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

Optimal. Leaf size=285 \[ -\frac {2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{d \sqrt {d x}}+\frac {b \sqrt [4]{c} \log \left (\sqrt {c} \sqrt {d} x-\sqrt {2} \sqrt [4]{c} \sqrt {d x}+\sqrt {d}\right )}{\sqrt {2} d^{3/2}}-\frac {b \sqrt [4]{c} \log \left (\sqrt {c} \sqrt {d} x+\sqrt {2} \sqrt [4]{c} \sqrt {d x}+\sqrt {d}\right )}{\sqrt {2} d^{3/2}}-\frac {2 b \sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {\sqrt {2} b \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {\sqrt {2} b \sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}+1\right )}{d^{3/2}}+\frac {2 b \sqrt [4]{c} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 285, normalized size of antiderivative = 1.00, 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, 300, 297, 1162, 617, 204, 1165, 628, 298, 205, 208} \[ -\frac {2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{d \sqrt {d x}}+\frac {b \sqrt [4]{c} \log \left (\sqrt {c} \sqrt {d} x-\sqrt {2} \sqrt [4]{c} \sqrt {d x}+\sqrt {d}\right )}{\sqrt {2} d^{3/2}}-\frac {b \sqrt [4]{c} \log \left (\sqrt {c} \sqrt {d} x+\sqrt {2} \sqrt [4]{c} \sqrt {d x}+\sqrt {d}\right )}{\sqrt {2} d^{3/2}}-\frac {2 b \sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {\sqrt {2} b \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {\sqrt {2} b \sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}+1\right )}{d^{3/2}}+\frac {2 b \sqrt [4]{c} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 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 300

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

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

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Mathematica [A]  time = 0.11, size = 268, normalized size = 0.94 \[ -\frac {x \left (4 a+4 b \tanh ^{-1}\left (c x^2\right )+2 b \sqrt [4]{c} \sqrt {x} \log \left (1-\sqrt [4]{c} \sqrt {x}\right )-2 b \sqrt [4]{c} \sqrt {x} \log \left (\sqrt [4]{c} \sqrt {x}+1\right )-\sqrt {2} b \sqrt [4]{c} \sqrt {x} \log \left (\sqrt {c} x-\sqrt {2} \sqrt [4]{c} \sqrt {x}+1\right )+\sqrt {2} b \sqrt [4]{c} \sqrt {x} \log \left (\sqrt {c} x+\sqrt {2} \sqrt [4]{c} \sqrt {x}+1\right )+2 \sqrt {2} b \sqrt [4]{c} \sqrt {x} \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{c} \sqrt {x}\right )-2 \sqrt {2} b \sqrt [4]{c} \sqrt {x} \tan ^{-1}\left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+1\right )+4 b \sqrt [4]{c} \sqrt {x} \tan ^{-1}\left (\sqrt [4]{c} \sqrt {x}\right )\right )}{2 (d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.83, size = 38, normalized size = 0.13 \[ -\frac {\sqrt {d x} {\left (b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a\right )}}{d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.85, size = 505, normalized size = 1.77 \[ -\frac {\frac {2 \, b \log \left (-\frac {c d^{2} x^{2} + d^{2}}{c d^{2} x^{2} - d^{2}}\right )}{\sqrt {d x}} + \frac {4 \, a}{\sqrt {d x}} - \frac {2 \, \sqrt {2} \left (c^{3} d^{2}\right )^{\frac {3}{4}} b \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{c^{2} d^{2}} - \frac {2 \, \sqrt {2} \left (c^{3} d^{2}\right )^{\frac {3}{4}} b \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{c^{2} d^{2}} - \frac {2 \, \sqrt {2} \left (-c^{3} d^{2}\right )^{\frac {3}{4}} b \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {d^{2}}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (-\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{c^{2} d^{2}} - \frac {2 \, \sqrt {2} \left (-c^{3} d^{2}\right )^{\frac {3}{4}} b \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {d^{2}}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (-\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{c^{2} d^{2}} + \frac {\sqrt {2} \left (c^{3} d^{2}\right )^{\frac {3}{4}} b \log \left (d x + \sqrt {2} \sqrt {d x} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {d^{2}}{c}}\right )}{c^{2} d^{2}} - \frac {\sqrt {2} \left (c^{3} d^{2}\right )^{\frac {3}{4}} b \log \left (d x - \sqrt {2} \sqrt {d x} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {d^{2}}{c}}\right )}{c^{2} d^{2}} + \frac {\sqrt {2} \left (-c^{3} d^{2}\right )^{\frac {3}{4}} b \log \left (d x + \sqrt {2} \sqrt {d x} \left (-\frac {d^{2}}{c}\right )^{\frac {1}{4}} + \sqrt {-\frac {d^{2}}{c}}\right )}{c^{2} d^{2}} - \frac {\sqrt {2} \left (-c^{3} d^{2}\right )^{\frac {3}{4}} b \log \left (d x - \sqrt {2} \sqrt {d x} \left (-\frac {d^{2}}{c}\right )^{\frac {1}{4}} + \sqrt {-\frac {d^{2}}{c}}\right )}{c^{2} d^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*log(-(c*d^2*x^2 + d^2)/(c*d^2*x^2 - d^2))/sqrt(d*x) + 4*a/sqrt(d*x) - 2*sqrt(2)*(c^3*d^2)^(3/4)*b*ar
ctan(1/2*sqrt(2)*(sqrt(2)*(d^2/c)^(1/4) + 2*sqrt(d*x))/(d^2/c)^(1/4))/(c^2*d^2) - 2*sqrt(2)*(c^3*d^2)^(3/4)*b*
arctan(-1/2*sqrt(2)*(sqrt(2)*(d^2/c)^(1/4) - 2*sqrt(d*x))/(d^2/c)^(1/4))/(c^2*d^2) - 2*sqrt(2)*(-c^3*d^2)^(3/4
)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(-d^2/c)^(1/4) + 2*sqrt(d*x))/(-d^2/c)^(1/4))/(c^2*d^2) - 2*sqrt(2)*(-c^3*d^2)
^(3/4)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*(-d^2/c)^(1/4) - 2*sqrt(d*x))/(-d^2/c)^(1/4))/(c^2*d^2) + sqrt(2)*(c^3*d
^2)^(3/4)*b*log(d*x + sqrt(2)*sqrt(d*x)*(d^2/c)^(1/4) + sqrt(d^2/c))/(c^2*d^2) - sqrt(2)*(c^3*d^2)^(3/4)*b*log
(d*x - sqrt(2)*sqrt(d*x)*(d^2/c)^(1/4) + sqrt(d^2/c))/(c^2*d^2) + sqrt(2)*(-c^3*d^2)^(3/4)*b*log(d*x + sqrt(2)
*sqrt(d*x)*(-d^2/c)^(1/4) + sqrt(-d^2/c))/(c^2*d^2) - sqrt(2)*(-c^3*d^2)^(3/4)*b*log(d*x - sqrt(2)*sqrt(d*x)*(
-d^2/c)^(1/4) + sqrt(-d^2/c))/(c^2*d^2))/d

