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

Optimal. Leaf size=332 \[ \frac{15 \log \left (-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \log \left (\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 \sqrt{2} (-a)^{13/8} c^{11/8}}+\frac{15 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}+1\right )}{256 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{15 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2} \]

[Out]

-x^(3/2)/(8*c*(a + c*x^4)^2) + (3*x^(3/2))/(64*a*c*(a + c*x^4)) - (15*ArcTan[1 - (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a
)^(1/8)])/(256*Sqrt[2]*(-a)^(13/8)*c^(11/8)) + (15*ArcTan[1 + (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(256*Sqrt
[2]*(-a)^(13/8)*c^(11/8)) - (15*ArcTan[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(256*(-a)^(13/8)*c^(11/8)) + (15*ArcTanh
[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(256*(-a)^(13/8)*c^(11/8)) + (15*Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*c^(1/8)*S
qrt[x] + c^(1/4)*x])/(512*Sqrt[2]*(-a)^(13/8)*c^(11/8)) - (15*Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt
[x] + c^(1/4)*x])/(512*Sqrt[2]*(-a)^(13/8)*c^(11/8))

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Rubi [A]  time = 0.304056, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.867, Rules used = {288, 290, 329, 300, 297, 1162, 617, 204, 1165, 628, 298, 205, 208} \[ \frac{15 \log \left (-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \log \left (\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 \sqrt{2} (-a)^{13/8} c^{11/8}}+\frac{15 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}+1\right )}{256 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{15 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(9/2)/(a + c*x^4)^3,x]

[Out]

-x^(3/2)/(8*c*(a + c*x^4)^2) + (3*x^(3/2))/(64*a*c*(a + c*x^4)) - (15*ArcTan[1 - (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a
)^(1/8)])/(256*Sqrt[2]*(-a)^(13/8)*c^(11/8)) + (15*ArcTan[1 + (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(256*Sqrt
[2]*(-a)^(13/8)*c^(11/8)) - (15*ArcTan[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(256*(-a)^(13/8)*c^(11/8)) + (15*ArcTanh
[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(256*(-a)^(13/8)*c^(11/8)) + (15*Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*c^(1/8)*S
qrt[x] + c^(1/4)*x])/(512*Sqrt[2]*(-a)^(13/8)*c^(11/8)) - (15*Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt
[x] + c^(1/4)*x])/(512*Sqrt[2]*(-a)^(13/8)*c^(11/8))

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 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 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 300

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(
a/b), 2]]}, Dist[r/(2*a), Int[x^m/(r + s*x^(n/2)), x], x] + Dist[r/(2*a), Int[x^m/(r - s*x^(n/2)), x], x]] /;
FreeQ[{a, b}, x] && IGtQ[n/4, 0] && IGtQ[m, 0] && LtQ[m, n/2] &&  !GtQ[a/b, 0]

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 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])

