3.8.59 \(\int \frac {(d x)^{11/2}}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [759]

Optimal. Leaf size=504 \[ -\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 d^5 \sqrt {d x} \left (a+b x^2\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 \sqrt [4]{a} d^{11/2} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{32 \sqrt {2} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {45 \sqrt [4]{a} d^{11/2} \left (a+b x^2\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{32 \sqrt {2} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 \sqrt [4]{a} d^{11/2} \left (a+b x^2\right ) \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{64 \sqrt {2} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {45 \sqrt [4]{a} d^{11/2} \left (a+b x^2\right ) \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{64 \sqrt {2} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

-9/16*d^3*(d*x)^(5/2)/b^2/((b*x^2+a)^2)^(1/2)-1/4*d*(d*x)^(9/2)/b/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+45/64*a^(1/4)*
d^(11/2)*(b*x^2+a)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/b^(13/4)*2^(1/2)/((b*x^2+a)^2)^(1/2)-
45/64*a^(1/4)*d^(11/2)*(b*x^2+a)*arctan(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/b^(13/4)*2^(1/2)/((b*x^
2+a)^2)^(1/2)+45/128*a^(1/4)*d^(11/2)*(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))/b^(13/4)*2^(1/2)/((b*x^2+a)^2)^(1/2)-45/128*a^(1/4)*d^(11/2)*(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))/b^(13/4)*2^(1/2)/((b*x^2+a)^2)^(1/2)+45/16*d^5*(b*x^2+a)*(d*x)^
(1/2)/b^3/((b*x^2+a)^2)^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 504, normalized size of antiderivative = 1.00, 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 = {1126, 294, 327, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {45 \sqrt [4]{a} d^{11/2} \left (a+b x^2\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{32 \sqrt {2} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {45 \sqrt [4]{a} d^{11/2} \left (a+b x^2\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{32 \sqrt {2} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 \sqrt [4]{a} d^{11/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {45 \sqrt [4]{a} d^{11/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 d^5 \sqrt {d x} \left (a+b x^2\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(11/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(-9*d^3*(d*x)^(5/2))/(16*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (d*(d*x)^(9/2))/(4*b*(a + b*x^2)*Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4]) + (45*d^5*Sqrt[d*x]*(a + b*x^2))/(16*b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (45*a^(1/4)*d^
(11/2)*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*b^(13/4)*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4]) - (45*a^(1/4)*d^(11/2)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[
d])])/(32*Sqrt[2]*b^(13/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (45*a^(1/4)*d^(11/2)*(a + b*x^2)*Log[Sqrt[a]*Sqr
t[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(64*Sqrt[2]*b^(13/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2
*x^4]) - (45*a^(1/4)*d^(11/2)*(a + b*x^2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sq
rt[d*x]])/(64*Sqrt[2]*b^(13/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

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 294

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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

