3.6.22 \(\int \frac {(d x)^{11/2}}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=333 \[ -\frac {15 d^{11/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{3/4} b^{13/4}}+\frac {15 d^{11/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{256 \sqrt {2} a^{3/4} b^{13/4}}-\frac {15 d^{11/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{3/4} b^{13/4}}+\frac {15 d^{11/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{3/4} b^{13/4}}-\frac {15 d^5 \sqrt {d x}}{64 b^3 \left (a+b x^2\right )}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3} \]

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Rubi [A]  time = 0.34, 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, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {15 d^{11/2} \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} a^{3/4} b^{13/4}}+\frac {15 d^{11/2} \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} a^{3/4} b^{13/4}}-\frac {15 d^{11/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{3/4} b^{13/4}}+\frac {15 d^{11/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{128 \sqrt {2} a^{3/4} b^{13/4}}-\frac {15 d^5 \sqrt {d x}}{64 b^3 \left (a+b x^2\right )}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(11/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

-(d*(d*x)^(9/2))/(6*b*(a + b*x^2)^3) - (3*d^3*(d*x)^(5/2))/(16*b^2*(a + b*x^2)^2) - (15*d^5*Sqrt[d*x])/(64*b^3
*(a + b*x^2)) - (15*d^(11/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(3/4)*b
^(13/4)) + (15*d^(11/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(3/4)*b^(13/
4)) - (15*d^(11/2)*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^(3/4)*b^(13/4)) + (15*d^(11/2)*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^(3/4)*b^(13/4))

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 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)^{11/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {(d x)^{11/2}}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3}+\frac {1}{4} \left (3 b^2 d^2\right ) \int \frac {(d x)^{7/2}}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}+\frac {1}{32} \left (15 d^4\right ) \int \frac {(d x)^{3/2}}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {15 d^5 \sqrt {d x}}{64 b^3 \left (a+b x^2\right )}+\frac {\left (15 d^6\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )} \, dx}{128 b^2}\\ &=-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {15 d^5 \sqrt {d x}}{64 b^3 \left (a+b x^2\right )}+\frac {\left (15 d^5\right ) \operatorname {Subst}\left (\int \frac {1}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{64 b^2}\\ &=-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {15 d^5 \sqrt {d x}}{64 b^3 \left (a+b x^2\right )}+\frac {\left (15 d^4\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 \sqrt {a} b^2}+\frac {\left (15 d^4\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 \sqrt {a} b^2}\\ &=-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {15 d^5 \sqrt {d x}}{64 b^3 \left (a+b x^2\right )}-\frac {\left (15 d^{11/2}\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} a^{3/4} b^{13/4}}-\frac {\left (15 d^{11/2}\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} a^{3/4} b^{13/4}}+\frac {\left (15 d^6\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 \sqrt {a} b^{7/2}}+\frac {\left (15 d^6\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 \sqrt {a} b^{7/2}}\\ &=-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {15 d^5 \sqrt {d x}}{64 b^3 \left (a+b x^2\right )}-\frac {15 d^{11/2} \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} a^{3/4} b^{13/4}}+\frac {15 d^{11/2} \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} a^{3/4} b^{13/4}}+\frac {\left (15 d^{11/2}\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} a^{3/4} b^{13/4}}-\frac {\left (15 d^{11/2}\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} a^{3/4} b^{13/4}}\\ &=-\frac {d (d x)^{9/2}}{6 b \left (a+b x^2\right )^3}-\frac {3 d^3 (d x)^{5/2}}{16 b^2 \left (a+b x^2\right )^2}-\frac {15 d^5 \sqrt {d x}}{64 b^3 \left (a+b x^2\right )}-\frac {15 d^{11/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{3/4} b^{13/4}}+\frac {15 d^{11/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{3/4} b^{13/4}}-\frac {15 d^{11/2} \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} a^{3/4} b^{13/4}}+\frac {15 d^{11/2} \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} a^{3/4} b^{13/4}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 299, normalized size = 0.90 \begin {gather*} \frac {d^5 \sqrt {d x} \left (-\frac {315 \sqrt {2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4} \sqrt {x}}+\frac {315 \sqrt {2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4} \sqrt {x}}-\frac {630 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{a^{3/4} \sqrt {x}}+\frac {630 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{a^{3/4} \sqrt {x}}-\frac {3840 a^2 \sqrt [4]{b}}{\left (a+b x^2\right )^3}-\frac {9216 a b^{5/4} x^2}{\left (a+b x^2\right )^3}-\frac {7168 b^{9/4} x^4}{\left (a+b x^2\right )^3}+\frac {840 \sqrt [4]{b}}{a+b x^2}+\frac {480 a \sqrt [4]{b}}{\left (a+b x^2\right )^2}\right )}{10752 b^{13/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(11/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(d^5*Sqrt[d*x]*((-3840*a^2*b^(1/4))/(a + b*x^2)^3 - (9216*a*b^(5/4)*x^2)/(a + b*x^2)^3 - (7168*b^(9/4)*x^4)/(a
+ b*x^2)^3 + (480*a*b^(1/4))/(a + b*x^2)^2 + (840*b^(1/4))/(a + b*x^2) - (630*Sqrt[2]*ArcTan[1 - (Sqrt[2]*b^(
1/4)*Sqrt[x])/a^(1/4)])/(a^(3/4)*Sqrt[x]) + (630*Sqrt[2]*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(a^(3/
4)*Sqrt[x]) - (315*Sqrt[2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(a^(3/4)*Sqrt[x]) + (31
5*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(a^(3/4)*Sqrt[x])))/(10752*b^(13/4))

