3.111 \(\int x^7 (a+b \tanh ^{-1}(c x^3)) \, dx\)

Optimal. Leaf size=176 \[ \frac{1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac{b \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{c} x}{\sqrt{3}}\right )}{16 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right )}{16 c^{8/3}}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac{3 b x^5}{40 c} \]

[Out]

(3*b*x^5)/(40*c) - (Sqrt[3]*b*ArcTan[1/Sqrt[3] - (2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) + (Sqrt[3]*b*ArcTan[1/Sq
rt[3] + (2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) - (b*ArcTanh[c^(1/3)*x])/(8*c^(8/3)) + (x^8*(a + b*ArcTanh[c*x^3]
))/8 + (b*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(32*c^(8/3)) - (b*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(32*c^(8/3))

________________________________________________________________________________________

Rubi [A]  time = 0.283719, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {6097, 321, 296, 634, 618, 204, 628, 206} \[ \frac{1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac{b \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{c} x}{\sqrt{3}}\right )}{16 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right )}{16 c^{8/3}}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac{3 b x^5}{40 c} \]

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*ArcTanh[c*x^3]),x]

[Out]

(3*b*x^5)/(40*c) - (Sqrt[3]*b*ArcTan[1/Sqrt[3] - (2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) + (Sqrt[3]*b*ArcTan[1/Sq
rt[3] + (2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) - (b*ArcTanh[c^(1/3)*x])/(8*c^(8/3)) + (x^8*(a + b*ArcTanh[c*x^3]
))/8 + (b*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(32*c^(8/3)) - (b*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(32*c^(8/3))

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 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 296

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[(2*k*m*Pi)/n] - s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi
)/n]*x + s^2*x^2), x] + Int[(r*Cos[(2*k*m*Pi)/n] + s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x
+ s^2*x^2), x]; (2*r^(m + 2)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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]

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^7 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\frac{1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{1}{8} (3 b c) \int \frac{x^{10}}{1-c^2 x^6} \, dx\\ &=\frac{3 b x^5}{40 c}+\frac{1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{(3 b) \int \frac{x^4}{1-c^2 x^6} \, dx}{8 c}\\ &=\frac{3 b x^5}{40 c}+\frac{1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{b \int \frac{1}{1-c^{2/3} x^2} \, dx}{8 c^{7/3}}-\frac{b \int \frac{-\frac{1}{2}-\frac{\sqrt [3]{c} x}{2}}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{7/3}}-\frac{b \int \frac{-\frac{1}{2}+\frac{\sqrt [3]{c} x}{2}}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{7/3}}\\ &=\frac{3 b x^5}{40 c}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac{1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \int \frac{-\sqrt [3]{c}+2 c^{2/3} x}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{8/3}}-\frac{b \int \frac{\sqrt [3]{c}+2 c^{2/3} x}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{8/3}}+\frac{(3 b) \int \frac{1}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{7/3}}+\frac{(3 b) \int \frac{1}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{7/3}}\\ &=\frac{3 b x^5}{40 c}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac{1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac{b \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{c} x\right )}{16 c^{8/3}}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{c} x\right )}{16 c^{8/3}}\\ &=\frac{3 b x^5}{40 c}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1-2 \sqrt [3]{c} x}{\sqrt{3}}\right )}{16 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1+2 \sqrt [3]{c} x}{\sqrt{3}}\right )}{16 c^{8/3}}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac{1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac{b \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}\\ \end{align*}

Mathematica [A]  time = 0.0302007, size = 198, normalized size = 1.12 \[ \frac{a x^8}{8}+\frac{b \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac{b \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{32 c^{8/3}}+\frac{b \log \left (1-\sqrt [3]{c} x\right )}{16 c^{8/3}}-\frac{b \log \left (\sqrt [3]{c} x+1\right )}{16 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x-1}{\sqrt{3}}\right )}{16 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x+1}{\sqrt{3}}\right )}{16 c^{8/3}}+\frac{3 b x^5}{40 c}+\frac{1}{8} b x^8 \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a + b*ArcTanh[c*x^3]),x]

[Out]

