3.687 \(\int \frac {(d x)^{9/2}}{a^2+2 a b x^2+b^2 x^4} \, dx\)

Optimal. Leaf size=298 \[ -\frac {7 a^{3/4} d^{9/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} b^{11/4}}+\frac {7 a^{3/4} d^{9/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} b^{11/4}}+\frac {7 a^{3/4} d^{9/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} b^{11/4}}-\frac {7 a^{3/4} d^{9/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} b^{11/4}}-\frac {d (d x)^{7/2}}{2 b \left (a+b x^2\right )}+\frac {7 d^3 (d x)^{3/2}}{6 b^2} \]

[Out]

7/6*d^3*(d*x)^(3/2)/b^2-1/2*d*(d*x)^(7/2)/b/(b*x^2+a)+7/8*a^(3/4)*d^(9/2)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)
/a^(1/4)/d^(1/2))/b^(11/4)*2^(1/2)-7/8*a^(3/4)*d^(9/2)*arctan(1+b^(1/4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/b
^(11/4)*2^(1/2)-7/16*a^(3/4)*d^(9/2)*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))
/b^(11/4)*2^(1/2)+7/16*a^(3/4)*d^(9/2)*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
))/b^(11/4)*2^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {28, 288, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {7 a^{3/4} d^{9/2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} b^{11/4}}+\frac {7 a^{3/4} d^{9/2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{8 \sqrt {2} b^{11/4}}+\frac {7 a^{3/4} d^{9/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{4 \sqrt {2} b^{11/4}}-\frac {7 a^{3/4} d^{9/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{4 \sqrt {2} b^{11/4}}-\frac {d (d x)^{7/2}}{2 b \left (a+b x^2\right )}+\frac {7 d^3 (d x)^{3/2}}{6 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(9/2)/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

(7*d^3*(d*x)^(3/2))/(6*b^2) - (d*(d*x)^(7/2))/(2*b*(a + b*x^2)) + (7*a^(3/4)*d^(9/2)*ArcTan[1 - (Sqrt[2]*b^(1/
4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*b^(11/4)) - (7*a^(3/4)*d^(9/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d
*x])/(a^(1/4)*Sqrt[d])])/(4*Sqrt[2]*b^(11/4)) - (7*a^(3/4)*d^(9/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - S
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*b^(11/4)) + (7*a^(3/4)*d^(9/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqr
t[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(8*Sqrt[2]*b^(11/4))

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

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Mathematica [C]  time = 0.02, size = 63, normalized size = 0.21 \[ -\frac {2 d^4 x \sqrt {d x} \left (7 \left (a+b x^2\right ) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {b x^2}{a}\right )-7 a-b x^2\right )}{3 b^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(9/2)/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

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

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fricas [A]  time = 0.86, size = 283, normalized size = 0.95 \[ \frac {84 \, \left (-\frac {a^{3} d^{18}}{b^{11}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )} \arctan \left (-\frac {\left (-\frac {a^{3} d^{18}}{b^{11}}\right )^{\frac {1}{4}} \sqrt {d x} a^{2} b^{3} d^{13} - \sqrt {a^{4} d^{27} x - \sqrt {-\frac {a^{3} d^{18}}{b^{11}}} a^{3} b^{5} d^{18}} \left (-\frac {a^{3} d^{18}}{b^{11}}\right )^{\frac {1}{4}} b^{3}}{a^{3} d^{18}}\right ) - 21 \, \left (-\frac {a^{3} d^{18}}{b^{11}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )} \log \left (343 \, \sqrt {d x} a^{2} d^{13} + 343 \, \left (-\frac {a^{3} d^{18}}{b^{11}}\right )^{\frac {3}{4}} b^{8}\right ) + 21 \, \left (-\frac {a^{3} d^{18}}{b^{11}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )} \log \left (343 \, \sqrt {d x} a^{2} d^{13} - 343 \, \left (-\frac {a^{3} d^{18}}{b^{11}}\right )^{\frac {3}{4}} b^{8}\right ) + 4 \, {\left (4 \, b d^{4} x^{3} + 7 \, a d^{4} x\right )} \sqrt {d x}}{24 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(9/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

