3.84 \(\int \sqrt{d x} (a+b \tanh ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=301 \[ \frac{2 (d x)^{3/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 d}+\frac{b \sqrt{d} \log \left (\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{3 \sqrt{2} c^{3/4}}-\frac{b \sqrt{d} \log \left (\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{3 \sqrt{2} c^{3/4}}+\frac{2 b \sqrt{d} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 c^{3/4}}-\frac{\sqrt{2} b \sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 c^{3/4}}+\frac{\sqrt{2} b \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}+1\right )}{3 c^{3/4}}-\frac{2 b \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 c^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.250824, antiderivative size = 301, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.722, Rules used = {6097, 16, 329, 301, 297, 1162, 617, 204, 1165, 628, 298, 205, 208} \[ \frac{2 (d x)^{3/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 d}+\frac{b \sqrt{d} \log \left (\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{3 \sqrt{2} c^{3/4}}-\frac{b \sqrt{d} \log \left (\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{3 \sqrt{2} c^{3/4}}+\frac{2 b \sqrt{d} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 c^{3/4}}-\frac{\sqrt{2} b \sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 c^{3/4}}+\frac{\sqrt{2} b \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}+1\right )}{3 c^{3/4}}-\frac{2 b \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 c^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*x]*(a + b*ArcTanh[c*x^2]),x]

[Out]

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

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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 301

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(
a/b), 2]]}, Dist[s/(2*b), Int[x^(m - n/2)/(r + s*x^(n/2)), x], x] - Dist[s/(2*b), Int[x^(m - n/2)/(r - s*x^(n/
2)), x], x]] /; FreeQ[{a, b}, x] && IGtQ[n/4, 0] && IGtQ[m, 0] && LeQ[n/2, m] && LtQ[m, n] &&  !GtQ[a/b, 0]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 205

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

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.071632, size = 227, normalized size = 0.75 \[ \frac{\sqrt{d x} \left (4 a c^{3/4} x^{3/2}+4 b c^{3/4} x^{3/2} \tanh ^{-1}\left (c x^2\right )+2 b \log \left (1-\sqrt [4]{c} \sqrt{x}\right )-2 b \log \left (\sqrt [4]{c} \sqrt{x}+1\right )+\sqrt{2} b \log \left (\sqrt{c} x-\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )-\sqrt{2} b \log \left (\sqrt{c} x+\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )-2 \sqrt{2} b \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{c} \sqrt{x}\right )+2 \sqrt{2} b \tan ^{-1}\left (\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )+4 b \tan ^{-1}\left (\sqrt [4]{c} \sqrt{x}\right )\right )}{6 c^{3/4} \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*x]*(a + b*ArcTanh[c*x^2]),x]

[Out]

(Sqrt[d*x]*(4*a*c^(3/4)*x^(3/2) - 2*Sqrt[2]*b*ArcTan[1 - Sqrt[2]*c^(1/4)*Sqrt[x]] + 2*Sqrt[2]*b*ArcTan[1 + Sqr
t[2]*c^(1/4)*Sqrt[x]] + 4*b*ArcTan[c^(1/4)*Sqrt[x]] + 4*b*c^(3/4)*x^(3/2)*ArcTanh[c*x^2] + 2*b*Log[1 - c^(1/4)
*Sqrt[x]] - 2*b*Log[1 + c^(1/4)*Sqrt[x]] + Sqrt[2]*b*Log[1 - Sqrt[2]*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - Sqrt[2]*b*
Log[1 + Sqrt[2]*c^(1/4)*Sqrt[x] + Sqrt[c]*x]))/(6*c^(3/4)*Sqrt[x])

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Maple [A]  time = 0.011, size = 280, normalized size = 0.9 \begin{align*}{\frac{2\,a}{3\,d} \left ( dx \right ) ^{{\frac{3}{2}}}}+{\frac{2\,b{\it Artanh} \left ( c{x}^{2} \right ) }{3\,d} \left ( dx \right ) ^{{\frac{3}{2}}}}+{\frac{db\sqrt{2}}{6\,c}\ln \left ({ \left ( dx-\sqrt [4]{{\frac{{d}^{2}}{c}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{{d}^{2}}{c}}} \right ) \left ( dx+\sqrt [4]{{\frac{{d}^{2}}{c}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{{d}^{2}}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}+{\frac{db\sqrt{2}}{3\,c}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}+{\frac{db\sqrt{2}}{3\,c}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}+{\frac{2\,db}{3\,c}\arctan \left ({\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}-{\frac{db}{3\,c}\ln \left ({ \left ( \sqrt{dx}+\sqrt [4]{{\frac{{d}^{2}}{c}}} \right ) \left ( \sqrt{dx}-\sqrt [4]{{\frac{{d}^{2}}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(1/2)*(a+b*arctanh(c*x^2)),x)

[Out]

2/3/d*(d*x)^(3/2)*a+2/3/d*b*(d*x)^(3/2)*arctanh(c*x^2)+1/6*d*b/c/(d^2/c)^(1/4)*2^(1/2)*ln((d*x-(d^2/c)^(1/4)*(
d*x)^(1/2)*2^(1/2)+(d^2/c)^(1/2))/(d*x+(d^2/c)^(1/4)*(d*x)^(1/2)*2^(1/2)+(d^2/c)^(1/2)))+1/3*d*b/c/(d^2/c)^(1/
4)*2^(1/2)*arctan(2^(1/2)/(d^2/c)^(1/4)*(d*x)^(1/2)+1)+1/3*d*b/c/(d^2/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2/c)^
(1/4)*(d*x)^(1/2)-1)+2/3*d*b/c/(d^2/c)^(1/4)*arctan((d*x)^(1/2)/(d^2/c)^(1/4))-1/3*d*b/c/(d^2/c)^(1/4)*ln(((d*
x)^(1/2)+(d^2/c)^(1/4))/((d*x)^(1/2)-(d^2/c)^(1/4)))

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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((d*x)^(1/2)*(a+b*arctanh(c*x^2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.15992, size = 80, normalized size = 0.27 \begin{align*} \frac{1}{3} \,{\left (b x \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a x\right )} \sqrt{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(1/2)*(a+b*arctanh(c*x^2)),x, algorithm="fricas")

[Out]

1/3*(b*x*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a*x)*sqrt(d*x)

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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((d*x)**(1/2)*(a+b*atanh(c*x**2)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d x}{\left (b \operatorname{artanh}\left (c x^{2}\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(1/2)*(a+b*arctanh(c*x^2)),x, algorithm="giac")

[Out]

integrate(sqrt(d*x)*(b*arctanh(c*x^2) + a), x)