(3*b*x^5)/(40*c) + (a*x^8)/8 + (Sqrt[3]*b*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) + (Sqrt[3]*b*ArcTan
[(1 + 2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) + (b*x^8*ArcTanh[c*x^3])/8 + (b*Log[1 - c^(1/3)*x])/(16*c^(8/3)) - (
b*Log[1 + c^(1/3)*x])/(16*c^(8/3)) + (b*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(32*c^(8/3)) - (b*Log[1 + c^(1/3)*x
+ c^(2/3)*x^2])/(32*c^(8/3))

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 186, normalized size = 1.1 \begin{align*}{\frac{{x}^{8}a}{8}}+{\frac{{x}^{8}b{\it Artanh} \left ( c{x}^{3} \right ) }{8}}+{\frac{3\,b{x}^{5}}{40\,c}}+{\frac{b}{16\,{c}^{3}}\ln \left ( x-\sqrt [3]{{c}^{-1}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}-{\frac{b}{32\,{c}^{3}}\ln \left ({x}^{2}+\sqrt [3]{{c}^{-1}}x+ \left ({c}^{-1} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}+{\frac{b\sqrt{3}}{16\,{c}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{x}{\sqrt [3]{{c}^{-1}}}}+1 \right ) } \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}-{\frac{b}{16\,{c}^{3}}\ln \left ( x+\sqrt [3]{{c}^{-1}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}+{\frac{b}{32\,{c}^{3}}\ln \left ({x}^{2}-\sqrt [3]{{c}^{-1}}x+ \left ({c}^{-1} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}+{\frac{b\sqrt{3}}{16\,{c}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{x}{\sqrt [3]{{c}^{-1}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a+b*arctanh(c*x^3)),x)

[Out]

1/8*x^8*a+1/8*x^8*b*arctanh(c*x^3)+3/40*b*x^5/c+1/16*b/c^3/(1/c)^(1/3)*ln(x-(1/c)^(1/3))-1/32*b/c^3/(1/c)^(1/3
)*ln(x^2+(1/c)^(1/3)*x+(1/c)^(2/3))+1/16*b/c^3*3^(1/2)/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))-1/1
6*b/c^3/(1/c)^(1/3)*ln(x+(1/c)^(1/3))+1/32*b/c^3/(1/c)^(1/3)*ln(x^2-(1/c)^(1/3)*x+(1/c)^(2/3))+1/16*b/c^3*3^(1
/2)/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x-1))

________________________________________________________________________________________

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^7*(a+b*arctanh(c*x^3)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.88365, size = 659, normalized size = 3.74 \begin{align*} \frac{10 \, b c^{4} x^{8} \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + 20 \, a c^{4} x^{8} + 12 \, b c^{3} x^{5} + 10 \, \sqrt{3} b c \sqrt{-\left (-c^{2}\right )^{\frac{1}{3}}} \arctan \left (\frac{\sqrt{3}{\left (2 \, c x + \left (-c^{2}\right )^{\frac{1}{3}}\right )} \sqrt{-\left (-c^{2}\right )^{\frac{1}{3}}}}{3 \, c}\right ) + 10 \, \sqrt{3} b{\left (c^{2}\right )}^{\frac{1}{6}} c \arctan \left (\frac{\sqrt{3}{\left (c^{2}\right )}^{\frac{1}{6}}{\left (2 \, c x +{\left (c^{2}\right )}^{\frac{1}{3}}\right )}}{3 \, c}\right ) + 5 \, \left (-c^{2}\right )^{\frac{2}{3}} b \log \left (c^{2} x^{2} + \left (-c^{2}\right )^{\frac{1}{3}} c x + \left (-c^{2}\right )^{\frac{2}{3}}\right ) - 5 \, b{\left (c^{2}\right )}^{\frac{2}{3}} \log \left (c^{2} x^{2} +{\left (c^{2}\right )}^{\frac{1}{3}} c x +{\left (c^{2}\right )}^{\frac{2}{3}}\right ) - 10 \, \left (-c^{2}\right )^{\frac{2}{3}} b \log \left (c x - \left (-c^{2}\right )^{\frac{1}{3}}\right ) + 10 \, b{\left (c^{2}\right )}^{\frac{2}{3}} \log \left (c x -{\left (c^{2}\right )}^{\frac{1}{3}}\right )}{160 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arctanh(c*x^3)),x, algorithm="fricas")