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maple [A]  time = 0.03, size = 272, normalized size = 0.95 \[ -\frac {2 a}{d \sqrt {d x}}-\frac {2 b \arctanh \left (c \,x^{2}\right )}{d \sqrt {d x}}-\frac {2 b \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{d \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}+\frac {b \ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{d \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}+\frac {b \sqrt {2}\, \ln \left (\frac {d x -\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {d^{2}}{c}}}{d x +\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {d^{2}}{c}}}\right )}{2 d \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}+\frac {b \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}+1\right )}{d \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}+\frac {b \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}-1\right )}{d \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 296, normalized size = 1.04 \[ -\frac {b {\left (\frac {4 \, \operatorname {artanh}\left (c x^{2}\right )}{\sqrt {d x}} - \frac {{\left (d^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} \sqrt {d} + 2 \, \sqrt {d x} \sqrt {c}\right )}}{2 \, \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} \sqrt {d} - 2 \, \sqrt {d x} \sqrt {c}\right )}}{2 \, \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {c} d x + \sqrt {2} \sqrt {d x} c^{\frac {1}{4}} \sqrt {d} + d\right )}{c^{\frac {3}{4}} \sqrt {d}} + \frac {\sqrt {2} \log \left (\sqrt {c} d x - \sqrt {2} \sqrt {d x} c^{\frac {1}{4}} \sqrt {d} + d\right )}{c^{\frac {3}{4}} \sqrt {d}}\right )} - 2 \, d^{2} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {d x} \sqrt {c}}{\sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d} \sqrt {c}} + \frac {\log \left (\frac {\sqrt {d x} \sqrt {c} - \sqrt {\sqrt {c} d}}{\sqrt {d x} \sqrt {c} + \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d} \sqrt {c}}\right )}\right )} c}{d^{2}}\right )} + \frac {4 \, a}{\sqrt {d x}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {atanh}\left (c\,x^2\right )}{{\left (d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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