3.215 \(\int x (a+b \tanh ^{-1}(c x^{3/2})) \, dx\)

Optimal. Leaf size=190 \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )+\frac {b \log \left (c^{2/3} x-\sqrt [3]{c} \sqrt {x}+1\right )}{8 c^{4/3}}-\frac {b \log \left (c^{2/3} x+\sqrt [3]{c} \sqrt {x}+1\right )}{8 c^{4/3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {1-2 \sqrt [3]{c} \sqrt {x}}{\sqrt {3}}\right )}{4 c^{4/3}}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {2 \sqrt [3]{c} \sqrt {x}+1}{\sqrt {3}}\right )}{4 c^{4/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt {x}\right )}{2 c^{4/3}}+\frac {3 b \sqrt {x}}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6097, 321, 329, 210, 634, 618, 204, 628, 206} \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )+\frac {b \log \left (c^{2/3} x-\sqrt [3]{c} \sqrt {x}+1\right )}{8 c^{4/3}}-\frac {b \log \left (c^{2/3} x+\sqrt [3]{c} \sqrt {x}+1\right )}{8 c^{4/3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {1-2 \sqrt [3]{c} \sqrt {x}}{\sqrt {3}}\right )}{4 c^{4/3}}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {2 \sqrt [3]{c} \sqrt {x}+1}{\sqrt {3}}\right )}{4 c^{4/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt {x}\right )}{2 c^{4/3}}+\frac {3 b \sqrt {x}}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*x^(3/2)]),x]

[Out]

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

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 210

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

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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

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Mathematica [A]  time = 0.04, size = 222, normalized size = 1.17 \[ \frac {a x^2}{2}+\frac {b \log \left (1-\sqrt [3]{c} \sqrt {x}\right )}{4 c^{4/3}}-\frac {b \log \left (\sqrt [3]{c} \sqrt {x}+1\right )}{4 c^{4/3}}+\frac {b \log \left (c^{2/3} x-\sqrt [3]{c} \sqrt {x}+1\right )}{8 c^{4/3}}-\frac {b \log \left (c^{2/3} x+\sqrt [3]{c} \sqrt {x}+1\right )}{8 c^{4/3}}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {2 \sqrt [3]{c} \sqrt {x}-1}{\sqrt {3}}\right )}{4 c^{4/3}}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {2 \sqrt [3]{c} \sqrt {x}+1}{\sqrt {3}}\right )}{4 c^{4/3}}+\frac {1}{2} b x^2 \tanh ^{-1}\left (c x^{3/2}\right )+\frac {3 b \sqrt {x}}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTanh[c*x^(3/2)]),x]

[Out]

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

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fricas [C]  time = 3.95, size = 1848, normalized size = 9.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(3/2))),x, algorithm="fricas")

[Out]

1/16*(8*a*c*x^2 + 4*sqrt(3)*sqrt(((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)
^2 - 4*((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)*b + 4*b^2)*c*arctan(-1/24
*(2*sqrt(3)*sqrt(((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2*c^2 + 4*b^2*c
^2 + 4*b^2*c*sqrt(x) + 4*b^2*x - 2*(2*b*c^2 + b*c*sqrt(x))*((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1
/3)*(I*sqrt(3) + 1) + 2*b))*(((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)*c^3
- 2*b*c^3)*sqrt(((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2 - 4*((1/2)^(1
/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)*b + 4*b^2) + sqrt(3)*(((1/2)^(1/3)*(b^3 -
(c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2*c^4 + 4*b^2*c^4 + 8*b^2*c^3*sqrt(x) - 4*(b*c^4 +
b*c^3*sqrt(x))*((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b))*sqrt(((1/2)^(1/3
)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2 - 4*((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4
+ b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)*b + 4*b^2))/b^3) - 2*((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)
^(1/3)*(I*sqrt(3) + 1) + 2*b)*c*log(1/2*((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1)
+ 2*b)*c - b*c + b*sqrt(x)) - 4*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) +
1) - b)*c*log((2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)*c + b*c + b
*sqrt(x)) - 8*sqrt(3*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)^2 +
6*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)*b + 3*b^2)*c*arctan(-1/
3*(2*sqrt((2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)^2*c^2 + b^2*c^2
- b^2*c*sqrt(x) + b^2*x + (2*b*c^2 - b*c*sqrt(x))*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(
1/3)*(I*sqrt(3) + 1) - b))*((2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) -
b)*c^3 + b*c^3)*sqrt(3*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)^2
+ 6*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)*b + 3*b^2) + ((2*(-1/
128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)^2*c^4 + b^2*c^4 - 2*b^2*c^3*sqrt
(x) + 2*(b*c^4 - b*c^3*sqrt(x))*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1
) - b))*sqrt(3*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)^2 + 6*(2*(
-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)*b + 3*b^2))/b^3) + (((1/2)^(1
/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)*c - 6*b*c)*log(((1/2)^(1/3)*(b^3 - (c^4 -
1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2*c^2 + 4*b^2*c^2 + 4*b^2*c*sqrt(x) + 4*b^2*x - 2*(2*b*c^2
+ b*c*sqrt(x))*((1/2)^(1/3)*(b^3 - (c^4 - 1)*b^3/c^4 + b^3/c^4)^(1/3)*(I*sqrt(3) + 1) + 2*b)) + 2*((2*(-1/128
*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)*c + 3*b*c)*log(4*(2*(-1/128*b^3 + 1
/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1) - b)^2*c^2 + 4*b^2*c^2 - 4*b^2*c*sqrt(x) + 4*b^2
*x + 4*(2*b*c^2 - b*c*sqrt(x))*(2*(-1/128*b^3 + 1/128*(c^4 - 1)*b^3/c^4 + 1/128*b^3/c^4)^(1/3)*(I*sqrt(3) + 1)
- b)) + 4*(b*c*x^2 - b*c)*log(-(c^2*x^3 + 2*c*x^(3/2) + 1)/(c^2*x^3 - 1)) + 24*b*sqrt(x))/c

