3.706 \(\int \frac {1}{\sqrt {d x} (a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=335 \[ -\frac {77 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}-\frac {77 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \sqrt {d x}}{192 a^3 d \left (a+b x^2\right )}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}+\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3} \]

[Out]

-77/256*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(15/4)/b^(1/4)*2^(1/2)/d^(1/2)+77/256*arctan(1
+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(15/4)/b^(1/4)*2^(1/2)/d^(1/2)-77/512*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))/a^(15/4)/b^(1/4)*2^(1/2)/d^(1/2)+77/512*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))/a^(15/4)/b^(1/4)*2^(1/2)/d^(1/2)+1/6*(d*x)^(1/2)/a/d/(b*
x^2+a)^3+11/48*(d*x)^(1/2)/a^2/d/(b*x^2+a)^2+77/192*(d*x)^(1/2)/a^3/d/(b*x^2+a)

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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, 211, 1165, 628, 1162, 617, 204} \[ \frac {77 \sqrt {d x}}{192 a^3 d \left (a+b x^2\right )}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}-\frac {77 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}-\frac {77 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*x]*(a^2 + 2*a*b*x^2 + b^2*x^4)^2),x]

[Out]

Sqrt[d*x]/(6*a*d*(a + b*x^2)^3) + (11*Sqrt[d*x])/(48*a^2*d*(a + b*x^2)^2) + (77*Sqrt[d*x])/(192*a^3*d*(a + b*x
^2)) - (77*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(15/4)*b^(1/4)*Sqrt[d]) +
(77*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(15/4)*b^(1/4)*Sqrt[d]) - (77*L
og[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*a^(15/4)*b^(1/4)*Sqr
t[d]) + (77*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^(15/4
)*b^(1/4)*Sqrt[d])

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 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 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 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 {1}{\sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3}+\frac {\left (11 b^3\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )^3} \, dx}{12 a}\\ &=\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}+\frac {\left (77 b^2\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )^2} \, dx}{96 a^2}\\ &=\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}+\frac {77 \sqrt {d x}}{192 a^3 d \left (a+b x^2\right )}+\frac {(77 b) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )} \, dx}{128 a^3}\\ &=\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}+\frac {77 \sqrt {d x}}{192 a^3 d \left (a+b x^2\right )}+\frac {(77 b) \operatorname {Subst}\left (\int \frac {1}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{64 a^3 d}\\ &=\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}+\frac {77 \sqrt {d x}}{192 a^3 d \left (a+b x^2\right )}+\frac {(77 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 )}{128 a^{7/2} d^2}+\frac {(77 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 )}{128 a^{7/2} d^2}\\ &=\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}+\frac {77 \sqrt {d x}}{192 a^3 d \left (a+b x^2\right )}+\frac {77 \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^{7/2} \sqrt {b}}+\frac {77 \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^{7/2} \sqrt {b}}-\frac {77 \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^{15/4} \sqrt [4]{b} \sqrt {d}}-\frac {77 \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^{15/4} \sqrt [4]{b} \sqrt {d}}\\ &=\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}+\frac {77 \sqrt {d x}}{192 a^3 d \left (a+b x^2\right )}-\frac {77 \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^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \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^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \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^{15/4} \sqrt [4]{b} \sqrt {d}}-\frac {77 \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^{15/4} \sqrt [4]{b} \sqrt {d}}\\ &=\frac {\sqrt {d x}}{6 a d \left (a+b x^2\right )^3}+\frac {11 \sqrt {d x}}{48 a^2 d \left (a+b x^2\right )^2}+\frac {77 \sqrt {d x}}{192 a^3 d \left (a+b x^2\right )}-\frac {77 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{15/4} \sqrt [4]{b} \sqrt {d}}-\frac {77 \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^{15/4} \sqrt [4]{b} \sqrt {d}}+\frac {77 \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^{15/4} \sqrt [4]{b} \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 253, normalized size = 0.76 \[ \frac {\sqrt {x} \left (\frac {256 a^{11/4} \sqrt {x}}{\left (a+b x^2\right )^3}+\frac {352 a^{7/4} \sqrt {x}}{\left (a+b x^2\right )^2}+\frac {616 a^{3/4} \sqrt {x}}{a+b x^2}-\frac {231 \sqrt {2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{\sqrt [4]{b}}+\frac {231 \sqrt {2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{\sqrt [4]{b}}-\frac {462 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{b}}+\frac {462 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{b}}\right )}{1536 a^{15/4} \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*x]*(a^2 + 2*a*b*x^2 + b^2*x^4)^2),x]

