3.766 \(\int \frac{1}{(d x)^{3/2} (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=506 \[ -\frac{45 \sqrt [4]{b} \left (a+b x^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{45 \sqrt [4]{b} \left (a+b x^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{45 \sqrt [4]{b} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{32 \sqrt{2} a^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \sqrt [4]{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{32 \sqrt{2} a^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \left (a+b x^2\right )}{16 a^3 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )} \]

[Out]

9/(16*a^2*d*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*d*Sqrt[d*x]*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) - (45*(a + b*x^2))/(16*a^3*d*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (45*b^(1/4)*(a + b*x^2)*A
rcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(13/4)*d^(3/2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) - (45*b^(1/4)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(
13/4)*d^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (45*b^(1/4)*(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^(13/4)*d^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (45*b
^(1/4)*(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
^(13/4)*d^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.384298, antiderivative size = 506, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1112, 290, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{45 \sqrt [4]{b} \left (a+b x^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{45 \sqrt [4]{b} \left (a+b x^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{45 \sqrt [4]{b} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{32 \sqrt{2} a^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \sqrt [4]{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{32 \sqrt{2} a^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \left (a+b x^2\right )}{16 a^3 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*x)^(3/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

9/(16*a^2*d*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*d*Sqrt[d*x]*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) - (45*(a + b*x^2))/(16*a^3*d*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (45*b^(1/4)*(a + b*x^2)*A
rcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(13/4)*d^(3/2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) - (45*b^(1/4)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(
13/4)*d^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (45*b^(1/4)*(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^(13/4)*d^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (45*b
^(1/4)*(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
^(13/4)*d^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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 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 325

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

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]

Rubi steps

\begin{align*} \int \frac{1}{(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{1}{(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{1}{4 a d \sqrt{d x} \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (9 b \left (a b+b^2 x^2\right )\right ) \int \frac{1}{(d x)^{3/2} \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{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (45 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{(d x)^{3/2} \left (a b+b^2 x^2\right )} \, dx}{32 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \left (a+b x^2\right )}{16 a^3 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (45 b \left (a b+b^2 x^2\right )\right ) \int \frac{\sqrt{d x}}{a b+b^2 x^2} \, dx}{32 a^3 d^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \left (a+b x^2\right )}{16 a^3 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (45 b \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^3 d^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \left (a+b x^2\right )}{16 a^3 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (45 \sqrt{b} \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 d^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (45 \sqrt{b} \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 d^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \left (a+b x^2\right )}{16 a^3 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (45 \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^{13/4} b^{3/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (45 \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^{13/4} b^{3/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (45 \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 b d \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (45 \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 b d \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \left (a+b x^2\right )}{16 a^3 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \sqrt [4]{b} \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^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{45 \sqrt [4]{b} \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^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (45 \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^{13/4} b^{3/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (45 \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^{13/4} b^{3/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{9}{16 a^2 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{4 a d \sqrt{d x} \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \left (a+b x^2\right )}{16 a^3 d \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{45 \sqrt [4]{b} \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^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \sqrt [4]{b} \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^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{45 \sqrt [4]{b} \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^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{45 \sqrt [4]{b} \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^{13/4} d^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0151114, size = 52, normalized size = 0.1 \[ -\frac{2 x \left (a+b x^2\right )^3 \, _2F_1\left (-\frac{1}{4},3;\frac{3}{4};-\frac{b x^2}{a}\right )}{a^3 (d x)^{3/2} \left (\left (a+b x^2\right )^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*x)^(3/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

