3.752 \(\int \frac{\sqrt{d x}}{\sqrt{a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=368 \[ \frac{\sqrt{d} \left (a+b x^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{2 \sqrt{2} \sqrt [4]{a} b^{3/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{d} \left (a+b x^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{2 \sqrt{2} \sqrt [4]{a} b^{3/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{d} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{\sqrt{2} \sqrt [4]{a} b^{3/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\sqrt{d} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{\sqrt{2} \sqrt [4]{a} b^{3/4} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.246074, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {1112, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{\sqrt{d} \left (a+b x^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{2 \sqrt{2} \sqrt [4]{a} b^{3/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{d} \left (a+b x^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{2 \sqrt{2} \sqrt [4]{a} b^{3/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{d} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{\sqrt{2} \sqrt [4]{a} b^{3/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\sqrt{d} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{\sqrt{2} \sqrt [4]{a} b^{3/4} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*x]/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0452284, size = 85, normalized size = 0.23 \[ \frac{\sqrt{d x} \left (a+b x^2\right ) \left (\tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{-a}}\right )+\tanh ^{-1}\left (\frac{a \sqrt [4]{b} \sqrt{x}}{(-a)^{5/4}}\right )\right )}{\sqrt [4]{-a} b^{3/4} \sqrt{x} \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*x]/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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Maple [A]  time = 0.225, size = 183, normalized size = 0.5 \begin{align*}{\frac{d \left ( b{x}^{2}+a \right ) \sqrt{2}}{4\,b} \left ( \ln \left ( -{ \left ( \sqrt [4]{{\frac{a{d}^{2}}{b}}}\sqrt{dx}\sqrt{2}-dx-\sqrt{{\frac{a{d}^{2}}{b}}} \right ) \left ( dx+\sqrt [4]{{\frac{a{d}^{2}}{b}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{a{d}^{2}}{b}}} \right ) ^{-1}} \right ) +2\,\arctan \left ({ \left ( \sqrt{2}\sqrt{dx}+\sqrt [4]{{\frac{a{d}^{2}}{b}}} \right ){\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}} \right ) +2\,\arctan \left ({ \left ( \sqrt{2}\sqrt{dx}-\sqrt [4]{{\frac{a{d}^{2}}{b}}} \right ){\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}} \right ) \right ){\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}{\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.62866, size = 390, normalized size = 1.06 \begin{align*} -2 \, \left (-\frac{d^{2}}{a b^{3}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{d x} b d \left (-\frac{d^{2}}{a b^{3}}\right )^{\frac{1}{4}} - \sqrt{-a b d^{2} \sqrt{-\frac{d^{2}}{a b^{3}}} + d^{3} x} b \left (-\frac{d^{2}}{a b^{3}}\right )^{\frac{1}{4}}}{d^{2}}\right ) + \frac{1}{2} \, \left (-\frac{d^{2}}{a b^{3}}\right )^{\frac{1}{4}} \log \left (a b^{2} \left (-\frac{d^{2}}{a b^{3}}\right )^{\frac{3}{4}} + \sqrt{d x} d\right ) - \frac{1}{2} \, \left (-\frac{d^{2}}{a b^{3}}\right )^{\frac{1}{4}} \log \left (-a b^{2} \left (-\frac{d^{2}}{a b^{3}}\right )^{\frac{3}{4}} + \sqrt{d x} d\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 152.039, size = 41, normalized size = 0.11 \begin{align*} 2 d \operatorname{RootSum}{\left (256 t^{4} a b^{3} d^{2} + 1, \left ( t \mapsto t \log{\left (64 t^{3} a b^{2} d^{2} + \sqrt{d x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*RootSum(256*_t**4*a*b**3*d**2 + 1, Lambda(_t, _t*log(64*_t**3*a*b**2*d**2 + sqrt(d*x))))

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Giac [A]  time = 1.28289, size = 339, normalized size = 0.92 \begin{align*} \frac{1}{4} \,{\left (\frac{2 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}} + 2 \, \sqrt{d x}\right )}}{2 \, \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}}}\right )}{a b^{3} d} + \frac{2 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}} - 2 \, \sqrt{d x}\right )}}{2 \, \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}}}\right )}{a b^{3} d} - \frac{\sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{3}{4}} \log \left (d x + \sqrt{2} \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}} \sqrt{d x} + \sqrt{\frac{a d^{2}}{b}}\right )}{a b^{3} d} + \frac{\sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{3}{4}} \log \left (d x - \sqrt{2} \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}} \sqrt{d x} + \sqrt{\frac{a d^{2}}{b}}\right )}{a b^{3} d}\right )} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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