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]

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

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Rubi [A]  time = 0.25, antiderivative size = 368, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.83, size = 173, normalized size = 0.47 \[ -2 \, \left (-\frac {d^{2}}{a b^{3}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} b d \left (-\frac {d^{2}}{a b^{3}}\right )^{\frac {1}{4}} - \sqrt {-a b d^{2} \sqrt {-\frac {d^{2}}{a b^{3}}} + d^{3} x} b \left (-\frac {d^{2}}{a b^{3}}\right )^{\frac {1}{4}}}{d^{2}}\right ) + \frac {1}{2} \, \left (-\frac {d^{2}}{a b^{3}}\right )^{\frac {1}{4}} \log \left (a b^{2} \left (-\frac {d^{2}}{a b^{3}}\right )^{\frac {3}{4}} + \sqrt {d x} d\right ) - \frac {1}{2} \, \left (-\frac {d^{2}}{a b^{3}}\right )^{\frac {1}{4}} \log \left (-a b^{2} \left (-\frac {d^{2}}{a b^{3}}\right )^{\frac {3}{4}} + \sqrt {d x} d\right ) \]

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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giac [A]  time = 0.19, size = 242, normalized size = 0.66 \[ \frac {{\left (\frac {2 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{3}} + \frac {2 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{3}} - \frac {\sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{3}} + \frac {\sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{3}}\right )} \mathrm {sgn}\left (b x^{2} + a\right )}{4 \, d} \]

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) + 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) - 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) +
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))*sgn(b*x^2 + a
)/d

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maple [A]  time = 0.01, size = 183, normalized size = 0.50 \[ \frac {\left (b \,x^{2}+a \right ) \sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+\ln \left (-\frac {-d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}-\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )\right ) d}{4 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b} \]

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

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maxima [A]  time = 3.00, size = 216, normalized size = 0.59 \[ \frac {1}{4} \, d {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )} \]

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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 57.26, size = 41, normalized size = 0.11 \[ 2 d \operatorname {RootSum} {\left (256 t^{4} a b^{3} d^{2} + 1, \left (t \mapsto t \log {\left (64 t^{3} a b^{2} d^{2} + \sqrt {d x} \right )} \right )\right )} \]

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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