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {artanh}\left (c x^{\frac {3}{2}}\right ) + a\right )} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(3/2))),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^(3/2)) + a)*x, x)

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maple [A]  time = 0.04, size = 194, normalized size = 1.02 \[ \frac {a \,x^{2}}{2}+\frac {b \,x^{2} \arctanh \left (c \,x^{\frac {3}{2}}\right )}{2}+\frac {3 b \sqrt {x}}{2 c}+\frac {b \ln \left (\sqrt {x}-\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{4 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}} \sqrt {x}+\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sqrt {x}}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{4 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (\sqrt {x}+\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{4 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}} \sqrt {x}+\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sqrt {x}}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{4 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x^(3/2))),x)

[Out]

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

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maxima [A]  time = 0.42, size = 172, normalized size = 0.91 \[ \frac {1}{2} \, a x^{2} + \frac {1}{8} \, {\left (4 \, x^{2} \operatorname {artanh}\left (c x^{\frac {3}{2}}\right ) - c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} \sqrt {x} + c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {7}{3}}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} \sqrt {x} - c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {7}{3}}} + \frac {\log \left (c^{\frac {2}{3}} x + c^{\frac {1}{3}} \sqrt {x} + 1\right )}{c^{\frac {7}{3}}} - \frac {\log \left (c^{\frac {2}{3}} x - c^{\frac {1}{3}} \sqrt {x} + 1\right )}{c^{\frac {7}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} \sqrt {x} + 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {7}{3}}} - \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} \sqrt {x} - 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {7}{3}}} - \frac {12 \, \sqrt {x}}{c^{2}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(3/2))),x, algorithm="maxima")

[Out]

1/2*a*x^2 + 1/8*(4*x^2*arctanh(c*x^(3/2)) - c*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*c^(2/3)*sqrt(x) + c^(1/3))/c^(1
/3))/c^(7/3) + 2*sqrt(3)*arctan(1/3*sqrt(3)*(2*c^(2/3)*sqrt(x) - c^(1/3))/c^(1/3))/c^(7/3) + log(c^(2/3)*x + c
^(1/3)*sqrt(x) + 1)/c^(7/3) - log(c^(2/3)*x - c^(1/3)*sqrt(x) + 1)/c^(7/3) + 2*log((c^(1/3)*sqrt(x) + 1)/c^(1/
3))/c^(7/3) - 2*log((c^(1/3)*sqrt(x) - 1)/c^(1/3))/c^(7/3) - 12*sqrt(x)/c^2))*b

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mupad [B]  time = 11.97, size = 247, normalized size = 1.30 \[ \frac {a\,x^2}{2}+\frac {3\,b\,\sqrt {x}}{2\,c}+\frac {b\,\ln \left (\frac {c^{1/3}\,\sqrt {x}-1}{c^{1/3}\,\sqrt {x}+1}\right )}{4\,c^{4/3}}+\frac {\ln \left (1-c\,x^{3/2}\right )\,\left (\frac {b\,x^2}{2}-\frac {b\,c^2\,x^5}{2}\right )}{2\,c^2\,x^3-2}+\frac {b\,x^2\,\ln \left (c\,x^{3/2}+1\right )}{4}+\frac {b\,\ln \left (\frac {\sqrt {3}\,c^{2/3}\,x+c^{2/3}\,x\,1{}\mathrm {i}-c^{1/3}\,\sqrt {x}\,4{}\mathrm {i}-\sqrt {3}+1{}\mathrm {i}}{2\,c^{2/3}\,x+1-\sqrt {3}\,1{}\mathrm {i}}\right )\,\sqrt {-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}{4\,c^{4/3}}+\frac {\sqrt {2}\,b\,\ln \left (\frac {2\,\sqrt {2}-c^{1/3}\,\sqrt {x}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}-\sqrt {2}\,c^{2/3}\,x+\sqrt {2}\,\sqrt {3}\,c^{2/3}\,x\,1{}\mathrm {i}}{2\,c^{2/3}\,x+1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\sqrt {1+\sqrt {3}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8\,c^{4/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c*x^(3/2))),x)

[Out]

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

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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**(3/2))),x)

[Out]

Timed out

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