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 631

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 642

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 1126

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 1176

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 1179

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)^{11/2}}{\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)^{11/2}}{\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)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (9 d^2 \left (a b+b^2 x^2\right )\right ) \int \frac {(d x)^{7/2}}{\left (a b+b^2 x^2\right )^2} \, dx}{8 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (45 d^4 \left (a b+b^2 x^2\right )\right ) \int \frac {(d x)^{3/2}}{a b+b^2 x^2} \, dx}{32 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 d^5 \sqrt {d x} \left (a+b x^2\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (45 a d^6 \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^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 d^5 \sqrt {d x} \left (a+b x^2\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (45 a d^5 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \frac {1}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 d^5 \sqrt {d x} \left (a+b x^2\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (45 \sqrt {a} d^4 \left (a b+b^2 x^2\right )\right ) \text {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 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (45 \sqrt {a} d^4 \left (a b+b^2 x^2\right )\right ) \text {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 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 d^5 \sqrt {d x} \left (a+b x^2\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (45 \sqrt [4]{a} d^{11/2} \left (a b+b^2 x^2\right )\right ) \text {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} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (45 \sqrt [4]{a} d^{11/2} \left (a b+b^2 x^2\right )\right ) \text {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} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (45 \sqrt {a} d^6 \left (a b+b^2 x^2\right )\right ) \text {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 b^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (45 \sqrt {a} d^6 \left (a b+b^2 x^2\right )\right ) \text {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 b^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 d^5 \sqrt {d x} \left (a+b x^2\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 \sqrt [4]{a} d^{11/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {45 \sqrt [4]{a} d^{11/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (45 \sqrt [4]{a} d^{11/2} \left (a b+b^2 x^2\right )\right ) \text {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} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (45 \sqrt [4]{a} d^{11/2} \left (a b+b^2 x^2\right )\right ) \text {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} b^{17/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {9 d^3 (d x)^{5/2}}{16 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d (d x)^{9/2}}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 d^5 \sqrt {d x} \left (a+b x^2\right )}{16 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 \sqrt [4]{a} d^{11/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {45 \sqrt [4]{a} d^{11/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {45 \sqrt [4]{a} d^{11/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {45 \sqrt [4]{a} d^{11/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 195, normalized size = 0.39 \begin {gather*} \frac {d^5 \sqrt {d x} \left (4 \sqrt [4]{b} \sqrt {x} \left (45 a^2+81 a b x^2+32 b^2 x^4\right )+45 \sqrt {2} \sqrt [4]{a} \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )-45 \sqrt {2} \sqrt [4]{a} \left (a+b x^2\right )^2 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )\right )}{64 b^{13/4} \sqrt {x} \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(11/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(d^5*Sqrt[d*x]*(4*b^(1/4)*Sqrt[x]*(45*a^2 + 81*a*b*x^2 + 32*b^2*x^4) + 45*Sqrt[2]*a^(1/4)*(a + b*x^2)^2*ArcTan
[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] - 45*Sqrt[2]*a^(1/4)*(a + b*x^2)^2*ArcTanh[(Sqrt[2]*
a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)]))/(64*b^(13/4)*Sqrt[x]*(a + b*x^2)*Sqrt[(a + b*x^2)^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(695\) vs. \(2(323)=646\).
time = 0.07, size = 696, normalized size = 1.38

method	result	size
risch	\(\frac {2 x \,d^{6} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b^{3} \sqrt {d x}\, \left (b \,x^{2}+a \right )}+\frac {\left (\frac {17 a d \left (d x \right )^{\frac {5}{2}}}{16 b^{2} \left (d^{2} x^{2} b +a \,d^{2}\right )^{2}}+\frac {13 a^{2} d^{3} \sqrt {d x}}{16 b^{3} \left (d^{2} x^{2} b +a \,d^{2}\right )^{2}}-\frac {45 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{128 b^{3} d}-\frac {45 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{64 b^{3} d}-\frac {45 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{64 b^{3} d}\right ) d^{6} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(299\)
default	\(\frac {\left (-45 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) b^{2} d^{2} x^{4}-90 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) b^{2} d^{2} x^{4}-90 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) b^{2} d^{2} x^{4}+256 \sqrt {d x}\, b^{2} d^{2} x^{4}-90 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) a b \,d^{2} x^{2}-180 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) a b \,d^{2} x^{2}-180 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) a b \,d^{2} x^{2}+136 \left (d x \right )^{\frac {5}{2}} a b +512 \sqrt {d x}\, a b \,d^{2} x^{2}-45 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) a^{2} d^{2}-90 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) a^{2} d^{2}-90 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {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 ) a^{2} d^{2}+360 \sqrt {d x}\, a^{2} d^{2}\right ) d^{3} \left (b \,x^{2}+a \right )}{128 b^{3} \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}\)	\(696\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(11/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/128*(-45*(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)))*b^2*d^2*x^4-90*(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))*b^2*d^2*x^4-90*(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))*b^2*d^2*x^4+256*(d*x)^(1/2)*b^2*d^2*x^4-90*(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*b*d^
2*x^2-180*(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*b*d^2*x^2-18
0*(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*b*d^2*x^2+136*(d*x)^
(5/2)*a*b+512*(d*x)^(1/2)*a*b*d^2*x^2-45*(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^2-90*(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^2-90*(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+360*(d*x)^(1/2)*a^2*d^2)*d^3*(b*x^2+a)/b^3/((b*x^2+a)
^2)^(3/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(11/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*a*d^(11/2)*x^(5/2)/(a*b^3*x^2 + a^2*b^2 + (b^4*x^2 + a*b^3)*x^2) + d^(11/2)*integrate(x^(3/2)/(b^3*x^2 + a
*b^2), x) - 13/128*(2*sqrt(2)*sqrt(a)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sq
rt(a)*sqrt(b)))/sqrt(sqrt(a)*sqrt(b)) + 2*sqrt(2)*sqrt(a)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqr
t(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/sqrt(sqrt(a)*sqrt(b)) + sqrt(2)*a^(1/4)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(
x) + sqrt(b)*x + sqrt(a))/b^(1/4) - sqrt(2)*a^(1/4)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a)
)/b^(1/4))*d^(11/2)/b^3 + 1/16*(9*a*b*d^(11/2)*x^(5/2) + 13*a^2*d^(11/2)*sqrt(x))/(b^5*x^4 + 2*a*b^4*x^2 + a^2
*b^3)

