3.6.90 \(\int \frac {1}{(d x)^{7/2} (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=553 \[ \frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b^{5/4} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{64 \sqrt {2} a^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 b^{5/4} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{64 \sqrt {2} a^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 b^{5/4} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{32 \sqrt {2} a^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b^{5/4} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{32 \sqrt {2} a^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b \left (a+b x^2\right )}{16 a^4 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.43, antiderivative size = 553, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1112, 290, 325, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {117 b \left (a+b x^2\right )}{16 a^4 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*x)^(7/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

13/(16*a^2*d*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*d*(d*x)^(5/2)*(a + b*x^2)*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4]) - (117*(a + b*x^2))/(80*a^3*d*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (117*b*(a + b*x^2
))/(16*a^4*d^3*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (117*b^(5/4)*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/
4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(17/4)*d^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (117*b^(5/4)
*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(32*Sqrt[2]*a^(17/4)*d^(7/2)*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4]) + (117*b^(5/4)*(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^(17/4)*d^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (117*b^(5/4)*(a + b*x^2)*Lo
g[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(64*Sqrt[2]*a^(17/4)*d^(7/2)*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 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 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 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 {1}{(d x)^{7/2} \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 {1}{(d x)^{7/2} \left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (13 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{(d x)^{7/2} \left (a b+b^2 x^2\right )^2} \, dx}{8 a \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{(d x)^{7/2} \left (a b+b^2 x^2\right )} \, dx}{32 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (117 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{(d x)^{3/2} \left (a b+b^2 x^2\right )} \, dx}{32 a^3 d^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b \left (a+b x^2\right )}{16 a^4 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {\sqrt {d x}}{a b+b^2 x^2} \, dx}{32 a^4 d^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b \left (a+b x^2\right )}{16 a^4 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 b^2 \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^4 d^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b \left (a+b x^2\right )}{16 a^4 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (117 b^{3/2} \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^4 d^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 b^{3/2} \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^4 d^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b \left (a+b x^2\right )}{16 a^4 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 \sqrt [4]{b} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 \sqrt [4]{b} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 \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^4 d^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 \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^4 d^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b \left (a+b x^2\right )}{16 a^4 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (117 \sqrt [4]{b} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (117 \sqrt [4]{b} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {13}{16 a^2 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a d (d x)^{5/2} \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 \left (a+b x^2\right )}{80 a^3 d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b \left (a+b x^2\right )}{16 a^4 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {117 b^{5/4} \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^{17/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 54, normalized size = 0.10 \begin {gather*} -\frac {2 x \left (a+b x^2\right )^3 \, _2F_1\left (-\frac {5}{4},3;-\frac {1}{4};-\frac {b x^2}{a}\right )}{5 a^3 (d x)^{7/2} \left (\left (a+b x^2\right )^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d*x)^(7/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

(-2*x*(a + b*x^2)^3*Hypergeometric2F1[-5/4, 3, -1/4, -((b*x^2)/a)])/(5*a^3*(d*x)^(7/2)*((a + b*x^2)^2)^(3/2))

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IntegrateAlgebraic [A]  time = 90.94, size = 269, normalized size = 0.49 \begin {gather*} \frac {\left (a d^2+b d^2 x^2\right ) \left (-\frac {117 b^{5/4} \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^{17/4} d^{7/2}}-\frac {117 b^{5/4} \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^{17/4} d^{7/2}}+\frac {-32 a^3 d^6+416 a^2 b d^6 x^2+1053 a b^2 d^6 x^4+585 b^3 d^6 x^6}{80 a^4 d^3 (d x)^{5/2} \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[1/((d*x)^(7/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

((a*d^2 + b*d^2*x^2)*((-32*a^3*d^6 + 416*a^2*b*d^6*x^2 + 1053*a*b^2*d^6*x^4 + 585*b^3*d^6*x^6)/(80*a^4*d^3*(d*
x)^(5/2)*(a*d^2 + b*d^2*x^2)^2) - (117*b^(5/4)*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^(17/4)*d^(7/2)) - (117*b^(5/4)*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^(17/4)*d^(7/2))))/(d^2*Sqrt[(a*d^2 + b*d^2*x^2)
^2/d^4])

