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

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

[Out]

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

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Rubi [A]  time = 0.24, 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, 301, 211, 1165, 628, 1162, 617, 204, 212, 208, 205} \[ \frac {2 \sqrt {d x} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{d}-\frac {b \log \left (\sqrt {c} \sqrt {d} x-\sqrt {2} \sqrt [4]{c} \sqrt {d x}+\sqrt {d}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt {d}}+\frac {b \log \left (\sqrt {c} \sqrt {d} x+\sqrt {2} \sqrt [4]{c} \sqrt {d x}+\sqrt {d}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt {d}}-\frac {2 b \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{c} \sqrt {d}}-\frac {\sqrt {2} b \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{c} \sqrt {d}}+\frac {\sqrt {2} b \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}+1\right )}{\sqrt [4]{c} \sqrt {d}}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{c} \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 227, normalized size = 0.80 \[ \frac {\sqrt {x} \left (4 a \sqrt [4]{c} \sqrt {x}+4 b \sqrt [4]{c} \sqrt {x} \tanh ^{-1}\left (c x^2\right )+2 b \log \left (1-\sqrt [4]{c} \sqrt {x}\right )-2 b \log \left (\sqrt [4]{c} \sqrt {x}+1\right )-\sqrt {2} b \log \left (\sqrt {c} x-\sqrt {2} \sqrt [4]{c} \sqrt {x}+1\right )+\sqrt {2} b \log \left (\sqrt {c} x+\sqrt {2} \sqrt [4]{c} \sqrt {x}+1\right )-2 \sqrt {2} b \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{c} \sqrt {x}\right )+2 \sqrt {2} b \tan ^{-1}\left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+1\right )-4 b \tan ^{-1}\left (\sqrt [4]{c} \sqrt {x}\right )\right )}{2 \sqrt [4]{c} \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(4*a*c^(1/4)*Sqrt[x] - 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^(1/4)*Sqrt[x]*ArcTanh[c*x^2] + 2*b*Log[1 - c^(1/4)*S
qrt[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*Lo
g[1 + Sqrt[2]*c^(1/4)*Sqrt[x] + Sqrt[c]*x]))/(2*c^(1/4)*Sqrt[d*x])

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fricas [A]  time = 1.96, size = 34, normalized size = 0.12 \[ \frac {\sqrt {d x} {\left (b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.16, size = 493, normalized size = 1.73 \[ \frac {{\left (c d^{2} {\left (\frac {2 \, \sqrt {2} \left (c^{3} d^{2}\right )^{\frac {1}{4}} \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 {1}{4}} \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 {1}{4}} \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 {1}{4}} \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 {1}{4}} \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 {1}{4}} \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 {1}{4}} \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 {1}{4}} \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}}\right )} + 2 \, \sqrt {d x} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )\right )} b + 4 \, \sqrt {d x} a}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 273, normalized size = 0.96 \[ \frac {2 a \sqrt {d x}}{d}+\frac {2 b \sqrt {d x}\, \arctanh \left (c \,x^{2}\right )}{d}-\frac {b \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{d}-\frac {2 b \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{d}+\frac {b \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d x +\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {d^{2}}{c}}}{d x -\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {d^{2}}{c}}}\right )}{2 d}+\frac {b \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}+1\right )}{d}+\frac {b \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}-1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a*(d*x)^(1/2)+2/d*b*(d*x)^(1/2)*arctanh(c*x^2)-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)))-2/d*b*(d^2/c)^(1/4)*arctan((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 = 0.42, size = 296, normalized size = 1.04 \[ \frac {{\left (4 \, \sqrt {d x} \operatorname {artanh}\left (c x^{2}\right ) + \frac {c {\left (\frac {\frac {2 \, \sqrt {2} d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} \sqrt {d} + 2 \, \sqrt {d x} \sqrt {c}\right )}}{2 \, \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d}} + \frac {2 \, \sqrt {2} d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} \sqrt {d} - 2 \, \sqrt {d x} \sqrt {c}\right )}}{2 \, \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d}} + \frac {\sqrt {2} d^{\frac {5}{2}} \log \left (\sqrt {c} d x + \sqrt {2} \sqrt {d x} c^{\frac {1}{4}} \sqrt {d} + d\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} d^{\frac {5}{2}} \log \left (\sqrt {c} d x - \sqrt {2} \sqrt {d x} c^{\frac {1}{4}} \sqrt {d} + d\right )}{c^{\frac {1}{4}}}}{c} - \frac {2 \, {\left (\frac {2 \, d^{3} \arctan \left (\frac {\sqrt {d x} \sqrt {c}}{\sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d}} - \frac {d^{3} \log \left (\frac {\sqrt {d x} \sqrt {c} - \sqrt {\sqrt {c} d}}{\sqrt {d x} \sqrt {c} + \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d}}\right )}}{c}\right )}}{d^{2}}\right )} b + 4 \, \sqrt {d x} a}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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