3.112 \(\int x^4 (a+b \tanh ^{-1}(c x^3)) \, dx\)

Optimal. Leaf size=117 \[ \frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (1-c^{2/3} x^2\right )}{10 c^{5/3}}-\frac{b \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{20 c^{5/3}}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 c^{2/3} x^2+1}{\sqrt{3}}\right )}{10 c^{5/3}}+\frac{3 b x^2}{10 c} \]

[Out]

(3*b*x^2)/(10*c) - (Sqrt[3]*b*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/(10*c^(5/3)) + (x^5*(a + b*ArcTanh[c*x^3]))
/5 + (b*Log[1 - c^(2/3)*x^2])/(10*c^(5/3)) - (b*Log[1 + c^(2/3)*x^2 + c^(4/3)*x^4])/(20*c^(5/3))

________________________________________________________________________________________

Rubi [A]  time = 0.0972647, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {6097, 275, 321, 200, 31, 634, 617, 204, 628} \[ \frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (1-c^{2/3} x^2\right )}{10 c^{5/3}}-\frac{b \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{20 c^{5/3}}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 c^{2/3} x^2+1}{\sqrt{3}}\right )}{10 c^{5/3}}+\frac{3 b x^2}{10 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTanh[c*x^3]),x]

[Out]

(3*b*x^2)/(10*c) - (Sqrt[3]*b*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/(10*c^(5/3)) + (x^5*(a + b*ArcTanh[c*x^3]))
/5 + (b*Log[1 - c^(2/3)*x^2])/(10*c^(5/3)) - (b*Log[1 + c^(2/3)*x^2 + c^(4/3)*x^4])/(20*c^(5/3))

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 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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int x^4 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{1}{5} (3 b c) \int \frac{x^7}{1-c^2 x^6} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{1}{10} (3 b c) \operatorname{Subst}\left (\int \frac{x^3}{1-c^2 x^3} \, dx,x,x^2\right )\\ &=\frac{3 b x^2}{10 c}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^3} \, dx,x,x^2\right )}{10 c}\\ &=\frac{3 b x^2}{10 c}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{b \operatorname{Subst}\left (\int \frac{1}{1-c^{2/3} x} \, dx,x,x^2\right )}{10 c}-\frac{b \operatorname{Subst}\left (\int \frac{2+c^{2/3} x}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )}{10 c}\\ &=\frac{3 b x^2}{10 c}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (1-c^{2/3} x^2\right )}{10 c^{5/3}}-\frac{b \operatorname{Subst}\left (\int \frac{c^{2/3}+2 c^{4/3} x}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )}{20 c^{5/3}}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )}{20 c}\\ &=\frac{3 b x^2}{10 c}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (1-c^{2/3} x^2\right )}{10 c^{5/3}}-\frac{b \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{20 c^{5/3}}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 c^{2/3} x^2\right )}{10 c^{5/3}}\\ &=\frac{3 b x^2}{10 c}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1+2 c^{2/3} x^2}{\sqrt{3}}\right )}{10 c^{5/3}}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (1-c^{2/3} x^2\right )}{10 c^{5/3}}-\frac{b \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{20 c^{5/3}}\\ \end{align*}

Mathematica [A]  time = 0.0282765, size = 198, normalized size = 1.69 \[ \frac{a x^5}{5}-\frac{b \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{20 c^{5/3}}-\frac{b \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{20 c^{5/3}}+\frac{b \log \left (1-\sqrt [3]{c} x\right )}{10 c^{5/3}}+\frac{b \log \left (\sqrt [3]{c} x+1\right )}{10 c^{5/3}}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x-1}{\sqrt{3}}\right )}{10 c^{5/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x+1}{\sqrt{3}}\right )}{10 c^{5/3}}+\frac{3 b x^2}{10 c}+\frac{1}{5} b x^5 \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcTanh[c*x^3]),x]

[Out]

(3*b*x^2)/(10*c) + (a*x^5)/5 - (Sqrt[3]*b*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]])/(10*c^(5/3)) + (Sqrt[3]*b*ArcTan
[(1 + 2*c^(1/3)*x)/Sqrt[3]])/(10*c^(5/3)) + (b*x^5*ArcTanh[c*x^3])/5 + (b*Log[1 - c^(1/3)*x])/(10*c^(5/3)) + (
b*Log[1 + c^(1/3)*x])/(10*c^(5/3)) - (b*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(20*c^(5/3)) - (b*Log[1 + c^(1/3)*x
+ c^(2/3)*x^2])/(20*c^(5/3))

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 114, normalized size = 1. \begin{align*}{\frac{a{x}^{5}}{5}}+{\frac{{x}^{5}b{\it Artanh} \left ( c{x}^{3} \right ) }{5}}+{\frac{3\,b{x}^{2}}{10\,c}}+{\frac{b}{10\,{c}^{3}}\ln \left ({x}^{2}-\sqrt [3]{{c}^{-2}} \right ) \left ({c}^{-2} \right ) ^{-{\frac{2}{3}}}}-{\frac{b}{20\,{c}^{3}}\ln \left ({x}^{4}+\sqrt [3]{{c}^{-2}}{x}^{2}+ \left ({c}^{-2} \right ) ^{{\frac{2}{3}}} \right ) \left ({c}^{-2} \right ) ^{-{\frac{2}{3}}}}-{\frac{b\sqrt{3}}{10\,{c}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{{x}^{2}}{\sqrt [3]{{c}^{-2}}}}+1 \right ) } \right ) \left ({c}^{-2} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c*x^3)),x)

[Out]

