\(\int \frac {1}{(d x)^{7/2} (a^2+2 a b x^2+b^2 x^4)} \, dx\) [695]

Optimal result

Integrand size = 28, antiderivative size = 318 \[ \int \frac {1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {9}{10 a^2 d (d x)^{5/2}}+\frac {9 b}{2 a^3 d^3 \sqrt {d x}}+\frac {1}{2 a d (d x)^{5/2} \left (a+b x^2\right )}-\frac {9 b^{5/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} a^{13/4} d^{7/2}}+\frac {9 b^{5/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} a^{13/4} d^{7/2}}+\frac {9 b^{5/4} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{8 \sqrt {2} a^{13/4} d^{7/2}}-\frac {9 b^{5/4} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{8 \sqrt {2} a^{13/4} d^{7/2}} \]

[Out]

-9/10/a^2/d/(d*x)^(5/2)+1/2/a/d/(d*x)^(5/2)/(b*x^2+a)-9/8*b^(5/4)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)
/d^(1/2))/a^(13/4)/d^(7/2)*2^(1/2)+9/8*b^(5/4)*arctan(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(13/4)/
d^(7/2)*2^(1/2)+9/16*b^(5/4)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1/2))/a^(13/4
)/d^(7/2)*2^(1/2)-9/16*b^(5/4)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1/2))/a^(13
/4)/d^(7/2)*2^(1/2)+9/2*b/a^3/d^3/(d*x)^(1/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {28, 296, 331, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {9 b^{5/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} a^{13/4} d^{7/2}}+\frac {9 b^{5/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} a^{13/4} d^{7/2}}+\frac {9 b^{5/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{13/4} d^{7/2}}-\frac {9 b^{5/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} a^{13/4} d^{7/2}}+\frac {9 b}{2 a^3 d^3 \sqrt {d x}}-\frac {9}{10 a^2 d (d x)^{5/2}}+\frac {1}{2 a d (d x)^{5/2} \left (a+b x^2\right )} \]

[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 210

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

Rule 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 303

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 331

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 335

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 631

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 642

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 1176

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 1179

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*} \text {integral}& = 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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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) \text {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) \text {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 ) \text {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 ) \text {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 [A] (verified)

Time = 0.07 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {\sqrt {d x} \left (4 \sqrt [4]{a} \left (4 a^2-36 a b x^2-45 b^2 x^4\right )+45 \sqrt {2} b^{5/4} x^{5/2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )+45 \sqrt {2} b^{5/4} x^{5/2} \left (a+b x^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )\right )}{40 a^{13/4} d^4 x^3 \left (a+b x^2\right )} \]

[In]

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

[Out]

-1/40*(Sqrt[d*x]*(4*a^(1/4)*(4*a^2 - 36*a*b*x^2 - 45*b^2*x^4) + 45*Sqrt[2]*b^(5/4)*x^(5/2)*(a + b*x^2)*ArcTan[
(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] + 45*Sqrt[2]*b^(5/4)*x^(5/2)*(a + b*x^2)*ArcTanh[(Sqr
t[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)]))/(a^(13/4)*d^4*x^3*(a + b*x^2))

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\frac {2 \left (-10 b \,x^{2}+a \right )}{5 a^{3} \sqrt {d x}\, x^{2} d^{3}}+\frac {b^{2} \left (\frac {\left (d x \right )^{\frac {3}{2}}}{2 b \,d^{2} x^{2}+2 a \,d^{2}}+\frac {9 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} d^{3}}\)	\(198\)
derivativedivides	\(2 d^{3} \left (-\frac {1}{5 a^{2} d^{4} \left (d x \right )^{\frac {5}{2}}}+\frac {2 b}{a^{3} d^{6} \sqrt {d x}}+\frac {b^{2} \left (\frac {\left (d x \right )^{\frac {3}{2}}}{4 b \,d^{2} x^{2}+4 a \,d^{2}}+\frac {9 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} d^{6}}\right )\)	\(206\)
default	\(2 d^{3} \left (-\frac {1}{5 a^{2} d^{4} \left (d x \right )^{\frac {5}{2}}}+\frac {2 b}{a^{3} d^{6} \sqrt {d x}}+\frac {b^{2} \left (\frac {\left (d x \right )^{\frac {3}{2}}}{4 b \,d^{2} x^{2}+4 a \,d^{2}}+\frac {9 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} d^{6}}\right )\)	\(206\)
pseudoelliptic	\(\frac {\frac {9 b \sqrt {2}\, \left (b \,x^{2}+a \right ) \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )\right ) \left (d x \right )^{\frac {5}{2}}}{16}-\frac {2 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} d^{2} \left (-\frac {45}{4} b^{2} x^{4}-9 a b \,x^{2}+a^{2}\right )}{5}}{d^{3} a^{3} \left (d x \right )^{\frac {5}{2}} \left (b \,x^{2}+a \right ) \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\)	\(227\)