1/24*(84*(-a^3*d^18/b^11)^(1/4)*(b^3*x^2 + a*b^2)*arctan(-((-a^3*d^18/b^11)^(1/4)*sqrt(d*x)*a^2*b^3*d^13 - sqr
t(a^4*d^27*x - sqrt(-a^3*d^18/b^11)*a^3*b^5*d^18)*(-a^3*d^18/b^11)^(1/4)*b^3)/(a^3*d^18)) - 21*(-a^3*d^18/b^11
)^(1/4)*(b^3*x^2 + a*b^2)*log(343*sqrt(d*x)*a^2*d^13 + 343*(-a^3*d^18/b^11)^(3/4)*b^8) + 21*(-a^3*d^18/b^11)^(
1/4)*(b^3*x^2 + a*b^2)*log(343*sqrt(d*x)*a^2*d^13 - 343*(-a^3*d^18/b^11)^(3/4)*b^8) + 4*(4*b*d^4*x^3 + 7*a*d^4
*x)*sqrt(d*x))/(b^3*x^2 + a*b^2)

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giac [A]  time = 0.19, size = 277, normalized size = 0.93 \[ \frac {1}{48} \, {\left (\frac {24 \, \sqrt {d x} a d^{2} x}{{\left (b d^{2} x^{2} + a d^{2}\right )} b^{2}} + \frac {32 \, \sqrt {d x} x}{b^{2}} - \frac {42 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{b^{5} d} - \frac {42 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{b^{5} d} + \frac {21 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{b^{5} d} - \frac {21 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{b^{5} d}\right )} d^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(9/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

1/48*(24*sqrt(d*x)*a*d^2*x/((b*d^2*x^2 + a*d^2)*b^2) + 32*sqrt(d*x)*x/b^2 - 42*sqrt(2)*(a*b^3*d^2)^(3/4)*arcta
n(1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4) + 2*sqrt(d*x))/(a*d^2/b)^(1/4))/(b^5*d) - 42*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))/(b^5*d) + 21*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))/(b^5*d) - 21*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))/(b^5*d))*d^4

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maple [A]  time = 0.02, size = 226, normalized size = 0.76 \[ \frac {\left (d x \right )^{\frac {3}{2}} a \,d^{5}}{2 \left (b \,d^{2} x^{2}+d^{2} a \right ) b^{2}}-\frac {7 \sqrt {2}\, a \,d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{3}}-\frac {7 \sqrt {2}\, a \,d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{3}}-\frac {7 \sqrt {2}\, a \,d^{5} \ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )}{16 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{3}}+\frac {2 \left (d x \right )^{\frac {3}{2}} d^{3}}{3 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(9/2)/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

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

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maxima [A]  time = 3.12, size = 273, normalized size = 0.92 \[ \frac {\frac {24 \, \left (d x\right )^{\frac {3}{2}} a d^{6}}{b^{3} d^{2} x^{2} + a b^{2} d^{2}} - \frac {21 \, a d^{6} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{b^{2}} + \frac {32 \, \left (d x\right )^{\frac {3}{2}} d^{4}}{b^{2}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(9/2)/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

1/48*(24*(d*x)^(3/2)*a*d^6/(b^3*d^2*x^2 + a*b^2*d^2) - 21*a*d^6*(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)*a
rctan(-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 - sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/
4)*b^(3/4)))/b^2 + 32*(d*x)^(3/2)*d^4/b^2)/d

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mupad [B]  time = 0.12, size = 112, normalized size = 0.38 \[ \frac {2\,d^3\,{\left (d\,x\right )}^{3/2}}{3\,b^2}+\frac {7\,{\left (-a\right )}^{3/4}\,d^{9/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{4\,b^{11/4}}+\frac {a\,d^5\,{\left (d\,x\right )}^{3/2}}{2\,\left (b^3\,d^2\,x^2+a\,b^2\,d^2\right )}+\frac {{\left (-a\right )}^{3/4}\,d^{9/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )\,7{}\mathrm {i}}{4\,b^{11/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(9/2)/(a^2 + b^2*x^4 + 2*a*b*x^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{\frac {9}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(9/2)/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

Integral((d*x)**(9/2)/(a + b*x**2)**2, x)

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