3.6.86 \(\int \frac {\sqrt {d x}}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=460 \[ \frac {5 (d x)^{3/2}}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \sqrt {d} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{64 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 \sqrt {d} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{64 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 \sqrt {d} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{32 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \sqrt {d} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{32 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.33, antiderivative size = 460, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1112, 290, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {5 (d x)^{3/2}}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \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 )}{64 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 \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 )}{64 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 \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 )}{32 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \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 )}{32 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*x]/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(5*(d*x)^(3/2))/(16*a^2*d*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (d*x)^(3/2)/(4*a*d*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x
^2 + b^2*x^4]) - (5*Sqrt[d]*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]
*a^(9/4)*b^(3/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*Sqrt[d]*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*
x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(9/4)*b^(3/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*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]])/(64*Sqrt[2]*a^(9/4)*b^(3/4)*Sqrt
[a^2 + 2*a*b*x^2 + b^2*x^4]) - (5*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]])/(64*Sqrt[2]*a^(9/4)*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 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 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}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {\sqrt {d x}}{\left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 b \left (a b+b^2 x^2\right )\right ) \int \frac {\sqrt {d x}}{\left (a b+b^2 x^2\right )^2} \, dx}{8 a \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5 (d x)^{3/2}}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 \left (a b+b^2 x^2\right )\right ) \int \frac {\sqrt {d x}}{a b+b^2 x^2} \, dx}{32 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5 (d x)^{3/2}}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 \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 )}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5 (d x)^{3/2}}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (5 \left (a b+b^2 x^2\right )\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 )}{32 a^2 \sqrt {b} d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 \left (a b+b^2 x^2\right )\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 )}{32 a^2 \sqrt {b} d \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5 (d x)^{3/2}}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 \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 )}{64 \sqrt {2} a^{9/4} b^{7/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 \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 )}{64 \sqrt {2} a^{9/4} b^{7/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 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 )}{64 a^2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 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 )}{64 a^2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5 (d x)^{3/2}}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \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 )}{64 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 \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 )}{64 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 \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 )}{32 \sqrt {2} a^{9/4} b^{7/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (5 \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 )}{32 \sqrt {2} a^{9/4} b^{7/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5 (d x)^{3/2}}{16 a^2 d \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 \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 )}{32 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \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 )}{32 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \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 )}{64 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 \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 )}{64 \sqrt {2} a^{9/4} b^{3/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.12 \begin {gather*} \frac {2 x \sqrt {d x} \left (a+b x^2\right )^3 \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};-\frac {b x^2}{a}\right )}{3 a^3 \left (\left (a+b x^2\right )^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*x]/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 77.41, size = 238, normalized size = 0.52 \begin {gather*} \frac {\left (a d^2+b d^2 x^2\right ) \left (-\frac {5 \sqrt {d} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{a} \sqrt {d}}{\sqrt {2} \sqrt [4]{b}}-\frac {\sqrt [4]{b} \sqrt {d} x}{\sqrt {2} \sqrt [4]{a}}}{\sqrt {d x}}\right )}{32 \sqrt {2} a^{9/4} b^{3/4}}-\frac {5 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d} \sqrt {d x}}{\sqrt {a} d+\sqrt {b} d x}\right )}{32 \sqrt {2} a^{9/4} b^{3/4}}+\frac {(d x)^{3/2} \left (9 a d^3+5 b d^3 x^2\right )}{16 a^2 \left (a d^2+b d^2 x^2\right )^2}\right )}{d^2 \sqrt {\frac {\left (a d^2+b d^2 x^2\right )^2}{d^4}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[d*x]/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

((a*d^2 + b*d^2*x^2)*(((d*x)^(3/2)*(9*a*d^3 + 5*b*d^3*x^2))/(16*a^2*(a*d^2 + b*d^2*x^2)^2) - (5*Sqrt[d]*ArcTan
[((a^(1/4)*Sqrt[d])/(Sqrt[2]*b^(1/4)) - (b^(1/4)*Sqrt[d]*x)/(Sqrt[2]*a^(1/4)))/Sqrt[d*x]])/(32*Sqrt[2]*a^(9/4)
*b^(3/4)) - (5*Sqrt[d]*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x])/(Sqrt[a]*d + Sqrt[b]*d*x)])/(32*Sqr
t[2]*a^(9/4)*b^(3/4))))/(d^2*Sqrt[(a*d^2 + b*d^2*x^2)^2/d^4])

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fricas [A]  time = 0.64, size = 304, normalized size = 0.66 \begin {gather*} -\frac {20 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \left (-\frac {d^{2}}{a^{9} b^{3}}\right )^{\frac {1}{4}} \arctan \left (-\frac {125 \, \sqrt {d x} a^{2} b d \left (-\frac {d^{2}}{a^{9} b^{3}}\right )^{\frac {1}{4}} - \sqrt {-15625 \, a^{5} b d^{2} \sqrt {-\frac {d^{2}}{a^{9} b^{3}}} + 15625 \, d^{3} x} a^{2} b \left (-\frac {d^{2}}{a^{9} b^{3}}\right )^{\frac {1}{4}}}{125 \, d^{2}}\right ) - 5 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \left (-\frac {d^{2}}{a^{9} b^{3}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} b^{2} \left (-\frac {d^{2}}{a^{9} b^{3}}\right )^{\frac {3}{4}} + 125 \, \sqrt {d x} d\right ) + 5 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \left (-\frac {d^{2}}{a^{9} b^{3}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} b^{2} \left (-\frac {d^{2}}{a^{9} b^{3}}\right )^{\frac {3}{4}} + 125 \, \sqrt {d x} d\right ) - 4 \, {\left (5 \, b x^{3} + 9 \, a x\right )} \sqrt {d x}}{64 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

