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

Optimal. Leaf size=316 \[ -\frac {9 a^{5/4} d^{11/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} b^{13/4}}+\frac {9 a^{5/4} d^{11/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} b^{13/4}}-\frac {9 a^{5/4} d^{11/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} b^{13/4}}+\frac {9 a^{5/4} d^{11/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} b^{13/4}}-\frac {9 a d^5 \sqrt {d x}}{2 b^3}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}+\frac {9 d^3 (d x)^{5/2}}{10 b^2} \]

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Rubi [A]  time = 0.38, antiderivative size = 316, 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, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {9 a^{5/4} 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 )}{8 \sqrt {2} b^{13/4}}+\frac {9 a^{5/4} 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 )}{8 \sqrt {2} b^{13/4}}-\frac {9 a^{5/4} d^{11/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} b^{13/4}}+\frac {9 a^{5/4} d^{11/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} b^{13/4}}-\frac {9 a d^5 \sqrt {d x}}{2 b^3}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}+\frac {9 d^3 (d x)^{5/2}}{10 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a*d^5*Sqrt[d*x])/(2*b^3) + (9*d^3*(d*x)^(5/2))/(10*b^2) - (d*(d*x)^(9/2))/(2*b*(a + b*x^2)) - (9*a^(5/4)*d
^(11/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*b^(13/4)) + (9*a^(5/4)*d^(11/2)*
ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*b^(13/4)) - (9*a^(5/4)*d^(11/2)*Log[Sqrt
[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*b^(13/4)) + (9*a^(5/4)*d^(11/
2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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}}{a^2+2 a b x^2+b^2 x^4} \, dx &=b^2 \int \frac {(d x)^{11/2}}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}+\frac {1}{4} \left (9 d^2\right ) \int \frac {(d x)^{7/2}}{a b+b^2 x^2} \, dx\\ &=\frac {9 d^3 (d x)^{5/2}}{10 b^2}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}-\frac {\left (9 a d^4\right ) \int \frac {(d x)^{3/2}}{a b+b^2 x^2} \, dx}{4 b}\\ &=-\frac {9 a d^5 \sqrt {d x}}{2 b^3}+\frac {9 d^3 (d x)^{5/2}}{10 b^2}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}+\frac {\left (9 a^2 d^6\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )} \, dx}{4 b^2}\\ &=-\frac {9 a d^5 \sqrt {d x}}{2 b^3}+\frac {9 d^3 (d x)^{5/2}}{10 b^2}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}+\frac {\left (9 a^2 d^5\right ) \operatorname {Subst}\left (\int \frac {1}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{2 b^2}\\ &=-\frac {9 a d^5 \sqrt {d x}}{2 b^3}+\frac {9 d^3 (d x)^{5/2}}{10 b^2}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}+\frac {\left (9 a^{3/2} 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 )}{4 b^2}+\frac {\left (9 a^{3/2} 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 )}{4 b^2}\\ &=-\frac {9 a d^5 \sqrt {d x}}{2 b^3}+\frac {9 d^3 (d x)^{5/2}}{10 b^2}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}-\frac {\left (9 a^{5/4} 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 )}{8 \sqrt {2} b^{13/4}}-\frac {\left (9 a^{5/4} 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 )}{8 \sqrt {2} b^{13/4}}+\frac {\left (9 a^{3/2} 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 )}{8 b^{7/2}}+\frac {\left (9 a^{3/2} 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 )}{8 b^{7/2}}\\ &=-\frac {9 a d^5 \sqrt {d x}}{2 b^3}+\frac {9 d^3 (d x)^{5/2}}{10 b^2}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}-\frac {9 a^{5/4} 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 )}{8 \sqrt {2} b^{13/4}}+\frac {9 a^{5/4} 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 )}{8 \sqrt {2} b^{13/4}}+\frac {\left (9 a^{5/4} 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 )}{4 \sqrt {2} b^{13/4}}-\frac {\left (9 a^{5/4} 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 )}{4 \sqrt {2} b^{13/4}}\\ &=-\frac {9 a d^5 \sqrt {d x}}{2 b^3}+\frac {9 d^3 (d x)^{5/2}}{10 b^2}-\frac {d (d x)^{9/2}}{2 b \left (a+b x^2\right )}-\frac {9 a^{5/4} d^{11/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} b^{13/4}}+\frac {9 a^{5/4} d^{11/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} b^{13/4}}-\frac {9 a^{5/4} 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 )}{8 \sqrt {2} b^{13/4}}+\frac {9 a^{5/4} 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 )}{8 \sqrt {2} b^{13/4}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 235, normalized size = 0.74 \begin {gather*} \frac {d^5 \sqrt {d x} \left (-45 \sqrt {2} a^{5/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )+45 \sqrt {2} a^{5/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )-90 \sqrt {2} a^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )+90 \sqrt {2} a^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )+\frac {8 \sqrt [4]{b} \sqrt {x} \left (-45 a^2-36 a b x^2+4 b^2 x^4\right )}{a+b x^2}\right )}{80 b^{13/4} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^5*Sqrt[d*x]*((8*b^(1/4)*Sqrt[x]*(-45*a^2 - 36*a*b*x^2 + 4*b^2*x^4))/(a + b*x^2) - 90*Sqrt[2]*a^(5/4)*ArcTan
[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)] + 90*Sqrt[2]*a^(5/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)] - 4
5*Sqrt[2]*a^(5/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] + 45*Sqrt[2]*a^(5/4)*Log[Sqrt[a]
+ Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x]))/(80*b^(13/4)*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.50, size = 218, normalized size = 0.69 \begin {gather*} -\frac {9 a^{5/4} 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 )}{4 \sqrt {2} b^{13/4}}+\frac {9 a^{5/4} 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 )}{4 \sqrt {2} b^{13/4}}+\frac {-45 a^2 d^7 \sqrt {d x}-36 a b d^5 (d x)^{5/2}+4 b^2 d^3 (d x)^{9/2}}{10 b^3 \left (a d^2+b d^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-45*a^2*d^7*Sqrt[d*x] - 36*a*b*d^5*(d*x)^(5/2) + 4*b^2*d^3*(d*x)^(9/2))/(10*b^3*(a*d^2 + b*d^2*x^2)) - (9*a^(
5/4)*d^(11/2)*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]])
/(4*Sqrt[2]*b^(13/4)) + (9*a^(5/4)*d^(11/2)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x])/(Sqrt[a]*d + S
qrt[b]*d*x)])/(4*Sqrt[2]*b^(13/4))

