3.688 \(\int \frac{(d x)^{7/2}}{a^2+2 a b x^2+b^2 x^4} \, dx\)

Optimal. Leaf size=298 \[ \frac{5 \sqrt [4]{a} d^{7/2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{8 \sqrt{2} b^{9/4}}-\frac{5 \sqrt [4]{a} d^{7/2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{8 \sqrt{2} b^{9/4}}+\frac{5 \sqrt [4]{a} d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{4 \sqrt{2} b^{9/4}}-\frac{5 \sqrt [4]{a} d^{7/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{4 \sqrt{2} b^{9/4}}-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}+\frac{5 d^3 \sqrt{d x}}{2 b^2} \]

[Out]

(5*d^3*Sqrt[d*x])/(2*b^2) - (d*(d*x)^(5/2))/(2*b*(a + b*x^2)) + (5*a^(1/4)*d^(7/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)
*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*b^(9/4)) - (5*a^(1/4)*d^(7/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x]
)/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*b^(9/4)) + (5*a^(1/4)*d^(7/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[
2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*b^(9/4)) - (5*a^(1/4)*d^(7/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*
x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*b^(9/4))

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Rubi [A]  time = 0.294855, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {28, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{5 \sqrt [4]{a} d^{7/2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{8 \sqrt{2} b^{9/4}}-\frac{5 \sqrt [4]{a} d^{7/2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{8 \sqrt{2} b^{9/4}}+\frac{5 \sqrt [4]{a} d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{4 \sqrt{2} b^{9/4}}-\frac{5 \sqrt [4]{a} d^{7/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{4 \sqrt{2} b^{9/4}}-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}+\frac{5 d^3 \sqrt{d x}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(7/2)/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

(5*d^3*Sqrt[d*x])/(2*b^2) - (d*(d*x)^(5/2))/(2*b*(a + b*x^2)) + (5*a^(1/4)*d^(7/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)
*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*b^(9/4)) - (5*a^(1/4)*d^(7/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x]
)/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*b^(9/4)) + (5*a^(1/4)*d^(7/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[
2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*b^(9/4)) - (5*a^(1/4)*d^(7/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*
x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*b^(9/4))

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 321

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 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 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 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]

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])