[Out]

(Sqrt[x]*((256*a^(11/4)*Sqrt[x])/(a + b*x^2)^3 + (352*a^(7/4)*Sqrt[x])/(a + b*x^2)^2 + (616*a^(3/4)*Sqrt[x])/(
a + b*x^2) - (462*Sqrt[2]*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/b^(1/4) + (462*Sqrt[2]*ArcTan[1 + (Sq
rt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/b^(1/4) - (231*Sqrt[2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b
]*x])/b^(1/4) + (231*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/b^(1/4)))/(1536*a^(15
/4)*Sqrt[d*x])

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fricas [A]  time = 1.09, size = 357, normalized size = 1.07 \[ \frac {924 \, {\left (a^{3} b^{3} d x^{6} + 3 \, a^{4} b^{2} d x^{4} + 3 \, a^{5} b d x^{2} + a^{6} d\right )} \left (-\frac {1}{a^{15} b d^{2}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {a^{8} d^{2} \sqrt {-\frac {1}{a^{15} b d^{2}}} + d x} a^{11} b d \left (-\frac {1}{a^{15} b d^{2}}\right )^{\frac {3}{4}} - \sqrt {d x} a^{11} b d \left (-\frac {1}{a^{15} b d^{2}}\right )^{\frac {3}{4}}\right ) + 231 \, {\left (a^{3} b^{3} d x^{6} + 3 \, a^{4} b^{2} d x^{4} + 3 \, a^{5} b d x^{2} + a^{6} d\right )} \left (-\frac {1}{a^{15} b d^{2}}\right )^{\frac {1}{4}} \log \left (a^{4} d \left (-\frac {1}{a^{15} b d^{2}}\right )^{\frac {1}{4}} + \sqrt {d x}\right ) - 231 \, {\left (a^{3} b^{3} d x^{6} + 3 \, a^{4} b^{2} d x^{4} + 3 \, a^{5} b d x^{2} + a^{6} d\right )} \left (-\frac {1}{a^{15} b d^{2}}\right )^{\frac {1}{4}} \log \left (-a^{4} d \left (-\frac {1}{a^{15} b d^{2}}\right )^{\frac {1}{4}} + \sqrt {d x}\right ) + 4 \, {\left (77 \, b^{2} x^{4} + 198 \, a b x^{2} + 153 \, a^{2}\right )} \sqrt {d x}}{768 \, {\left (a^{3} b^{3} d x^{6} + 3 \, a^{4} b^{2} d x^{4} + 3 \, a^{5} b d x^{2} + a^{6} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(924*(a^3*b^3*d*x^6 + 3*a^4*b^2*d*x^4 + 3*a^5*b*d*x^2 + a^6*d)*(-1/(a^15*b*d^2))^(1/4)*arctan(sqrt(a^8*d
^2*sqrt(-1/(a^15*b*d^2)) + d*x)*a^11*b*d*(-1/(a^15*b*d^2))^(3/4) - sqrt(d*x)*a^11*b*d*(-1/(a^15*b*d^2))^(3/4))
+ 231*(a^3*b^3*d*x^6 + 3*a^4*b^2*d*x^4 + 3*a^5*b*d*x^2 + a^6*d)*(-1/(a^15*b*d^2))^(1/4)*log(a^4*d*(-1/(a^15*b
*d^2))^(1/4) + sqrt(d*x)) - 231*(a^3*b^3*d*x^6 + 3*a^4*b^2*d*x^4 + 3*a^5*b*d*x^2 + a^6*d)*(-1/(a^15*b*d^2))^(1
/4)*log(-a^4*d*(-1/(a^15*b*d^2))^(1/4) + sqrt(d*x)) + 4*(77*b^2*x^4 + 198*a*b*x^2 + 153*a^2)*sqrt(d*x))/(a^3*b
^3*d*x^6 + 3*a^4*b^2*d*x^4 + 3*a^5*b*d*x^2 + a^6*d)

