3.699 \(\int \frac {(d x)^{13/2}}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=333 \[ \frac {77 d^{13/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {77 d^{13/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {77 d^{13/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} \sqrt [4]{a} b^{15/4}}+\frac {77 d^{13/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {77 d^5 (d x)^{3/2}}{192 b^3 \left (a+b x^2\right )}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3} \]

[Out]

-1/6*d*(d*x)^(11/2)/b/(b*x^2+a)^3-11/48*d^3*(d*x)^(7/2)/b^2/(b*x^2+a)^2-77/192*d^5*(d*x)^(3/2)/b^3/(b*x^2+a)-7
7/256*d^(13/2)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(1/4)/b^(15/4)*2^(1/2)+77/256*d^(13/2)*
arctan(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(1/4)/b^(15/4)*2^(1/2)+77/512*d^(13/2)*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^(1/4)/b^(15/4)*2^(1/2)-77/512*d^(13/2)*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^(1/4)/b^(15/4)*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {28, 288, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {77 d^5 (d x)^{3/2}}{192 b^3 \left (a+b x^2\right )}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}+\frac {77 d^{13/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {77 d^{13/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {77 d^{13/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} \sqrt [4]{a} b^{15/4}}+\frac {77 d^{13/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(13/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

-(d*(d*x)^(11/2))/(6*b*(a + b*x^2)^3) - (11*d^3*(d*x)^(7/2))/(48*b^2*(a + b*x^2)^2) - (77*d^5*(d*x)^(3/2))/(19
2*b^3*(a + b*x^2)) - (77*d^(13/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(1
/4)*b^(15/4)) + (77*d^(13/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(1/4)*b
^(15/4)) + (77*d^(13/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqr
t[2]*a^(1/4)*b^(15/4)) - (77*d^(13/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d
*x]])/(256*Sqrt[2]*a^(1/4)*b^(15/4))

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 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 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 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 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 {(d x)^{13/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {(d x)^{13/2}}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}+\frac {1}{12} \left (11 b^2 d^2\right ) \int \frac {(d x)^{9/2}}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}+\frac {1}{96} \left (77 d^4\right ) \int \frac {(d x)^{5/2}}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {77 d^5 (d x)^{3/2}}{192 b^3 \left (a+b x^2\right )}+\frac {\left (77 d^6\right ) \int \frac {\sqrt {d x}}{a b+b^2 x^2} \, dx}{128 b^2}\\ &=-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {77 d^5 (d x)^{3/2}}{192 b^3 \left (a+b x^2\right )}+\frac {\left (77 d^5\right ) \operatorname {Subst}\left (\int \frac {x^2}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{64 b^2}\\ &=-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {77 d^5 (d x)^{3/2}}{192 b^3 \left (a+b x^2\right )}-\frac {\left (77 d^5\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 )}{128 b^{5/2}}+\frac {\left (77 d^5\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 )}{128 b^{5/2}}\\ &=-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {77 d^5 (d x)^{3/2}}{192 b^3 \left (a+b x^2\right )}+\frac {\left (77 d^{13/2}\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 )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}+\frac {\left (77 d^{13/2}\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 )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}+\frac {\left (77 d^7\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 )}{256 b^4}+\frac {\left (77 d^7\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 )}{256 b^4}\\ &=-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {77 d^5 (d x)^{3/2}}{192 b^3 \left (a+b x^2\right )}+\frac {77 d^{13/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {77 d^{13/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}+\frac {\left (77 d^{13/2}\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 )}{128 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {\left (77 d^{13/2}\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 )}{128 \sqrt {2} \sqrt [4]{a} b^{15/4}}\\ &=-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}-\frac {11 d^3 (d x)^{7/2}}{48 b^2 \left (a+b x^2\right )^2}-\frac {77 d^5 (d x)^{3/2}}{192 b^3 \left (a+b x^2\right )}-\frac {77 d^{13/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} \sqrt [4]{a} b^{15/4}}+\frac {77 d^{13/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} \sqrt [4]{a} b^{15/4}}+\frac {77 d^{13/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}-\frac {77 d^{13/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} \sqrt [4]{a} b^{15/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 83, normalized size = 0.25 \[ \frac {2 d^6 x \sqrt {d x} \left (77 \left (a+b x^2\right )^3 \, _2F_1\left (\frac {3}{4},4;\frac {7}{4};-\frac {b x^2}{a}\right )-a \left (77 a^2+99 a b x^2+45 b^2 x^4\right )\right )}{45 a b^3 \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(13/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(2*d^6*x*Sqrt[d*x]*(-(a*(77*a^2 + 99*a*b*x^2 + 45*b^2*x^4)) + 77*(a + b*x^2)^3*Hypergeometric2F1[3/4, 4, 7/4,
-((b*x^2)/a)]))/(45*a*b^3*(a + b*x^2)^3)

