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

Optimal. Leaf size=335 \[ \frac {5 d^{5/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}-\frac {5 d^{5/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}-\frac {5 d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {5 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3} \]

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Rubi [A]  time = 0.35, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {28, 288, 290, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {5 d^{5/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}-\frac {5 d^{5/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}-\frac {5 d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {5 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(d*x)^(3/2))/(6*b*(a + b*x^2)^3) + (d*(d*x)^(3/2))/(16*a*b*(a + b*x^2)^2) + (5*d*(d*x)^(3/2))/(64*a^2*b*(a
+ b*x^2)) - (5*d^(5/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(9/4)*b^(7/4
)) + (5*d^(5/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(9/4)*b^(7/4)) + (5*
d^(5/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*a^(9/4)*b^(
7/4)) - (5*d^(5/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*
a^(9/4)*b^(7/4))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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 )^2} \, dx &=b^4 \int \frac {(d x)^{5/2}}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {1}{4} \left (b^2 d^2\right ) \int \frac {\sqrt {d x}}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}+\frac {\left (5 b d^2\right ) \int \frac {\sqrt {d x}}{\left (a b+b^2 x^2\right )^2} \, dx}{32 a}\\ &=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac {\left (5 d^2\right ) \int \frac {\sqrt {d x}}{a b+b^2 x^2} \, dx}{128 a^2}\\ &=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {x^2}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{64 a^2}\\ &=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}-\frac {(5 d) \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 )}{128 a^2 \sqrt {b}}+\frac {(5 d) \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 )}{128 a^2 \sqrt {b}}\\ &=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac {\left (5 d^{5/2}\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 )}{256 \sqrt {2} a^{9/4} b^{7/4}}+\frac {\left (5 d^{5/2}\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 )}{256 \sqrt {2} a^{9/4} b^{7/4}}+\frac {\left (5 d^3\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 )}{256 a^2 b^2}+\frac {\left (5 d^3\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 )}{256 a^2 b^2}\\ &=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac {5 d^{5/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}-\frac {5 d^{5/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}+\frac {\left (5 d^{5/2}\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 )}{128 \sqrt {2} a^{9/4} b^{7/4}}-\frac {\left (5 d^{5/2}\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 )}{128 \sqrt {2} a^{9/4} b^{7/4}}\\ &=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}-\frac {5 d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {5 d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {5 d^{5/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}-\frac {5 d^{5/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 60, normalized size = 0.18 \begin {gather*} \frac {2 d (d x)^{3/2} \left (\left (a+b x^2\right )^3 \, _2F_1\left (\frac {3}{4},4;\frac {7}{4};-\frac {b x^2}{a}\right )-a^3\right )}{9 a^3 b \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(d*x)^(3/2)*(-a^3 + (a + b*x^2)^3*Hypergeometric2F1[3/4, 4, 7/4, -((b*x^2)/a)]))/(9*a^3*b*(a + b*x^2)^3)

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IntegrateAlgebraic [A]  time = 0.79, size = 213, normalized size = 0.64 \begin {gather*} -\frac {5 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 )}{128 \sqrt {2} a^{9/4} b^{7/4}}-\frac {5 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 )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(d x)^{3/2} \left (-5 a^2 d^7+42 a b d^7 x^2+15 b^2 d^7 x^4\right )}{192 a^2 b \left (a d^2+b d^2 x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^(3/2)*(-5*a^2*d^7 + 42*a*b*d^7*x^2 + 15*b^2*d^7*x^4))/(192*a^2*b*(a*d^2 + b*d^2*x^2)^3) - (5*d^(5/2)*Ar
cTan[((a^(1/4)*Sqrt[d])/(Sqrt[2]*b^(1/4)) - (b^(1/4)*Sqrt[d]*x)/(Sqrt[2]*a^(1/4)))/Sqrt[d*x]])/(128*Sqrt[2]*a^
(9/4)*b^(7/4)) - (5*d^(5/2)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x])/(Sqrt[a]*d + Sqrt[b]*d*x)])/(1
28*Sqrt[2]*a^(9/4)*b^(7/4))

