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

Optimal. Leaf size=190 \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )+\frac{b \log \left (c^{2/3} x-\sqrt [3]{c} \sqrt{x}+1\right )}{16 c^{8/3}}-\frac{b \log \left (c^{2/3} x+\sqrt [3]{c} \sqrt{x}+1\right )}{16 c^{8/3}}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1-2 \sqrt [3]{c} \sqrt{x}}{\sqrt{3}}\right )}{8 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} \sqrt{x}+1}{\sqrt{3}}\right )}{8 c^{8/3}}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt{x}\right )}{4 c^{8/3}}+\frac{3 b x^{5/2}}{20 c} \]

[Out]

(3*b*x^(5/2))/(20*c) - (Sqrt[3]*b*ArcTan[(1 - 2*c^(1/3)*Sqrt[x])/Sqrt[3]])/(8*c^(8/3)) + (Sqrt[3]*b*ArcTan[(1
+ 2*c^(1/3)*Sqrt[x])/Sqrt[3]])/(8*c^(8/3)) - (b*ArcTanh[c^(1/3)*Sqrt[x]])/(4*c^(8/3)) + (x^4*(a + b*ArcTanh[c*
x^(3/2)]))/4 + (b*Log[1 - c^(1/3)*Sqrt[x] + c^(2/3)*x])/(16*c^(8/3)) - (b*Log[1 + c^(1/3)*Sqrt[x] + c^(2/3)*x]
)/(16*c^(8/3))

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Rubi [A]  time = 0.300586, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.562, Rules used = {6097, 321, 329, 296, 634, 618, 204, 628, 206} \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )+\frac{b \log \left (c^{2/3} x-\sqrt [3]{c} \sqrt{x}+1\right )}{16 c^{8/3}}-\frac{b \log \left (c^{2/3} x+\sqrt [3]{c} \sqrt{x}+1\right )}{16 c^{8/3}}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1-2 \sqrt [3]{c} \sqrt{x}}{\sqrt{3}}\right )}{8 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} \sqrt{x}+1}{\sqrt{3}}\right )}{8 c^{8/3}}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt{x}\right )}{4 c^{8/3}}+\frac{3 b x^{5/2}}{20 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x^(3/2)]),x]

[Out]

(3*b*x^(5/2))/(20*c) - (Sqrt[3]*b*ArcTan[(1 - 2*c^(1/3)*Sqrt[x])/Sqrt[3]])/(8*c^(8/3)) + (Sqrt[3]*b*ArcTan[(1
+ 2*c^(1/3)*Sqrt[x])/Sqrt[3]])/(8*c^(8/3)) - (b*ArcTanh[c^(1/3)*Sqrt[x]])/(4*c^(8/3)) + (x^4*(a + b*ArcTanh[c*
x^(3/2)]))/4 + (b*Log[1 - c^(1/3)*Sqrt[x] + c^(2/3)*x])/(16*c^(8/3)) - (b*Log[1 + c^(1/3)*Sqrt[x] + c^(2/3)*x]
)/(16*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 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 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^3 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac{1}{8} (3 b c) \int \frac{x^{9/2}}{1-c^2 x^3} \, dx\\ &=\frac{3 b x^{5/2}}{20 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac{(3 b) \int \frac{x^{3/2}}{1-c^2 x^3} \, dx}{8 c}\\ &=\frac{3 b x^{5/2}}{20 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac{(3 b) \operatorname{Subst}\left (\int \frac{x^4}{1-c^2 x^6} \, dx,x,\sqrt{x}\right )}{4 c}\\ &=\frac{3 b x^{5/2}}{20 c}+\frac{1}{4} x^4 \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 )}{4 c^{7/3}}-\frac{b \operatorname{Subst}\left (\int \frac{-\frac{1}{2}-\frac{\sqrt [3]{c} x}{2}}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx,x,\sqrt{x}\right )}{4 c^{7/3}}-\frac{b \operatorname{Subst}\left (\int \frac{-\frac{1}{2}+\frac{\sqrt [3]{c} x}{2}}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx,x,\sqrt{x}\right )}{4 c^{7/3}}\\ &=\frac{3 b x^{5/2}}{20 c}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt{x}\right )}{4 c^{8/3}}+\frac{1}{4} x^4 \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 )}{16 c^{8/3}}-\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 )}{16 c^{8/3}}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx,x,\sqrt{x}\right )}{16 c^{7/3}}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx,x,\sqrt{x}\right )}{16 c^{7/3}}\\ &=\frac{3 b x^{5/2}}{20 c}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt{x}\right )}{4 c^{8/3}}+\frac{1}{4} x^4 \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 )}{16 c^{8/3}}-\frac{b \log \left (1+\sqrt [3]{c} \sqrt{x}+c^{2/3} 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} \sqrt{x}\right )}{8 c^{8/3}}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{c} \sqrt{x}\right )}{8 c^{8/3}}\\ &=\frac{3 b x^{5/2}}{20 c}-\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1-2 \sqrt [3]{c} \sqrt{x}}{\sqrt{3}}\right )}{8 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1+2 \sqrt [3]{c} \sqrt{x}}{\sqrt{3}}\right )}{8 c^{8/3}}-\frac{b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt{x}\right )}{4 c^{8/3}}+\frac{1}{4} x^4 \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 )}{16 c^{8/3}}-\frac{b \log \left (1+\sqrt [3]{c} \sqrt{x}+c^{2/3} x\right )}{16 c^{8/3}}\\ \end{align*}

