3.761 \(\int \frac{(d x)^{7/2}}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=458 \[ -\frac{5 d^3 \sqrt{d x}}{16 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 d^{7/2} \left (a+b x^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 d^{7/2} \left (a+b x^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 d^{7/2} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 d^{7/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{32 \sqrt{2} a^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

(-5*d^3*Sqrt[d*x])/(16*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (d*(d*x)^(5/2))/(4*b*(a + b*x^2)*Sqrt[a^2 + 2*a*
b*x^2 + b^2*x^4]) - (5*d^(7/2)*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt
[2]*a^(3/4)*b^(9/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*d^(7/2)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt
[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(3/4)*b^(9/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (5*d^(7/2)*(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^(3/4)*b^(9/4)*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*d^(7/2)*(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^(3/4)*b^(9/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.322187, antiderivative size = 458, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1112, 288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{5 d^3 \sqrt{d x}}{16 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 d^{7/2} \left (a+b x^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 d^{7/2} \left (a+b x^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 d^{7/2} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 d^{7/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{32 \sqrt{2} a^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d^3*Sqrt[d*x])/(16*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (d*(d*x)^(5/2))/(4*b*(a + b*x^2)*Sqrt[a^2 + 2*a*
b*x^2 + b^2*x^4]) - (5*d^(7/2)*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt
[2]*a^(3/4)*b^(9/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*d^(7/2)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt
[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(3/4)*b^(9/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (5*d^(7/2)*(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^(3/4)*b^(9/4)*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*d^(7/2)*(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^(3/4)*b^(9/4)*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 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 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 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]

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])

Rubi steps

\begin{align*} \int \frac{(d x)^{7/2}}{\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)^{7/2}}{\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)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (5 d^2 \left (a b+b^2 x^2\right )\right ) \int \frac{(d x)^{3/2}}{\left (a b+b^2 x^2\right )^2} \, dx}{8 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{5 d^3 \sqrt{d x}}{16 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (5 d^4 \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 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{5 d^3 \sqrt{d x}}{16 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (5 d^3 \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 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{5 d^3 \sqrt{d x}}{16 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (5 d^2 \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 \sqrt{a} b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (5 d^2 \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 \sqrt{a} b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{5 d^3 \sqrt{d x}}{16 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (5 d^{7/2} \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^{3/4} b^{13/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (5 d^{7/2} \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^{3/4} b^{13/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (5 d^4 \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 \sqrt{a} b^{7/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (5 d^4 \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 \sqrt{a} b^{7/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{5 d^3 \sqrt{d x}}{16 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 d^{7/2} \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^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 d^{7/2} \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^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (5 d^{7/2} \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^{3/4} b^{13/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (5 d^{7/2} \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^{3/4} b^{13/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{5 d^3 \sqrt{d x}}{16 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d (d x)^{5/2}}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 d^{7/2} \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^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 d^{7/2} \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^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 d^{7/2} \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^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 d^{7/2} \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^{3/4} b^{9/4} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.150843, size = 447, normalized size = 0.98 \[ -\frac{5 (d x)^{7/2} \left (a+b x^2\right )^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4} x^{7/2} \left (\left (a+b x^2\right )^2\right )^{3/2}}+\frac{5 (d x)^{7/2} \left (a+b x^2\right )^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4} x^{7/2} \left (\left (a+b x^2\right )^2\right )^{3/2}}-\frac{5 (d x)^{7/2} \left (a+b x^2\right )^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4} x^{7/2} \left (\left (a+b x^2\right )^2\right )^{3/2}}+\frac{5 (d x)^{7/2} \left (a+b x^2\right )^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{3/4} b^{9/4} x^{7/2} \left (\left (a+b x^2\right )^2\right )^{3/2}}+\frac{5 (d x)^{7/2} \left (a+b x^2\right )^2}{48 b^2 x^3 \left (\left (a+b x^2\right )^2\right )^{3/2}}-\frac{5 a (d x)^{7/2} \left (a+b x^2\right )}{12 b^2 x^3 \left (\left (a+b x^2\right )^2\right )^{3/2}}-\frac{2 (d x)^{7/2} \left (a+b x^2\right )}{3 b x \left (\left (a+b x^2\right )^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*(d*x)^(7/2)*(a + b*x^2))/(12*b^2*x^3*((a + b*x^2)^2)^(3/2)) - (2*(d*x)^(7/2)*(a + b*x^2))/(3*b*x*((a + b
*x^2)^2)^(3/2)) + (5*(d*x)^(7/2)*(a + b*x^2)^2)/(48*b^2*x^3*((a + b*x^2)^2)^(3/2)) - (5*(d*x)^(7/2)*(a + b*x^2
)^3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b^(9/4)*x^(7/2)*((a + b*x^2)^2)^(3/2))
+ (5*(d*x)^(7/2)*(a + b*x^2)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b^(9/4)*x^(7
/2)*((a + b*x^2)^2)^(3/2)) - (5*(d*x)^(7/2)*(a + b*x^2)^3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt
[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(9/4)*x^(7/2)*((a + b*x^2)^2)^(3/2)) + (5*(d*x)^(7/2)*(a + b*x^2)^3*Log[Sqrt[a]
+ Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(9/4)*x^(7/2)*((a + b*x^2)^2)^(3/2))