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Fricas [A]
time = 0.37, size = 305, normalized size = 0.61 \begin {gather*} -\frac {180 \, \left (-\frac {a d^{22}}{b^{13}}\right )^{\frac {1}{4}} {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )} \arctan \left (-\frac {\left (-\frac {a d^{22}}{b^{13}}\right )^{\frac {3}{4}} \sqrt {d x} b^{10} d^{5} - \sqrt {d^{11} x + \sqrt {-\frac {a d^{22}}{b^{13}}} b^{6}} \left (-\frac {a d^{22}}{b^{13}}\right )^{\frac {3}{4}} b^{10}}{a d^{22}}\right ) + 45 \, \left (-\frac {a d^{22}}{b^{13}}\right )^{\frac {1}{4}} {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )} \log \left (45 \, \sqrt {d x} d^{5} + 45 \, \left (-\frac {a d^{22}}{b^{13}}\right )^{\frac {1}{4}} b^{3}\right ) - 45 \, \left (-\frac {a d^{22}}{b^{13}}\right )^{\frac {1}{4}} {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )} \log \left (45 \, \sqrt {d x} d^{5} - 45 \, \left (-\frac {a d^{22}}{b^{13}}\right )^{\frac {1}{4}} b^{3}\right ) - 4 \, {\left (32 \, b^{2} d^{5} x^{4} + 81 \, a b d^{5} x^{2} + 45 \, a^{2} d^{5}\right )} \sqrt {d x}}{64 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(11/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/64*(180*(-a*d^22/b^13)^(1/4)*(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*arctan(-((-a*d^22/b^13)^(3/4)*sqrt(d*x)*b^10
*d^5 - sqrt(d^11*x + sqrt(-a*d^22/b^13)*b^6)*(-a*d^22/b^13)^(3/4)*b^10)/(a*d^22)) + 45*(-a*d^22/b^13)^(1/4)*(b
^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*log(45*sqrt(d*x)*d^5 + 45*(-a*d^22/b^13)^(1/4)*b^3) - 45*(-a*d^22/b^13)^(1/4)*
(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*log(45*sqrt(d*x)*d^5 - 45*(-a*d^22/b^13)^(1/4)*b^3) - 4*(32*b^2*d^5*x^4 + 81
*a*b*d^5*x^2 + 45*a^2*d^5)*sqrt(d*x))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d x\right )^{\frac {11}{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)**(11/2)/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral((d*x)**(11/2)/((a + b*x**2)**2)**(3/2), x)

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Giac [A]
time = 3.62, size = 343, normalized size = 0.68 \begin {gather*} -\frac {1}{128} \, d^{5} {\left (\frac {90 \, \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 )}{b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {90 \, \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 )}{b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {45 \, \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 )}{b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {45 \, \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 )}{b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {256 \, \sqrt {d x}}{b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {8 \, {\left (17 \, \sqrt {d x} a b d^{4} x^{2} + 13 \, \sqrt {d x} a^{2} d^{4}\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{2} b^{3} \mathrm {sgn}\left (b x^{2} + a\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(11/2)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

-1/128*d^5*(90*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))/(b^4*sgn(b*x^2 + a)) + 90*sqrt(2)*(a*b^3*d^2)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4) - 2*sq
rt(d*x))/(a*d^2/b)^(1/4))/(b^4*sgn(b*x^2 + a)) + 45*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))/(b^4*sgn(b*x^2 + a)) - 45*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))/(b^4*sgn(b*x^2 + a)) - 256*sqrt(d*x)/(b^3*sgn(b*x^2 + a)) - 8*(17*sqrt(d*x)*a*
b*d^4*x^2 + 13*sqrt(d*x)*a^2*d^4)/((b*d^2*x^2 + a*d^2)^2*b^3*sgn(b*x^2 + a)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d\,x\right )}^{11/2}}{{\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)^(11/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int((d*x)^(11/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)

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