Mathematica [A]  time = 0.0534598, size = 222, normalized size = 1.17 \[ \frac{a x^4}{4}+\frac{b \log \left (1-\sqrt [3]{c} \sqrt{x}\right )}{8 c^{8/3}}-\frac{b \log \left (\sqrt [3]{c} \sqrt{x}+1\right )}{8 c^{8/3}}+\frac{b \log \left (c^{2/3} x-\sqrt [3]{c} \sqrt{x}+1\right )}{16 c^{8/3}}-\frac{b \log \left (c^{2/3} x+\sqrt [3]{c} \sqrt{x}+1\right )}{16 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} \sqrt{x}-1}{\sqrt{3}}\right )}{8 c^{8/3}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 \sqrt [3]{c} \sqrt{x}+1}{\sqrt{3}}\right )}{8 c^{8/3}}+\frac{3 b x^{5/2}}{20 c}+\frac{1}{4} b x^4 \tanh ^{-1}\left (c x^{3/2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*x^(5/2))/(20*c) + (a*x^4)/4 + (Sqrt[3]*b*ArcTan[(-1 + 2*c^(1/3)*Sqrt[x])/Sqrt[3]])/(8*c^(8/3)) + (Sqrt[3]
*b*ArcTan[(1 + 2*c^(1/3)*Sqrt[x])/Sqrt[3]])/(8*c^(8/3)) + (b*x^4*ArcTanh[c*x^(3/2)])/4 + (b*Log[1 - c^(1/3)*Sq
rt[x]])/(8*c^(8/3)) - (b*Log[1 + c^(1/3)*Sqrt[x]])/(8*c^(8/3)) + (b*Log[1 - c^(1/3)*Sqrt[x] + c^(2/3)*x])/(16*
c^(8/3)) - (b*Log[1 + c^(1/3)*Sqrt[x] + c^(2/3)*x])/(16*c^(8/3))

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Maple [A]  time = 0.036, size = 194, normalized size = 1. \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{{x}^{4}b}{4}{\it Artanh} \left ( c{x}^{{\frac{3}{2}}} \right ) }+{\frac{3\,b}{20\,c}{x}^{{\frac{5}{2}}}}+{\frac{b}{8\,{c}^{3}}\ln \left ( \sqrt{x}-\sqrt [3]{{c}^{-1}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}-{\frac{b}{16\,{c}^{3}}\ln \left ( x+\sqrt [3]{{c}^{-1}}\sqrt{x}+ \left ({c}^{-1} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}+{\frac{b\sqrt{3}}{8\,{c}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{\sqrt{x}}{\sqrt [3]{{c}^{-1}}}}+1 \right ) } \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}-{\frac{b}{8\,{c}^{3}}\ln \left ( \sqrt{x}+\sqrt [3]{{c}^{-1}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}+{\frac{b}{16\,{c}^{3}}\ln \left ( x-\sqrt [3]{{c}^{-1}}\sqrt{x}+ \left ({c}^{-1} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}+{\frac{b\sqrt{3}}{8\,{c}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{\sqrt{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^3*(a+b*arctanh(c*x^(3/2))),x)

