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

Optimal. Leaf size=165 \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{4 c^{2/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}} \]

[Out]

-1/2*b*arctanh(c^(1/3)*x)/c^(2/3)+1/2*x^2*(a+b*arctanh(c*x^3))+1/8*b*ln(1-c^(1/3)*x+c^(2/3)*x^2)/c^(2/3)-1/8*b
*ln(1+c^(1/3)*x+c^(2/3)*x^2)/c^(2/3)+1/4*b*arctan(-1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(1/2)/c^(2/3)+1/4*b*ar
ctan(1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(1/2)/c^(2/3)

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Rubi [A]  time = 0.25, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6097, 296, 634, 618, 204, 628, 206} \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{4 c^{2/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*x^3]),x]

[Out]

-(Sqrt[3]*b*ArcTan[1/Sqrt[3] - (2*c^(1/3)*x)/Sqrt[3]])/(4*c^(2/3)) + (Sqrt[3]*b*ArcTan[1/Sqrt[3] + (2*c^(1/3)*
x)/Sqrt[3]])/(4*c^(2/3)) - (b*ArcTanh[c^(1/3)*x])/(2*c^(2/3)) + (x^2*(a + b*ArcTanh[c*x^3]))/2 + (b*Log[1 - c^
(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3)) - (b*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3))

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 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])

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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 187, normalized size = 1.13 \[ \frac {a x^2}{2}+\frac {b \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{8 c^{2/3}}+\frac {b \log \left (1-\sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b \log \left (\sqrt [3]{c} x+1\right )}{4 c^{2/3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x-1}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x+1}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {1}{2} b x^2 \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTanh[c*x^3]),x]

[Out]

(a*x^2)/2 + (Sqrt[3]*b*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]])/(4*c^(2/3)) + (Sqrt[3]*b*ArcTan[(1 + 2*c^(1/3)*x)/S
qrt[3]])/(4*c^(2/3)) + (b*x^2*ArcTanh[c*x^3])/2 + (b*Log[1 - c^(1/3)*x])/(4*c^(2/3)) - (b*Log[1 + c^(1/3)*x])/
(4*c^(2/3)) + (b*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3)) - (b*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/
3))

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fricas [A]  time = 0.61, size = 238, normalized size = 1.44 \[ \frac {2 \, b c^{2} x^{2} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 4 \, a c^{2} x^{2} + 2 \, \sqrt {3} b c \sqrt {-\left (-c^{2}\right )^{\frac {1}{3}}} \arctan \left (\frac {\sqrt {3} {\left (2 \, c x + \left (-c^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\left (-c^{2}\right )^{\frac {1}{3}}}}{3 \, c}\right ) + 2 \, \sqrt {3} b {\left (c^{2}\right )}^{\frac {1}{6}} c \arctan \left (\frac {\sqrt {3} {\left (c^{2}\right )}^{\frac {1}{6}} {\left (2 \, c x + {\left (c^{2}\right )}^{\frac {1}{3}}\right )}}{3 \, c}\right ) + \left (-c^{2}\right )^{\frac {2}{3}} b \log \left (c^{2} x^{2} + \left (-c^{2}\right )^{\frac {1}{3}} c x + \left (-c^{2}\right )^{\frac {2}{3}}\right ) - b {\left (c^{2}\right )}^{\frac {2}{3}} \log \left (c^{2} x^{2} + {\left (c^{2}\right )}^{\frac {1}{3}} c x + {\left (c^{2}\right )}^{\frac {2}{3}}\right ) - 2 \, \left (-c^{2}\right )^{\frac {2}{3}} b \log \left (c x - \left (-c^{2}\right )^{\frac {1}{3}}\right ) + 2 \, b {\left (c^{2}\right )}^{\frac {2}{3}} \log \left (c x - {\left (c^{2}\right )}^{\frac {1}{3}}\right )}{8 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*b*c^2*x^2*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 4*a*c^2*x^2 + 2*sqrt(3)*b*c*sqrt(-(-c^2)^(1/3))*arctan(1/3*sq
rt(3)*(2*c*x + (-c^2)^(1/3))*sqrt(-(-c^2)^(1/3))/c) + 2*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) + (-c^2)^(2/3)*b*log(c^2*x^2 + (-c^2)^(1/3)*c*x + (-c^2)^(2/3)) - b*(c^2)^(2/3)*log(
c^2*x^2 + (c^2)^(1/3)*c*x + (c^2)^(2/3)) - 2*(-c^2)^(2/3)*b*log(c*x - (-c^2)^(1/3)) + 2*b*(c^2)^(2/3)*log(c*x
- (c^2)^(1/3)))/c^2