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fricas [A]  time = 2.07, size = 283, normalized size = 0.90 \begin {gather*} \frac {180 \, \left (-\frac {a^{5} d^{22}}{b^{13}}\right )^{\frac {1}{4}} {\left (b^{4} x^{2} + a b^{3}\right )} \arctan \left (-\frac {\left (-\frac {a^{5} d^{22}}{b^{13}}\right )^{\frac {3}{4}} \sqrt {d x} a b^{10} d^{5} - \left (-\frac {a^{5} d^{22}}{b^{13}}\right )^{\frac {3}{4}} \sqrt {a^{2} d^{11} x + \sqrt {-\frac {a^{5} d^{22}}{b^{13}}} b^{6}} b^{10}}{a^{5} d^{22}}\right ) + 45 \, \left (-\frac {a^{5} d^{22}}{b^{13}}\right )^{\frac {1}{4}} {\left (b^{4} x^{2} + a b^{3}\right )} \log \left (9 \, \sqrt {d x} a d^{5} + 9 \, \left (-\frac {a^{5} d^{22}}{b^{13}}\right )^{\frac {1}{4}} b^{3}\right ) - 45 \, \left (-\frac {a^{5} d^{22}}{b^{13}}\right )^{\frac {1}{4}} {\left (b^{4} x^{2} + a b^{3}\right )} \log \left (9 \, \sqrt {d x} a d^{5} - 9 \, \left (-\frac {a^{5} d^{22}}{b^{13}}\right )^{\frac {1}{4}} b^{3}\right ) + 4 \, {\left (4 \, b^{2} d^{5} x^{4} - 36 \, a b d^{5} x^{2} - 45 \, a^{2} d^{5}\right )} \sqrt {d x}}{40 \, {\left (b^{4} x^{2} + a 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),x, algorithm="fricas")

[Out]

1/40*(180*(-a^5*d^22/b^13)^(1/4)*(b^4*x^2 + a*b^3)*arctan(-((-a^5*d^22/b^13)^(3/4)*sqrt(d*x)*a*b^10*d^5 - (-a^
5*d^22/b^13)^(3/4)*sqrt(a^2*d^11*x + sqrt(-a^5*d^22/b^13)*b^6)*b^10)/(a^5*d^22)) + 45*(-a^5*d^22/b^13)^(1/4)*(
b^4*x^2 + a*b^3)*log(9*sqrt(d*x)*a*d^5 + 9*(-a^5*d^22/b^13)^(1/4)*b^3) - 45*(-a^5*d^22/b^13)^(1/4)*(b^4*x^2 +
a*b^3)*log(9*sqrt(d*x)*a*d^5 - 9*(-a^5*d^22/b^13)^(1/4)*b^3) + 4*(4*b^2*d^5*x^4 - 36*a*b*d^5*x^2 - 45*a^2*d^5)
*sqrt(d*x))/(b^4*x^2 + a*b^3)

