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

Optimal. Leaf size=335 \[ \frac {15 \sqrt {d} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{13/4} b^{3/4}}-\frac {15 \sqrt {d} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{13/4} b^{3/4}}-\frac {15 \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{13/4} b^{3/4}}+\frac {15 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{13/4} b^{3/4}}+\frac {15 (d x)^{3/2}}{64 a^3 d \left (a+b x^2\right )}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3} \]

[Out]

1/6*(d*x)^(3/2)/a/d/(b*x^2+a)^3+3/16*(d*x)^(3/2)/a^2/d/(b*x^2+a)^2+15/64*(d*x)^(3/2)/a^3/d/(b*x^2+a)-15/256*ar
ctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))*d^(1/2)/a^(13/4)/b^(3/4)*2^(1/2)+15/256*arctan(1+b^(1/4)*2
^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))*d^(1/2)/a^(13/4)/b^(3/4)*2^(1/2)+15/512*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^(13/4)/b^(3/4)*2^(1/2)-15/512*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^(13/4)/b^(3/4)*2^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ \frac {15 \sqrt {d} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{13/4} b^{3/4}}-\frac {15 \sqrt {d} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{13/4} b^{3/4}}-\frac {15 \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{13/4} b^{3/4}}+\frac {15 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{13/4} b^{3/4}}+\frac {15 (d x)^{3/2}}{64 a^3 d \left (a+b x^2\right )}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x)^(3/2)/(6*a*d*(a + b*x^2)^3) + (3*(d*x)^(3/2))/(16*a^2*d*(a + b*x^2)^2) + (15*(d*x)^(3/2))/(64*a^3*d*(a +
b*x^2)) - (15*Sqrt[d]*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(13/4)*b^(3/4
)) + (15*Sqrt[d]*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(13/4)*b^(3/4)) + (
15*Sqrt[d]*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*a^(13/4)
*b^(3/4)) - (15*Sqrt[d]*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqr
t[2]*a^(13/4)*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}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {\sqrt {d x}}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3}+\frac {\left (3 b^3\right ) \int \frac {\sqrt {d x}}{\left (a b+b^2 x^2\right )^3} \, dx}{4 a}\\ &=\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {\left (15 b^2\right ) \int \frac {\sqrt {d x}}{\left (a b+b^2 x^2\right )^2} \, dx}{32 a^2}\\ &=\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {15 (d x)^{3/2}}{64 a^3 d \left (a+b x^2\right )}+\frac {(15 b) \int \frac {\sqrt {d x}}{a b+b^2 x^2} \, dx}{128 a^3}\\ &=\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {15 (d x)^{3/2}}{64 a^3 d \left (a+b x^2\right )}+\frac {(15 b) \operatorname {Subst}\left (\int \frac {x^2}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{64 a^3 d}\\ &=\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {15 (d x)^{3/2}}{64 a^3 d \left (a+b x^2\right )}-\frac {\left (15 \sqrt {b}\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 )}{128 a^3 d}+\frac {\left (15 \sqrt {b}\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 )}{128 a^3 d}\\ &=\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {15 (d x)^{3/2}}{64 a^3 d \left (a+b x^2\right )}+\frac {\left (15 \sqrt {d}\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 )}{256 \sqrt {2} a^{13/4} b^{3/4}}+\frac {\left (15 \sqrt {d}\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 )}{256 \sqrt {2} a^{13/4} b^{3/4}}+\frac {(15 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 )}{256 a^3 b}+\frac {(15 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 )}{256 a^3 b}\\ &=\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {15 (d x)^{3/2}}{64 a^3 d \left (a+b x^2\right )}+\frac {15 \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 )}{256 \sqrt {2} a^{13/4} b^{3/4}}-\frac {15 \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 )}{256 \sqrt {2} a^{13/4} b^{3/4}}+\frac {\left (15 \sqrt {d}\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 )}{128 \sqrt {2} a^{13/4} b^{3/4}}-\frac {\left (15 \sqrt {d}\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 )}{128 \sqrt {2} a^{13/4} b^{3/4}}\\ &=\frac {(d x)^{3/2}}{6 a d \left (a+b x^2\right )^3}+\frac {3 (d x)^{3/2}}{16 a^2 d \left (a+b x^2\right )^2}+\frac {15 (d x)^{3/2}}{64 a^3 d \left (a+b x^2\right )}-\frac {15 \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{13/4} b^{3/4}}+\frac {15 \sqrt {d} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{13/4} b^{3/4}}+\frac {15 \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 )}{256 \sqrt {2} a^{13/4} b^{3/4}}-\frac {15 \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 )}{256 \sqrt {2} a^{13/4} b^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.10 \[ \frac {2 x \sqrt {d x} \, _2F_1\left (\frac {3}{4},4;\frac {7}{4};-\frac {b x^2}{a}\right )}{3 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.01, size = 359, normalized size = 1.07 \[ -\frac {180 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )} \left (-\frac {d^{2}}{a^{13} b^{3}}\right )^{\frac {1}{4}} \arctan \left (-\frac {3375 \, \sqrt {d x} a^{3} b d \left (-\frac {d^{2}}{a^{13} b^{3}}\right )^{\frac {1}{4}} - \sqrt {-11390625 \, a^{7} b d^{2} \sqrt {-\frac {d^{2}}{a^{13} b^{3}}} + 11390625 \, d^{3} x} a^{3} b \left (-\frac {d^{2}}{a^{13} b^{3}}\right )^{\frac {1}{4}}}{3375 \, d^{2}}\right ) - 45 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )} \left (-\frac {d^{2}}{a^{13} b^{3}}\right )^{\frac {1}{4}} \log \left (3375 \, a^{10} b^{2} \left (-\frac {d^{2}}{a^{13} b^{3}}\right )^{\frac {3}{4}} + 3375 \, \sqrt {d x} d\right ) + 45 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )} \left (-\frac {d^{2}}{a^{13} b^{3}}\right )^{\frac {1}{4}} \log \left (-3375 \, a^{10} b^{2} \left (-\frac {d^{2}}{a^{13} b^{3}}\right )^{\frac {3}{4}} + 3375 \, \sqrt {d x} d\right ) - 4 \, {\left (45 \, b^{2} x^{5} + 126 \, a b x^{3} + 113 \, a^{2} x\right )} \sqrt {d x}}{768 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} \]

