3.773 \(\int \frac {(d x)^{15/2}}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=554 \[ -\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {195 d^7 \sqrt {d x}}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {195 d^{15/2} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{4096 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {195 d^{15/2} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{4096 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {195 d^{15/2} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{2048 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {195 d^{15/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{2048 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

-1/8*d*(d*x)^(13/2)/b/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2)-13/96*d^3*(d*x)^(9/2)/b^2/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2
)-39/256*d^5*(d*x)^(5/2)/b^3/(b*x^2+a)/((b*x^2+a)^2)^(1/2)-195/4096*d^(15/2)*(b*x^2+a)*arctan(1-b^(1/4)*2^(1/2
)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(3/4)/b^(17/4)*2^(1/2)/((b*x^2+a)^2)^(1/2)+195/4096*d^(15/2)*(b*x^2+a)*arctan
(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(3/4)/b^(17/4)*2^(1/2)/((b*x^2+a)^2)^(1/2)-195/8192*d^(15/2)
*(b*x^2+a)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1/2))/a^(3/4)/b^(17/4)*2^(1/2)/
((b*x^2+a)^2)^(1/2)+195/8192*d^(15/2)*(b*x^2+a)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*(
d*x)^(1/2))/a^(3/4)/b^(17/4)*2^(1/2)/((b*x^2+a)^2)^(1/2)-195/1024*d^7*(d*x)^(1/2)/b^4/((b*x^2+a)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.42, antiderivative size = 554, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1112, 288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {195 d^7 \sqrt {d x}}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {195 d^{15/2} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{4096 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {195 d^{15/2} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{4096 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {195 d^{15/2} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{2048 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {195 d^{15/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{2048 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(15/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(-195*d^7*Sqrt[d*x])/(1024*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (d*(d*x)^(13/2))/(8*b*(a + b*x^2)^3*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4]) - (13*d^3*(d*x)^(9/2))/(96*b^2*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (39*d
^5*(d*x)^(5/2))/(256*b^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (195*d^(15/2)*(a + b*x^2)*ArcTan[1 - (
Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(2048*Sqrt[2]*a^(3/4)*b^(17/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
+ (195*d^(15/2)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(2048*Sqrt[2]*a^(3/4)*
b^(17/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (195*d^(15/2)*(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]])/(4096*Sqrt[2]*a^(3/4)*b^(17/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (195*d
^(15/2)*(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]])/(4096*Sqrt[2
]*a^(3/4)*b^(17/4)*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 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)^{15/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x^2\right )\right ) \int \frac {(d x)^{15/2}}{\left (a b+b^2 x^2\right )^5} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (13 b^2 d^2 \left (a b+b^2 x^2\right )\right ) \int \frac {(d x)^{11/2}}{\left (a b+b^2 x^2\right )^4} \, dx}{16 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (39 d^4 \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}{64 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (195 d^6 \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}{512 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {195 d^7 \sqrt {d x}}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (195 d^8 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )} \, dx}{2048 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {195 d^7 \sqrt {d x}}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (195 d^7 \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 )}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {195 d^7 \sqrt {d x}}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (195 d^6 \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 )}{2048 \sqrt {a} b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (195 d^6 \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 )}{2048 \sqrt {a} b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {195 d^7 \sqrt {d x}}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (195 d^{15/2} \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 )}{4096 \sqrt {2} a^{3/4} b^{21/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (195 d^{15/2} \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 )}{4096 \sqrt {2} a^{3/4} b^{21/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (195 d^8 \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 )}{4096 \sqrt {a} b^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (195 d^8 \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 )}{4096 \sqrt {a} b^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {195 d^7 \sqrt {d x}}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {195 d^{15/2} \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 )}{4096 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {195 d^{15/2} \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 )}{4096 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (195 d^{15/2} \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 )}{2048 \sqrt {2} a^{3/4} b^{21/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (195 d^{15/2} \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 )}{2048 \sqrt {2} a^{3/4} b^{21/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {195 d^7 \sqrt {d x}}{1024 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{13/2}}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {13 d^3 (d x)^{9/2}}{96 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {39 d^5 (d x)^{5/2}}{256 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {195 d^{15/2} \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 )}{2048 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {195 d^{15/2} \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 )}{2048 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {195 d^{15/2} \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 )}{4096 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {195 d^{15/2} \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 )}{4096 \sqrt {2} a^{3/4} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 366, normalized size = 0.66 \[ \frac {(d x)^{15/2} \left (a+b x^2\right ) \left (-\frac {45045 \sqrt {2} \left (a+b x^2\right )^4 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4}}+\frac {45045 \sqrt {2} \left (a+b x^2\right )^4 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4}}-\frac {90090 \sqrt {2} \left (a+b x^2\right )^4 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac {90090 \sqrt {2} \left (a+b x^2\right )^4 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{a^{3/4}}-599040 a^3 \sqrt [4]{b} \sqrt {x}-1916928 a^2 b^{5/4} x^{5/2}+49920 a^2 \sqrt [4]{b} \sqrt {x} \left (a+b x^2\right )-2342912 a b^{9/4} x^{9/2}+120120 \sqrt [4]{b} \sqrt {x} \left (a+b x^2\right )^3+68640 a \sqrt [4]{b} \sqrt {x} \left (a+b x^2\right )^2-1261568 b^{13/4} x^{13/2}\right )}{1892352 b^{17/4} x^{15/2} \left (\left (a+b x^2\right )^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(15/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

