3.747 \(\int \frac{x^{11/2}}{(a+c x^4)^2} \, dx\)

Optimal. Leaf size=308 \[ -\frac{5 \log \left (-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}+\frac{5 \log \left (\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{3/8} c^{13/8}}+\frac{5 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}+1\right )}{16 \sqrt{2} (-a)^{3/8} c^{13/8}}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{x^{5/2}}{4 c \left (a+c x^4\right )} \]

[Out]

-x^(5/2)/(4*c*(a + c*x^4)) - (5*ArcTan[1 - (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(3/8)*c^(13
/8)) + (5*ArcTan[1 + (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(3/8)*c^(13/8)) - (5*ArcTan[(c^(1
/8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(3/8)*c^(13/8)) - (5*ArcTanh[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(3/8)*
c^(13/8)) - (5*Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(3/8)*c^(13/
8)) + (5*Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(3/8)*c^(13/8))

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Rubi [A]  time = 0.25079, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {288, 329, 301, 211, 1165, 628, 1162, 617, 204, 212, 208, 205} \[ -\frac{5 \log \left (-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}+\frac{5 \log \left (\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{3/8} c^{13/8}}+\frac{5 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}+1\right )}{16 \sqrt{2} (-a)^{3/8} c^{13/8}}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{x^{5/2}}{4 c \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^(11/2)/(a + c*x^4)^2,x]

[Out]

-x^(5/2)/(4*c*(a + c*x^4)) - (5*ArcTan[1 - (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(3/8)*c^(13
/8)) + (5*ArcTan[1 + (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(3/8)*c^(13/8)) - (5*ArcTan[(c^(1
/8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(3/8)*c^(13/8)) - (5*ArcTanh[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(3/8)*
c^(13/8)) - (5*Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(3/8)*c^(13/
8)) + (5*Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(3/8)*c^(13/8))

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 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 301

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

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])

Rule 212

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

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]

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]

Rubi steps

\begin{align*} \int \frac{x^{11/2}}{\left (a+c x^4\right )^2} \, dx &=-\frac{x^{5/2}}{4 c \left (a+c x^4\right )}+\frac{5 \int \frac{x^{3/2}}{a+c x^4} \, dx}{8 c}\\ &=-\frac{x^{5/2}}{4 c \left (a+c x^4\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{a+c x^8} \, dx,x,\sqrt{x}\right )}{4 c}\\ &=-\frac{x^{5/2}}{4 c \left (a+c x^4\right )}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a}-\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{8 c^{3/2}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{8 c^{3/2}}\\ &=-\frac{x^{5/2}}{4 c \left (a+c x^4\right )}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{-a}-\sqrt [4]{c} x^2} \, dx,x,\sqrt{x}\right )}{16 \sqrt [4]{-a} c^{3/2}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{-a}+\sqrt [4]{c} x^2} \, dx,x,\sqrt{x}\right )}{16 \sqrt [4]{-a} c^{3/2}}+\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt [4]{-a}-\sqrt [4]{c} x^2}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{16 \sqrt [4]{-a} c^{3/2}}+\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt [4]{-a}+\sqrt [4]{c} x^2}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{16 \sqrt [4]{-a} c^{3/2}}\\ &=-\frac{x^{5/2}}{4 c \left (a+c x^4\right )}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}+\frac{5 \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 )}{32 \sqrt [4]{-a} c^{7/4}}+\frac{5 \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 )}{32 \sqrt [4]{-a} c^{7/4}}-\frac{5 \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 )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}-\frac{5 \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 )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}\\ &=-\frac{x^{5/2}}{4 c \left (a+c x^4\right )}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{5 \log \left (\sqrt [4]{-a}-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}+\frac{5 \log \left (\sqrt [4]{-a}+\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{3/8} c^{13/8}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{3/8} c^{13/8}}\\ &=-\frac{x^{5/2}}{4 c \left (a+c x^4\right )}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{3/8} c^{13/8}}+\frac{5 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{3/8} c^{13/8}}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{3/8} c^{13/8}}-\frac{5 \log \left (\sqrt [4]{-a}-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}+\frac{5 \log \left (\sqrt [4]{-a}+\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{3/8} c^{13/8}}\\ \end{align*}

Mathematica [C]  time = 0.0117147, size = 54, normalized size = 0.18 \[ \frac{2 x^{5/2} \, _2F_1\left (\frac{5}{8},2;\frac{13}{8};-\frac{c x^4}{a}\right )}{3 a c}-\frac{2 x^{5/2}}{3 c \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(11/2)/(a + c*x^4)^2,x]