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)^2,x, algorithm="fricas")

[Out]

-1/768*(180*(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6)*(-d^2/(a^13*b^3))^(1/4)*arctan(-1/3375*(3375*sqr
t(d*x)*a^3*b*d*(-d^2/(a^13*b^3))^(1/4) - sqrt(-11390625*a^7*b*d^2*sqrt(-d^2/(a^13*b^3)) + 11390625*d^3*x)*a^3*
b*(-d^2/(a^13*b^3))^(1/4))/d^2) - 45*(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6)*(-d^2/(a^13*b^3))^(1/4)
*log(3375*a^10*b^2*(-d^2/(a^13*b^3))^(3/4) + 3375*sqrt(d*x)*d) + 45*(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2
+ a^6)*(-d^2/(a^13*b^3))^(1/4)*log(-3375*a^10*b^2*(-d^2/(a^13*b^3))^(3/4) + 3375*sqrt(d*x)*d) - 4*(45*b^2*x^5
+ 126*a*b*x^3 + 113*a^2*x)*sqrt(d*x))/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6)

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giac [A]  time = 0.22, size = 302, normalized size = 0.90 \[ \frac {\frac {90 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{4} b^{3}} + \frac {90 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{4} b^{3}} - \frac {45 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{4} b^{3}} + \frac {45 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{4} b^{3}} + \frac {8 \, {\left (45 \, \sqrt {d x} b^{2} d^{7} x^{5} + 126 \, \sqrt {d x} a b d^{7} x^{3} + 113 \, \sqrt {d x} a^{2} d^{7} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} a^{3}}}{1536 \, d} \]

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)^2,x, algorithm="giac")

[Out]

1/1536*(90*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^4*b^3) + 90*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^4*b^3) - 45*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^4*b^3) + 45*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^4*
b^3) + 8*(45*sqrt(d*x)*b^2*d^7*x^5 + 126*sqrt(d*x)*a*b*d^7*x^3 + 113*sqrt(d*x)*a^2*d^7*x)/((b*d^2*x^2 + a*d^2)
^3*a^3))/d

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maple [A]  time = 0.02, size = 272, normalized size = 0.81 \[ \frac {113 \left (d x \right )^{\frac {3}{2}} d^{5}}{192 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a}+\frac {21 \left (d x \right )^{\frac {7}{2}} b \,d^{3}}{32 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a^{2}}+\frac {15 \left (d x \right )^{\frac {11}{2}} b^{2} d}{64 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a^{3}}+\frac {15 \sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{3} b}+\frac {15 \sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{3} b}+\frac {15 \sqrt {2}\, d \ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{512 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{3} b} \]

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)^2,x)

[Out]

