3.695 \(\int \frac{1}{(d x)^{7/2} (a^2+2 a b x^2+b^2 x^4)} \, dx\)

Optimal. Leaf size=318 \[ \frac{9 b^{5/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{8 \sqrt{2} a^{13/4} d^{7/2}}-\frac{9 b^{5/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{8 \sqrt{2} a^{13/4} d^{7/2}}-\frac{9 b^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{4 \sqrt{2} a^{13/4} d^{7/2}}+\frac{9 b^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{4 \sqrt{2} a^{13/4} d^{7/2}}+\frac{9 b}{2 a^3 d^3 \sqrt{d x}}-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )} \]

[Out]

-9/(10*a^2*d*(d*x)^(5/2)) + (9*b)/(2*a^3*d^3*Sqrt[d*x]) + 1/(2*a*d*(d*x)^(5/2)*(a + b*x^2)) - (9*b^(5/4)*ArcTa
n[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*a^(13/4)*d^(7/2)) + (9*b^(5/4)*ArcTan[1 + (Sq
rt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*a^(13/4)*d^(7/2)) + (9*b^(5/4)*Log[Sqrt[a]*Sqrt[d] + S
qrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*a^(13/4)*d^(7/2)) - (9*b^(5/4)*Log[Sqrt[a]*S
qrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*a^(13/4)*d^(7/2))

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Rubi [A]  time = 0.342626, antiderivative size = 318, 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, 290, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{9 b^{5/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{8 \sqrt{2} a^{13/4} d^{7/2}}-\frac{9 b^{5/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{8 \sqrt{2} a^{13/4} d^{7/2}}-\frac{9 b^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{4 \sqrt{2} a^{13/4} d^{7/2}}+\frac{9 b^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{4 \sqrt{2} a^{13/4} d^{7/2}}+\frac{9 b}{2 a^3 d^3 \sqrt{d x}}-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*x)^(7/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

-9/(10*a^2*d*(d*x)^(5/2)) + (9*b)/(2*a^3*d^3*Sqrt[d*x]) + 1/(2*a*d*(d*x)^(5/2)*(a + b*x^2)) - (9*b^(5/4)*ArcTa
n[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*a^(13/4)*d^(7/2)) + (9*b^(5/4)*ArcTan[1 + (Sq
rt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*a^(13/4)*d^(7/2)) + (9*b^(5/4)*Log[Sqrt[a]*Sqrt[d] + S
qrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*a^(13/4)*d^(7/2)) - (9*b^(5/4)*Log[Sqrt[a]*S
qrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*a^(13/4)*d^(7/2))

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 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -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{1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx &=b^2 \int \frac{1}{(d x)^{7/2} \left (a b+b^2 x^2\right )^2} \, dx\\ &=\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}+\frac{(9 b) \int \frac{1}{(d x)^{7/2} \left (a b+b^2 x^2\right )} \, dx}{4 a}\\ &=-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}-\frac{\left (9 b^2\right ) \int \frac{1}{(d x)^{3/2} \left (a b+b^2 x^2\right )} \, dx}{4 a^2 d^2}\\ &=-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{9 b}{2 a^3 d^3 \sqrt{d x}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}+\frac{\left (9 b^3\right ) \int \frac{\sqrt{d x}}{a b+b^2 x^2} \, dx}{4 a^3 d^4}\\ &=-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{9 b}{2 a^3 d^3 \sqrt{d x}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}+\frac{\left (9 b^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{a b+\frac{b^2 x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{2 a^3 d^5}\\ &=-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{9 b}{2 a^3 d^3 \sqrt{d x}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}-\frac{\left (9 b^{5/2}\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 a^3 d^5}+\frac{\left (9 b^{5/2}\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 a^3 d^5}\\ &=-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{9 b}{2 a^3 d^3 \sqrt{d x}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}+\frac{\left (9 b^{5/4}\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} a^{13/4} d^{7/2}}+\frac{\left (9 b^{5/4}\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} a^{13/4} d^{7/2}}+\frac{(9 b) \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 a^3 d^3}+\frac{(9 b) \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 a^3 d^3}\\ &=-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{9 b}{2 a^3 d^3 \sqrt{d x}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}+\frac{9 b^{5/4} \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} a^{13/4} d^{7/2}}-\frac{9 b^{5/4} \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} a^{13/4} d^{7/2}}+\frac{\left (9 b^{5/4}\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} a^{13/4} d^{7/2}}-\frac{\left (9 b^{5/4}\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} a^{13/4} d^{7/2}}\\ &=-\frac{9}{10 a^2 d (d x)^{5/2}}+\frac{9 b}{2 a^3 d^3 \sqrt{d x}}+\frac{1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}-\frac{9 b^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{4 \sqrt{2} a^{13/4} d^{7/2}}+\frac{9 b^{5/4} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{4 \sqrt{2} a^{13/4} d^{7/2}}+\frac{9 b^{5/4} \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} a^{13/4} d^{7/2}}-\frac{9 b^{5/4} \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} a^{13/4} d^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0112613, size = 37, normalized size = 0.12 \[ -\frac{2 \sqrt{d x} \, _2F_1\left (-\frac{5}{4},2;-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 a^2 d^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*x)^(7/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

(-2*Sqrt[d*x]*Hypergeometric2F1[-5/4, 2, -1/4, -((b*x^2)/a)])/(5*a^2*d^4*x^3)