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giac [A]  time = 0.44, size = 179, normalized size = 1.08 \[ \frac {1}{4} \, b x^{2} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + \frac {1}{2} \, a x^{2} + \frac {\sqrt {3} b {\left | c \right |}^{\frac {1}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{4 \, c} + \frac {\sqrt {3} b {\left | c \right |}^{\frac {1}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{4 \, c} - \frac {b c \log \left (x^{2} + \frac {x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{8 \, {\left | c \right |}^{\frac {5}{3}}} + \frac {b c \log \left (x^{2} - \frac {x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{8 \, {\left | c \right |}^{\frac {5}{3}}} - \frac {b c \log \left ({\left | x + \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right )}{4 \, {\left | c \right |}^{\frac {5}{3}}} + \frac {b {\left | c \right |}^{\frac {1}{3}} \log \left ({\left | x - \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right )}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 177, normalized size = 1.07 \[ \frac {a \,x^{2}}{2}+\frac {b \,x^{2} \arctanh \left (c \,x^{3}\right )}{2}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{8 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{8 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2+1/2*b*x^2*arctanh(c*x^3)+1/4*b/c/(1/c)^(1/3)*ln(x-(1/c)^(1/3))-1/8*b/c/(1/c)^(1/3)*ln(x^2+(1/c)^(1/3
)*x+(1/c)^(2/3))+1/4*b*3^(1/2)/c/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))-1/4*b/c/(1/c)^(1/3)*ln(x+
(1/c)^(1/3))+1/8*b/c/(1/c)^(1/3)*ln(x^2-(1/c)^(1/3)*x+(1/c)^(2/3))+1/4*b*3^(1/2)/c/(1/c)^(1/3)*arctan(1/3*3^(1
/2)*(2/(1/c)^(1/3)*x-1))

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maxima [A]  time = 0.41, size = 155, normalized size = 0.94 \[ \frac {1}{2} \, a x^{2} + \frac {1}{8} \, {\left (4 \, x^{2} \operatorname {artanh}\left (c x^{3}\right ) + c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x + c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x - c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} - \frac {\log \left (c^{\frac {2}{3}} x^{2} + c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} + \frac {\log \left (c^{\frac {2}{3}} x^{2} - c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} x + 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} x - 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.25, size = 118, normalized size = 0.72 \[ \frac {a\,x^2}{2}+\frac {b\,\left (-\frac {\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}-\mathrm {i}\right )}{2}\right )}{2}+\frac {\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}+1{}\mathrm {i}\right )}{2}\right )}{2}+\mathrm {atan}\left (c^{1/3}\,x\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{2\,c^{2/3}}+\frac {b\,x^2\,\ln \left (c\,x^3+1\right )}{4}-\frac {b\,x^2\,\ln \left (1-c\,x^3\right )}{4}+\frac {\sqrt {3}\,b\,\left (\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}-\mathrm {i}\right )}{2}\right )+\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}+1{}\mathrm {i}\right )}{2}\right )\right )}{4\,c^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c*x^3)),x)

[Out]

(a*x^2)/2 + (b*(atan((c^(1/3)*x*(3^(1/2) + 1i))/2)/2 - atan((c^(1/3)*x*(3^(1/2) - 1i))/2)/2 + atan(c^(1/3)*x*1
i))*1i)/(2*c^(2/3)) + (b*x^2*log(c*x^3 + 1))/4 - (b*x^2*log(1 - c*x^3))/4 + (3^(1/2)*b*(atan((c^(1/3)*x*(3^(1/
2) - 1i))/2) + atan((c^(1/3)*x*(3^(1/2) + 1i))/2)))/(4*c^(2/3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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