(-2*x*(a + b*x^2)^3*Hypergeometric2F1[-1/4, 3, 3/4, -((b*x^2)/a)])/(a^3*(d*x)^(3/2)*((a + b*x^2)^2)^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.237, size = 645, normalized size = 1.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128/d*(45*2^(1/2)*ln(-((a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)-d*x-(a*d^2/b)^(1/2))/(d*x+(a*d^2/b)^(1/4)*(d*x)^
(1/2)*2^(1/2)+(a*d^2/b)^(1/2)))*(d*x)^(1/2)*x^4*b^2+90*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a*d^2/b)^(1/4))/(a
*d^2/b)^(1/4))*(d*x)^(1/2)*x^4*b^2+90*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4))*(d
*x)^(1/2)*x^4*b^2+360*(a*d^2/b)^(1/4)*x^4*b^2+90*2^(1/2)*ln(-((a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)-d*x-(a*d^2/b
)^(1/2))/(d*x+(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2)))*(d*x)^(1/2)*x^2*a*b+180*2^(1/2)*arctan((2^
(1/2)*(d*x)^(1/2)+(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4))*(d*x)^(1/2)*x^2*a*b+180*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2
)-(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4))*(d*x)^(1/2)*x^2*a*b+648*(a*d^2/b)^(1/4)*x^2*a*b+45*2^(1/2)*ln(-((a*d^2/b)^
(1/4)*(d*x)^(1/2)*2^(1/2)-d*x-(a*d^2/b)^(1/2))/(d*x+(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2)))*(d*x
)^(1/2)*a^2+90*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4))*(d*x)^(1/2)*a^2+90*2^(1/2
)*arctan((2^(1/2)*(d*x)^(1/2)-(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4))*(d*x)^(1/2)*a^2+256*(a*d^2/b)^(1/4)*a^2)*(b*x^
2+a)/(a*d^2/b)^(1/4)/(d*x)^(1/2)/a^3/((b*x^2+a)^2)^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(3/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77116, size = 837, normalized size = 1.65 \begin{align*} \frac{180 \,{\left (a^{3} b^{2} d^{2} x^{5} + 2 \, a^{4} b d^{2} x^{3} + a^{5} d^{2} x\right )} \left (-\frac{b}{a^{13} d^{6}}\right )^{\frac{1}{4}} \arctan \left (-\frac{91125 \, \sqrt{d x} a^{3} b d \left (-\frac{b}{a^{13} d^{6}}\right )^{\frac{1}{4}} - \sqrt{-8303765625 \, a^{7} b d^{4} \sqrt{-\frac{b}{a^{13} d^{6}}} + 8303765625 \, b^{2} d x} a^{3} d \left (-\frac{b}{a^{13} d^{6}}\right )^{\frac{1}{4}}}{91125 \, b}\right ) - 45 \,{\left (a^{3} b^{2} d^{2} x^{5} + 2 \, a^{4} b d^{2} x^{3} + a^{5} d^{2} x\right )} \left (-\frac{b}{a^{13} d^{6}}\right )^{\frac{1}{4}} \log \left (91125 \, a^{10} d^{5} \left (-\frac{b}{a^{13} d^{6}}\right )^{\frac{3}{4}} + 91125 \, \sqrt{d x} b\right ) + 45 \,{\left (a^{3} b^{2} d^{2} x^{5} + 2 \, a^{4} b d^{2} x^{3} + a^{5} d^{2} x\right )} \left (-\frac{b}{a^{13} d^{6}}\right )^{\frac{1}{4}} \log \left (-91125 \, a^{10} d^{5} \left (-\frac{b}{a^{13} d^{6}}\right )^{\frac{3}{4}} + 91125 \, \sqrt{d x} b\right ) - 4 \,{\left (45 \, b^{2} x^{4} + 81 \, a b x^{2} + 32 \, a^{2}\right )} \sqrt{d x}}{64 \,{\left (a^{3} b^{2} d^{2} x^{5} + 2 \, a^{4} b d^{2} x^{3} + a^{5} d^{2} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(3/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/64*(180*(a^3*b^2*d^2*x^5 + 2*a^4*b*d^2*x^3 + a^5*d^2*x)*(-b/(a^13*d^6))^(1/4)*arctan(-1/91125*(91125*sqrt(d*
x)*a^3*b*d*(-b/(a^13*d^6))^(1/4) - sqrt(-8303765625*a^7*b*d^4*sqrt(-b/(a^13*d^6)) + 8303765625*b^2*d*x)*a^3*d*
(-b/(a^13*d^6))^(1/4))/b) - 45*(a^3*b^2*d^2*x^5 + 2*a^4*b*d^2*x^3 + a^5*d^2*x)*(-b/(a^13*d^6))^(1/4)*log(91125
*a^10*d^5*(-b/(a^13*d^6))^(3/4) + 91125*sqrt(d*x)*b) + 45*(a^3*b^2*d^2*x^5 + 2*a^4*b*d^2*x^3 + a^5*d^2*x)*(-b/
(a^13*d^6))^(1/4)*log(-91125*a^10*d^5*(-b/(a^13*d^6))^(3/4) + 91125*sqrt(d*x)*b) - 4*(45*b^2*x^4 + 81*a*b*x^2
+ 32*a^2)*sqrt(d*x))/(a^3*b^2*d^2*x^5 + 2*a^4*b*d^2*x^3 + a^5*d^2*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d x\right )^{\frac{3}{2}} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((d*x)**(3/2)*((a + b*x**2)**2)**(3/2)), x)

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Giac [A]  time = 1.40361, size = 554, normalized size = 1.09 \begin{align*} -\frac{\frac{256}{\sqrt{d x} a^{3} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac{8 \,{\left (13 \, \sqrt{d x} b^{2} d^{3} x^{3} + 17 \, \sqrt{d x} a b d^{3} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{2} a^{3} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \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^{2} d^{2} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \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^{2} d^{2} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} - \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^{2} d^{2} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \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^{2} d^{2} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )}}{128 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(3/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

-1/128*(256/(sqrt(d*x)*a^3*sgn(b*d^4*x^2 + a*d^4)) + 8*(13*sqrt(d*x)*b^2*d^3*x^3 + 17*sqrt(d*x)*a*b*d^3*x)/((b
*d^2*x^2 + a*d^2)^2*a^3*sgn(b*d^4*x^2 + a*d^4)) + 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^2*d^2*sgn(b*d^4*x^2 + a*d^4)) + 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^2*d^2*sgn(b*d^4*x^2 + a
*d^4)) - 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^2*d^
2*sgn(b*d^4*x^2 + a*d^4)) + 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^2*d^2*sgn(b*d^4*x^2 + a*d^4)))/d