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

Optimal. Leaf size=283 \[ \frac {\sqrt {d} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}-\frac {\sqrt {d} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}-\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )} \]

[Out]

1/2*(d*x)^(3/2)/a/d/(b*x^2+a)-1/8*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))*d^(1/2)/a^(5/4)/b^(3/4
)*2^(1/2)+1/8*arctan(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))*d^(1/2)/a^(5/4)/b^(3/4)*2^(1/2)+1/16*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^(5/4)/b^(3/4)*2^(1/2)-1/16*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^(5/4)/b^(3/4)*2^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {28, 290, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {\sqrt {d} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}-\frac {\sqrt {d} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{5/4} b^{3/4}}-\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} a^{5/4} b^{3/4}}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x)^(3/2)/(2*a*d*(a + b*x^2)) - (Sqrt[d]*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[
2]*a^(5/4)*b^(3/4)) + (Sqrt[d]*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*a^(5/4)*b
^(3/4)) + (Sqrt[d]*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*a^
(5/4)*b^(3/4)) - (Sqrt[d]*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqr
t[2]*a^(5/4)*b^(3/4))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.11 \[ \frac {2 x \sqrt {d x} \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {b x^2}{a}\right )}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[d*x]*Hypergeometric2F1[3/4, 2, 7/4, -((b*x^2)/a)])/(3*a^2)

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fricas [A]  time = 1.13, size = 232, normalized size = 0.82 \[ -\frac {4 \, {\left (a b x^{2} + a^{2}\right )} \left (-\frac {d^{2}}{a^{5} b^{3}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a b d \left (-\frac {d^{2}}{a^{5} b^{3}}\right )^{\frac {1}{4}} - \sqrt {-a^{3} b d^{2} \sqrt {-\frac {d^{2}}{a^{5} b^{3}}} + d^{3} x} a b \left (-\frac {d^{2}}{a^{5} b^{3}}\right )^{\frac {1}{4}}}{d^{2}}\right ) - {\left (a b x^{2} + a^{2}\right )} \left (-\frac {d^{2}}{a^{5} b^{3}}\right )^{\frac {1}{4}} \log \left (a^{4} b^{2} \left (-\frac {d^{2}}{a^{5} b^{3}}\right )^{\frac {3}{4}} + \sqrt {d x} d\right ) + {\left (a b x^{2} + a^{2}\right )} \left (-\frac {d^{2}}{a^{5} b^{3}}\right )^{\frac {1}{4}} \log \left (-a^{4} b^{2} \left (-\frac {d^{2}}{a^{5} b^{3}}\right )^{\frac {3}{4}} + \sqrt {d x} d\right ) - 4 \, \sqrt {d x} x}{8 \, {\left (a b x^{2} + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(1/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

-1/8*(4*(a*b*x^2 + a^2)*(-d^2/(a^5*b^3))^(1/4)*arctan(-(sqrt(d*x)*a*b*d*(-d^2/(a^5*b^3))^(1/4) - sqrt(-a^3*b*d
^2*sqrt(-d^2/(a^5*b^3)) + d^3*x)*a*b*(-d^2/(a^5*b^3))^(1/4))/d^2) - (a*b*x^2 + a^2)*(-d^2/(a^5*b^3))^(1/4)*log
(a^4*b^2*(-d^2/(a^5*b^3))^(3/4) + sqrt(d*x)*d) + (a*b*x^2 + a^2)*(-d^2/(a^5*b^3))^(1/4)*log(-a^4*b^2*(-d^2/(a^
5*b^3))^(3/4) + sqrt(d*x)*d) - 4*sqrt(d*x)*x)/(a*b*x^2 + a^2)

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giac [A]  time = 0.19, size = 264, normalized size = 0.93 \[ \frac {\frac {8 \, \sqrt {d x} d^{3} x}{{\left (b d^{2} x^{2} + a d^{2}\right )} a} + \frac {2 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{2} b^{3}} + \frac {2 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{2} b^{3}} - \frac {\sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{2} b^{3}} + \frac {\sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{2} b^{3}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(1/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

1/16*(8*sqrt(d*x)*d^3*x/((b*d^2*x^2 + a*d^2)*a) + 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^2*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^2*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^2*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^2*b^3))/d

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maple [A]  time = 0.01, size = 210, normalized size = 0.74 \[ \frac {\sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a b}+\frac {\sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a b}+\frac {\sqrt {2}\, d \ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{16 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a b}+\frac {\left (d x \right )^{\frac {3}{2}} d}{2 \left (b \,d^{2} x^{2}+d^{2} a \right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(1/2)/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

1/2*d*(d*x)^(3/2)/a/(b*d^2*x^2+a*d^2)+1/16*d/a/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)))+1/8*d/a/b/(a/b*d^2)^(1/4)*2
^(1/2)*arctan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)+1)+1/8*d/a/b/(a/b*d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b*d^2
)^(1/4)*(d*x)^(1/2)-1)

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maxima [A]  time = 3.06, size = 255, normalized size = 0.90 \[ \frac {\frac {8 \, \left (d x\right )^{\frac {3}{2}} d^{2}}{a b d^{2} x^{2} + a^{2} d^{2}} + \frac {d^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{a}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(1/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

1/16*(8*(d*x)^(3/2)*d^2/(a*b*d^2*x^2 + a^2*d^2) + d^2*(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)
))/a)/d

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mupad [B]  time = 0.11, size = 90, normalized size = 0.32 \[ \frac {\sqrt {d}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{4\,{\left (-a\right )}^{5/4}\,b^{3/4}}-\frac {\sqrt {d}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{4\,{\left (-a\right )}^{5/4}\,b^{3/4}}+\frac {d\,{\left (d\,x\right )}^{3/2}}{2\,a\,\left (b\,d^2\,x^2+a\,d^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(1/2)/(a^2 + b^2*x^4 + 2*a*b*x^2),x)

[Out]

(d^(1/2)*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(4*(-a)^(5/4)*b^(3/4)) - (d^(1/2)*atan((b^(1/4)*(d
*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(4*(-a)^(5/4)*b^(3/4)) + (d*(d*x)^(3/2))/(2*a*(a*d^2 + b*d^2*x^2))

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sympy [A]  time = 6.67, size = 78, normalized size = 0.28 \[ \frac {2 d^{3} \left (d x\right )^{\frac {3}{2}}}{4 a^{2} d^{4} + 4 a b d^{4} x^{2}} + 2 d^{3} \operatorname {RootSum} {\left (65536 t^{4} a^{5} b^{3} d^{10} + 1, \left (t \mapsto t \log {\left (4096 t^{3} a^{4} b^{2} d^{8} + \sqrt {d x} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(1/2)/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

2*d**3*(d*x)**(3/2)/(4*a**2*d**4 + 4*a*b*d**4*x**2) + 2*d**3*RootSum(65536*_t**4*a**5*b**3*d**10 + 1, Lambda(_
t, _t*log(4096*_t**3*a**4*b**2*d**8 + sqrt(d*x))))

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