3.341 \(\int \frac{x^{19/2}}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=239 \[ -\frac{5 \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{9/4}}+\frac{5 \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{9/4}}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{3/4} c^{9/4}}+\frac{5 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{3/4} c^{9/4}}-\frac{5 \sqrt{x}}{16 c^2 \left (b+c x^2\right )}-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2} \]

[Out]

-x^(5/2)/(4*c*(b + c*x^2)^2) - (5*Sqrt[x])/(16*c^2*(b + c*x^2)) - (5*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1
/4)])/(32*Sqrt[2]*b^(3/4)*c^(9/4)) + (5*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(3/4)*c^(
9/4)) - (5*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(3/4)*c^(9/4)) + (5*Log[S
qrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(3/4)*c^(9/4))

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Rubi [A]  time = 0.189325, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {1584, 288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{5 \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{9/4}}+\frac{5 \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{9/4}}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{3/4} c^{9/4}}+\frac{5 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{3/4} c^{9/4}}-\frac{5 \sqrt{x}}{16 c^2 \left (b+c x^2\right )}-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(19/2)/(b*x^2 + c*x^4)^3,x]

[Out]

-x^(5/2)/(4*c*(b + c*x^2)^2) - (5*Sqrt[x])/(16*c^2*(b + c*x^2)) - (5*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1
/4)])/(32*Sqrt[2]*b^(3/4)*c^(9/4)) + (5*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(3/4)*c^(
9/4)) - (5*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(3/4)*c^(9/4)) + (5*Log[S
qrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(3/4)*c^(9/4))

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]

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 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 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 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 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 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])

Rubi steps

\begin{align*} \int \frac{x^{19/2}}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{x^{7/2}}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2}+\frac{5 \int \frac{x^{3/2}}{\left (b+c x^2\right )^2} \, dx}{8 c}\\ &=-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2}-\frac{5 \sqrt{x}}{16 c^2 \left (b+c x^2\right )}+\frac{5 \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{32 c^2}\\ &=-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2}-\frac{5 \sqrt{x}}{16 c^2 \left (b+c x^2\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{16 c^2}\\ &=-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2}-\frac{5 \sqrt{x}}{16 c^2 \left (b+c x^2\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 \sqrt{b} c^2}+\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 \sqrt{b} c^2}\\ &=-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2}-\frac{5 \sqrt{x}}{16 c^2 \left (b+c x^2\right )}+\frac{5 \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 \sqrt{b} c^{5/2}}+\frac{5 \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 \sqrt{b} c^{5/2}}-\frac{5 \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^{3/4} c^{9/4}}-\frac{5 \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^{3/4} c^{9/4}}\\ &=-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2}-\frac{5 \sqrt{x}}{16 c^2 \left (b+c x^2\right )}-\frac{5 \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{9/4}}+\frac{5 \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{9/4}}+\frac{5 \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^{3/4} c^{9/4}}-\frac{5 \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^{3/4} c^{9/4}}\\ &=-\frac{x^{5/2}}{4 c \left (b+c x^2\right )^2}-\frac{5 \sqrt{x}}{16 c^2 \left (b+c x^2\right )}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{3/4} c^{9/4}}+\frac{5 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{3/4} c^{9/4}}-\frac{5 \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{9/4}}+\frac{5 \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.111162, size = 242, normalized size = 1.01 \[ \frac{-\frac{15 \sqrt{2} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{b^{3/4}}+\frac{15 \sqrt{2} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{b^{3/4}}-\frac{30 \sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{b^{3/4}}+\frac{30 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{b^{3/4}}-\frac{256 c^{5/4} x^{5/2}}{\left (b+c x^2\right )^2}+\frac{40 \sqrt [4]{c} \sqrt{x}}{b+c x^2}-\frac{160 b \sqrt [4]{c} \sqrt{x}}{\left (b+c x^2\right )^2}}{384 c^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(19/2)/(b*x^2 + c*x^4)^3,x]

[Out]

((-160*b*c^(1/4)*Sqrt[x])/(b + c*x^2)^2 - (256*c^(5/4)*x^(5/2))/(b + c*x^2)^2 + (40*c^(1/4)*Sqrt[x])/(b + c*x^
2) - (30*Sqrt[2]*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/b^(3/4) + (30*Sqrt[2]*ArcTan[1 + (Sqrt[2]*c^(1
/4)*Sqrt[x])/b^(1/4)])/b^(3/4) - (15*Sqrt[2]*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/b^(3/
4) + (15*Sqrt[2]*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/b^(3/4))/(384*c^(9/4))

