3.3.28 \(\int \frac {x^{9/2}}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=251 \[ -\frac {45 \sqrt [4]{c} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{13/4}}+\frac {45 \sqrt [4]{c} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{13/4}}+\frac {45 \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{13/4}}-\frac {45 \sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{13/4}}-\frac {45}{16 b^3 \sqrt {x}}+\frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}+\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2} \]

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Rubi [A]  time = 0.22, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1584, 290, 325, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}-\frac {45 \sqrt [4]{c} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{13/4}}+\frac {45 \sqrt [4]{c} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{13/4}}+\frac {45 \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{13/4}}-\frac {45 \sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{13/4}}-\frac {45}{16 b^3 \sqrt {x}}+\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(9/2)/(b*x^2 + c*x^4)^3,x]

[Out]

-45/(16*b^3*Sqrt[x]) + 1/(4*b*Sqrt[x]*(b + c*x^2)^2) + 9/(16*b^2*Sqrt[x]*(b + c*x^2)) + (45*c^(1/4)*ArcTan[1 -
(Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(13/4)) - (45*c^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/
b^(1/4)])/(32*Sqrt[2]*b^(13/4)) - (45*c^(1/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*
Sqrt[2]*b^(13/4)) + (45*c^(1/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(13/
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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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]

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 )^3} \, dx &=\int \frac {1}{x^{3/2} \left (b+c x^2\right )^3} \, dx\\ &=\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2}+\frac {9 \int \frac {1}{x^{3/2} \left (b+c x^2\right )^2} \, dx}{8 b}\\ &=\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2}+\frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}+\frac {45 \int \frac {1}{x^{3/2} \left (b+c x^2\right )} \, dx}{32 b^2}\\ &=-\frac {45}{16 b^3 \sqrt {x}}+\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2}+\frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}-\frac {(45 c) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{32 b^3}\\ &=-\frac {45}{16 b^3 \sqrt {x}}+\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2}+\frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}-\frac {(45 c) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^3}\\ &=-\frac {45}{16 b^3 \sqrt {x}}+\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2}+\frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}+\frac {\left (45 \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^3}-\frac {\left (45 \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^3}\\ &=-\frac {45}{16 b^3 \sqrt {x}}+\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2}+\frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}-\frac {45 \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 )}{64 b^3}-\frac {45 \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 )}{64 b^3}-\frac {\left (45 \sqrt [4]{c}\right ) \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 )}{64 \sqrt {2} b^{13/4}}-\frac {\left (45 \sqrt [4]{c}\right ) \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 )}{64 \sqrt {2} b^{13/4}}\\ &=-\frac {45}{16 b^3 \sqrt {x}}+\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2}+\frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}-\frac {45 \sqrt [4]{c} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{13/4}}+\frac {45 \sqrt [4]{c} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{13/4}}-\frac {\left (45 \sqrt [4]{c}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{13/4}}+\frac {\left (45 \sqrt [4]{c}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{13/4}}\\ &=-\frac {45}{16 b^3 \sqrt {x}}+\frac {1}{4 b \sqrt {x} \left (b+c x^2\right )^2}+\frac {9}{16 b^2 \sqrt {x} \left (b+c x^2\right )}+\frac {45 \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{13/4}}-\frac {45 \sqrt [4]{c} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{13/4}}-\frac {45 \sqrt [4]{c} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{13/4}}+\frac {45 \sqrt [4]{c} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{13/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.11 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {1}{4},3;\frac {3}{4};-\frac {c x^2}{b}\right )}{b^3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(9/2)/(b*x^2 + c*x^4)^3,x]

[Out]

(-2*Hypergeometric2F1[-1/4, 3, 3/4, -((c*x^2)/b)])/(b^3*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.49, size = 160, normalized size = 0.64 \begin {gather*} \frac {45 \sqrt [4]{c} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2} \sqrt [4]{c}}-\frac {\sqrt [4]{c} x}{\sqrt {2} \sqrt [4]{b}}}{\sqrt {x}}\right )}{32 \sqrt {2} b^{13/4}}+\frac {45 \sqrt [4]{c} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{13/4}}+\frac {-32 b^2-81 b c x^2-45 c^2 x^4}{16 b^3 \sqrt {x} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(9/2)/(b*x^2 + c*x^4)^3,x]

[Out]

(-32*b^2 - 81*b*c*x^2 - 45*c^2*x^4)/(16*b^3*Sqrt[x]*(b + c*x^2)^2) + (45*c^(1/4)*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1
/4)) - (c^(1/4)*x)/(Sqrt[2]*b^(1/4)))/Sqrt[x]])/(32*Sqrt[2]*b^(13/4)) + (45*c^(1/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c
^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqrt[2]*b^(13/4))

