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

Optimal. Leaf size=459 \[ \frac {d \sqrt {d x}}{16 a b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \sqrt {d x}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 d^{3/2} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{64 \sqrt {2} a^{7/4} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 d^{3/2} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{64 \sqrt {2} a^{7/4} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 d^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{32 \sqrt {2} a^{7/4} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 d^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{32 \sqrt {2} a^{7/4} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.33, antiderivative size = 459, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1112, 288, 290, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {3 d^{3/2} \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^{7/4} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 d^{3/2} \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^{7/4} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 d^{3/2} \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^{7/4} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 d^{3/2} \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^{7/4} b^{5/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {d \sqrt {d x}}{16 a b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \sqrt {d x}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[d*x])/(16*a*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (d*Sqrt[d*x])/(4*b*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) - (3*d^(3/2)*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(
7/4)*b^(5/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (3*d^(3/2)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/
(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(7/4)*b^(5/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (3*d^(3/2)*(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^(7/4)*b^(5/4)*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4]) + (3*d^(3/2)*(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^(7/4)*b^(5/4)*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 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 288

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

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

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Mathematica [A]  time = 0.21, size = 272, normalized size = 0.59 \begin {gather*} \frac {(d x)^{3/2} \left (a+b x^2\right ) \left (8 a^{3/4} \sqrt [4]{b} \sqrt {x} \left (a+b x^2\right )-32 a^{7/4} \sqrt [4]{b} \sqrt {x}-3 \sqrt {2} \left (a+b x^2\right )^2 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )+3 \sqrt {2} \left (a+b x^2\right )^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )-6 \sqrt {2} \left (a+b x^2\right )^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )+6 \sqrt {2} \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )\right )}{128 a^{7/4} b^{5/4} x^{3/2} \left (\left (a+b x^2\right )^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^(3/2)*(a + b*x^2)*(-32*a^(7/4)*b^(1/4)*Sqrt[x] + 8*a^(3/4)*b^(1/4)*Sqrt[x]*(a + b*x^2) - 6*Sqrt[2]*(a +
b*x^2)^2*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)] + 6*Sqrt[2]*(a + b*x^2)^2*ArcTan[1 + (Sqrt[2]*b^(1/4)*
Sqrt[x])/a^(1/4)] - 3*Sqrt[2]*(a + b*x^2)^2*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] + 3*Sqr
t[2]*(a + b*x^2)^2*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x]))/(128*a^(7/4)*b^(5/4)*x^(3/2)*(
(a + b*x^2)^2)^(3/2))

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IntegrateAlgebraic [A]  time = 73.26, size = 244, normalized size = 0.53 \begin {gather*} \frac {\left (a d^2+b d^2 x^2\right ) \left (-\frac {3 d^{3/2} \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^{7/4} b^{5/4}}+\frac {3 d^{3/2} \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^{7/4} b^{5/4}}+\frac {b d^3 (d x)^{5/2}-3 a d^5 \sqrt {d x}}{16 a b \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[(d*x)^(3/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

((a*d^2 + b*d^2*x^2)*((-3*a*d^5*Sqrt[d*x] + b*d^3*(d*x)^(5/2))/(16*a*b*(a*d^2 + b*d^2*x^2)^2) - (3*d^(3/2)*Arc
Tan[((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^(7
/4)*b^(5/4)) + (3*d^(3/2)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x])/(Sqrt[a]*d + Sqrt[b]*d*x)])/(32*
Sqrt[2]*a^(7/4)*b^(5/4))))/(d^2*Sqrt[(a*d^2 + b*d^2*x^2)^2/d^4])

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fricas [A]  time = 1.07, size = 308, normalized size = 0.67 \begin {gather*} \frac {12 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac {d^{6}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{5} b^{4} d \left (-\frac {d^{6}}{a^{7} b^{5}}\right )^{\frac {3}{4}} - \sqrt {a^{4} b^{2} \sqrt {-\frac {d^{6}}{a^{7} b^{5}}} + d^{3} x} a^{5} b^{4} \left (-\frac {d^{6}}{a^{7} b^{5}}\right )^{\frac {3}{4}}}{d^{6}}\right ) + 3 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac {d^{6}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \log \left (3 \, a^{2} b \left (-\frac {d^{6}}{a^{7} b^{5}}\right )^{\frac {1}{4}} + 3 \, \sqrt {d x} d\right ) - 3 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac {d^{6}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \log \left (-3 \, a^{2} b \left (-\frac {d^{6}}{a^{7} b^{5}}\right )^{\frac {1}{4}} + 3 \, \sqrt {d x} d\right ) + 4 \, {\left (b d x^{2} - 3 \, a d\right )} \sqrt {d x}}{64 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/64*(12*(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)*(-d^6/(a^7*b^5))^(1/4)*arctan(-(sqrt(d*x)*a^5*b^4*d*(-d^6/(a^7*b^
5))^(3/4) - sqrt(a^4*b^2*sqrt(-d^6/(a^7*b^5)) + d^3*x)*a^5*b^4*(-d^6/(a^7*b^5))^(3/4))/d^6) + 3*(a*b^3*x^4 + 2
*a^2*b^2*x^2 + a^3*b)*(-d^6/(a^7*b^5))^(1/4)*log(3*a^2*b*(-d^6/(a^7*b^5))^(1/4) + 3*sqrt(d*x)*d) - 3*(a*b^3*x^
4 + 2*a^2*b^2*x^2 + a^3*b)*(-d^6/(a^7*b^5))^(1/4)*log(-3*a^2*b*(-d^6/(a^7*b^5))^(1/4) + 3*sqrt(d*x)*d) + 4*(b*
d*x^2 - 3*a*d)*sqrt(d*x))/(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)