-1/64*(20*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)*(-d^2/(a^9*b^3))^(1/4)*arctan(-1/125*(125*sqrt(d*x)*a^2*b*d*(-d^2/
(a^9*b^3))^(1/4) - sqrt(-15625*a^5*b*d^2*sqrt(-d^2/(a^9*b^3)) + 15625*d^3*x)*a^2*b*(-d^2/(a^9*b^3))^(1/4))/d^2
) - 5*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)*(-d^2/(a^9*b^3))^(1/4)*log(125*a^7*b^2*(-d^2/(a^9*b^3))^(3/4) + 125*sq
rt(d*x)*d) + 5*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)*(-d^2/(a^9*b^3))^(1/4)*log(-125*a^7*b^2*(-d^2/(a^9*b^3))^(3/4
) + 125*sqrt(d*x)*d) - 4*(5*b*x^3 + 9*a*x)*sqrt(d*x))/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)

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giac [A]  time = 0.34, size = 368, normalized size = 0.80 \begin {gather*} \frac {\frac {8 \, {\left (5 \, \sqrt {d x} b d^{5} x^{3} + 9 \, \sqrt {d x} a d^{5} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{2} a^{2} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {10 \, \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^{3} b^{3} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {10 \, \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^{3} b^{3} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} - \frac {5 \, \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^{3} b^{3} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {5 \, \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^{3} b^{3} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )}}{128 \, d} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

1/128*(8*(5*sqrt(d*x)*b*d^5*x^3 + 9*sqrt(d*x)*a*d^5*x)/((b*d^2*x^2 + a*d^2)^2*a^2*sgn(b*d^4*x^2 + a*d^4)) + 10
*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^3*b^
3*sgn(b*d^4*x^2 + a*d^4)) + 10*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^3*b^3*sgn(b*d^4*x^2 + a*d^4)) - 5*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^3*b^3*sgn(b*d^4*x^2 + a*d^4)) + 5*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^3*b^3*sgn(b*d^4*x^2 + a*d^4)))/d

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maple [B]  time = 0.01, size = 617, normalized size = 1.34 \begin {gather*} \frac {\left (10 \sqrt {2}\, b^{2} d^{2} x^{4} \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 )+10 \sqrt {2}\, b^{2} d^{2} x^{4} \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 )+5 \sqrt {2}\, b^{2} d^{2} x^{4} \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 )+20 \sqrt {2}\, a b \,d^{2} x^{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 )+20 \sqrt {2}\, a b \,d^{2} x^{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 )+10 \sqrt {2}\, a b \,d^{2} x^{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 )+10 \sqrt {2}\, a^{2} d^{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 )+10 \sqrt {2}\, a^{2} d^{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 )+5 \sqrt {2}\, a^{2} d^{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 )+40 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (d x \right )^{\frac {3}{2}} b^{2} x^{2}+72 \left (d x \right )^{\frac {3}{2}} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a b \right ) \left (b \,x^{2}+a \right )}{128 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} a^{2} b d} \end {gather*}

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)^(3/2),x)

[Out]

1/128*(5*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)))*x^4*b^2*d^2+10*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/
4))*x^4*b^2*d^2+10*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))*x^4*b^2*d^2+40*(a/b*d
^2)^(1/4)*(d*x)^(3/2)*x^2*b^2+10*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)))*x^2*a*b*d^2+20*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d
^2)^(1/4))/(a/b*d^2)^(1/4))*x^2*a*b*d^2+20*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4
))*x^2*a*b*d^2+72*(d*x)^(3/2)*a*b*(a/b*d^2)^(1/4)+5*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)))*a^2*d^2+10*2^(1/2)*arctan((2^(1/2)*(d*
x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))*a^2*d^2+10*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/
b*d^2)^(1/4))*a^2*d^2)/d*(b*x^2+a)/(a/b*d^2)^(1/4)/b/a^2/((b*x^2+a)^2)^(3/2)

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maxima [A]  time = 3.21, size = 265, normalized size = 0.58 \begin {gather*} \frac {\sqrt {d} x^{\frac {3}{2}}}{2 \, {\left (a^{2} b x^{2} + a^{3} + {\left (a b^{2} x^{2} + a^{2} b\right )} x^{2}\right )}} + \frac {5 \, b \sqrt {d} x^{\frac {7}{2}} + a \sqrt {d} x^{\frac {3}{2}}}{16 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} + \frac {5 \, \sqrt {d} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{128 \, a^{2}} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

1/2*sqrt(d)*x^(3/2)/(a^2*b*x^2 + a^3 + (a*b^2*x^2 + a^2*b)*x^2) + 1/16*(5*b*sqrt(d)*x^(7/2) + a*sqrt(d)*x^(3/2
))/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4) + 5/128*sqrt(d)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) +
2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqr
t(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log
(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/
4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/a^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {d\,x}}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \end {gather*}

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)^(3/2),x)

[Out]

int((d*x)^(1/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d x}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)**(3/2),x)

[Out]

Integral(sqrt(d*x)/((a + b*x**2)**2)**(3/2), x)

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