15/64*d/(b*d^2*x^2+a*d^2)^3/a^3*b^2*(d*x)^(11/2)+21/32*d^3/(b*d^2*x^2+a*d^2)^3/a^2*b*(d*x)^(7/2)+113/192*d^5/(
b*d^2*x^2+a*d^2)^3/a*(d*x)^(3/2)+15/512*d/a^3/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)))+15/256*d/a^3/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)+15/256*d/a^3/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 = 2.98, size = 317, normalized size = 0.95 \[ \frac {\frac {8 \, {\left (45 \, \left (d x\right )^{\frac {11}{2}} b^{2} d^{2} + 126 \, \left (d x\right )^{\frac {7}{2}} a b d^{4} + 113 \, \left (d x\right )^{\frac {3}{2}} a^{2} d^{6}\right )}}{a^{3} b^{3} d^{6} x^{6} + 3 \, a^{4} b^{2} d^{6} x^{4} + 3 \, a^{5} b d^{6} x^{2} + a^{6} d^{6}} + \frac {45 \, d^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{a^{3}}}{1536 \, d} \]

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)^2,x, algorithm="maxima")

[Out]

1/1536*(8*(45*(d*x)^(11/2)*b^2*d^2 + 126*(d*x)^(7/2)*a*b*d^4 + 113*(d*x)^(3/2)*a^2*d^6)/(a^3*b^3*d^6*x^6 + 3*a
^4*b^2*d^6*x^4 + 3*a^5*b*d^6*x^2 + a^6*d^6) + 45*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^3)/d

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mupad [B]  time = 0.10, size = 150, normalized size = 0.45 \[ \frac {\frac {113\,d^5\,{\left (d\,x\right )}^{3/2}}{192\,a}+\frac {21\,b\,d^3\,{\left (d\,x\right )}^{7/2}}{32\,a^2}+\frac {15\,b^2\,d\,{\left (d\,x\right )}^{11/2}}{64\,a^3}}{a^3\,d^6+3\,a^2\,b\,d^6\,x^2+3\,a\,b^2\,d^6\,x^4+b^3\,d^6\,x^6}-\frac {15\,\sqrt {d}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{13/4}\,b^{3/4}}+\frac {15\,\sqrt {d}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{13/4}\,b^{3/4}} \]

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)^2,x)

[Out]

((113*d^5*(d*x)^(3/2))/(192*a) + (21*b*d^3*(d*x)^(7/2))/(32*a^2) + (15*b^2*d*(d*x)^(11/2))/(64*a^3))/(a^3*d^6
+ b^3*d^6*x^6 + 3*a^2*b*d^6*x^2 + 3*a*b^2*d^6*x^4) - (15*d^(1/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2
))))/(128*(-a)^(13/4)*b^(3/4)) + (15*d^(1/2)*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*(-a)^(13/
4)*b^(3/4))

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sympy [A]  time = 28.99, size = 252, normalized size = 0.75 \[ \frac {226 a^{2} d^{11} \left (d x\right )^{\frac {3}{2}}}{384 a^{6} d^{12} + 1152 a^{5} b d^{12} x^{2} + 1152 a^{4} b^{2} d^{12} x^{4} + 384 a^{3} b^{3} d^{12} x^{6}} + \frac {252 a b d^{9} \left (d x\right )^{\frac {7}{2}}}{384 a^{6} d^{12} + 1152 a^{5} b d^{12} x^{2} + 1152 a^{4} b^{2} d^{12} x^{4} + 384 a^{3} b^{3} d^{12} x^{6}} + \frac {90 b^{2} d^{7} \left (d x\right )^{\frac {11}{2}}}{384 a^{6} d^{12} + 1152 a^{5} b d^{12} x^{2} + 1152 a^{4} b^{2} d^{12} x^{4} + 384 a^{3} b^{3} d^{12} x^{6}} + 2 d^{7} \operatorname {RootSum} {\left (68719476736 t^{4} a^{13} b^{3} d^{26} + 50625, \left (t \mapsto t \log {\left (\frac {134217728 t^{3} a^{10} b^{2} d^{20}}{3375} + \sqrt {d x} \right )} \right )\right )} \]

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)**2,x)

[Out]

226*a**2*d**11*(d*x)**(3/2)/(384*a**6*d**12 + 1152*a**5*b*d**12*x**2 + 1152*a**4*b**2*d**12*x**4 + 384*a**3*b*
*3*d**12*x**6) + 252*a*b*d**9*(d*x)**(7/2)/(384*a**6*d**12 + 1152*a**5*b*d**12*x**2 + 1152*a**4*b**2*d**12*x**
4 + 384*a**3*b**3*d**12*x**6) + 90*b**2*d**7*(d*x)**(11/2)/(384*a**6*d**12 + 1152*a**5*b*d**12*x**2 + 1152*a**
4*b**2*d**12*x**4 + 384*a**3*b**3*d**12*x**6) + 2*d**7*RootSum(68719476736*_t**4*a**13*b**3*d**26 + 50625, Lam
bda(_t, _t*log(134217728*_t**3*a**10*b**2*d**20/3375 + sqrt(d*x))))

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