3.762 \(\int \frac{(d x)^{5/2}}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.333142, antiderivative size = 459, normalized size of antiderivative = 1., 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, 297, 1162, 617, 204, 1165, 628} \[ \frac{3 d^{5/2} \left (a+b x^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 d^{5/2} \left (a+b x^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 d^{5/2} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{32 \sqrt{2} a^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 d^{5/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{32 \sqrt{2} a^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 d (d x)^{3/2}}{16 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{3/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(5/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(3*d*(d*x)^(3/2))/(16*a*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (d*(d*x)^(3/2))/(4*b*(a + b*x^2)*Sqrt[a^2 + 2*a*b
*x^2 + b^2*x^4]) - (3*d^(5/2)*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[
2]*a^(5/4)*b^(7/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (3*d^(5/2)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[
d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(5/4)*b^(7/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (3*d^(5/2)*(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^(5/4)*b^(7/4)*Sq
rt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (3*d^(5/2)*(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^(5/4)*b^(7/4)*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 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 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{(d x)^{5/2}}{\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)^{5/2}}{\left (a b+b^2 x^2\right )^3} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{d (d x)^{3/2}}{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{\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{3 d (d x)^{3/2}}{16 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{3/2}}{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{\sqrt{d x}}{a b+b^2 x^2} \, dx}{32 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{3 d (d x)^{3/2}}{16 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{3/2}}{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{x^2}{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{3 d (d x)^{3/2}}{16 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{3/2}}{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{\sqrt{a} d-\sqrt{b} x^2}{a b+\frac{b^2 x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{32 a b^{3/2} \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{\sqrt{a} d+\sqrt{b} x^2}{a b+\frac{b^2 x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{32 a b^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{3 d (d x)^{3/2}}{16 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{3/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (3 d^{5/2} \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^{5/4} b^{11/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (3 d^{5/2} \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^{5/4} b^{11/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (3 d^3 \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 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (3 d^3 \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 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{3 d (d x)^{3/2}}{16 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{3/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 d^{5/2} \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^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 d^{5/2} \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^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (3 d^{5/2} \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^{5/4} b^{11/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (3 d^{5/2} \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^{5/4} b^{11/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{3 d (d x)^{3/2}}{16 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{3/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 d^{5/2} \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^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 d^{5/2} \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^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 d^{5/2} \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^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 d^{5/2} \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^{5/4} b^{7/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.027012, size = 73, normalized size = 0.16 \[ \frac{2 d (d x)^{3/2} \left (\left (a+b x^2\right )^2 \, _2F_1\left (\frac{3}{4},3;\frac{7}{4};-\frac{b x^2}{a}\right )-a^2\right )}{5 a^2 b \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(5/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(2*d*(d*x)^(3/2)*(-a^2 + (a + b*x^2)^2*Hypergeometric2F1[3/4, 3, 7/4, -((b*x^2)/a)]))/(5*a^2*b*(a + b*x^2)*Sqr
