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

Optimal. Leaf size=317 \[ \frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}+\frac{b d^{3/2} \log \left (\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{5 \sqrt{2} c^{5/4}}-\frac{b d^{3/2} \log \left (\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{5 \sqrt{2} c^{5/4}}-\frac{2 b d^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{\sqrt{2} b d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}-\frac{\sqrt{2} b d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}+1\right )}{5 c^{5/4}}-\frac{2 b d^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{8 b d \sqrt{d x}}{5 c} \]

[Out]

(8*b*d*Sqrt[d*x])/(5*c) - (2*b*d^(3/2)*ArcTan[(c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (Sqrt[2]*b*d^(3/2)*A
rcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) - (Sqrt[2]*b*d^(3/2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*S
qrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (2*(d*x)^(5/2)*(a + b*ArcTanh[c*x^2]))/(5*d) - (2*b*d^(3/2)*ArcTanh[(c^(1/4)
*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (b*d^(3/2)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x - Sqrt[2]*c^(1/4)*Sqrt[d*x]])/(
5*Sqrt[2]*c^(5/4)) - (b*d^(3/2)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x + Sqrt[2]*c^(1/4)*Sqrt[d*x]])/(5*Sqrt[2]*c^(5/
4))

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Rubi [A]  time = 0.299853, antiderivative size = 317, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {6097, 16, 321, 329, 214, 212, 208, 205, 211, 1165, 628, 1162, 617, 204} \[ \frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}+\frac{b d^{3/2} \log \left (\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{5 \sqrt{2} c^{5/4}}-\frac{b d^{3/2} \log \left (\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{5 \sqrt{2} c^{5/4}}-\frac{2 b d^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{\sqrt{2} b d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}-\frac{\sqrt{2} b d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}+1\right )}{5 c^{5/4}}-\frac{2 b d^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{8 b d \sqrt{d x}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d*Sqrt[d*x])/(5*c) - (2*b*d^(3/2)*ArcTan[(c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (Sqrt[2]*b*d^(3/2)*A
rcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) - (Sqrt[2]*b*d^(3/2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*S
qrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (2*(d*x)^(5/2)*(a + b*ArcTanh[c*x^2]))/(5*d) - (2*b*d^(3/2)*ArcTanh[(c^(1/4)
*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (b*d^(3/2)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x - Sqrt[2]*c^(1/4)*Sqrt[d*x]])/(
5*Sqrt[2]*c^(5/4)) - (b*d^(3/2)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x + Sqrt[2]*c^(1/4)*Sqrt[d*x]])/(5*Sqrt[2]*c^(5/
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 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 214

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

Rule 212

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

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]

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 211

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

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 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])

Rubi steps

\begin{align*} \int (d x)^{3/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{(4 b c) \int \frac{x (d x)^{5/2}}{1-c^2 x^4} \, dx}{5 d}\\ &=\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{(4 b c) \int \frac{(d x)^{7/2}}{1-c^2 x^4} \, dx}{5 d^2}\\ &=\frac{8 b d \sqrt{d x}}{5 c}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{\left (4 b d^2\right ) \int \frac{1}{\sqrt{d x} \left (1-c^2 x^4\right )} \, dx}{5 c}\\ &=\frac{8 b d \sqrt{d x}}{5 c}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{(8 b d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{c^2 x^8}{d^4}} \, dx,x,\sqrt{d x}\right )}{5 c}\\ &=\frac{8 b d \sqrt{d x}}{5 c}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{\left (4 b d^3\right ) \operatorname{Subst}\left (\int \frac{1}{d^2-c x^4} \, dx,x,\sqrt{d x}\right )}{5 c}-\frac{\left (4 b d^3\right ) \operatorname{Subst}\left (\int \frac{1}{d^2+c x^4} \, dx,x,\sqrt{d x}\right )}{5 c}\\ &=\frac{8 b d \sqrt{d x}}{5 c}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{\left (2 b d^2\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{c} x^2} \, dx,x,\sqrt{d x}\right )}{5 c}-\frac{\left (2 b d^2\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{c} x^2} \, dx,x,\sqrt{d x}\right )}{5 c}-\frac{\left (2 b d^2\right ) \operatorname{Subst}\left (\int \frac{d-\sqrt{c} x^2}{d^2+c x^4} \, dx,x,\sqrt{d x}\right )}{5 c}-\frac{\left (2 b d^2\right ) \operatorname{Subst}\left (\int \frac{d+\sqrt{c} x^2}{d^2+c x^4} \, dx,x,\sqrt{d x}\right )}{5 c}\\ &=\frac{8 b d \sqrt{d x}}{5 c}-\frac{2 b d^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{2 b d^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{\left (b d^{3/2}\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 )}{5 \sqrt{2} c^{5/4}}+\frac{\left (b d^{3/2}\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 )}{5 \sqrt{2} c^{5/4}}-\frac{\left (b d^2\right ) \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 )}{5 c^{3/2}}-\frac{\left (b d^2\right ) \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 )}{5 c^{3/2}}\\ &=\frac{8 b d \sqrt{d x}}{5 c}-\frac{2 b d^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{2 b d^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{b d^{3/2} \log \left (\sqrt{d}+\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}\right )}{5 \sqrt{2} c^{5/4}}-\frac{b d^{3/2} \log \left (\sqrt{d}+\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}\right )}{5 \sqrt{2} c^{5/4}}-\frac{\left (\sqrt{2} b d^{3/2}\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 )}{5 c^{5/4}}+\frac{\left (\sqrt{2} b d^{3/2}\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 )}{5 c^{5/4}}\\ &=\frac{8 b d \sqrt{d x}}{5 c}-\frac{2 b d^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{\sqrt{2} b d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}-\frac{\sqrt{2} b d^{3/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac{2 b d^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/4}}+\frac{b d^{3/2} \log \left (\sqrt{d}+\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}\right )}{5 \sqrt{2} c^{5/4}}-\frac{b d^{3/2} \log \left (\sqrt{d}+\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}\right )}{5 \sqrt{2} c^{5/4}}\\ \end{align*}