[In]

int(1/(d*x)^(7/2)/(b^2*x^4+2*a*b*x^2+a^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {45 \, {\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}} \log \left (729 \, a^{10} d^{11} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {3}{4}} + 729 \, \sqrt {d x} b^{4}\right ) - 45 \, {\left (i \, a^{3} b d^{4} x^{5} + i \, a^{4} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}} \log \left (729 i \, a^{10} d^{11} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {3}{4}} + 729 \, \sqrt {d x} b^{4}\right ) - 45 \, {\left (-i \, a^{3} b d^{4} x^{5} - i \, a^{4} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}} \log \left (-729 i \, a^{10} d^{11} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {3}{4}} + 729 \, \sqrt {d x} b^{4}\right ) - 45 \, {\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}} \log \left (-729 \, a^{10} d^{11} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {3}{4}} + 729 \, \sqrt {d x} b^{4}\right ) + 4 \, {\left (45 \, b^{2} x^{4} + 36 \, a b x^{2} - 4 \, a^{2}\right )} \sqrt {d x}}{40 \, {\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )}} \]

[In]

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

[Out]

1/40*(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) + 7
29*sqrt(d*x)*b^4) - 45*(I*a^3*b*d^4*x^5 + I*a^4*d^4*x^3)*(-b^5/(a^13*d^14))^(1/4)*log(729*I*a^10*d^11*(-b^5/(a
^13*d^14))^(3/4) + 729*sqrt(d*x)*b^4) - 45*(-I*a^3*b*d^4*x^5 - I*a^4*d^4*x^3)*(-b^5/(a^13*d^14))^(1/4)*log(-72
9*I*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^1
4))^(1/4)*log(-729*a^10*d^11*(-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)

Sympy [F]

\[ \int \frac {1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\int \frac {1}{\left (d x\right )^{\frac {7}{2}} \left (a + b x^{2}\right )^{2}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {\frac {8 \, {\left (45 \, b^{2} d^{4} x^{4} + 36 \, a b d^{4} x^{2} - 4 \, a^{2} d^{4}\right )}}{\left (d x\right )^{\frac {9}{2}} a^{3} b d^{2} + \left (d x\right )^{\frac {5}{2}} a^{4} d^{4}} + \frac {45 \, b^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{a^{3} d^{2}}}{80 \, d} \]

[In]

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

[Out]

1/80*(8*(45*b^2*d^4*x^4 + 36*a*b*d^4*x^2 - 4*a^2*d^4)/((d*x)^(9/2)*a^3*b*d^2 + (d*x)^(5/2)*a^4*d^4) + 45*b^2*(
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))/(s
qrt(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)*s
qrt(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^3*d^2))/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 307, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {\sqrt {d x} b^{2} x}{2 \, {\left (b d^{2} x^{2} + a d^{2}\right )} a^{3} d^{2}} + \frac {9 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{4} b d^{5}} + \frac {9 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{4} b d^{5}} - \frac {9 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{16 \, a^{4} b d^{5}} + \frac {9 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{16 \, a^{4} b d^{5}} + \frac {2 \, {\left (10 \, b d^{2} x^{2} - a d^{2}\right )}}{5 \, \sqrt {d x} a^{3} d^{5} x^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 14.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.36 \[ \int \frac {1}{(d x)^{7/2} \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {\frac {9\,b^2\,d\,x^4}{2\,a^3}-\frac {2\,d}{5\,a}+\frac {18\,b\,d\,x^2}{5\,a^2}}{b\,{\left (d\,x\right )}^{9/2}+a\,d^2\,{\left (d\,x\right )}^{5/2}}-\frac {9\,{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {d\,x}}{a^{1/4}\,\sqrt {d}}\right )}{4\,a^{13/4}\,d^{7/2}}+\frac {9\,{\left (-b\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {d\,x}}{a^{1/4}\,\sqrt {d}}\right )}{4\,a^{13/4}\,d^{7/2}} \]

[In]

int(1/((d*x)^(7/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)),x)

[Out]

((9*b^2*d*x^4)/(2*a^3) - (2*d)/(5*a) + (18*b*d*x^2)/(5*a^2))/(b*(d*x)^(9/2) + a*d^2*(d*x)^(5/2)) - (9*(-b)^(5/
4)*atan(((-b)^(1/4)*(d*x)^(1/2))/(a^(1/4)*d^(1/2))))/(4*a^(13/4)*d^(7/2)) + (9*(-b)^(5/4)*atanh(((-b)^(1/4)*(d
*x)^(1/2))/(a^(1/4)*d^(1/2))))/(4*a^(13/4)*d^(7/2))