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giac [A]  time = 0.21, size = 297, normalized size = 0.94 \begin {gather*} -\frac {1}{80} \, d^{5} {\left (\frac {40 \, \sqrt {d x} a^{2} d^{2}}{{\left (b d^{2} x^{2} + a d^{2}\right )} b^{3}} - \frac {90 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} a \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 )}{b^{4}} - \frac {90 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} a \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 )}{b^{4}} - \frac {45 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} a \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 )}{b^{4}} + \frac {45 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} a \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 )}{b^{4}} - \frac {32 \, {\left (\sqrt {d x} b^{8} d^{10} x^{2} - 10 \, \sqrt {d x} a b^{7} d^{10}\right )}}{b^{10} d^{10}}\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),x, algorithm="giac")

[Out]

-1/80*d^5*(40*sqrt(d*x)*a^2*d^2/((b*d^2*x^2 + a*d^2)*b^3) - 90*sqrt(2)*(a*b^3*d^2)^(1/4)*a*arctan(1/2*sqrt(2)*
(sqrt(2)*(a*d^2/b)^(1/4) + 2*sqrt(d*x))/(a*d^2/b)^(1/4))/b^4 - 90*sqrt(2)*(a*b^3*d^2)^(1/4)*a*arctan(-1/2*sqrt
(2)*(sqrt(2)*(a*d^2/b)^(1/4) - 2*sqrt(d*x))/(a*d^2/b)^(1/4))/b^4 - 45*sqrt(2)*(a*b^3*d^2)^(1/4)*a*log(d*x + sq
rt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/b^4 + 45*sqrt(2)*(a*b^3*d^2)^(1/4)*a*log(d*x - sqrt(2)*(a*d^2
/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/b^4 - 32*(sqrt(d*x)*b^8*d^10*x^2 - 10*sqrt(d*x)*a*b^7*d^10)/(b^10*d^10))

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maple [A]  time = 0.02, size = 242, normalized size = 0.77 \begin {gather*} -\frac {\sqrt {d x}\, a^{2} d^{7}}{2 \left (b \,d^{2} x^{2}+d^{2} a \right ) b^{3}}+\frac {9 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \,d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{8 b^{3}}+\frac {9 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \,d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{8 b^{3}}+\frac {9 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \,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 )}{16 b^{3}}-\frac {4 \sqrt {d x}\, a \,d^{5}}{b^{3}}+\frac {2 \left (d x \right )^{\frac {5}{2}} d^{3}}{5 b^{2}} \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),x)

[Out]

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

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maxima [A]  time = 3.04, size = 300, normalized size = 0.95 \begin {gather*} -\frac {\frac {40 \, \sqrt {d x} a^{2} d^{8}}{b^{4} d^{2} x^{2} + a b^{3} d^{2}} - \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 )} a^{2}}{b^{3}} - \frac {32 \, {\left (\left (d x\right )^{\frac {5}{2}} b d^{4} - 10 \, \sqrt {d x} a d^{6}\right )}}{b^{3}}}{80 \, 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),x, algorithm="maxima")

[Out]

-1/80*(40*sqrt(d*x)*a^2*d^8/(b^4*d^2*x^2 + a*b^3*d^2) - 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)))*a^2/b^3 - 32*((d*x)^(5/2)*b*d^4 - 10*sqrt(d*x)*a*d^6)/b^3)/d

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mupad [B]  time = 4.27, size = 129, normalized size = 0.41 \begin {gather*} \frac {2\,d^3\,{\left (d\,x\right )}^{5/2}}{5\,b^2}-\frac {9\,{\left (-a\right )}^{5/4}\,d^{11/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{4\,b^{13/4}}-\frac {a^2\,d^7\,\sqrt {d\,x}}{2\,\left (b^4\,d^2\,x^2+a\,b^3\,d^2\right )}-\frac {4\,a\,d^5\,\sqrt {d\,x}}{b^3}+\frac {{\left (-a\right )}^{5/4}\,d^{11/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )\,9{}\mathrm {i}}{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),x)

[Out]

(2*d^3*(d*x)^(5/2))/(5*b^2) - (9*(-a)^(5/4)*d^(11/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(4*b^(1
3/4)) + ((-a)^(5/4)*d^(11/2)*atan((b^(1/4)*(d*x)^(1/2)*1i)/((-a)^(1/4)*d^(1/2)))*9i)/(4*b^(13/4)) - (a^2*d^7*(
d*x)^(1/2))/(2*(a*b^3*d^2 + b^4*d^2*x^2)) - (4*a*d^5*(d*x)^(1/2))/b^3

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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 )^{2}}\, 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),x)

[Out]

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

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