[Out]

1/160*(10*b*c^4*x^8*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 20*a*c^4*x^8 + 12*b*c^3*x^5 + 10*sqrt(3)*b*c*sqrt(-(-c^2)^
(1/3))*arctan(1/3*sqrt(3)*(2*c*x + (-c^2)^(1/3))*sqrt(-(-c^2)^(1/3))/c) + 10*sqrt(3)*b*(c^2)^(1/6)*c*arctan(1/
3*sqrt(3)*(c^2)^(1/6)*(2*c*x + (c^2)^(1/3))/c) + 5*(-c^2)^(2/3)*b*log(c^2*x^2 + (-c^2)^(1/3)*c*x + (-c^2)^(2/3
)) - 5*b*(c^2)^(2/3)*log(c^2*x^2 + (c^2)^(1/3)*c*x + (c^2)^(2/3)) - 10*(-c^2)^(2/3)*b*log(c*x - (-c^2)^(1/3))
+ 10*b*(c^2)^(2/3)*log(c*x - (c^2)^(1/3)))/c^4

________________________________________________________________________________________

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**7*(a+b*atanh(c*x**3)),x)

[Out]

Exception raised: KeyError

________________________________________________________________________________________

Giac [A]  time = 1.49266, size = 281, normalized size = 1.6 \begin{align*} -\frac{1}{32} \, b c^{15}{\left (\frac{2 \, \left (-\frac{1}{c}\right )^{\frac{2}{3}} \log \left ({\left | x - \left (-\frac{1}{c}\right )^{\frac{1}{3}} \right |}\right )}{c^{17}} - \frac{2 \, \sqrt{3}{\left | c \right |}^{\frac{4}{3}} \arctan \left (\frac{1}{3} \, \sqrt{3} c^{\frac{1}{3}}{\left (2 \, x + \frac{1}{c^{\frac{1}{3}}}\right )}\right )}{c^{19}} + \frac{{\left | c \right |}^{\frac{4}{3}} \log \left (x^{2} + \frac{x}{c^{\frac{1}{3}}} + \frac{1}{c^{\frac{2}{3}}}\right )}{c^{19}} - \frac{2 \, \log \left ({\left | x - \frac{1}{c^{\frac{1}{3}}} \right |}\right )}{c^{\frac{53}{3}}} + \frac{2 \, \sqrt{3} \left (-c^{2}\right )^{\frac{2}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \, x + \left (-\frac{1}{c}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{1}{c}\right )^{\frac{1}{3}}}\right )}{c^{19}} - \frac{\left (-c^{2}\right )^{\frac{2}{3}} \log \left (x^{2} + x \left (-\frac{1}{c}\right )^{\frac{1}{3}} + \left (-\frac{1}{c}\right )^{\frac{2}{3}}\right )}{c^{19}}\right )} + \frac{1}{16} \, b x^{8} \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + \frac{1}{8} \, a x^{8} + \frac{3 \, b x^{5}}{40 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arctanh(c*x^3)),x, algorithm="giac")

[Out]

-1/32*b*c^15*(2*(-1/c)^(2/3)*log(abs(x - (-1/c)^(1/3)))/c^17 - 2*sqrt(3)*abs(c)^(4/3)*arctan(1/3*sqrt(3)*c^(1/
3)*(2*x + 1/c^(1/3)))/c^19 + abs(c)^(4/3)*log(x^2 + x/c^(1/3) + 1/c^(2/3))/c^19 - 2*log(abs(x - 1/c^(1/3)))/c^
(53/3) + 2*sqrt(3)*(-c^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x + (-1/c)^(1/3))/(-1/c)^(1/3))/c^19 - (-c^2)^(2/3)*log(
x^2 + x*(-1/c)^(1/3) + (-1/c)^(2/3))/c^19) + 1/16*b*x^8*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 1/8*a*x^8 + 3/40*b*x^5
/c