((d*x)^(15/2)*(a + b*x^2)*(-599040*a^3*b^(1/4)*Sqrt[x] - 1916928*a^2*b^(5/4)*x^(5/2) - 2342912*a*b^(9/4)*x^(9/
2) - 1261568*b^(13/4)*x^(13/2) + 49920*a^2*b^(1/4)*Sqrt[x]*(a + b*x^2) + 68640*a*b^(1/4)*Sqrt[x]*(a + b*x^2)^2
+ 120120*b^(1/4)*Sqrt[x]*(a + b*x^2)^3 - (90090*Sqrt[2]*(a + b*x^2)^4*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^
(1/4)])/a^(3/4) + (90090*Sqrt[2]*(a + b*x^2)^4*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) - (45045
*Sqrt[2]*(a + b*x^2)^4*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4) + (45045*Sqrt[2]*(a
+ b*x^2)^4*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4)))/(1892352*b^(17/4)*x^(15/2)*(
(a + b*x^2)^2)^(5/2))

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fricas [A]  time = 1.07, size = 431, normalized size = 0.78 \[ \frac {2340 \, {\left (b^{8} x^{8} + 4 \, a b^{7} x^{6} + 6 \, a^{2} b^{6} x^{4} + 4 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )} \left (-\frac {d^{30}}{a^{3} b^{17}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\left (-\frac {d^{30}}{a^{3} b^{17}}\right )^{\frac {3}{4}} \sqrt {d x} a^{2} b^{13} d^{7} - \sqrt {d^{15} x + \sqrt {-\frac {d^{30}}{a^{3} b^{17}}} a^{2} b^{8}} \left (-\frac {d^{30}}{a^{3} b^{17}}\right )^{\frac {3}{4}} a^{2} b^{13}}{d^{30}}\right ) + 585 \, {\left (b^{8} x^{8} + 4 \, a b^{7} x^{6} + 6 \, a^{2} b^{6} x^{4} + 4 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )} \left (-\frac {d^{30}}{a^{3} b^{17}}\right )^{\frac {1}{4}} \log \left (195 \, \sqrt {d x} d^{7} + 195 \, \left (-\frac {d^{30}}{a^{3} b^{17}}\right )^{\frac {1}{4}} a b^{4}\right ) - 585 \, {\left (b^{8} x^{8} + 4 \, a b^{7} x^{6} + 6 \, a^{2} b^{6} x^{4} + 4 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )} \left (-\frac {d^{30}}{a^{3} b^{17}}\right )^{\frac {1}{4}} \log \left (195 \, \sqrt {d x} d^{7} - 195 \, \left (-\frac {d^{30}}{a^{3} b^{17}}\right )^{\frac {1}{4}} a b^{4}\right ) - 4 \, {\left (1853 \, b^{3} d^{7} x^{6} + 3107 \, a b^{2} d^{7} x^{4} + 2223 \, a^{2} b d^{7} x^{2} + 585 \, a^{3} d^{7}\right )} \sqrt {d x}}{12288 \, {\left (b^{8} x^{8} + 4 \, a b^{7} x^{6} + 6 \, a^{2} b^{6} x^{4} + 4 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(15/2)/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/12288*(2340*(b^8*x^8 + 4*a*b^7*x^6 + 6*a^2*b^6*x^4 + 4*a^3*b^5*x^2 + a^4*b^4)*(-d^30/(a^3*b^17))^(1/4)*arcta
n(-((-d^30/(a^3*b^17))^(3/4)*sqrt(d*x)*a^2*b^13*d^7 - sqrt(d^15*x + sqrt(-d^30/(a^3*b^17))*a^2*b^8)*(-d^30/(a^
3*b^17))^(3/4)*a^2*b^13)/d^30) + 585*(b^8*x^8 + 4*a*b^7*x^6 + 6*a^2*b^6*x^4 + 4*a^3*b^5*x^2 + a^4*b^4)*(-d^30/
(a^3*b^17))^(1/4)*log(195*sqrt(d*x)*d^7 + 195*(-d^30/(a^3*b^17))^(1/4)*a*b^4) - 585*(b^8*x^8 + 4*a*b^7*x^6 + 6
*a^2*b^6*x^4 + 4*a^3*b^5*x^2 + a^4*b^4)*(-d^30/(a^3*b^17))^(1/4)*log(195*sqrt(d*x)*d^7 - 195*(-d^30/(a^3*b^17)
)^(1/4)*a*b^4) - 4*(1853*b^3*d^7*x^6 + 3107*a*b^2*d^7*x^4 + 2223*a^2*b*d^7*x^2 + 585*a^3*d^7)*sqrt(d*x))/(b^8*
x^8 + 4*a*b^7*x^6 + 6*a^2*b^6*x^4 + 4*a^3*b^5*x^2 + a^4*b^4)