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^{9/2}}{\left (a+c x^4\right )^3} \, dx &=-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2}+\frac{3 \int \frac{\sqrt{x}}{\left (a+c x^4\right )^2} \, dx}{16 c}\\ &=-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}+\frac{15 \int \frac{\sqrt{x}}{a+c x^4} \, dx}{128 a c}\\ &=-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{a+c x^8} \, dx,x,\sqrt{x}\right )}{64 a c}\\ &=-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{-a}-\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{128 (-a)^{3/2} c}+\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{128 (-a)^{3/2} c}\\ &=-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{-a}-\sqrt [4]{c} x^2} \, dx,x,\sqrt{x}\right )}{256 (-a)^{3/2} c^{5/4}}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{-a}+\sqrt [4]{c} x^2} \, dx,x,\sqrt{x}\right )}{256 (-a)^{3/2} c^{5/4}}-\frac{15 \operatorname{Subst}\left (\int \frac{\sqrt [4]{-a}-\sqrt [4]{c} x^2}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{256 (-a)^{3/2} c^{5/4}}+\frac{15 \operatorname{Subst}\left (\int \frac{\sqrt [4]{-a}+\sqrt [4]{c} x^2}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{256 (-a)^{3/2} c^{5/4}}\\ &=-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}-\frac{15 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{15 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt [4]{-a}}{\sqrt [4]{c}}-\frac{\sqrt{2} \sqrt [8]{-a} x}{\sqrt [8]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{512 (-a)^{3/2} c^{3/2}}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt [4]{-a}}{\sqrt [4]{c}}+\frac{\sqrt{2} \sqrt [8]{-a} x}{\sqrt [8]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{512 (-a)^{3/2} c^{3/2}}+\frac{15 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [8]{-a}}{\sqrt [8]{c}}+2 x}{-\frac{\sqrt [4]{-a}}{\sqrt [4]{c}}-\frac{\sqrt{2} \sqrt [8]{-a} x}{\sqrt [8]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}+\frac{15 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [8]{-a}}{\sqrt [8]{c}}-2 x}{-\frac{\sqrt [4]{-a}}{\sqrt [4]{c}}+\frac{\sqrt{2} \sqrt [8]{-a} x}{\sqrt [8]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}\\ &=-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}-\frac{15 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{15 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{15 \log \left (\sqrt [4]{-a}-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \log \left (\sqrt [4]{-a}+\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 \sqrt{2} (-a)^{13/8} c^{11/8}}\\ &=-\frac{x^{3/2}}{8 c \left (a+c x^4\right )^2}+\frac{3 x^{3/2}}{64 a c \left (a+c x^4\right )}-\frac{15 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 \sqrt{2} (-a)^{13/8} c^{11/8}}+\frac{15 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{15 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{256 (-a)^{13/8} c^{11/8}}+\frac{15 \log \left (\sqrt [4]{-a}-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}-\frac{15 \log \left (\sqrt [4]{-a}+\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{512 \sqrt{2} (-a)^{13/8} c^{11/8}}\\ \end{align*}

Mathematica [C]  time = 0.0157076, size = 45, normalized size = 0.14 \[ \frac{2 x^{3/2} \left (\frac{\, _2F_1\left (\frac{3}{8},3;\frac{11}{8};-\frac{c x^4}{a}\right )}{a^2}-\frac{1}{\left (a+c x^4\right )^2}\right )}{13 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(9/2)/(a + c*x^4)^3,x]

[Out]

(2*x^(3/2)*(-(a + c*x^4)^(-2) + Hypergeometric2F1[3/8, 3, 11/8, -((c*x^4)/a)]/a^2))/(13*c)

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Maple [C]  time = 0.018, size = 61, normalized size = 0.2 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{4}+a \right ) ^{2}} \left ( -{\frac{5\,{x}^{3/2}}{128\,c}}+{\frac{3\,{x}^{11/2}}{128\,a}} \right ) }+{\frac{15}{512\,{c}^{2}a}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+a \right ) }{\frac{1}{{{\it \_R}}^{5}}\ln \left ( \sqrt{x}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-5/128*x^(3/2)/c+3/128/a*x^(11/2))/(c*x^4+a)^2+15/512/c^2/a*sum(1/_R^5*ln(x^(1/2)-_R),_R=RootOf(_Z^8*c+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3 \, c x^{\frac{11}{2}} - 5 \, a x^{\frac{3}{2}}}{64 \,{\left (a c^{3} x^{8} + 2 \, a^{2} c^{2} x^{4} + a^{3} c\right )}} + 15 \, \int \frac{\sqrt{x}}{128 \,{\left (a c^{2} x^{4} + a^{2} c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(3*c*x^(11/2) - 5*a*x^(3/2))/(a*c^3*x^8 + 2*a^2*c^2*x^4 + a^3*c) + 15*integrate(1/128*sqrt(x)/(a*c^2*x^4
+ a^2*c), x)

