3.332 \(\int \frac {x^{9/2}}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=218 \[ \frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}-\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{3/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{5/4} c^{3/4}}+\frac {x^{3/2}}{2 b \left (b+c x^2\right )} \]

[Out]

1/2*x^(3/2)/b/(c*x^2+b)-1/8*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(5/4)/c^(3/4)*2^(1/2)+1/8*arctan(1+c^(
1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(5/4)/c^(3/4)*2^(1/2)+1/16*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2
))/b^(5/4)/c^(3/4)*2^(1/2)-1/16*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(5/4)/c^(3/4)*2^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1584, 290, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}-\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{3/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{5/4} c^{3/4}}+\frac {x^{3/2}}{2 b \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^(9/2)/(b*x^2 + c*x^4)^2,x]

[Out]

x^(3/2)/(2*b*(b + c*x^2)) - ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(4*Sqrt[2]*b^(5/4)*c^(3/4)) + ArcTan
[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(4*Sqrt[2]*b^(5/4)*c^(3/4)) + Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sq
rt[x] + Sqrt[c]*x]/(8*Sqrt[2]*b^(5/4)*c^(3/4)) - Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]/(8
*Sqrt[2]*b^(5/4)*c^(3/4))

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 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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {\sqrt {x}}{\left (b+c x^2\right )^2} \, dx\\ &=\frac {x^{3/2}}{2 b \left (b+c x^2\right )}+\frac {\int \frac {\sqrt {x}}{b+c x^2} \, dx}{4 b}\\ &=\frac {x^{3/2}}{2 b \left (b+c x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 b}\\ &=\frac {x^{3/2}}{2 b \left (b+c x^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b \sqrt {c}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b \sqrt {c}}\\ &=\frac {x^{3/2}}{2 b \left (b+c x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{8 b c}+\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{8 b c}+\frac {\operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}\\ &=\frac {x^{3/2}}{2 b \left (b+c x^2\right )}+\frac {\log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}-\frac {\log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{3/4}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{3/4}}\\ &=\frac {x^{3/2}}{2 b \left (b+c x^2\right )}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{3/4}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{3/4}}+\frac {\log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}-\frac {\log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.13 \[ \frac {2 x^{3/2} \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)])/(3*b^2)

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fricas [A]  time = 0.86, size = 182, normalized size = 0.83 \[ -\frac {4 \, {\left (b c x^{2} + b^{2}\right )} \left (-\frac {1}{b^{5} c^{3}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {-b^{3} c \sqrt {-\frac {1}{b^{5} c^{3}}} + x} b c \left (-\frac {1}{b^{5} c^{3}}\right )^{\frac {1}{4}} - b c \sqrt {x} \left (-\frac {1}{b^{5} c^{3}}\right )^{\frac {1}{4}}\right ) - {\left (b c x^{2} + b^{2}\right )} \left (-\frac {1}{b^{5} c^{3}}\right )^{\frac {1}{4}} \log \left (b^{4} c^{2} \left (-\frac {1}{b^{5} c^{3}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) + {\left (b c x^{2} + b^{2}\right )} \left (-\frac {1}{b^{5} c^{3}}\right )^{\frac {1}{4}} \log \left (-b^{4} c^{2} \left (-\frac {1}{b^{5} c^{3}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 4 \, x^{\frac {3}{2}}}{8 \, {\left (b c x^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*(b*c*x^2 + b^2)*(-1/(b^5*c^3))^(1/4)*arctan(sqrt(-b^3*c*sqrt(-1/(b^5*c^3)) + x)*b*c*(-1/(b^5*c^3))^(1/
4) - b*c*sqrt(x)*(-1/(b^5*c^3))^(1/4)) - (b*c*x^2 + b^2)*(-1/(b^5*c^3))^(1/4)*log(b^4*c^2*(-1/(b^5*c^3))^(3/4)
+ sqrt(x)) + (b*c*x^2 + b^2)*(-1/(b^5*c^3))^(1/4)*log(-b^4*c^2*(-1/(b^5*c^3))^(3/4) + sqrt(x)) - 4*x^(3/2))/(
b*c*x^2 + b^2)

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giac [A]  time = 0.18, size = 199, normalized size = 0.91 \[ \frac {x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} b} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{2} c^{3}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{2} c^{3}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{2} c^{3}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{2} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(3/2)/((c*x^2 + b)*b) + 1/8*sqrt(2)*(b*c^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(
b/c)^(1/4))/(b^2*c^3) + 1/8*sqrt(2)*(b*c^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^
(1/4))/(b^2*c^3) - 1/16*sqrt(2)*(b*c^3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^3) + 1/1
6*sqrt(2)*(b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^3)

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maple [A]  time = 0.01, size = 158, normalized size = 0.72 \[ \frac {x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) b}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c}+\frac {\sqrt {2}\, \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(3/2)/b/(c*x^2+b)+1/16/b/c/(b/c)^(1/4)*2^(1/2)*ln((x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x+(b/c)^(
1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2)))+1/8/b/c/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+1/8/b/c/(
b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 3.04, size = 194, normalized size = 0.89 \[ \frac {x^{\frac {3}{2}}}{2 \, {\left (b c x^{2} + b^{2}\right )}} + \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(3/2)/(b*c*x^2 + b^2) + 1/16*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))
/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/
4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c
^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(
c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/b

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mupad [B]  time = 4.34, size = 64, normalized size = 0.29 \[ \frac {x^{3/2}}{2\,b\,\left (c\,x^2+b\right )}-\frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{4\,{\left (-b\right )}^{5/4}\,c^{3/4}}+\frac {\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{4\,{\left (-b\right )}^{5/4}\,c^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)/(b*x^2 + c*x^4)^2,x)

[Out]

x^(3/2)/(2*b*(b + c*x^2)) - atan((c^(1/4)*x^(1/2))/(-b)^(1/4))/(4*(-b)^(5/4)*c^(3/4)) + atanh((c^(1/4)*x^(1/2)
)/(-b)^(1/4))/(4*(-b)^(5/4)*c^(3/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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