[Out]

1/4*x^4*a+1/4*x^4*b*arctanh(c*x^(3/2))+3/20*b*x^(5/2)/c+1/8*b/c^3/(1/c)^(1/3)*ln(x^(1/2)-(1/c)^(1/3))-1/16*b/c
^3/(1/c)^(1/3)*ln(x+(1/c)^(1/3)*x^(1/2)+(1/c)^(2/3))+1/8*b/c^3*3^(1/2)/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)
^(1/3)*x^(1/2)+1))-1/8*b/c^3/(1/c)^(1/3)*ln(x^(1/2)+(1/c)^(1/3))+1/16*b/c^3/(1/c)^(1/3)*ln(x-(1/c)^(1/3)*x^(1/
2)+(1/c)^(2/3))+1/8*b/c^3*3^(1/2)/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x^(1/2)-1))

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 10.9902, size = 4629, normalized size = 24.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160*(40*a*c*x^4 + 24*b*x^(5/2) - 20*sqrt(3)*sqrt(((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*s
qrt(3) + 1) + 2*b)^2 - 4*((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)*b + 4*b
^2)*c*arctan(1/24*(4*sqrt(3)*sqrt(((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b
)^2*b^2*c^5*sqrt(x) + 4*b^4*c^5*sqrt(x) + 4*b^4*c^2 + 4*b^4*x - 2*(2*b^3*c^5*sqrt(x) + b^3*c^2)*((1/2)^(1/3)*(
b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b))*sqrt(((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 +
b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2 - 4*((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3)
+ 1) + 2*b)*b + 4*b^2)*c^3 - sqrt(3)*(((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) +
2*b)^2*c^8 - 4*((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)*b*c^8 + 4*b^2*c^
8 + 8*b^2*c^3*sqrt(x))*sqrt(((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2 -
4*((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)*b + 4*b^2))/b^3) - 10*((1/2)^(
1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)*c*log(-1/4*((1/2)^(1/3)*(b^3 - (c^8 - 1)
*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2*c^5 + ((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3
)*(I*sqrt(3) + 1) + 2*b)*b*c^5 - b^2*c^5 + b^2*sqrt(x)) - 20*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/10
24*b^3/c^8)^(1/3)*(I*sqrt(3) + 1) - b)*c*log((4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3
)*(I*sqrt(3) + 1) - b)^2*c^5 + 2*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sqrt(3)
+ 1) - b)*b*c^5 + b^2*c^5 + b^2*sqrt(x)) - 40*sqrt(3*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/
c^8)^(1/3)*(I*sqrt(3) + 1) - b)^2 + 6*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sq
rt(3) + 1) - b)*b + 3*b^2)*c*arctan(1/3*((4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I
*sqrt(3) + 1) - b)^2*c^8 + 2*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sqrt(3) + 1
) - b)*b*c^8 + b^2*c^8 - 2*b^2*c^3*sqrt(x) + sqrt(-4*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c
^8)^(1/3)*(I*sqrt(3) + 1) - b)^2*b^2*c^5*sqrt(x) - 4*b^4*c^5*sqrt(x) + 4*b^4*c^2 + 4*b^4*x - 4*(2*b^3*c^5*sqrt
(x) - b^3*c^2)*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sqrt(3) + 1) - b))*c^3)*s
qrt(3*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sqrt(3) + 1) - b)^2 + 6*(4*(-1/102
4*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sqrt(3) + 1) - b)*b + 3*b^2)/b^3) + 5*(((1/2)^(1/3
)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)*c - 6*b*c)*log(((1/2)^(1/3)*(b^3 - (c^8 - 1
)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b)^2*b^2*c^5*sqrt(x) + 4*b^4*c^5*sqrt(x) + 4*b^4*c^2 + 4*b^4*x
- 2*(2*b^3*c^5*sqrt(x) + b^3*c^2)*((1/2)^(1/3)*(b^3 - (c^8 - 1)*b^3/c^8 + b^3/c^8)^(1/3)*(I*sqrt(3) + 1) + 2*b
)) + 10*((4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sqrt(3) + 1) - b)*c + 3*b*c)*lo
g(-4*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sqrt(3) + 1) - b)^2*b^2*c^5*sqrt(x)
- 4*b^4*c^5*sqrt(x) + 4*b^4*c^2 + 4*b^4*x - 4*(2*b^3*c^5*sqrt(x) - b^3*c^2)*(4*(-1/1024*b^3 + 1/1024*(c^8 - 1
)*b^3/c^8 + 1/1024*b^3/c^8)^(1/3)*(I*sqrt(3) + 1) - b)) + 20*(b*c*x^4 - b*c)*log(-(c^2*x^3 + 2*c*x^(3/2) + 1)/
(c^2*x^3 - 1)))/c

