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)+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]  time = 0.34, antiderivative size = 318, normalized size of antiderivative = 1.00, 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 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 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 )} \, 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 = 0.01, 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)

________________________________________________________________________________________

fricas [A]  time = 1.06, size = 323, normalized size = 1.02 \[ -\frac {180 \, {\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}} \arctan \left (-\frac {729 \, \sqrt {d x} a^{3} b^{4} d^{3} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}} - \sqrt {-531441 \, a^{7} b^{5} d^{8} \sqrt {-\frac {b^{5}}{a^{13} d^{14}}} + 531441 \, b^{8} d x} a^{3} d^{3} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}}}{729 \, b^{5}}\right ) - 45 \, {\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}} \log \left (729 \, a^{10} d^{11} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {3}{4}} + 729 \, \sqrt {d x} b^{4}\right ) + 45 \, {\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {1}{4}} \log \left (-729 \, a^{10} d^{11} \left (-\frac {b^{5}}{a^{13} d^{14}}\right )^{\frac {3}{4}} + 729 \, \sqrt {d x} b^{4}\right ) - 4 \, {\left (45 \, b^{2} x^{4} + 36 \, a b x^{2} - 4 \, a^{2}\right )} \sqrt {d x}}{40 \, {\left (a^{3} b d^{4} x^{5} + a^{4} d^{4} x^{3}\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 307, normalized size = 0.97 \[ \frac {\sqrt {d x} b^{2} x}{2 \, {\left (b d^{2} x^{2} + a d^{2}\right )} a^{3} d^{2}} + \frac {9 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{4} b d^{5}} + \frac {9 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{4} b d^{5}} - \frac {9 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{16 \, a^{4} b d^{5}} + \frac {9 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{16 \, a^{4} b d^{5}} + \frac {2 \, {\left (10 \, b d^{2} x^{2} - a d^{2}\right )}}{5 \, \sqrt {d x} a^{3} d^{5} x^{2}} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 242, normalized size = 0.76 \[ -\frac {2}{5 \left (d x \right )^{\frac {5}{2}} a^{2} d}+\frac {\left (d x \right )^{\frac {3}{2}} b^{2}}{2 \left (b \,d^{2} x^{2}+d^{2} a \right ) a^{3} d^{3}}+\frac {9 \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{3} d^{3}}+\frac {9 \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{3} d^{3}}+\frac {9 \sqrt {2}\, b \ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{16 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a^{3} d^{3}}+\frac {4 b}{\sqrt {d x}\, a^{3} d^{3}} \]

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]

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/b*d^2)^(1/4)*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)))+9/8/d^3*b/a^3/(
a/b*d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)+1)+9/8/d^3*b/a^3/(a/b*d^2)^(1/4)*2^(1/2)*arc
tan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)-1)-2/5/a^2/d/(d*x)^(5/2)+4*b/a^3/d^3/(d*x)^(1/2)

________________________________________________________________________________________

maxima [A]  time = 3.01, size = 290, normalized size = 0.91 \[ \frac {\frac {8 \, {\left (45 \, b^{2} d^{4} x^{4} + 36 \, a b d^{4} x^{2} - 4 \, a^{2} d^{4}\right )}}{\left (d x\right )^{\frac {9}{2}} a^{3} b d^{2} + \left (d x\right )^{\frac {5}{2}} a^{4} d^{4}} + \frac {45 \, b^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{a^{3} d^{2}}}{80 \, d} \]

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]

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

________________________________________________________________________________________

mupad [B]  time = 4.35, size = 113, normalized size = 0.36 \[ \frac {\frac {9\,b^2\,d\,x^4}{2\,a^3}-\frac {2\,d}{5\,a}+\frac {18\,b\,d\,x^2}{5\,a^2}}{b\,{\left (d\,x\right )}^{9/2}+a\,d^2\,{\left (d\,x\right )}^{5/2}}-\frac {9\,{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {d\,x}}{a^{1/4}\,\sqrt {d}}\right )}{4\,a^{13/4}\,d^{7/2}}+\frac {9\,{\left (-b\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {d\,x}}{a^{1/4}\,\sqrt {d}}\right )}{4\,a^{13/4}\,d^{7/2}} \]

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)),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))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d x\right )^{\frac {7}{2}} \left (a + b x^{2}\right )^{2}}\, dx \]

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)

________________________________________________________________________________________