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fricas [A]  time = 0.88, size = 396, normalized size = 1.18 \begin {gather*} -\frac {60 \, {\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \arctan \left (-\frac {125 \, \sqrt {d x} a^{2} b^{2} d^{7} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}} - \sqrt {-15625 \, a^{5} b^{3} d^{10} \sqrt {-\frac {d^{10}}{a^{9} b^{7}}} + 15625 \, d^{15} x} a^{2} b^{2} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}}}{125 \, d^{10}}\right ) - 15 \, {\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} b^{5} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + 125 \, \sqrt {d x} d^{7}\right ) + 15 \, {\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} b^{5} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + 125 \, \sqrt {d x} d^{7}\right ) - 4 \, {\left (15 \, b^{2} d^{2} x^{5} + 42 \, a b d^{2} x^{3} - 5 \, a^{2} d^{2} x\right )} \sqrt {d x}}{768 \, {\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} 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)^2,x, algorithm="fricas")

[Out]

-1/768*(60*(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5*b)*(-d^10/(a^9*b^7))^(1/4)*arctan(-1/125*(125*sq
rt(d*x)*a^2*b^2*d^7*(-d^10/(a^9*b^7))^(1/4) - sqrt(-15625*a^5*b^3*d^10*sqrt(-d^10/(a^9*b^7)) + 15625*d^15*x)*a
^2*b^2*(-d^10/(a^9*b^7))^(1/4))/d^10) - 15*(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5*b)*(-d^10/(a^9*b
^7))^(1/4)*log(125*a^7*b^5*(-d^10/(a^9*b^7))^(3/4) + 125*sqrt(d*x)*d^7) + 15*(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*
a^4*b^2*x^2 + a^5*b)*(-d^10/(a^9*b^7))^(1/4)*log(-125*a^7*b^5*(-d^10/(a^9*b^7))^(3/4) + 125*sqrt(d*x)*d^7) - 4
*(15*b^2*d^2*x^5 + 42*a*b*d^2*x^3 - 5*a^2*d^2*x)*sqrt(d*x))/(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5
*b)

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giac [A]  time = 0.22, size = 317, normalized size = 0.95 \begin {gather*} \frac {1}{1536} \, d^{2} {\left (\frac {8 \, {\left (15 \, \sqrt {d x} b^{2} d^{6} x^{5} + 42 \, \sqrt {d x} a b d^{6} x^{3} - 5 \, \sqrt {d x} a^{2} d^{6} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} a^{2} b} + \frac {30 \, \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^{3} b^{4} d} + \frac {30 \, \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^{3} b^{4} d} - \frac {15 \, \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^{3} b^{4} d} + \frac {15 \, \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^{3} b^{4} d}\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)^2,x, algorithm="giac")

[Out]

1/1536*d^2*(8*(15*sqrt(d*x)*b^2*d^6*x^5 + 42*sqrt(d*x)*a*b*d^6*x^3 - 5*sqrt(d*x)*a^2*d^6*x)/((b*d^2*x^2 + a*d^
2)^3*a^2*b) + 30*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^3*b^4*d) + 30*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^3*b^4*d) - 15*sqrt(2)*(a*b^3*d^2)^(3/4)*log(d*x + sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sq
rt(a*d^2/b))/(a^3*b^4*d) + 15*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^3*b^4*d))

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maple [A]  time = 0.02, size = 277, normalized size = 0.83 \begin {gather*} -\frac {5 \left (d x \right )^{\frac {3}{2}} d^{7}}{192 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b}+\frac {7 \left (d x \right )^{\frac {7}{2}} d^{5}}{32 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a}+\frac {5 \left (d x \right )^{\frac {11}{2}} b \,d^{3}}{64 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} a^{2}}+\frac {5 \sqrt {2}\, d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{2} b^{2}}+\frac {5 \sqrt {2}\, d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{2} b^{2}}+\frac {5 \sqrt {2}\, d^{3} \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 )}{512 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{2} b^{2}} \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)^2,x)