Mathematica [A]  time = 0.0933528, size = 240, normalized size = 0.76 \[ \frac{(d x)^{3/2} \left (4 a c^{5/4} x^{5/2}+4 b c^{5/4} x^{5/2} \tanh ^{-1}\left (c x^2\right )+16 b \sqrt [4]{c} \sqrt{x}+2 b \log \left (1-\sqrt [4]{c} \sqrt{x}\right )-2 b \log \left (\sqrt [4]{c} \sqrt{x}+1\right )+\sqrt{2} b \log \left (\sqrt{c} x-\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )-\sqrt{2} b \log \left (\sqrt{c} x+\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )+2 \sqrt{2} b \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{c} \sqrt{x}\right )-2 \sqrt{2} b \tan ^{-1}\left (\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )-4 b \tan ^{-1}\left (\sqrt [4]{c} \sqrt{x}\right )\right )}{10 c^{5/4} x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^(3/2)*(16*b*c^(1/4)*Sqrt[x] + 4*a*c^(5/4)*x^(5/2) + 2*Sqrt[2]*b*ArcTan[1 - Sqrt[2]*c^(1/4)*Sqrt[x]] - 2
*Sqrt[2]*b*ArcTan[1 + Sqrt[2]*c^(1/4)*Sqrt[x]] - 4*b*ArcTan[c^(1/4)*Sqrt[x]] + 4*b*c^(5/4)*x^(5/2)*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]))/(10*c^(5/4)*x^(3/2))

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Maple [A]  time = 0.016, size = 292, normalized size = 0.9 \begin{align*}{\frac{2\,a}{5\,d} \left ( dx \right ) ^{{\frac{5}{2}}}}+{\frac{2\,b{\it Artanh} \left ( c{x}^{2} \right ) }{5\,d} \left ( dx \right ) ^{{\frac{5}{2}}}}+{\frac{8\,bd}{5\,c}\sqrt{dx}}-{\frac{bd\sqrt{2}}{10\,c}\sqrt [4]{{\frac{{d}^{2}}{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{bd\sqrt{2}}{5\,c}\sqrt [4]{{\frac{{d}^{2}}{c}}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}+1 \right ) }-{\frac{bd\sqrt{2}}{5\,c}\sqrt [4]{{\frac{{d}^{2}}{c}}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}-1 \right ) }-{\frac{bd}{5\,c}\sqrt [4]{{\frac{{d}^{2}}{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{2\,bd}{5\,c}\sqrt [4]{{\frac{{d}^{2}}{c}}}\arctan \left ({\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/d*(d*x)^(5/2)*a+2/5/d*b*(d*x)^(5/2)*arctanh(c*x^2)+8/5*b*d*(d*x)^(1/2)/c-1/10*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))
)-1/5*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)-1/5*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)-1/5*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)))-2/5*d*b/c*(d^2/c)^(1/4)*arctan((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)^(3/2)*(a+b*arctanh(c*x^2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.27177, size = 109, normalized size = 0.34 \begin{align*} \frac{{\left (b c d x^{2} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a c d x^{2} + 8 \, b d\right )} \sqrt{d x}}{5 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b*c*d*x^2*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a*c*d*x^2 + 8*b*d)*sqrt(d*x)/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((d*x)**(3/2)*(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 \left (d x\right )^{\frac{3}{2}}{\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)^(3/2)*(a+b*arctanh(c*x^2)),x, algorithm="giac")

[Out]

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