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giac [A]  time = 0.27, size = 308, normalized size = 0.92 \[ \frac {77 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{256 \, a^{4} b d} + \frac {77 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{256 \, a^{4} b d} + \frac {77 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{512 \, a^{4} b d} - \frac {77 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{512 \, a^{4} b d} + \frac {77 \, \sqrt {d x} b^{2} d^{5} x^{4} + 198 \, \sqrt {d x} a b d^{5} x^{2} + 153 \, \sqrt {d x} a^{2} d^{5}}{192 \, {\left (b d^{2} x^{2} + a d^{2}\right )}^{3} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

77/256*sqrt(2)*(a*b^3*d^2)^(1/4)*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*d) + 77/256*sqrt(2)*(a*b^3*d^2)^(1/4)*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*d) + 77/512*sqrt(2)*(a*b^3*d^2)^(1/4)*log(d*x + sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d
^2/b))/(a^4*b*d) - 77/512*sqrt(2)*(a*b^3*d^2)^(1/4)*log(d*x - sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b
))/(a^4*b*d) + 1/192*(77*sqrt(d*x)*b^2*d^5*x^4 + 198*sqrt(d*x)*a*b*d^5*x^2 + 153*sqrt(d*x)*a^2*d^5)/((b*d^2*x^
2 + a*d^2)^3*a^3)

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maple [A]  time = 0.02, size = 269, normalized size = 0.80 \[ \frac {51 \sqrt {d x}\, d^{5}}{64 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a}+\frac {33 \left (d x \right )^{\frac {5}{2}} b \,d^{3}}{32 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a^{2}}+\frac {77 \left (d x \right )^{\frac {9}{2}} b^{2} d}{192 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a^{3}}+\frac {77 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{256 a^{4} d}+\frac {77 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{256 a^{4} d}+\frac {77 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{512 a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

77/192*d/(b*d^2*x^2+a*d^2)^3/a^3*b^2*(d*x)^(9/2)+33/32*d^3/(b*d^2*x^2+a*d^2)^3/a^2*b*(d*x)^(5/2)+51/64*d^5/(b*
d^2*x^2+a*d^2)^3/a*(d*x)^(1/2)+77/512/d/a^4*(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)))+77/256/d/a^4*(a/b*d^2)^(1/4)*2^(
1/2)*arctan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)+1)+77/256/d/a^4*(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 = 3.09, size = 322, normalized size = 0.96 \[ \frac {\frac {8 \, {\left (77 \, \left (d x\right )^{\frac {9}{2}} b^{2} d^{2} + 198 \, \left (d x\right )^{\frac {5}{2}} a b d^{4} + 153 \, \sqrt {d x} a^{2} d^{6}\right )}}{a^{3} b^{3} d^{6} x^{6} + 3 \, a^{4} b^{2} d^{6} x^{4} + 3 \, a^{5} b d^{6} x^{2} + a^{6} d^{6}} + \frac {231 \, {\left (\frac {\sqrt {2} d^{2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} d^{2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {3}{4}} b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a}} + \frac {2 \, \sqrt {2} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {a}}\right )}}{a^{3}}}{1536 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*(8*(77*(d*x)^(9/2)*b^2*d^2 + 198*(d*x)^(5/2)*a*b*d^4 + 153*sqrt(d*x)*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) + 231*(sqrt(2)*d^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)^(3/4)*b^(1/4)) - sqrt(2)*d^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)^(3/4)*b^(1/4)) + 2*sqrt(2)*d*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(a)) + 2*sqrt(2)*d*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)*sq
rt(a)))/a^3)/d

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mupad [B]  time = 4.28, size = 150, normalized size = 0.45 \[ \frac {\frac {51\,d^5\,\sqrt {d\,x}}{64\,a}+\frac {33\,b\,d^3\,{\left (d\,x\right )}^{5/2}}{32\,a^2}+\frac {77\,b^2\,d\,{\left (d\,x\right )}^{9/2}}{192\,a^3}}{a^3\,d^6+3\,a^2\,b\,d^6\,x^2+3\,a\,b^2\,d^6\,x^4+b^3\,d^6\,x^6}+\frac {77\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{15/4}\,b^{1/4}\,\sqrt {d}}+\frac {77\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{15/4}\,b^{1/4}\,\sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((51*d^5*(d*x)^(1/2))/(64*a) + (33*b*d^3*(d*x)^(5/2))/(32*a^2) + (77*b^2*d*(d*x)^(9/2))/(192*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) + (77*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*
(-a)^(15/4)*b^(1/4)*d^(1/2)) + (77*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*(-a)^(15/4)*b^(1/4)
*d^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d x} \left (a + b x^{2}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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