[Out]

5/64*d^3/(b*d^2*x^2+a*d^2)^3/a^2*b*(d*x)^(11/2)+7/32*d^5/(b*d^2*x^2+a*d^2)^3/a*(d*x)^(7/2)-5/192*d^7/(b*d^2*x^
2+a*d^2)^3/b*(d*x)^(3/2)+5/512*d^3/a^2/b^2/(a/b*d^2)^(1/4)*2^(1/2)*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)))+5/256*d^3/a^2/b^2/(a/b*d^2)^(1/4)
*2^(1/2)*arctan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)+1)+5/256*d^3/a^2/b^2/(a/b*d^2)^(1/4)*2^(1/2)*arctan(2^(1/2
)/(a/b*d^2)^(1/4)*(d*x)^(1/2)-1)

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maxima [A]  time = 3.02, size = 323, normalized size = 0.96 \begin {gather*} \frac {\frac {15 \, d^{4} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{a^{2} b} + \frac {8 \, {\left (15 \, \left (d x\right )^{\frac {11}{2}} b^{2} d^{4} + 42 \, \left (d x\right )^{\frac {7}{2}} a b d^{6} - 5 \, \left (d x\right )^{\frac {3}{2}} a^{2} d^{8}\right )}}{a^{2} b^{4} d^{6} x^{6} + 3 \, a^{3} b^{3} d^{6} x^{4} + 3 \, a^{4} b^{2} d^{6} x^{2} + a^{5} b d^{6}}}{1536 \, d} \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)^2,x, algorithm="maxima")

[Out]

1/1536*(15*d^4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) + 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a
)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4)
- 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sqrt(b)) - sqrt(2)*log(sqrt(b)*d*x +
sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)) + sqrt(2)*log(sqrt(b)*d*x - sqrt
(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)))/(a^2*b) + 8*(15*(d*x)^(11/2)*b^2*d^4
+ 42*(d*x)^(7/2)*a*b*d^6 - 5*(d*x)^(3/2)*a^2*d^8)/(a^2*b^4*d^6*x^6 + 3*a^3*b^3*d^6*x^4 + 3*a^4*b^2*d^6*x^2 +
a^5*b*d^6))/d

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mupad [B]  time = 4.23, size = 149, normalized size = 0.44 \begin {gather*} \frac {\frac {7\,d^5\,{\left (d\,x\right )}^{7/2}}{32\,a}-\frac {5\,d^7\,{\left (d\,x\right )}^{3/2}}{192\,b}+\frac {5\,b\,d^3\,{\left (d\,x\right )}^{11/2}}{64\,a^2}}{a^3\,d^6+3\,a^2\,b\,d^6\,x^2+3\,a\,b^2\,d^6\,x^4+b^3\,d^6\,x^6}+\frac {5\,d^{5/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{9/4}\,b^{7/4}}-\frac {5\,d^{5/2}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{9/4}\,b^{7/4}} \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)^2,x)

[Out]

((7*d^5*(d*x)^(7/2))/(32*a) - (5*d^7*(d*x)^(3/2))/(192*b) + (5*b*d^3*(d*x)^(11/2))/(64*a^2))/(a^3*d^6 + b^3*d^
6*x^6 + 3*a^2*b*d^6*x^2 + 3*a*b^2*d^6*x^4) + (5*d^(5/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128
*(-a)^(9/4)*b^(7/4)) - (5*d^(5/2)*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*(-a)^(9/4)*b^(7/4))

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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 (a + b x^{2}\right )^{4}}\, 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)**2,x)

[Out]

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

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