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giac [A]  time = 0.35, size = 405, normalized size = 0.73 \[ \frac {1}{24576} \, d^{7} {\left (\frac {1170 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{5} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {1170 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{5} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {585 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{5} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} - \frac {585 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{5} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} - \frac {8 \, {\left (1853 \, \sqrt {d x} b^{3} d^{8} x^{6} + 3107 \, \sqrt {d x} a b^{2} d^{8} x^{4} + 2223 \, \sqrt {d x} a^{2} b d^{8} x^{2} + 585 \, \sqrt {d x} a^{3} d^{8}\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{4} b^{4} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(15/2)/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/24576*d^7*(1170*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*b^5*sgn(b*d^4*x^2 + a*d^4)) + 1170*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*b^5*sgn(b*d^4*x^2 + a*d^4)) + 585*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*b^5*sgn(b*d^4*x^2 + a*d^4)) - 585*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*b^5*sgn(b*d^4*x^2 + a*d^4)) - 8*(1853*
sqrt(d*x)*b^3*d^8*x^6 + 3107*sqrt(d*x)*a*b^2*d^8*x^4 + 2223*sqrt(d*x)*a^2*b*d^8*x^2 + 585*sqrt(d*x)*a^3*d^8)/(
(b*d^2*x^2 + a*d^2)^4*b^4*sgn(b*d^4*x^2 + a*d^4)))

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maple [B]  time = 0.02, size = 1134, normalized size = 2.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(15/2)/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/24576*(585*(a/b*d^2)^(1/4)*2^(1/2)*b^4*d^6*x^8*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)))+1170*(a/b*d^2)^(1/4)*2^(1/2)*b^4*d^6*x^8*arctan((2^
(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+1170*(a/b*d^2)^(1/4)*2^(1/2)*b^4*d^6*x^8*arctan((2^(1/2)*(
d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+2340*(a/b*d^2)^(1/4)*2^(1/2)*a*b^3*d^6*x^6*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)))+4680*(a/b*d^
2)^(1/4)*2^(1/2)*a*b^3*d^6*x^6*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+4680*(a/b*d^2)^(1
/4)*2^(1/2)*a*b^3*d^6*x^6*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+3510*(a/b*d^2)^(1/4)*2
^(1/2)*a^2*b^2*d^6*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)))+7020*(a/b*d^2)^(1/4)*2^(1/2)*a^2*b^2*d^6*x^4*arctan((2^(1/2)*(d*x)^(1/2)+(a/b
*d^2)^(1/4))/(a/b*d^2)^(1/4))+7020*(a/b*d^2)^(1/4)*2^(1/2)*a^2*b^2*d^6*x^4*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^
2)^(1/4))/(a/b*d^2)^(1/4))-14824*(d*x)^(13/2)*a*b^3+2340*(a/b*d^2)^(1/4)*2^(1/2)*a^3*b*d^6*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)))+4680*
(a/b*d^2)^(1/4)*2^(1/2)*a^3*b*d^6*x^2*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+4680*(a/b*
d^2)^(1/4)*2^(1/2)*a^3*b*d^6*x^2*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))-24856*(d*x)^(9/
2)*a^2*b^2*d^2+585*(a/b*d^2)^(1/4)*2^(1/2)*a^4*d^6*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)))+1170*(a/b*d^2)^(1/4)*2^(1/2)*a^4*d^6*arctan((2^(1
/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+1170*(a/b*d^2)^(1/4)*2^(1/2)*a^4*d^6*arctan((2^(1/2)*(d*x)^(
1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))-17784*(d*x)^(5/2)*a^3*b*d^4-4680*(d*x)^(1/2)*a^4*d^6)*d*(b*x^2+a)/a/b^4
/((b*x^2+a)^2)^(5/2)