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Fricas [B]  time = 1.77149, size = 1801, normalized size = 5.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1024*(60*sqrt(2)*(a*c^3*x^8 + 2*a^2*c^2*x^4 + a^3*c)*(-1/(a^13*c^11))^(1/8)*arctan(sqrt(2)*sqrt(a^10*c^8*(-
1/(a^13*c^11))^(3/4) + sqrt(2)*a^5*c^4*sqrt(x)*(-1/(a^13*c^11))^(3/8) + x)*a^8*c^7*(-1/(a^13*c^11))^(5/8) - sq
rt(2)*a^8*c^7*sqrt(x)*(-1/(a^13*c^11))^(5/8) + 1) + 60*sqrt(2)*(a*c^3*x^8 + 2*a^2*c^2*x^4 + a^3*c)*(-1/(a^13*c
^11))^(1/8)*arctan(sqrt(2)*sqrt(a^10*c^8*(-1/(a^13*c^11))^(3/4) - sqrt(2)*a^5*c^4*sqrt(x)*(-1/(a^13*c^11))^(3/
8) + x)*a^8*c^7*(-1/(a^13*c^11))^(5/8) - sqrt(2)*a^8*c^7*sqrt(x)*(-1/(a^13*c^11))^(5/8) - 1) - 15*sqrt(2)*(a*c
^3*x^8 + 2*a^2*c^2*x^4 + a^3*c)*(-1/(a^13*c^11))^(1/8)*log(a^10*c^8*(-1/(a^13*c^11))^(3/4) + sqrt(2)*a^5*c^4*s
qrt(x)*(-1/(a^13*c^11))^(3/8) + x) + 15*sqrt(2)*(a*c^3*x^8 + 2*a^2*c^2*x^4 + a^3*c)*(-1/(a^13*c^11))^(1/8)*log
(a^10*c^8*(-1/(a^13*c^11))^(3/4) - sqrt(2)*a^5*c^4*sqrt(x)*(-1/(a^13*c^11))^(3/8) + x) - 120*(a*c^3*x^8 + 2*a^
2*c^2*x^4 + a^3*c)*(-1/(a^13*c^11))^(1/8)*arctan(sqrt(a^10*c^8*(-1/(a^13*c^11))^(3/4) + x)*a^8*c^7*(-1/(a^13*c
^11))^(5/8) - a^8*c^7*sqrt(x)*(-1/(a^13*c^11))^(5/8)) + 30*(a*c^3*x^8 + 2*a^2*c^2*x^4 + a^3*c)*(-1/(a^13*c^11)
)^(1/8)*log(a^5*c^4*(-1/(a^13*c^11))^(3/8) + sqrt(x)) - 30*(a*c^3*x^8 + 2*a^2*c^2*x^4 + a^3*c)*(-1/(a^13*c^11)
)^(1/8)*log(-a^5*c^4*(-1/(a^13*c^11))^(3/8) + sqrt(x)) - 16*(3*c*x^5 - 5*a*x)*sqrt(x))/(a*c^3*x^8 + 2*a^2*c^2*
x^4 + a^3*c)

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

[Out]

Timed out

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Giac [B]  time = 1.52469, size = 663, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/512*sqrt(-sqrt(2) + 2)*(a/c)^(3/8)*arctan((sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + 2*sqrt(x))/(sqrt(sqrt(2) + 2)*
(a/c)^(1/8)))/(a^2*c) - 15/512*sqrt(-sqrt(2) + 2)*(a/c)^(3/8)*arctan(-(sqrt(-sqrt(2) + 2)*(a/c)^(1/8) - 2*sqrt
(x))/(sqrt(sqrt(2) + 2)*(a/c)^(1/8)))/(a^2*c) + 15/512*sqrt(sqrt(2) + 2)*(a/c)^(3/8)*arctan((sqrt(sqrt(2) + 2)
*(a/c)^(1/8) + 2*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/(a^2*c) + 15/512*sqrt(sqrt(2) + 2)*(a/c)^(3/8)*arc
tan(-(sqrt(sqrt(2) + 2)*(a/c)^(1/8) - 2*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/(a^2*c) + 15/1024*sqrt(-sqr
t(2) + 2)*(a/c)^(3/8)*log(sqrt(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a^2*c) - 15/1024*sqrt(-sqr
t(2) + 2)*(a/c)^(3/8)*log(-sqrt(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a^2*c) - 15/1024*sqrt(sqr
t(2) + 2)*(a/c)^(3/8)*log(sqrt(x)*sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a^2*c) + 15/1024*sqrt(sqr
t(2) + 2)*(a/c)^(3/8)*log(-sqrt(x)*sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a^2*c) + 1/64*(3*c*x^(11
/2) - 5*a*x^(3/2))/((c*x^4 + a)^2*a*c)