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Maple [A]  time = 0.063, size = 242, normalized size = 0.8 \begin{align*} -{\frac{2}{5\,{a}^{2}d} \left ( dx \right ) ^{-{\frac{5}{2}}}}+4\,{\frac{b}{{a}^{3}{d}^{3}\sqrt{dx}}}+{\frac{{b}^{2}}{2\,{a}^{3}{d}^{3} \left ( b{d}^{2}{x}^{2}+a{d}^{2} \right ) } \left ( dx \right ) ^{{\frac{3}{2}}}}+{\frac{9\,b\sqrt{2}}{16\,{a}^{3}{d}^{3}}\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{9\,b\sqrt{2}}{8\,{a}^{3}{d}^{3}}\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{9\,b\sqrt{2}}{8\,{a}^{3}{d}^{3}}\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(1/(d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

-2/5/a^2/d/(d*x)^(5/2)+4*b/a^3/d^3/(d*x)^(1/2)+1/2/d^3*b^2/a^3*(d*x)^(3/2)/(b*d^2*x^2+a*d^2)+9/16/d^3*b/a^3/(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)))+9/8/d^3*b/a^3/(a*d^2/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a*d^2/b)^(1/4)*(d*x)^(1/
2)+1)+9/8/d^3*b/a^3/(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(1/(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 = 1.3864, size = 767, normalized size = 2.41 \begin{align*} -\frac{180 \,{\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )} \left (-\frac{b^{5}}{a^{13} d^{14}}\right )^{\frac{1}{4}} \arctan \left (-\frac{729 \, \sqrt{d x} a^{3} b^{4} d^{3} \left (-\frac{b^{5}}{a^{13} d^{14}}\right )^{\frac{1}{4}} - \sqrt{-531441 \, a^{7} b^{5} d^{8} \sqrt{-\frac{b^{5}}{a^{13} d^{14}}} + 531441 \, b^{8} d x} a^{3} d^{3} \left (-\frac{b^{5}}{a^{13} d^{14}}\right )^{\frac{1}{4}}}{729 \, b^{5}}\right ) - 45 \,{\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )} \left (-\frac{b^{5}}{a^{13} d^{14}}\right )^{\frac{1}{4}} \log \left (729 \, a^{10} d^{11} \left (-\frac{b^{5}}{a^{13} d^{14}}\right )^{\frac{3}{4}} + 729 \, \sqrt{d x} b^{4}\right ) + 45 \,{\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )} \left (-\frac{b^{5}}{a^{13} d^{14}}\right )^{\frac{1}{4}} \log \left (-729 \, a^{10} d^{11} \left (-\frac{b^{5}}{a^{13} d^{14}}\right )^{\frac{3}{4}} + 729 \, \sqrt{d x} b^{4}\right ) - 4 \,{\left (45 \, b^{2} x^{4} + 36 \, a b x^{2} - 4 \, a^{2}\right )} \sqrt{d x}}{40 \,{\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

-1/40*(180*(a^3*b*d^4*x^5 + a^4*d^4*x^3)*(-b^5/(a^13*d^14))^(1/4)*arctan(-1/729*(729*sqrt(d*x)*a^3*b^4*d^3*(-b
^5/(a^13*d^14))^(1/4) - sqrt(-531441*a^7*b^5*d^8*sqrt(-b^5/(a^13*d^14)) + 531441*b^8*d*x)*a^3*d^3*(-b^5/(a^13*
d^14))^(1/4))/b^5) - 45*(a^3*b*d^4*x^5 + a^4*d^4*x^3)*(-b^5/(a^13*d^14))^(1/4)*log(729*a^10*d^11*(-b^5/(a^13*d
^14))^(3/4) + 729*sqrt(d*x)*b^4) + 45*(a^3*b*d^4*x^5 + a^4*d^4*x^3)*(-b^5/(a^13*d^14))^(1/4)*log(-729*a^10*d^1
1*(-b^5/(a^13*d^14))^(3/4) + 729*sqrt(d*x)*b^4) - 4*(45*b^2*x^4 + 36*a*b*x^2 - 4*a^2)*sqrt(d*x))/(a^3*b*d^4*x^
5 + a^4*d^4*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\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(1/(d*x)**(7/2)/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

Integral(1/((d*x)**(7/2)*(a + b*x**2)**2), x)

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Giac [A]  time = 1.25903, size = 414, normalized size = 1.3 \begin{align*} \frac{\sqrt{d x} b^{2} x}{2 \,{\left (b d^{2} x^{2} + a d^{2}\right )} a^{3} d^{2}} + \frac{9 \, \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 )}{8 \, a^{4} b d^{5}} + \frac{9 \, \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 )}{8 \, a^{4} b d^{5}} - \frac{9 \, \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 )}{16 \, a^{4} b d^{5}} + \frac{9 \, \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 )}{16 \, a^{4} b d^{5}} + \frac{2 \,{\left (10 \, b d^{2} x^{2} - a d^{2}\right )}}{5 \, \sqrt{d x} a^{3} d^{5} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

1/2*sqrt(d*x)*b^2*x/((b*d^2*x^2 + a*d^2)*a^3*d^2) + 9/8*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^4*b*d^5) + 9/8*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^4*b*d^5) - 9/16*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^4*b*d^5) + 9/16*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^4*b*d^5) + 2/5*(10*b*d^2*x^2 - a*d^2)/(sqrt(d*x)*a^3*d^5
*x^2)