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maxima [A]  time = 3.76, size = 583, normalized size = 1.05 \[ \frac {195 \, d^{7} {\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 )}}{8192 \, b^{4}} - \frac {15 \, b^{3} d^{\frac {15}{2}} x^{\frac {13}{2}} + 65 \, a b^{2} d^{\frac {15}{2}} x^{\frac {9}{2}} + 117 \, a^{2} b d^{\frac {15}{2}} x^{\frac {5}{2}} + 195 \, a^{3} d^{\frac {15}{2}} \sqrt {x}}{1024 \, {\left (b^{8} x^{8} + 4 \, a b^{7} x^{6} + 6 \, a^{2} b^{6} x^{4} + 4 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}} - \frac {{\left (113 \, b^{4} d^{\frac {15}{2}} x^{5} + 282 \, a b^{3} d^{\frac {15}{2}} x^{3} + 201 \, a^{2} b^{2} d^{\frac {15}{2}} x\right )} x^{\frac {11}{2}} + 2 \, {\left (63 \, a b^{3} d^{\frac {15}{2}} x^{5} + 174 \, a^{2} b^{2} d^{\frac {15}{2}} x^{3} + 143 \, a^{3} b d^{\frac {15}{2}} x\right )} x^{\frac {7}{2}} + {\left (45 \, a^{2} b^{2} d^{\frac {15}{2}} x^{5} + 130 \, a^{3} b d^{\frac {15}{2}} x^{3} + 117 \, a^{4} d^{\frac {15}{2}} x\right )} x^{\frac {3}{2}}}{192 \, {\left (a^{3} b^{6} x^{6} + 3 \, a^{4} b^{5} x^{4} + 3 \, a^{5} b^{4} x^{2} + a^{6} b^{3} + {\left (b^{9} x^{6} + 3 \, a b^{8} x^{4} + 3 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )} x^{6} + 3 \, {\left (a b^{8} x^{6} + 3 \, a^{2} b^{7} x^{4} + 3 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )} x^{4} + 3 \, {\left (a^{2} b^{7} x^{6} + 3 \, a^{3} b^{6} x^{4} + 3 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(15/2)/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

195/8192*d^7*(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)))/b^4 - 1/1024*(15*b^3*d^(15/2)*x^(13/2) + 65*a*b^2*d^(15/2)*x^(9/2
) + 117*a^2*b*d^(15/2)*x^(5/2) + 195*a^3*d^(15/2)*sqrt(x))/(b^8*x^8 + 4*a*b^7*x^6 + 6*a^2*b^6*x^4 + 4*a^3*b^5*
x^2 + a^4*b^4) - 1/192*((113*b^4*d^(15/2)*x^5 + 282*a*b^3*d^(15/2)*x^3 + 201*a^2*b^2*d^(15/2)*x)*x^(11/2) + 2*
(63*a*b^3*d^(15/2)*x^5 + 174*a^2*b^2*d^(15/2)*x^3 + 143*a^3*b*d^(15/2)*x)*x^(7/2) + (45*a^2*b^2*d^(15/2)*x^5 +
130*a^3*b*d^(15/2)*x^3 + 117*a^4*d^(15/2)*x)*x^(3/2))/(a^3*b^6*x^6 + 3*a^4*b^5*x^4 + 3*a^5*b^4*x^2 + a^6*b^3
+ (b^9*x^6 + 3*a*b^8*x^4 + 3*a^2*b^7*x^2 + a^3*b^6)*x^6 + 3*(a*b^8*x^6 + 3*a^2*b^7*x^4 + 3*a^3*b^6*x^2 + a^4*b
^5)*x^4 + 3*(a^2*b^7*x^6 + 3*a^3*b^6*x^4 + 3*a^4*b^5*x^2 + a^5*b^4)*x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d\,x\right )}^{15/2}}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(15/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int((d*x)^(15/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(15/2)/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Timed out

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