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Maple [B]  time = 0.238, size = 666, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/128*(5*(a*d^2/b)^(1/4)*2^(1/2)*ln((d*x+(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(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^2+10*(a*d^2/b)^(1/4)*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^2+10*(a*d^2/b)^(1/4)*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^2+10*(a*d^2/b)^(1/4)*2^(1/2)*ln((d*x+(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(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^2+20*(a*d^2/b)^(1/4)*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^2+20*(a*d^2/b)^(1/4)*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^2+5*(a*d^2/b)^(1/4)*2^(1/2)*ln((d*x+(a*d^2/b
)^(1/4)*(d*x)^(1/2)*2^(1/2)+(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^
2+10*(a*d^2/b)^(1/4)*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^2+10*(a*d^2/b
)^(1/4)*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^2-72*(d*x)^(5/2)*a*b-40*(d
*x)^(1/2)*a^2*d^2)*d*(b*x^2+a)/a/b^2/((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)^(7/2)/(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 = 1.67061, size = 695, normalized size = 1.52 \begin{align*} \frac{20 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac{d^{14}}{a^{3} b^{9}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\left (-\frac{d^{14}}{a^{3} b^{9}}\right )^{\frac{3}{4}} \sqrt{d x} a^{2} b^{7} d^{3} - \sqrt{d^{7} x + \sqrt{-\frac{d^{14}}{a^{3} b^{9}}} a^{2} b^{4}} \left (-\frac{d^{14}}{a^{3} b^{9}}\right )^{\frac{3}{4}} a^{2} b^{7}}{d^{14}}\right ) + 5 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac{d^{14}}{a^{3} b^{9}}\right )^{\frac{1}{4}} \log \left (5 \, \sqrt{d x} d^{3} + 5 \, \left (-\frac{d^{14}}{a^{3} b^{9}}\right )^{\frac{1}{4}} a b^{2}\right ) - 5 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac{d^{14}}{a^{3} b^{9}}\right )^{\frac{1}{4}} \log \left (5 \, \sqrt{d x} d^{3} - 5 \, \left (-\frac{d^{14}}{a^{3} b^{9}}\right )^{\frac{1}{4}} a b^{2}\right ) - 4 \,{\left (9 \, b d^{3} x^{2} + 5 \, a d^{3}\right )} \sqrt{d x}}{64 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/64*(20*(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)*(-d^14/(a^3*b^9))^(1/4)*arctan(-((-d^14/(a^3*b^9))^(3/4)*sqrt(d*x)*
a^2*b^7*d^3 - sqrt(d^7*x + sqrt(-d^14/(a^3*b^9))*a^2*b^4)*(-d^14/(a^3*b^9))^(3/4)*a^2*b^7)/d^14) + 5*(b^4*x^4
+ 2*a*b^3*x^2 + a^2*b^2)*(-d^14/(a^3*b^9))^(1/4)*log(5*sqrt(d*x)*d^3 + 5*(-d^14/(a^3*b^9))^(1/4)*a*b^2) - 5*(b
^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)*(-d^14/(a^3*b^9))^(1/4)*log(5*sqrt(d*x)*d^3 - 5*(-d^14/(a^3*b^9))^(1/4)*a*b^2)
- 4*(9*b*d^3*x^2 + 5*a*d^3)*sqrt(d*x))/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{\frac{7}{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)**(7/2)/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral((d*x)**(7/2)/((a + b*x**2)**2)**(3/2), x)

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Giac [A]  time = 1.38142, size = 501, normalized size = 1.09 \begin{align*} \frac{1}{128} \, d^{2}{\left (\frac{10 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{1}{4}} d \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 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{1}{4}} d \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 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{1}{4}} d \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 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{1}{4}} d \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 b^{3} \mathrm{sgn}\left (b d^{4} x^{2} + a d^{4}\right )} - \frac{8 \,{\left (9 \, \sqrt{d x} b d^{5} x^{2} + 5 \, \sqrt{d x} a d^{5}\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{2} b^{2} \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)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

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