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IntegrateAlgebraic [A]  time = 0.81, size = 213, normalized size = 0.64 \begin {gather*} -\frac {15 d^{11/2} \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 )}{128 \sqrt {2} a^{3/4} b^{13/4}}+\frac {15 d^{11/2} \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 )}{128 \sqrt {2} a^{3/4} b^{13/4}}-\frac {d^7 \sqrt {d x} \left (45 a^2 d^4+126 a b d^4 x^2+113 b^2 d^4 x^4\right )}{192 b^3 \left (a d^2+b d^2 x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d*x)^(11/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

-1/192*(d^7*Sqrt[d*x]*(45*a^2*d^4 + 126*a*b*d^4*x^2 + 113*b^2*d^4*x^4))/(b^3*(a*d^2 + b*d^2*x^2)^3) - (15*d^(1
1/2)*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]])/(128*Sqr
t[2]*a^(3/4)*b^(13/4)) + (15*d^(11/2)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x])/(Sqrt[a]*d + Sqrt[b]
*d*x)])/(128*Sqrt[2]*a^(3/4)*b^(13/4))

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fricas [A]  time = 1.77, size = 373, normalized size = 1.12 \begin {gather*} \frac {180 \, {\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^{22}}{a^{3} b^{13}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\left (-\frac {d^{22}}{a^{3} b^{13}}\right )^{\frac {3}{4}} \sqrt {d x} a^{2} b^{10} d^{5} - \sqrt {d^{11} x + \sqrt {-\frac {d^{22}}{a^{3} b^{13}}} a^{2} b^{6}} \left (-\frac {d^{22}}{a^{3} b^{13}}\right )^{\frac {3}{4}} a^{2} b^{10}}{d^{22}}\right ) + 45 \, {\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^{22}}{a^{3} b^{13}}\right )^{\frac {1}{4}} \log \left (15 \, \sqrt {d x} d^{5} + 15 \, \left (-\frac {d^{22}}{a^{3} b^{13}}\right )^{\frac {1}{4}} a b^{3}\right ) - 45 \, {\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^{22}}{a^{3} b^{13}}\right )^{\frac {1}{4}} \log \left (15 \, \sqrt {d x} d^{5} - 15 \, \left (-\frac {d^{22}}{a^{3} b^{13}}\right )^{\frac {1}{4}} a b^{3}\right ) - 4 \, {\left (113 \, b^{2} d^{5} x^{4} + 126 \, a b d^{5} x^{2} + 45 \, a^{2} d^{5}\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(11/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

1/768*(180*(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2*b^4*x^2 + a^3*b^3)*(-d^22/(a^3*b^13))^(1/4)*arctan(-((-d^22/(a^3*b^1
3))^(3/4)*sqrt(d*x)*a^2*b^10*d^5 - sqrt(d^11*x + sqrt(-d^22/(a^3*b^13))*a^2*b^6)*(-d^22/(a^3*b^13))^(3/4)*a^2*
b^10)/d^22) + 45*(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2*b^4*x^2 + a^3*b^3)*(-d^22/(a^3*b^13))^(1/4)*log(15*sqrt(d*x)*d
^5 + 15*(-d^22/(a^3*b^13))^(1/4)*a*b^3) - 45*(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2*b^4*x^2 + a^3*b^3)*(-d^22/(a^3*b^1
3))^(1/4)*log(15*sqrt(d*x)*d^5 - 15*(-d^22/(a^3*b^13))^(1/4)*a*b^3) - 4*(113*b^2*d^5*x^4 + 126*a*b*d^5*x^2 + 4
5*a^2*d^5)*sqrt(d*x))/(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2*b^4*x^2 + a^3*b^3)

