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

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

[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))

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Rubi [A]  time = 0.348355, antiderivative size = 335, normalized size of antiderivative = 1., 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} \[ \frac{5 d^{5/2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{256 \sqrt{2} a^{9/4} b^{7/4}}-\frac{5 d^{5/2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{256 \sqrt{2} a^{9/4} b^{7/4}}-\frac{5 d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{128 \sqrt{2} a^{9/4} b^{7/4}}+\frac{5 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{128 \sqrt{2} a^{9/4} b^{7/4}}+\frac{5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac{d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}-\frac{d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3} \]

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 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 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 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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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*}

Mathematica [C]  time = 0.020821, size = 60, normalized size = 0.18 \[ \frac{2 d (d x)^{3/2} \left (\left (a+b x^2\right )^3 \, _2F_1\left (\frac{3}{4},4;\frac{7}{4};-\frac{b x^2}{a}\right )-a^3\right )}{9 a^3 b \left (a+b x^2\right )^3} \]

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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Maple [A]  time = 0.065, size = 277, normalized size = 0.8 \begin{align*}{\frac{5\,{d}^{3}b}{64\, \left ( b{d}^{2}{x}^{2}+a{d}^{2} \right ) ^{3}{a}^{2}} \left ( dx \right ) ^{{\frac{11}{2}}}}+{\frac{7\,{d}^{5}}{32\, \left ( b{d}^{2}{x}^{2}+a{d}^{2} \right ) ^{3}a} \left ( dx \right ) ^{{\frac{7}{2}}}}-{\frac{5\,{d}^{7}}{192\, \left ( b{d}^{2}{x}^{2}+a{d}^{2} \right ) ^{3}b} \left ( dx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{d}^{3}\sqrt{2}}{512\,{b}^{2}{a}^{2}}\ln \left ({ \left ( dx-\sqrt [4]{{\frac{a{d}^{2}}{b}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{a{d}^{2}}{b}}} \right ) \left ( dx+\sqrt [4]{{\frac{a{d}^{2}}{b}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{a{d}^{2}}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}+{\frac{5\,{d}^{3}\sqrt{2}}{256\,{b}^{2}{a}^{2}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}+{\frac{5\,{d}^{3}\sqrt{2}}{256\,{b}^{2}{a}^{2}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}} \end{align*}

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*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)))+5/256*d^3/a^2/b^2/(a*d^2/b)^(1/4)
*2^(1/2)*arctan(2^(1/2)/(a*d^2/b)^(1/4)*(d*x)^(1/2)+1)+5/256*d^3/a^2/b^2/(a*d^2/b)^(1/4)*2^(1/2)*arctan(2^(1/2
)/(a*d^2/b)^(1/4)*(d*x)^(1/2)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.41831, size = 898, normalized size = 2.68 \begin{align*} -\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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{\frac{5}{2}}}{\left (a + b x^{2}\right )^{4}}\, dx \end{align*}

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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Giac [A]  time = 1.28677, size = 409, normalized size = 1.22 \begin{align*} \frac{1}{1536} \, d{\left (\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}} + \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}} - \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}} + \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}} + \frac{8 \,{\left (15 \, \sqrt{d x} b^{2} d^{7} x^{5} + 42 \, \sqrt{d x} a b d^{7} x^{3} - 5 \, \sqrt{d x} a^{2} d^{7} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} a^{2} b}\right )} \end{align*}

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*(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) + 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) - 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) + 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) + 8*(15*sqrt(d*x)*b^2*d^7*x^5 + 42*sqrt(d*x)*a*b*d^7*x^3 - 5*sqrt(d*x)*a^2*d^7*x)/((b*d^2*x^2 + a*d^2)^
3*a^2*b))