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [C]  time = 1.49211, size = 296, normalized size = 1.56 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{320} \,{\left (40 \, x^{4} \log \left (-\frac{c x^{\frac{3}{2}} + 1}{c x^{\frac{3}{2}} - 1}\right ) + c{\left (\frac{48 \, x^{\frac{5}{2}}}{c^{2}} - \frac{10 \, \sqrt{3}{\left (-i \, \sqrt{3} - 1\right )}^{2} \arctan \left (\frac{\sqrt{3}{\left (2 \, \sqrt{x} + \left (-\frac{1}{c}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{1}{c}\right )^{\frac{1}{3}}}\right )}{c^{\frac{11}{3}}} + \frac{5 \,{\left (-i \, \sqrt{3} - 1\right )}^{2} \log \left (x + \sqrt{x} \left (-\frac{1}{c}\right )^{\frac{1}{3}} + \left (-\frac{1}{c}\right )^{\frac{2}{3}}\right )}{c^{\frac{11}{3}}} - \frac{40 \, \left (-\frac{1}{c}\right )^{\frac{2}{3}} \log \left ({\left | \sqrt{x} - \left (-\frac{1}{c}\right )^{\frac{1}{3}} \right |}\right )}{c^{3}} + \frac{40 \, \sqrt{3}{\left | c \right |}^{\frac{4}{3}} \arctan \left (\frac{1}{3} \, \sqrt{3} c^{\frac{1}{3}}{\left (2 \, \sqrt{x} + \frac{1}{c^{\frac{1}{3}}}\right )}\right )}{c^{5}} - \frac{20 \,{\left | c \right |}^{\frac{4}{3}} \log \left (x + \frac{\sqrt{x}}{c^{\frac{1}{3}}} + \frac{1}{c^{\frac{2}{3}}}\right )}{c^{5}} + \frac{40 \, \log \left ({\left | \sqrt{x} - \frac{1}{c^{\frac{1}{3}}} \right |}\right )}{c^{\frac{11}{3}}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4 + 1/320*(40*x^4*log(-(c*x^(3/2) + 1)/(c*x^(3/2) - 1)) + c*(48*x^(5/2)/c^2 - 10*sqrt(3)*(-I*sqrt(3) -
1)^2*arctan(1/3*sqrt(3)*(2*sqrt(x) + (-1/c)^(1/3))/(-1/c)^(1/3))/c^(11/3) + 5*(-I*sqrt(3) - 1)^2*log(x + sqrt
(x)*(-1/c)^(1/3) + (-1/c)^(2/3))/c^(11/3) - 40*(-1/c)^(2/3)*log(abs(sqrt(x) - (-1/c)^(1/3)))/c^3 + 40*sqrt(3)*
abs(c)^(4/3)*arctan(1/3*sqrt(3)*c^(1/3)*(2*sqrt(x) + 1/c^(1/3)))/c^5 - 20*abs(c)^(4/3)*log(x + sqrt(x)/c^(1/3)
+ 1/c^(2/3))/c^5 + 40*log(abs(sqrt(x) - 1/c^(1/3)))/c^(11/3)))*b