1/5*a*x^5+1/5*x^5*b*arctanh(c*x^3)+3/10*b*x^2/c+1/10*b/c^3/(1/c^2)^(2/3)*ln(x^2-(1/c^2)^(1/3))-1/20*b/c^3/(1/c
^2)^(2/3)*ln(x^4+(1/c^2)^(1/3)*x^2+(1/c^2)^(2/3))-1/10*b/c^3/(1/c^2)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c^
2)^(1/3)*x^2+1))

________________________________________________________________________________________

Maxima [A]  time = 1.50079, size = 161, normalized size = 1.38 \begin{align*} \frac{1}{5} \, a x^{5} + \frac{1}{20} \,{\left (4 \, x^{5} \operatorname{artanh}\left (c x^{3}\right ) + c{\left (\frac{6 \, x^{2}}{c^{2}} - \frac{2 \, \sqrt{3}{\left (c^{2}\right )}^{\frac{2}{3}} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{3}}{\left (2 \, x^{2} + \frac{1}{c^{2}}^{\frac{1}{3}}\right )}\right )}{c^{4}} - \frac{{\left (c^{2}\right )}^{\frac{2}{3}} \log \left (x^{4} + \frac{1}{c^{2}}^{\frac{1}{3}} x^{2} + \frac{1}{c^{2}}^{\frac{2}{3}}\right )}{c^{4}} + \frac{2 \,{\left (c^{2}\right )}^{\frac{2}{3}} \log \left (x^{2} - \frac{1}{c^{2}}^{\frac{1}{3}}\right )}{c^{4}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x^3)),x, algorithm="maxima")

[Out]

1/5*a*x^5 + 1/20*(4*x^5*arctanh(c*x^3) + c*(6*x^2/c^2 - 2*sqrt(3)*(c^2)^(2/3)*arctan(1/3*sqrt(3)*(c^2)^(1/3)*(
2*x^2 + (c^(-2))^(1/3)))/c^4 - (c^2)^(2/3)*log(x^4 + (c^(-2))^(1/3)*x^2 + (c^(-2))^(2/3))/c^4 + 2*(c^2)^(2/3)*
log(x^2 - (c^(-2))^(1/3))/c^4))*b

________________________________________________________________________________________

Fricas [A]  time = 1.79069, size = 414, normalized size = 3.54 \begin{align*} \frac{2 \, b c^{3} x^{5} \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + 4 \, a c^{3} x^{5} + 6 \, b c^{2} x^{2} - 2 \, \sqrt{3} b{\left (c^{2}\right )}^{\frac{1}{6}} c \arctan \left (-\frac{\sqrt{3}{\left (4 \, c^{2} x^{4} - 2 \,{\left (c^{2}\right )}^{\frac{2}{3}} x^{2} +{\left (c^{2}\right )}^{\frac{1}{3}}\right )}{\left (c^{2}\right )}^{\frac{1}{6}}}{8 \, c^{3} x^{6} + c}\right ) - b{\left (c^{2}\right )}^{\frac{2}{3}} \log \left (c^{2} x^{4} +{\left (c^{2}\right )}^{\frac{2}{3}} x^{2} +{\left (c^{2}\right )}^{\frac{1}{3}}\right ) + 2 \, b{\left (c^{2}\right )}^{\frac{2}{3}} \log \left (c^{2} x^{2} -{\left (c^{2}\right )}^{\frac{2}{3}}\right )}{20 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x^3)),x, algorithm="fricas")

[Out]

1/20*(2*b*c^3*x^5*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 4*a*c^3*x^5 + 6*b*c^2*x^2 - 2*sqrt(3)*b*(c^2)^(1/6)*c*arctan
(-sqrt(3)*(4*c^2*x^4 - 2*(c^2)^(2/3)*x^2 + (c^2)^(1/3))*(c^2)^(1/6)/(8*c^3*x^6 + c)) - b*(c^2)^(2/3)*log(c^2*x
^4 + (c^2)^(2/3)*x^2 + (c^2)^(1/3)) + 2*b*(c^2)^(2/3)*log(c^2*x^2 - (c^2)^(2/3)))/c^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atanh(c*x**3)),x)

[Out]

Exception raised: KeyError

________________________________________________________________________________________

Giac [A]  time = 1.18607, size = 170, normalized size = 1.45 \begin{align*} -\frac{1}{20} \, b c^{9}{\left (\frac{2 \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + \frac{1}{{\left | c \right |}^{\frac{2}{3}}}\right )}{\left | c \right |}^{\frac{2}{3}}\right )}{c^{10}{\left | c \right |}^{\frac{2}{3}}} + \frac{\log \left (x^{4} + \frac{x^{2}}{{\left | c \right |}^{\frac{2}{3}}} + \frac{1}{{\left | c \right |}^{\frac{4}{3}}}\right )}{c^{10}{\left | c \right |}^{\frac{2}{3}}} - \frac{2 \, \log \left ({\left | x^{2} - \frac{1}{{\left | c \right |}^{\frac{2}{3}}} \right |}\right )}{c^{10}{\left | c \right |}^{\frac{2}{3}}}\right )} + \frac{1}{10} \, b x^{5} \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + \frac{1}{5} \, a x^{5} + \frac{3 \, b x^{2}}{10 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x^3)),x, algorithm="giac")

[Out]

-1/20*b*c^9*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1/abs(c)^(2/3))*abs(c)^(2/3))/(c^10*abs(c)^(2/3)) + log(x^4
+ x^2/abs(c)^(2/3) + 1/abs(c)^(4/3))/(c^10*abs(c)^(2/3)) - 2*log(abs(x^2 - 1/abs(c)^(2/3)))/(c^10*abs(c)^(2/3
))) + 1/10*b*x^5*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 1/5*a*x^5 + 3/10*b*x^2/c