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 370, normalized size = 1.11 \[ -\frac {924 \, {\left (b^{6} x^{6} + 3 \, a b^{5} x^{4} + 3 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )} \left (-\frac {d^{26}}{a b^{15}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\left (-\frac {d^{26}}{a b^{15}}\right )^{\frac {1}{4}} \sqrt {d x} b^{4} d^{19} - \sqrt {d^{39} x - \sqrt {-\frac {d^{26}}{a b^{15}}} a b^{7} d^{26}} \left (-\frac {d^{26}}{a b^{15}}\right )^{\frac {1}{4}} b^{4}}{d^{26}}\right ) - 231 \, {\left (b^{6} x^{6} + 3 \, a b^{5} x^{4} + 3 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )} \left (-\frac {d^{26}}{a b^{15}}\right )^{\frac {1}{4}} \log \left (456533 \, \sqrt {d x} d^{19} + 456533 \, \left (-\frac {d^{26}}{a b^{15}}\right )^{\frac {3}{4}} a b^{11}\right ) + 231 \, {\left (b^{6} x^{6} + 3 \, a b^{5} x^{4} + 3 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )} \left (-\frac {d^{26}}{a b^{15}}\right )^{\frac {1}{4}} \log \left (456533 \, \sqrt {d x} d^{19} - 456533 \, \left (-\frac {d^{26}}{a b^{15}}\right )^{\frac {3}{4}} a b^{11}\right ) + 4 \, {\left (153 \, b^{2} d^{6} x^{5} + 198 \, a b d^{6} x^{3} + 77 \, a^{2} d^{6} x\right )} \sqrt {d x}}{768 \, {\left (b^{6} x^{6} + 3 \, a b^{5} x^{4} + 3 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(13/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

-1/768*(924*(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2*b^4*x^2 + a^3*b^3)*(-d^26/(a*b^15))^(1/4)*arctan(-((-d^26/(a*b^15))
^(1/4)*sqrt(d*x)*b^4*d^19 - sqrt(d^39*x - sqrt(-d^26/(a*b^15))*a*b^7*d^26)*(-d^26/(a*b^15))^(1/4)*b^4)/d^26) -
231*(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2*b^4*x^2 + a^3*b^3)*(-d^26/(a*b^15))^(1/4)*log(456533*sqrt(d*x)*d^19 + 4565
33*(-d^26/(a*b^15))^(3/4)*a*b^11) + 231*(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2*b^4*x^2 + a^3*b^3)*(-d^26/(a*b^15))^(1/
4)*log(456533*sqrt(d*x)*d^19 - 456533*(-d^26/(a*b^15))^(3/4)*a*b^11) + 4*(153*b^2*d^6*x^5 + 198*a*b*d^6*x^3 +
77*a^2*d^6*x)*sqrt(d*x))/(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2*b^4*x^2 + a^3*b^3)

________________________________________________________________________________________

giac [A]  time = 0.20, size = 314, normalized size = 0.94 \[ -\frac {1}{1536} \, d^{6} {\left (\frac {8 \, {\left (153 \, \sqrt {d x} b^{2} d^{6} x^{5} + 198 \, \sqrt {d x} a b d^{6} x^{3} + 77 \, \sqrt {d x} a^{2} d^{6} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} b^{3}} - \frac {462 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{6} d} - \frac {462 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{6} d} + \frac {231 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{6} d} - \frac {231 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{6} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(13/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

