-c)*x) - 3*b*sqrt(-c)*log((c*x^2 - 2*sqrt(-c)*x - 1)/(c
*x^2 + 1)))/c^3]

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Sympy [A]  time = 19.3632, size = 185, normalized size = 2.85 \begin{align*} \begin{cases} \frac{a x^{5}}{5} + \frac{b x^{5} \operatorname{atanh}{\left (c x^{2} \right )}}{5} + \frac{2 b x^{3}}{15 c} - \frac{b \log{\left (x - i \sqrt{\frac{1}{c}} \right )}}{10 c^{3} \sqrt{\frac{1}{c}}} - \frac{i b \log{\left (x - i \sqrt{\frac{1}{c}} \right )}}{10 c^{3} \sqrt{\frac{1}{c}}} - \frac{b \log{\left (x + i \sqrt{\frac{1}{c}} \right )}}{10 c^{3} \sqrt{\frac{1}{c}}} + \frac{i b \log{\left (x + i \sqrt{\frac{1}{c}} \right )}}{10 c^{3} \sqrt{\frac{1}{c}}} + \frac{b \log{\left (x - \sqrt{\frac{1}{c}} \right )}}{5 c^{3} \sqrt{\frac{1}{c}}} + \frac{b \operatorname{atanh}{\left (c x^{2} \right )}}{5 c^{3} \sqrt{\frac{1}{c}}} & \text{for}\: c \neq 0 \\\frac{a x^{5}}{5} & \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 + b*x**5*atanh(c*x**2)/5 + 2*b*x**3/(15*c) - b*log(x - I*sqrt(1/c))/(10*c**3*sqrt(1/c)) -
I*b*log(x - I*sqrt(1/c))/(10*c**3*sqrt(1/c)) - b*log(x + I*sqrt(1/c))/(10*c**3*sqrt(1/c)) + I*b*log(x + I*sqrt
(1/c))/(10*c**3*sqrt(1/c)) + b*log(x - sqrt(1/c))/(5*c**3*sqrt(1/c)) + b*atanh(c*x**2)/(5*c**3*sqrt(1/c)), Ne(
c, 0)), (a*x**5/5, True))

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Giac [A]  time = 1.27424, size = 103, normalized size = 1.58 \begin{align*} \frac{1}{5} \, b c^{9}{\left (\frac{\arctan \left (\sqrt{c} x\right )}{c^{\frac{23}{2}}} + \frac{\arctan \left (\frac{c x}{\sqrt{-c}}\right )}{\sqrt{-c} c^{11}}\right )} + \frac{1}{10} \, b x^{5} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + \frac{1}{5} \, a x^{5} + \frac{2 \, b x^{3}}{15 \, c} \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]

1/5*b*c^9*(arctan(sqrt(c)*x)/c^(23/2) + arctan(c*x/sqrt(-c))/(sqrt(-c)*c^11)) + 1/10*b*x^5*log(-(c*x^2 + 1)/(c
*x^2 - 1)) + 1/5*a*x^5 + 2/15*b*x^3/c