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giac [A]  time = 0.21, size = 301, normalized size = 0.90 \begin {gather*} \frac {1}{1536} \, d^{5} {\left (\frac {90 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{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^{4}} + \frac {90 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{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^{4}} + \frac {45 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{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^{4}} - \frac {45 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{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^{4}} - \frac {8 \, {\left (113 \, \sqrt {d x} b^{2} d^{6} x^{4} + 126 \, \sqrt {d x} a b d^{6} x^{2} + 45 \, \sqrt {d x} a^{2} d^{6}\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} b^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(11/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

1/1536*d^5*(90*sqrt(2)*(a*b^3*d^2)^(1/4)*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^4) + 90*sqrt(2)*(a*b^3*d^2)^(1/4)*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^4) + 45*sqrt(2)*(a*b^3*d^2)^(1/4)*log(d*x + sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b
))/(a*b^4) - 45*sqrt(2)*(a*b^3*d^2)^(1/4)*log(d*x - sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/(a*b^4)
- 8*(113*sqrt(d*x)*b^2*d^6*x^4 + 126*sqrt(d*x)*a*b*d^6*x^2 + 45*sqrt(d*x)*a^2*d^6)/((b*d^2*x^2 + a*d^2)^3*b^3
))

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maple [A]  time = 0.02, size = 280, normalized size = 0.84 \begin {gather*} -\frac {15 \sqrt {d x}\, a^{2} d^{11}}{64 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b^{3}}-\frac {21 \left (d x \right )^{\frac {5}{2}} a \,d^{9}}{32 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b^{2}}-\frac {113 \left (d x \right )^{\frac {9}{2}} d^{7}}{192 \left (b \,d^{2} x^{2}+d^{2} a \right )^{3} b}+\frac {15 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{256 a \,b^{3}}+\frac {15 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{256 a \,b^{3}}+\frac {15 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{5} \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 a \,b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(11/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

-113/192*d^7/(b*d^2*x^2+a*d^2)^3/b*(d*x)^(9/2)-21/32*d^9/(b*d^2*x^2+a*d^2)^3/b^2*a*(d*x)^(5/2)-15/64*d^11/(b*d
^2*x^2+a*d^2)^3/b^3*a^2*(d*x)^(1/2)+15/512*d^5/b^3*(a/b*d^2)^(1/4)/a*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)))+15/256*d^5/b^3*(a/b*d^2
)^(1/4)/a*2^(1/2)*arctan(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)+1)+15/256*d^5/b^3*(a/b*d^2)^(1/4)/a*2^(1/2)*arcta
n(2^(1/2)/(a/b*d^2)^(1/4)*(d*x)^(1/2)-1)

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maxima [A]  time = 3.13, size = 326, normalized size = 0.98 \begin {gather*} -\frac {\frac {8 \, {\left (113 \, \left (d x\right )^{\frac {9}{2}} b^{2} d^{8} + 126 \, \left (d x\right )^{\frac {5}{2}} a b d^{10} + 45 \, \sqrt {d x} 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}} - \frac {45 \, {\left (\frac {\sqrt {2} d^{8} \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 {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} d^{8} \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 {3}{4}} b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} d^{7} \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 {a}} + \frac {2 \, \sqrt {2} d^{7} \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 {a}}\right )}}{b^{3}}}{1536 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(11/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

-1/1536*(8*(113*(d*x)^(9/2)*b^2*d^8 + 126*(d*x)^(5/2)*a*b*d^10 + 45*sqrt(d*x)*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) - 45*(sqrt(2)*d^8*log(sqrt(b)*d*x + sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x
)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(3/4)*b^(1/4)) - sqrt(2)*d^8*log(sqrt(b)*d*x - sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)
*b^(1/4) + sqrt(a)*d)/((a*d^2)^(3/4)*b^(1/4)) + 2*sqrt(2)*d^7*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(a)) + 2*sqrt(2)*d^7*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(a)))/b^3)/d

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mupad [B]  time = 4.29, size = 153, normalized size = 0.46 \begin {gather*} -\frac {\frac {113\,d^7\,{\left (d\,x\right )}^{9/2}}{192\,b}+\frac {15\,a^2\,d^{11}\,\sqrt {d\,x}}{64\,b^3}+\frac {21\,a\,d^9\,{\left (d\,x\right )}^{5/2}}{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 {15\,d^{11/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{3/4}\,b^{13/4}}-\frac {15\,d^{11/2}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{3/4}\,b^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(11/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

- ((113*d^7*(d*x)^(9/2))/(192*b) + (15*a^2*d^11*(d*x)^(1/2))/(64*b^3) + (21*a*d^9*(d*x)^(5/2))/(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) - (15*d^(11/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d
^(1/2))))/(128*(-a)^(3/4)*b^(13/4)) - (15*d^(11/2)*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*(-a
)^(3/4)*b^(13/4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d x\right )^{\frac {11}{2}}}{\left (a + b x^{2}\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(11/2)/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

Integral((d*x)**(11/2)/(a + b*x**2)**4, x)

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