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

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

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Rubi [A]  time = 0.33, antiderivative size = 459, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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 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 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)^{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*}

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Mathematica [C]  time = 0.03, size = 73, normalized size = 0.16 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 70.17, size = 245, normalized size = 0.53 \begin {gather*} \frac {\left (a d^2+b d^2 x^2\right ) \left (-\frac {3 d^{5/2} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{a} \sqrt {d}}{\sqrt {2} \sqrt [4]{b}}-\frac {\sqrt [4]{b} \sqrt {d} x}{\sqrt {2} \sqrt [4]{a}}}{\sqrt {d x}}\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}-\frac {3 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d} \sqrt {d x}}{\sqrt {a} d+\sqrt {b} d x}\right )}{32 \sqrt {2} a^{5/4} b^{7/4}}+\frac {3 b d^3 (d x)^{7/2}-a d^5 (d x)^{3/2}}{16 a b \left (a d^2+b d^2 x^2\right )^2}\right )}{d^2 \sqrt {\frac {\left (a d^2+b d^2 x^2\right )^2}{d^4}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 326, normalized size = 0.71 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.32, size = 383, normalized size = 0.83 \begin {gather*} \frac {1}{128} \, d^{2} {\left (\frac {8 \, {\left (3 \, \sqrt {d x} b d^{4} x^{3} - \sqrt {d x} a d^{4} 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} d \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} d \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} d \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} d \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )}\right )} \end {gather*}

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^2*(8*(3*sqrt(d*x)*b*d^4*x^3 - sqrt(d*x)*a*d^4*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*d*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*sq
rt(d*x))/(a*d^2/b)^(1/4))/(a^2*b^4*d*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*d*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*d*sgn(b*d^4*x^2 + a*d^4)))

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maple [B]  time = 0.02, size = 617, normalized size = 1.34 \begin {gather*} \frac {\left (6 \sqrt {2}\, b^{2} d^{4} x^{4} \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 )+6 \sqrt {2}\, b^{2} d^{4} x^{4} \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 )+3 \sqrt {2}\, b^{2} d^{4} x^{4} \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 )+12 \sqrt {2}\, a b \,d^{4} x^{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 )+12 \sqrt {2}\, a b \,d^{4} x^{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 )+6 \sqrt {2}\, a b \,d^{4} x^{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 )+6 \sqrt {2}\, a^{2} d^{4} \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 )+6 \sqrt {2}\, a^{2} d^{4} \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 )+3 \sqrt {2}\, a^{2} d^{4} \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 )-8 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (d x \right )^{\frac {3}{2}} a b \,d^{2}+24 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (d x \right )^{\frac {7}{2}} b^{2}\right ) \left (b \,x^{2}+a \right )}{128 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} a \,b^{2} d} \end {gather*}

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)*b^2*d^4*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)))+6*2^(1/2)*b^2*d^4*x^4*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a
/b*d^2)^(1/4))+6*2^(1/2)*b^2*d^4*x^4*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+24*(a/b*d^2
)^(1/4)*(d*x)^(7/2)*b^2+6*2^(1/2)*a*b*d^4*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)))+12*2^(1/2)*a*b*d^4*x^2*arctan((2^(1/2)*(d*x)^(1/2)+(
a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+12*2^(1/2)*a*b*d^4*x^2*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)
^(1/4))-8*(a/b*d^2)^(1/4)*(d*x)^(3/2)*a*b*d^2+3*2^(1/2)*a^2*d^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)))+6*2^(1/2)*a^2*d^4*arctan((2^(1/2)*
(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+6*2^(1/2)*a^2*d^4*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(
a/b*d^2)^(1/4)))/d*(b*x^2+a)/(a/b*d^2)^(1/4)/b^2/a/((b*x^2+a)^2)^(3/2)

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maxima [A]  time = 3.19, size = 272, normalized size = 0.59 \begin {gather*} -\frac {d^{\frac {5}{2}} x^{\frac {3}{2}}}{2 \, {\left (a b^{2} x^{2} + a^{2} b + {\left (b^{3} x^{2} + a b^{2}\right )} x^{2}\right )}} + \frac {3 \, d^{\frac {5}{2}} {\left (\frac {2 \, \sqrt {2} \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 {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \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 {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{128 \, a b} + \frac {3 \, b d^{\frac {5}{2}} x^{\frac {7}{2}} + 7 \, a d^{\frac {5}{2}} x^{\frac {3}{2}}}{16 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} \end {gather*}

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]

-1/2*d^(5/2)*x^(3/2)/(a*b^2*x^2 + a^2*b + (b^3*x^2 + a*b^2)*x^2) + 3/128*d^(5/2)*(2*sqrt(2)*arctan(1/2*sqrt(2)
*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt
(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqr
t(b))*sqrt(b)) - sqrt(2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2
)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/(a*b) + 1/16*(3*b*d^(5/2)*x^(
7/2) + 7*a*d^(5/2)*x^(3/2))/(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d\,x\right )}^{5/2}}{{\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)^(5/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int((d*x)^(5/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d x\right )^{\frac {5}{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)**(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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