t[(a + b*x^2)^2])

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Maple [B]  time = 0.24, size = 617, 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((d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/128*(3*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)))*x^4*b^2*d^4+6*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4)
)*x^4*b^2*d^4+6*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4))*x^4*b^2*d^4+24*(a*d^2/b)
^(1/4)*(d*x)^(7/2)*b^2+6*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)))*x^2*a*b*d^4+12*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a*d^2/b)^(1/4)
)/(a*d^2/b)^(1/4))*x^2*a*b*d^4+12*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4))*x^2*a*
b*d^4-8*(a*d^2/b)^(1/4)*(d*x)^(3/2)*a*b*d^2+3*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)))*a^2*d^4+6*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)
+(a*d^2/b)^(1/4))/(a*d^2/b)^(1/4))*a^2*d^4+6*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a*d^2/b)^(1/4))/(a*d^2/b)^(1
/4))*a^2*d^4)/d*(b*x^2+a)/(a*d^2/b)^(1/4)/b^2/a/((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((d*x)^(5/2)/(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 = 1.82999, size = 745, normalized size = 1.62 \begin{align*} -\frac{12 \,{\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac{d^{10}}{a^{5} b^{7}}\right )^{\frac{1}{4}} \arctan \left (-\frac{27 \, \sqrt{d x} a b^{2} d^{7} \left (-\frac{d^{10}}{a^{5} b^{7}}\right )^{\frac{1}{4}} - \sqrt{-729 \, a^{3} b^{3} d^{10} \sqrt{-\frac{d^{10}}{a^{5} b^{7}}} + 729 \, d^{15} x} a b^{2} \left (-\frac{d^{10}}{a^{5} b^{7}}\right )^{\frac{1}{4}}}{27 \, d^{10}}\right ) - 3 \,{\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac{d^{10}}{a^{5} b^{7}}\right )^{\frac{1}{4}} \log \left (27 \, a^{4} b^{5} \left (-\frac{d^{10}}{a^{5} b^{7}}\right )^{\frac{3}{4}} + 27 \, \sqrt{d x} d^{7}\right ) + 3 \,{\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )} \left (-\frac{d^{10}}{a^{5} b^{7}}\right )^{\frac{1}{4}} \log \left (-27 \, a^{4} b^{5} \left (-\frac{d^{10}}{a^{5} b^{7}}\right )^{\frac{3}{4}} + 27 \, \sqrt{d x} d^{7}\right ) - 4 \,{\left (3 \, b d^{2} x^{3} - a d^{2} x\right )} \sqrt{d x}}{64 \,{\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)/(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^10/(a^5*b^7))^(1/4)*arctan(-1/27*(27*sqrt(d*x)*a*b^2*d^7*(-d
^10/(a^5*b^7))^(1/4) - sqrt(-729*a^3*b^3*d^10*sqrt(-d^10/(a^5*b^7)) + 729*d^15*x)*a*b^2*(-d^10/(a^5*b^7))^(1/4
))/d^10) - 3*(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)*(-d^10/(a^5*b^7))^(1/4)*log(27*a^4*b^5*(-d^10/(a^5*b^7))^(3/4
) + 27*sqrt(d*x)*d^7) + 3*(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)*(-d^10/(a^5*b^7))^(1/4)*log(-27*a^4*b^5*(-d^10/(
a^5*b^7))^(3/4) + 27*sqrt(d*x)*d^7) - 4*(3*b*d^2*x^3 - a*d^2*x)*sqrt(d*x))/(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{\frac{5}{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((d*x)**(5/2)/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral((d*x)**(5/2)/((a + b*x**2)**2)**(3/2), x)

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Giac [A]  time = 1.38369, size = 498, normalized size = 1.08 \begin{align*} \frac{1}{128} \, d{\left (\frac{8 \,{\left (3 \, \sqrt{d x} b d^{5} x^{3} - \sqrt{d x} a d^{5} x\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 )} + \frac{6 \, \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^{2} b^{4} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac{6 \, \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^{2} b^{4} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} - \frac{3 \, \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^{2} b^{4} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac{3 \, \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^{2} b^{4} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/128*d*(8*(3*sqrt(d*x)*b*d^5*x^3 - sqrt(d*x)*a*d^5*x)/((b*d^2*x^2 + a*d^2)^2*a*b*sgn(b*d^4*x^2 + a*d^4)) + 6*
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^2*b^4
*sgn(b*d^4*x^2 + a*d^4)) + 6*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^2*b^4*sgn(b*d^4*x^2 + a*d^4)) - 3*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^2*b^4*sgn(b*d^4*x^2 + a*d^4)) + 3*sqrt(2)*(a*b^3*d^2)^(3/4)*log(d*x - s
qrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/(a^2*b^4*sgn(b*d^4*x^2 + a*d^4)))