Rubi steps

\begin{align*} \int \frac{(d x)^{7/2}}{a^2+2 a b x^2+b^2 x^4} \, dx &=b^2 \int \frac{(d x)^{7/2}}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}+\frac{1}{4} \left (5 d^2\right ) \int \frac{(d x)^{3/2}}{a b+b^2 x^2} \, dx\\ &=\frac{5 d^3 \sqrt{d x}}{2 b^2}-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}-\frac{\left (5 a d^4\right ) \int \frac{1}{\sqrt{d x} \left (a b+b^2 x^2\right )} \, dx}{4 b}\\ &=\frac{5 d^3 \sqrt{d x}}{2 b^2}-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}-\frac{\left (5 a d^3\right ) \operatorname{Subst}\left (\int \frac{1}{a b+\frac{b^2 x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{2 b}\\ &=\frac{5 d^3 \sqrt{d x}}{2 b^2}-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}-\frac{\left (5 \sqrt{a} d^2\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 )}{4 b}-\frac{\left (5 \sqrt{a} d^2\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 )}{4 b}\\ &=\frac{5 d^3 \sqrt{d x}}{2 b^2}-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}+\frac{\left (5 \sqrt [4]{a} d^{7/2}\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 )}{8 \sqrt{2} b^{9/4}}+\frac{\left (5 \sqrt [4]{a} d^{7/2}\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 )}{8 \sqrt{2} b^{9/4}}-\frac{\left (5 \sqrt{a} d^4\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 )}{8 b^{5/2}}-\frac{\left (5 \sqrt{a} d^4\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 )}{8 b^{5/2}}\\ &=\frac{5 d^3 \sqrt{d x}}{2 b^2}-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}+\frac{5 \sqrt [4]{a} d^{7/2} \log \left (\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}\right )}{8 \sqrt{2} b^{9/4}}-\frac{5 \sqrt [4]{a} d^{7/2} \log \left (\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}\right )}{8 \sqrt{2} b^{9/4}}-\frac{\left (5 \sqrt [4]{a} d^{7/2}\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 )}{4 \sqrt{2} b^{9/4}}+\frac{\left (5 \sqrt [4]{a} d^{7/2}\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 )}{4 \sqrt{2} b^{9/4}}\\ &=\frac{5 d^3 \sqrt{d x}}{2 b^2}-\frac{d (d x)^{5/2}}{2 b \left (a+b x^2\right )}+\frac{5 \sqrt [4]{a} d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{4 \sqrt{2} b^{9/4}}-\frac{5 \sqrt [4]{a} d^{7/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{4 \sqrt{2} b^{9/4}}+\frac{5 \sqrt [4]{a} d^{7/2} \log \left (\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}\right )}{8 \sqrt{2} b^{9/4}}-\frac{5 \sqrt [4]{a} d^{7/2} \log \left (\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}\right )}{8 \sqrt{2} b^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.154868, size = 244, normalized size = 0.82 \[ \frac{d^3 \sqrt{d x} \left (\frac{32 b^{5/4} x^2}{a+b x^2}+\frac{40 a \sqrt [4]{b}}{a+b x^2}+\frac{5 \sqrt{2} \sqrt [4]{a} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{\sqrt{x}}-\frac{5 \sqrt{2} \sqrt [4]{a} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{\sqrt{x}}+\frac{10 \sqrt{2} \sqrt [4]{a} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{x}}-\frac{10 \sqrt{2} \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{x}}\right )}{16 b^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(7/2)/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

(d^3*Sqrt[d*x]*((40*a*b^(1/4))/(a + b*x^2) + (32*b^(5/4)*x^2)/(a + b*x^2) + (10*Sqrt[2]*a^(1/4)*ArcTan[1 - (Sq
rt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/Sqrt[x] - (10*Sqrt[2]*a^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/
Sqrt[x] + (5*Sqrt[2]*a^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/Sqrt[x] - (5*Sqrt[2]*
a^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/Sqrt[x]))/(16*b^(9/4))

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Maple [A]  time = 0.059, size = 223, normalized size = 0.8 \begin{align*} 2\,{\frac{{d}^{3}\sqrt{dx}}{{b}^{2}}}+{\frac{{d}^{5}a}{2\,{b}^{2} \left ( b{d}^{2}{x}^{2}+a{d}^{2} \right ) }\sqrt{dx}}-{\frac{5\,{d}^{3}\sqrt{2}}{16\,{b}^{2}}\sqrt [4]{{\frac{a{d}^{2}}{b}}}\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{5\,{d}^{3}\sqrt{2}}{8\,{b}^{2}}\sqrt [4]{{\frac{a{d}^{2}}{b}}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}+1 \right ) }-{\frac{5\,{d}^{3}\sqrt{2}}{8\,{b}^{2}}\sqrt [4]{{\frac{a{d}^{2}}{b}}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