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giac [A]  time = 0.34, size = 367, normalized size = 0.80 \begin {gather*} \frac {1}{128} \, d {\left (\frac {6 \, \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 )}{a^{2} b^{2} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {6 \, \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 )}{a^{2} b^{2} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {3 \, \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 )}{a^{2} b^{2} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} - \frac {3 \, \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 )}{a^{2} b^{2} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {8 \, {\left (\sqrt {d x} b d^{4} x^{2} - 3 \, \sqrt {d x} a d^{4}\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{2} a b \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/128*d*(6*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^2*b^2*sgn(b*d^4*x^2 + a*d^4)) + 6*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^2*b^2*sgn(b*d^4*x^2 + a*d^4)) + 3*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^2*b^2*sgn(b*d^4*x^2 + a*d^4)) - 3*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^2*b^2*sgn(b*d^4*x^2 + a*d^4)) + 8*(sqrt(d*x)*b
*d^4*x^2 - 3*sqrt(d*x)*a*d^4)/((b*d^2*x^2 + a*d^2)^2*a*b*sgn(b*d^4*x^2 + a*d^4)))

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maple [B]  time = 0.02, size = 668, normalized size = 1.46 \begin {gather*} \frac {\left (6 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )+6 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )+3 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )+12 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )+12 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )+6 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )+6 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )+6 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )+3 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \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 )-24 \sqrt {d x}\, a^{2} d^{2}+8 \left (d x \right )^{\frac {5}{2}} a b \right ) \left (b \,x^{2}+a \right )}{128 \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)^(3/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/128*(3*(a/b*d^2)^(1/4)*2^(1/2)*b^2*d^2*x^4*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)))+6*(a/b*d^2)^(1/4)*2^(1/2)*b^2*d^2*x^4*arctan((2^(1/2)*(
d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+6*(a/b*d^2)^(1/4)*2^(1/2)*b^2*d^2*x^4*arctan((2^(1/2)*(d*x)^(1/2)
-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+6*(a/b*d^2)^(1/4)*2^(1/2)*a*b*d^2*x^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)))+12*(a/b*d^2)^(1/4)*2^(1/2)*
a*b*d^2*x^2*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+12*(a/b*d^2)^(1/4)*2^(1/2)*a*b*d^2*x
^2*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+3*(a/b*d^2)^(1/4)*2^(1/2)*a^2*d^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)))+6
*(a/b*d^2)^(1/4)*2^(1/2)*a^2*d^2*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+6*(a/b*d^2)^(1/
4)*2^(1/2)*a^2*d^2*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+8*(d*x)^(5/2)*a*b-24*(d*x)^(1
/2)*a^2*d^2)/d*(b*x^2+a)/b/a^2/((b*x^2+a)^2)^(3/2)

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maxima [A]  time = 3.22, size = 281, normalized size = 0.61 \begin {gather*} \frac {d^{\frac {3}{2}} x^{\frac {5}{2}}}{2 \, {\left (a^{2} b x^{2} + a^{3} + {\left (a b^{2} x^{2} + a^{2} b\right )} x^{2}\right )}} - \frac {7 \, b d^{\frac {3}{2}} x^{\frac {5}{2}} + 3 \, a d^{\frac {3}{2}} \sqrt {x}}{16 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} + \frac {3 \, d {\left (\frac {2 \, \sqrt {2} \sqrt {d} \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 {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} \sqrt {d} \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 {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} \sqrt {d} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} \sqrt {d} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}\right )}}{128 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

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

[Out]

int((d*x)^(3/2)/(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 {\left (d x\right )^{\frac {3}{2}}}{\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)**(3/2)/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

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