-1/1536*d^6*(8*(153*sqrt(d*x)*b^2*d^6*x^5 + 198*sqrt(d*x)*a*b*d^6*x^3 + 77*sqrt(d*x)*a^2*d^6*x)/((b*d^2*x^2 +
a*d^2)^3*b^3) - 462*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*b^6*d) - 462*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*b^6*d) + 231*sqrt(2)*(a*b^3*d^2)^(3/4)*log(d*x + sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + s
qrt(a*d^2/b))/(a*b^6*d) - 231*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*b^6*d))

________________________________________________________________________________________

maple [A]  time = 0.02, size = 271, normalized size = 0.81 \[ -\frac {77 \left (d x \right )^{\frac {3}{2}} a^{2} d^{11}}{192 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b^{3}}-\frac {33 \left (d x \right )^{\frac {7}{2}} a \,d^{9}}{32 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b^{2}}-\frac {51 \left (d x \right )^{\frac {11}{2}} d^{7}}{64 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b}+\frac {77 \sqrt {2}\, d^{7} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{4}}+\frac {77 \sqrt {2}\, d^{7} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{256 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{4}}+\frac {77 \sqrt {2}\, d^{7} \ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{512 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(13/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

-51/64*d^7/(b*d^2*x^2+a*d^2)^3/b*(d*x)^(11/2)-33/32*d^9/(b*d^2*x^2+a*d^2)^3/b^2*a*(d*x)^(7/2)-77/192*d^11/(b*d
^2*x^2+a*d^2)^3/b^3*a^2*(d*x)^(3/2)+77/512*d^7/b^4/(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)))+77/256*d^7/b^4/(a/b*d^2)^
(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)+1)+77/256*d^7/b^4/(a/b*d^2)^(1/4)*2^(1/2)*arctan(2^(1
/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)-1)

________________________________________________________________________________________

maxima [A]  time = 3.44, size = 317, normalized size = 0.95 \[ \frac {\frac {231 \, d^{8} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{b^{3}} - \frac {8 \, {\left (153 \, \left (d x\right )^{\frac {11}{2}} b^{2} d^{8} + 198 \, \left (d x\right )^{\frac {7}{2}} a b d^{10} + 77 \, \left (d x\right )^{\frac {3}{2}} a^{2} d^{12}\right )}}{b^{6} d^{6} x^{6} + 3 \, a b^{5} d^{6} x^{4} + 3 \, a^{2} b^{4} d^{6} x^{2} + a^{3} b^{3} d^{6}}}{1536 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(13/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

1/1536*(231*d^8*(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))/(sqrt(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)*sqrt(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 - sqr
t(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)))/b^3 - 8*(153*(d*x)^(11/2)*b^2*d^8 +
198*(d*x)^(7/2)*a*b*d^10 + 77*(d*x)^(3/2)*a^2*d^12)/(b^6*d^6*x^6 + 3*a*b^5*d^6*x^4 + 3*a^2*b^4*d^6*x^2 + a^3*
b^3*d^6))/d

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 153, normalized size = 0.46 \[ \frac {77\,d^{13/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{1/4}\,b^{15/4}}-\frac {\frac {51\,d^7\,{\left (d\,x\right )}^{11/2}}{64\,b}+\frac {77\,a^2\,d^{11}\,{\left (d\,x\right )}^{3/2}}{192\,b^3}+\frac {33\,a\,d^9\,{\left (d\,x\right )}^{7/2}}{32\,b^2}}{a^3\,d^6+3\,a^2\,b\,d^6\,x^2+3\,a\,b^2\,d^6\,x^4+b^3\,d^6\,x^6}-\frac {77\,d^{13/2}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{1/4}\,b^{15/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(13/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

(77*d^(13/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*(-a)^(1/4)*b^(15/4)) - ((51*d^7*(d*x)^(11/
2))/(64*b) + (77*a^2*d^11*(d*x)^(3/2))/(192*b^3) + (33*a*d^9*(d*x)^(7/2))/(32*b^2))/(a^3*d^6 + b^3*d^6*x^6 + 3
*a^2*b*d^6*x^2 + 3*a*b^2*d^6*x^4) - (77*d^(13/2)*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*(-a)^
(1/4)*b^(15/4))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{\frac {13}{2}}}{\left (a + b x^{2}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(13/2)/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

Integral((d*x)**(13/2)/(a + b*x**2)**4, x)

________________________________________________________________________________________