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fricas [A]  time = 1.81, size = 390, normalized size = 0.71 \begin {gather*} -\frac {2340 \, {\left (a^{4} b^{2} d^{4} x^{7} + 2 \, a^{5} b d^{4} x^{5} + a^{6} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{17} d^{14}}\right )^{\frac {1}{4}} \arctan \left (-\frac {1601613 \, \sqrt {d x} a^{4} b^{4} d^{3} \left (-\frac {b^{5}}{a^{17} d^{14}}\right )^{\frac {1}{4}} - \sqrt {-2565164201769 \, a^{9} b^{5} d^{8} \sqrt {-\frac {b^{5}}{a^{17} d^{14}}} + 2565164201769 \, b^{8} d x} a^{4} d^{3} \left (-\frac {b^{5}}{a^{17} d^{14}}\right )^{\frac {1}{4}}}{1601613 \, b^{5}}\right ) - 585 \, {\left (a^{4} b^{2} d^{4} x^{7} + 2 \, a^{5} b d^{4} x^{5} + a^{6} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{17} d^{14}}\right )^{\frac {1}{4}} \log \left (1601613 \, a^{13} d^{11} \left (-\frac {b^{5}}{a^{17} d^{14}}\right )^{\frac {3}{4}} + 1601613 \, \sqrt {d x} b^{4}\right ) + 585 \, {\left (a^{4} b^{2} d^{4} x^{7} + 2 \, a^{5} b d^{4} x^{5} + a^{6} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{17} d^{14}}\right )^{\frac {1}{4}} \log \left (-1601613 \, a^{13} d^{11} \left (-\frac {b^{5}}{a^{17} d^{14}}\right )^{\frac {3}{4}} + 1601613 \, \sqrt {d x} b^{4}\right ) - 4 \, {\left (585 \, b^{3} x^{6} + 1053 \, a b^{2} x^{4} + 416 \, a^{2} b x^{2} - 32 \, a^{3}\right )} \sqrt {d x}}{320 \, {\left (a^{4} b^{2} d^{4} x^{7} + 2 \, a^{5} b d^{4} x^{5} + a^{6} d^{4} x^{3}\right )}} \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

-1/320*(2340*(a^4*b^2*d^4*x^7 + 2*a^5*b*d^4*x^5 + a^6*d^4*x^3)*(-b^5/(a^17*d^14))^(1/4)*arctan(-1/1601613*(160
1613*sqrt(d*x)*a^4*b^4*d^3*(-b^5/(a^17*d^14))^(1/4) - sqrt(-2565164201769*a^9*b^5*d^8*sqrt(-b^5/(a^17*d^14)) +
2565164201769*b^8*d*x)*a^4*d^3*(-b^5/(a^17*d^14))^(1/4))/b^5) - 585*(a^4*b^2*d^4*x^7 + 2*a^5*b*d^4*x^5 + a^6*
d^4*x^3)*(-b^5/(a^17*d^14))^(1/4)*log(1601613*a^13*d^11*(-b^5/(a^17*d^14))^(3/4) + 1601613*sqrt(d*x)*b^4) + 58
5*(a^4*b^2*d^4*x^7 + 2*a^5*b*d^4*x^5 + a^6*d^4*x^3)*(-b^5/(a^17*d^14))^(1/4)*log(-1601613*a^13*d^11*(-b^5/(a^1
7*d^14))^(3/4) + 1601613*sqrt(d*x)*b^4) - 4*(585*b^3*x^6 + 1053*a*b^2*x^4 + 416*a^2*b*x^2 - 32*a^3)*sqrt(d*x))
/(a^4*b^2*d^4*x^7 + 2*a^5*b*d^4*x^5 + a^6*d^4*x^3)

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giac [A]  time = 0.34, size = 432, normalized size = 0.78 \begin {gather*} \frac {21 \, \sqrt {d x} b^{3} d^{3} x^{3} + 25 \, \sqrt {d x} a b^{2} d^{3} x}{16 \, {\left (b d^{2} x^{2} + a d^{2}\right )}^{2} a^{4} d^{3} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {117 \, \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 )}{64 \, a^{5} b d^{5} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {117 \, \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 )}{64 \, a^{5} b d^{5} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} - \frac {117 \, \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 )}{128 \, a^{5} b d^{5} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {117 \, \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 )}{128 \, a^{5} b d^{5} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} + \frac {2 \, {\left (15 \, b d^{2} x^{2} - a d^{2}\right )}}{5 \, \sqrt {d x} a^{4} d^{5} x^{2} \mathrm {sgn}\left (b d^{4} x^{2} + a d^{4}\right )} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

1/16*(21*sqrt(d*x)*b^3*d^3*x^3 + 25*sqrt(d*x)*a*b^2*d^3*x)/((b*d^2*x^2 + a*d^2)^2*a^4*d^3*sgn(b*d^4*x^2 + a*d^
4)) + 117/64*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^5*b*d^5*sgn(b*d^4*x^2 + a*d^4)) + 117/64*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^5*b*d^5*sgn(b*d^4*x^2 + a*d^4)) - 117/128*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^5*b*d^5*sgn(b*d^4*x^2 + a*d^4)) + 117/128*s
qrt(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^5*b*d^5*sgn(b*d^4*x^2
+ a*d^4)) + 2/5*(15*b*d^2*x^2 - a*d^2)/(sqrt(d*x)*a^4*d^5*x^2*sgn(b*d^4*x^2 + a*d^4))

