3.3.16 \(\int (a+b \tanh ^{-1}(c x^{3/2})) \, dx\) [216]

Optimal. Leaf size=170 \[ a x-\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1-2 \sqrt [3]{c} \sqrt {x}}{\sqrt {3}}\right )}{2 c^{2/3}}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1+2 \sqrt [3]{c} \sqrt {x}}{\sqrt {3}}\right )}{2 c^{2/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt {x}\right )}{c^{2/3}}+b x \tanh ^{-1}\left (c x^{3/2}\right )+\frac {b \log \left (1-\sqrt [3]{c} \sqrt {x}+c^{2/3} x\right )}{4 c^{2/3}}-\frac {b \log \left (1+\sqrt [3]{c} \sqrt {x}+c^{2/3} x\right )}{4 c^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6021, 335, 302, 648, 632, 210, 642, 212} \begin {gather*} a x-\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1-2 \sqrt [3]{c} \sqrt {x}}{\sqrt {3}}\right )}{2 c^{2/3}}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {2 \sqrt [3]{c} \sqrt {x}+1}{\sqrt {3}}\right )}{2 c^{2/3}}+\frac {b \log \left (c^{2/3} x-\sqrt [3]{c} \sqrt {x}+1\right )}{4 c^{2/3}}-\frac {b \log \left (c^{2/3} x+\sqrt [3]{c} \sqrt {x}+1\right )}{4 c^{2/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} \sqrt {x}\right )}{c^{2/3}}+b x \tanh ^{-1}\left (c x^{3/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[a + b*ArcTanh[c*x^(3/2)],x]

[Out]

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

Rule 210

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

Rule 212

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

Rule 302

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)/(a*n*s^m))*Int[1/(r^2 - s^2*x^2), x] + 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 335

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 632

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 642

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 648

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 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 114, normalized size = 0.67 \begin {gather*} a x+b x \tanh ^{-1}\left (c x^{3/2}\right )-\frac {b \left (\sqrt {3} \left (\text {ArcTan}\left (\frac {1-2 \sqrt [3]{c} \sqrt {x}}{\sqrt {3}}\right )-\text {ArcTan}\left (\frac {1+2 \sqrt [3]{c} \sqrt {x}}{\sqrt {3}}\right )\right )+2 \tanh ^{-1}\left (\sqrt [3]{c} \sqrt {x}\right )+\tanh ^{-1}\left (\frac {\sqrt [3]{c} \sqrt {x}}{1+c^{2/3} x}\right )\right )}{2 c^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcTanh[c*x^(3/2)],x]

[Out]

a*x + b*x*ArcTanh[c*x^(3/2)] - (b*(Sqrt[3]*(ArcTan[(1 - 2*c^(1/3)*Sqrt[x])/Sqrt[3]] - ArcTan[(1 + 2*c^(1/3)*Sq
rt[x])/Sqrt[3]]) + 2*ArcTanh[c^(1/3)*Sqrt[x]] + ArcTanh[(c^(1/3)*Sqrt[x])/(1 + c^(2/3)*x)]))/(2*c^(2/3))

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Maple [A]
time = 0.06, size = 179, normalized size = 1.05

method	result	size
derivativedivides	\(a x +b x \arctanh \left (c \,x^{\frac {3}{2}}\right )+\frac {b \ln \left (\sqrt {x}-\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}} \sqrt {x}+\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sqrt {x}}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (\sqrt {x}+\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}} \sqrt {x}+\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sqrt {x}}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}\)	\(179\)
default	\(a x +b x \arctanh \left (c \,x^{\frac {3}{2}}\right )+\frac {b \ln \left (\sqrt {x}-\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}} \sqrt {x}+\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sqrt {x}}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (\sqrt {x}+\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}} \sqrt {x}+\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sqrt {x}}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}\)	\(179\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 158, normalized size = 0.93 \begin {gather*} \frac {1}{4} \, {\left (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 {5}{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 {5}{3}}} - \frac {\log \left (c^{\frac {2}{3}} x + c^{\frac {1}{3}} \sqrt {x} + 1\right )}{c^{\frac {5}{3}}} + \frac {\log \left (c^{\frac {2}{3}} x - c^{\frac {1}{3}} \sqrt {x} + 1\right )}{c^{\frac {5}{3}}} - \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} \sqrt {x} + 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} \sqrt {x} - 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}}\right )} + 4 \, x \operatorname {artanh}\left (c x^{\frac {3}{2}}\right )\right )} b + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 1.22, size = 1682, normalized size = 9.89 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.43, size = 186, normalized size = 1.09 \begin {gather*} \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {3} {\left | c \right |}^{\frac {1}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \sqrt {x} + \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{c^{2}} + \frac {2 \, \sqrt {3} {\left | c \right |}^{\frac {1}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \sqrt {x} - \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{c^{2}} - \frac {{\left | c \right |}^{\frac {1}{3}} \log \left (x + \frac {\sqrt {x}}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{c^{2}} + \frac {{\left | c \right |}^{\frac {1}{3}} \log \left (x - \frac {\sqrt {x}}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{c^{2}} - \frac {2 \, {\left | c \right |}^{\frac {1}{3}} \log \left (\sqrt {x} + \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )}{c^{2}} + \frac {2 \, {\left | c \right |}^{\frac {1}{3}} \log \left ({\left | \sqrt {x} - \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right )}{c^{2}}\right )} + 2 \, x \log \left (-\frac {c x^{\frac {3}{2}} + 1}{c x^{\frac {3}{2}} - 1}\right )\right )} b + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.10, size = 107, normalized size = 0.63 \begin {gather*} a\,x+b\,x\,\mathrm {atanh}\left (c\,x^{3/2}\right )-\frac {b\,\mathrm {atanh}\left (c^{1/3}\,\sqrt {x}\right )}{c^{2/3}}+\frac {b\,\mathrm {atanh}\left (\frac {486\,c^8\,\sqrt {x}}{-243\,c^{23/3}+\sqrt {3}\,c^{23/3}\,243{}\mathrm {i}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,c^{2/3}}+\frac {b\,\mathrm {atanh}\left (\frac {486\,c^8\,\sqrt {x}}{243\,c^{23/3}+\sqrt {3}\,c^{23/3}\,243{}\mathrm {i}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,c^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*atanh(c*x^(3/2)),x)

[Out]

a*x + b*x*atanh(c*x^(3/2)) - (b*atanh(c^(1/3)*x^(1/2)))/c^(2/3) + (b*atanh((486*c^8*x^(1/2))/(3^(1/2)*c^(23/3)
*243i - 243*c^(23/3)))*(3^(1/2)*1i + 1))/(2*c^(2/3)) + (b*atanh((486*c^8*x^(1/2))/(3^(1/2)*c^(23/3)*243i + 243
*c^(23/3)))*(3^(1/2)*1i - 1))/(2*c^(2/3))

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