2*d^3*(d*x)^(1/2)/b^2+1/2*d^5/b^2*a*(d*x)^(1/2)/(b*d^2*x^2+a*d^2)-5/16*d^3/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)
))-5/8*d^3/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)-5/8*d^3/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)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38641, size = 549, normalized size = 1.84 \begin{align*} -\frac{20 \, \left (-\frac{a d^{14}}{b^{9}}\right )^{\frac{1}{4}}{\left (b^{3} x^{2} + a b^{2}\right )} \arctan \left (-\frac{\left (-\frac{a d^{14}}{b^{9}}\right )^{\frac{3}{4}} \sqrt{d x} b^{7} d^{3} - \sqrt{d^{7} x + \sqrt{-\frac{a d^{14}}{b^{9}}} b^{4}} \left (-\frac{a d^{14}}{b^{9}}\right )^{\frac{3}{4}} b^{7}}{a d^{14}}\right ) + 5 \, \left (-\frac{a d^{14}}{b^{9}}\right )^{\frac{1}{4}}{\left (b^{3} x^{2} + a b^{2}\right )} \log \left (5 \, \sqrt{d x} d^{3} + 5 \, \left (-\frac{a d^{14}}{b^{9}}\right )^{\frac{1}{4}} b^{2}\right ) - 5 \, \left (-\frac{a d^{14}}{b^{9}}\right )^{\frac{1}{4}}{\left (b^{3} x^{2} + a b^{2}\right )} \log \left (5 \, \sqrt{d x} d^{3} - 5 \, \left (-\frac{a d^{14}}{b^{9}}\right )^{\frac{1}{4}} b^{2}\right ) - 4 \,{\left (4 \, b d^{3} x^{2} + 5 \, a d^{3}\right )} \sqrt{d x}}{8 \,{\left (b^{3} x^{2} + a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

-1/8*(20*(-a*d^14/b^9)^(1/4)*(b^3*x^2 + a*b^2)*arctan(-((-a*d^14/b^9)^(3/4)*sqrt(d*x)*b^7*d^3 - sqrt(d^7*x + s
qrt(-a*d^14/b^9)*b^4)*(-a*d^14/b^9)^(3/4)*b^7)/(a*d^14)) + 5*(-a*d^14/b^9)^(1/4)*(b^3*x^2 + a*b^2)*log(5*sqrt(
d*x)*d^3 + 5*(-a*d^14/b^9)^(1/4)*b^2) - 5*(-a*d^14/b^9)^(1/4)*(b^3*x^2 + a*b^2)*log(5*sqrt(d*x)*d^3 - 5*(-a*d^
14/b^9)^(1/4)*b^2) - 4*(4*b*d^3*x^2 + 5*a*d^3)*sqrt(d*x))/(b^3*x^2 + a*b^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{\frac{7}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(7/2)/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

Integral((d*x)**(7/2)/(a + b*x**2)**2, x)

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Giac [A]  time = 1.15487, size = 362, normalized size = 1.21 \begin{align*} \frac{1}{16} \,{\left (\frac{8 \, \sqrt{d x} a d^{3}}{{\left (b d^{2} x^{2} + a d^{2}\right )} b^{2}} - \frac{10 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{1}{4}} d \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^{3}} - \frac{10 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{1}{4}} d \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^{3}} - \frac{5 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{1}{4}} d \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^{3}} + \frac{5 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{1}{4}} d \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^{3}} + \frac{32 \, \sqrt{d x} d}{b^{2}}\right )} d^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

1/16*(8*sqrt(d*x)*a*d^3/((b*d^2*x^2 + a*d^2)*b^2) - 10*sqrt(2)*(a*b^3*d^2)^(1/4)*d*arctan(1/2*sqrt(2)*(sqrt(2)
*(a*d^2/b)^(1/4) + 2*sqrt(d*x))/(a*d^2/b)^(1/4))/b^3 - 10*sqrt(2)*(a*b^3*d^2)^(1/4)*d*arctan(-1/2*sqrt(2)*(sqr
t(2)*(a*d^2/b)^(1/4) - 2*sqrt(d*x))/(a*d^2/b)^(1/4))/b^3 - 5*sqrt(2)*(a*b^3*d^2)^(1/4)*d*log(d*x + sqrt(2)*(a*
d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/b^3 + 5*sqrt(2)*(a*b^3*d^2)^(1/4)*d*log(d*x - sqrt(2)*(a*d^2/b)^(1/4)*
sqrt(d*x) + sqrt(a*d^2/b))/b^3 + 32*sqrt(d*x)*d/b^2)*d^2