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Maple [A]  time = 0.06, size = 170, normalized size = 0.7 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{2}+b \right ) ^{2}} \left ( -{\frac{9\,{x}^{5/2}}{32\,c}}-{\frac{5\,b\sqrt{x}}{32\,{c}^{2}}} \right ) }+{\frac{5\,\sqrt{2}}{128\,b{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }+{\frac{5\,\sqrt{2}}{64\,b{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{5\,\sqrt{2}}{64\,b{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(19/2)/(c*x^4+b*x^2)^3,x)

[Out]

2*(-9/32*x^(5/2)/c-5/32*b*x^(1/2)/c^2)/(c*x^2+b)^2+5/128/c^2*(b/c)^(1/4)/b*2^(1/2)*ln((x+(b/c)^(1/4)*x^(1/2)*2
^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2)))+5/64/c^2*(b/c)^(1/4)/b*2^(1/2)*arctan(2^(1/2)
/(b/c)^(1/4)*x^(1/2)+1)+5/64/c^2*(b/c)^(1/4)/b*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.40115, size = 599, normalized size = 2.51 \begin{align*} \frac{20 \,{\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )} \left (-\frac{1}{b^{3} c^{9}}\right )^{\frac{1}{4}} \arctan \left (\sqrt{b^{2} c^{4} \sqrt{-\frac{1}{b^{3} c^{9}}} + x} b^{2} c^{7} \left (-\frac{1}{b^{3} c^{9}}\right )^{\frac{3}{4}} - b^{2} c^{7} \sqrt{x} \left (-\frac{1}{b^{3} c^{9}}\right )^{\frac{3}{4}}\right ) + 5 \,{\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )} \left (-\frac{1}{b^{3} c^{9}}\right )^{\frac{1}{4}} \log \left (b c^{2} \left (-\frac{1}{b^{3} c^{9}}\right )^{\frac{1}{4}} + \sqrt{x}\right ) - 5 \,{\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )} \left (-\frac{1}{b^{3} c^{9}}\right )^{\frac{1}{4}} \log \left (-b c^{2} \left (-\frac{1}{b^{3} c^{9}}\right )^{\frac{1}{4}} + \sqrt{x}\right ) - 4 \,{\left (9 \, c x^{2} + 5 \, b\right )} \sqrt{x}}{64 \,{\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/64*(20*(c^4*x^4 + 2*b*c^3*x^2 + b^2*c^2)*(-1/(b^3*c^9))^(1/4)*arctan(sqrt(b^2*c^4*sqrt(-1/(b^3*c^9)) + x)*b^
2*c^7*(-1/(b^3*c^9))^(3/4) - b^2*c^7*sqrt(x)*(-1/(b^3*c^9))^(3/4)) + 5*(c^4*x^4 + 2*b*c^3*x^2 + b^2*c^2)*(-1/(
b^3*c^9))^(1/4)*log(b*c^2*(-1/(b^3*c^9))^(1/4) + sqrt(x)) - 5*(c^4*x^4 + 2*b*c^3*x^2 + b^2*c^2)*(-1/(b^3*c^9))
^(1/4)*log(-b*c^2*(-1/(b^3*c^9))^(1/4) + sqrt(x)) - 4*(9*c*x^2 + 5*b)*sqrt(x))/(c^4*x^4 + 2*b*c^3*x^2 + b^2*c^
2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(19/2)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.14332, size = 282, normalized size = 1.18 \begin{align*} \frac{5 \, \sqrt{2} \left (b c^{3}\right )^{\frac{1}{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 c^{3}} + \frac{5 \, \sqrt{2} \left (b c^{3}\right )^{\frac{1}{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 c^{3}} + \frac{5 \, \sqrt{2} \left (b c^{3}\right )^{\frac{1}{4}} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b c^{3}} - \frac{5 \, \sqrt{2} \left (b c^{3}\right )^{\frac{1}{4}} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b c^{3}} - \frac{9 \, c x^{\frac{5}{2}} + 5 \, b \sqrt{x}}{16 \,{\left (c x^{2} + b\right )}^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

5/64*sqrt(2)*(b*c^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b*c^3) + 5/64*sq
rt(2)*(b*c^3)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b*c^3) + 5/128*sqrt(2)
*(b*c^3)^(1/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^3) - 5/128*sqrt(2)*(b*c^3)^(1/4)*log(-sqr
t(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^3) - 1/16*(9*c*x^(5/2) + 5*b*sqrt(x))/((c*x^2 + b)^2*c^2)