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fricas [A]  time = 3.19, size = 263, normalized size = 1.05 \begin {gather*} \frac {180 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )} \left (-\frac {c}{b^{13}}\right )^{\frac {1}{4}} \arctan \left (-\frac {91125 \, b^{3} c \sqrt {x} \left (-\frac {c}{b^{13}}\right )^{\frac {1}{4}} - \sqrt {-8303765625 \, b^{7} c \sqrt {-\frac {c}{b^{13}}} + 8303765625 \, c^{2} x} b^{3} \left (-\frac {c}{b^{13}}\right )^{\frac {1}{4}}}{91125 \, c}\right ) - 45 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )} \left (-\frac {c}{b^{13}}\right )^{\frac {1}{4}} \log \left (91125 \, b^{10} \left (-\frac {c}{b^{13}}\right )^{\frac {3}{4}} + 91125 \, c \sqrt {x}\right ) + 45 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )} \left (-\frac {c}{b^{13}}\right )^{\frac {1}{4}} \log \left (-91125 \, b^{10} \left (-\frac {c}{b^{13}}\right )^{\frac {3}{4}} + 91125 \, c \sqrt {x}\right ) - 4 \, {\left (45 \, c^{2} x^{4} + 81 \, b c x^{2} + 32 \, b^{2}\right )} \sqrt {x}}{64 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/64*(180*(b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x)*(-c/b^13)^(1/4)*arctan(-1/91125*(91125*b^3*c*sqrt(x)*(-c/b^13)^(
1/4) - sqrt(-8303765625*b^7*c*sqrt(-c/b^13) + 8303765625*c^2*x)*b^3*(-c/b^13)^(1/4))/c) - 45*(b^3*c^2*x^5 + 2*
b^4*c*x^3 + b^5*x)*(-c/b^13)^(1/4)*log(91125*b^10*(-c/b^13)^(3/4) + 91125*c*sqrt(x)) + 45*(b^3*c^2*x^5 + 2*b^4
*c*x^3 + b^5*x)*(-c/b^13)^(1/4)*log(-91125*b^10*(-c/b^13)^(3/4) + 91125*c*sqrt(x)) - 4*(45*c^2*x^4 + 81*b*c*x^
2 + 32*b^2)*sqrt(x))/(b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x)

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giac [A]  time = 0.19, size = 220, normalized size = 0.88 \begin {gather*} -\frac {2}{b^{3} \sqrt {x}} - \frac {45 \, \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 )}{64 \, b^{4} c^{2}} - \frac {45 \, \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 )}{64 \, b^{4} c^{2}} + \frac {45 \, \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 )}{128 \, b^{4} c^{2}} - \frac {45 \, \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 )}{128 \, b^{4} c^{2}} - \frac {13 \, c^{2} x^{\frac {7}{2}} + 17 \, b c x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-2/(b^3*sqrt(x)) - 45/64*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^4*c^2) - 45/64*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^4*c^2) + 45/128*sqrt(2)*(b*c^3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c^2) - 45/128*s
qrt(2)*(b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c^2) - 1/16*(13*c^2*x^(7/2) + 17*b
*c*x^(3/2))/((c*x^2 + b)^2*b^3)

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maple [A]  time = 0.02, size = 178, normalized size = 0.71 \begin {gather*} -\frac {13 c^{2} x^{\frac {7}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {17 c \,x^{\frac {3}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{2}}-\frac {45 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {45 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {45 \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 )}{128 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {2}{b^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)/(c*x^4+b*x^2)^3,x)

[Out]

-13/16*c^2/b^3/(c*x^2+b)^2*x^(7/2)-17/16*c/b^2/(c*x^2+b)^2*x^(3/2)-45/128/b^3/(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)))-45/64/b^3/(b/c)^(1/4)*2^(1/2)*
arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-45/64/b^3/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-2/b^
3/x^(1/2)

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maxima [A]  time = 3.14, size = 230, normalized size = 0.92 \begin {gather*} -\frac {45 \, c^{2} x^{4} + 81 \, b c x^{2} + 32 \, b^{2}}{16 \, {\left (b^{3} c^{2} x^{\frac {9}{2}} + 2 \, b^{4} c x^{\frac {5}{2}} + b^{5} \sqrt {x}\right )}} - \frac {45 \, c {\left (\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}}}\right )}}{128 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

-1/16*(45*c^2*x^4 + 81*b*c*x^2 + 32*b^2)/(b^3*c^2*x^(9/2) + 2*b^4*c*x^(5/2) + b^5*sqrt(x)) - 45/128*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)) + 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^3

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mupad [B]  time = 4.37, size = 99, normalized size = 0.39 \begin {gather*} \frac {45\,{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{13/4}}-\frac {45\,{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{13/4}}-\frac {\frac {2}{b}+\frac {81\,c\,x^2}{16\,b^2}+\frac {45\,c^2\,x^4}{16\,b^3}}{b^2\,\sqrt {x}+c^2\,x^{9/2}+2\,b\,c\,x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)/(b*x^2 + c*x^4)^3,x)

[Out]

(45*(-c)^(1/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4)))/(32*b^(13/4)) - (45*(-c)^(1/4)*atan(((-c)^(1/4)*x^(1/2))/b
^(1/4)))/(32*b^(13/4)) - (2/b + (81*c*x^2)/(16*b^2) + (45*c^2*x^4)/(16*b^3))/(b^2*x^(1/2) + c^2*x^(9/2) + 2*b*
c*x^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(9/2)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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