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maple [A]  time = 0.03, size = 687, normalized size = 1.24 \begin {gather*} \frac {\left (4680 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{3} d^{2} x^{6}+8424 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a \,b^{2} d^{2} x^{4}+1170 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} b^{3} 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 )+1170 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} b^{3} 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 )+585 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} b^{3} 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 )+3328 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{2} b \,d^{2} x^{2}+2340 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} a \,b^{2} 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 )+2340 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} a \,b^{2} 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 )+1170 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} a \,b^{2} 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 )-256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{3} d^{2}+1170 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} a^{2} b \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 )+1170 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} a^{2} b \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 )+585 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} a^{2} b \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 )\right ) \left (b \,x^{2}+a \right )}{640 \left (d x \right )^{\frac {5}{2}} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} a^{4} d^{3}} \end {gather*}

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)^(3/2),x)

[Out]

1/640/d^3*(585*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)))*(d*x)^(5/2)*x^4*b^3+1170*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4
))/(a/b*d^2)^(1/4))*(d*x)^(5/2)*x^4*b^3+1170*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1
/4))*(d*x)^(5/2)*x^4*b^3+4680*(a/b*d^2)^(1/4)*x^6*b^3*d^2+1170*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)))*(d*x)^(5/2)*x^2*a*b^2+2340*
2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))*(d*x)^(5/2)*x^2*a*b^2+2340*2^(1/2)*arcta
n((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))*(d*x)^(5/2)*x^2*a*b^2+8424*(a/b*d^2)^(1/4)*x^4*a*b^2*
d^2+585*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)))*(d*x)^(5/2)*a^2*b+1170*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)+(a/b*d^2)^(1/4))/(a/b*d
^2)^(1/4))*(d*x)^(5/2)*a^2*b+1170*2^(1/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))*(d*x)^
(5/2)*a^2*b+3328*(a/b*d^2)^(1/4)*x^2*a^2*b*d^2-256*(a/b*d^2)^(1/4)*a^3*d^2)*(b*x^2+a)/(d*x)^(5/2)/(a/b*d^2)^(1
/4)/a^4/((b*x^2+a)^2)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {b^{2} x^{\frac {3}{2}}}{2 \, {\left (a^{4} b d^{\frac {7}{2}} x^{2} + a^{5} d^{\frac {7}{2}} + {\left (a^{3} b^{2} d^{\frac {7}{2}} x^{2} + a^{4} b d^{\frac {7}{2}}\right )} x^{2}\right )}} - 2 \, b \int \frac {1}{{\left (a^{3} b d^{\frac {7}{2}} x^{2} + a^{4} d^{\frac {7}{2}}\right )} x^{\frac {3}{2}}}\,{d x} + \frac {21 \, b^{3} x^{\frac {7}{2}} + 17 \, a b^{2} x^{\frac {3}{2}}}{16 \, {\left (a^{4} b^{2} d^{\frac {7}{2}} x^{4} + 2 \, a^{5} b d^{\frac {7}{2}} x^{2} + a^{6} d^{\frac {7}{2}}\right )}} + \frac {21 \, b^{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^{4} d^{\frac {7}{2}}} + \int \frac {1}{{\left (a^{2} b d^{\frac {7}{2}} x^{2} + a^{3} d^{\frac {7}{2}}\right )} x^{\frac {7}{2}}}\,{d x} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

1/2*b^2*x^(3/2)/(a^4*b*d^(7/2)*x^2 + a^5*d^(7/2) + (a^3*b^2*d^(7/2)*x^2 + a^4*b*d^(7/2))*x^2) - 2*b*integrate(
1/((a^3*b*d^(7/2)*x^2 + a^4*d^(7/2))*x^(3/2)), x) + 1/16*(21*b^3*x^(7/2) + 17*a*b^2*x^(3/2))/(a^4*b^2*d^(7/2)*
x^4 + 2*a^5*b*d^(7/2)*x^2 + a^6*d^(7/2)) + 21/128*b^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)*(sq
rt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*lo
g(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^4*d^(7/2)) + integrate(1/((a^2*b*d^(7/2)*x^2 + a^3*d^
(7/2))*x^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (d\,x\right )}^{7/2}\,{\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(1/((d*x)^(7/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)),x)

[Out]

int(1/((d*x)^(7/2)*(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 {1}{\left (d x\right )^{\frac {7}{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(1/(d*x)**(7/2)/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(1/((d*x)**(7/2)*((a + b*x**2)**2)**(3/2)), x)

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