[Out]

(-2*x^(5/2))/(3*c*(a + c*x^4)) + (2*x^(5/2)*Hypergeometric2F1[5/8, 2, 13/8, -((c*x^4)/a)])/(3*a*c)

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Maple [C]  time = 0.011, size = 47, normalized size = 0.2 \begin{align*} -{\frac{1}{4\,c \left ( c{x}^{4}+a \right ) }{x}^{{\frac{5}{2}}}}+{\frac{5}{32\,{c}^{2}}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+a \right ) }{\frac{1}{{{\it \_R}}^{3}}\ln \left ( \sqrt{x}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)/(c*x^4+a)^2,x)

[Out]

-1/4*x^(5/2)/c/(c*x^4+a)+5/32/c^2*sum(1/_R^3*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{x^{\frac{5}{2}}}{4 \,{\left (c^{2} x^{4} + a c\right )}} + 5 \, \int \frac{x^{\frac{3}{2}}}{8 \,{\left (c^{2} x^{4} + a c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

-1/4*x^(5/2)/(c^2*x^4 + a*c) + 5*integrate(1/8*x^(3/2)/(c^2*x^4 + a*c), x)

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Fricas [B]  time = 1.70398, size = 1488, normalized size = 4.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

1/64*(20*sqrt(2)*(c^2*x^4 + a*c)*(-1/(a^3*c^13))^(1/8)*arctan(sqrt(2)*sqrt(sqrt(2)*a^2*c^8*sqrt(x)*(-1/(a^3*c^
13))^(5/8) - a*c^3*(-1/(a^3*c^13))^(1/4) + x)*a*c^5*(-1/(a^3*c^13))^(3/8) - sqrt(2)*a*c^5*sqrt(x)*(-1/(a^3*c^1
3))^(3/8) + 1) + 20*sqrt(2)*(c^2*x^4 + a*c)*(-1/(a^3*c^13))^(1/8)*arctan(sqrt(2)*sqrt(-sqrt(2)*a^2*c^8*sqrt(x)
*(-1/(a^3*c^13))^(5/8) - a*c^3*(-1/(a^3*c^13))^(1/4) + x)*a*c^5*(-1/(a^3*c^13))^(3/8) - sqrt(2)*a*c^5*sqrt(x)*
(-1/(a^3*c^13))^(3/8) - 1) + 5*sqrt(2)*(c^2*x^4 + a*c)*(-1/(a^3*c^13))^(1/8)*log(sqrt(2)*a^2*c^8*sqrt(x)*(-1/(
a^3*c^13))^(5/8) - a*c^3*(-1/(a^3*c^13))^(1/4) + x) - 5*sqrt(2)*(c^2*x^4 + a*c)*(-1/(a^3*c^13))^(1/8)*log(-sqr
t(2)*a^2*c^8*sqrt(x)*(-1/(a^3*c^13))^(5/8) - a*c^3*(-1/(a^3*c^13))^(1/4) + x) - 16*x^(5/2) - 40*(c^2*x^4 + a*c
)*(-1/(a^3*c^13))^(1/8)*arctan(sqrt(-a*c^3*(-1/(a^3*c^13))^(1/4) + x)*a*c^5*(-1/(a^3*c^13))^(3/8) - a*c^5*sqrt
(x)*(-1/(a^3*c^13))^(3/8)) - 10*(c^2*x^4 + a*c)*(-1/(a^3*c^13))^(1/8)*log(a^2*c^8*(-1/(a^3*c^13))^(5/8) + sqrt
(x)) + 10*(c^2*x^4 + a*c)*(-1/(a^3*c^13))^(1/8)*log(-a^2*c^8*(-1/(a^3*c^13))^(5/8) + sqrt(x)))/(c^2*x^4 + a*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**(11/2)/(c*x**4+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.94351, size = 645, normalized size = 2.09 \begin{align*} -\frac{x^{\frac{5}{2}}}{4 \,{\left (c x^{4} + a\right )} c} - \frac{5 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \arctan \left (\frac{\sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + 2 \, \sqrt{x}}{\sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}}}\right )}{32 \, a c} - \frac{5 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \arctan \left (-\frac{\sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} - 2 \, \sqrt{x}}{\sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}}}\right )}{32 \, a c} + \frac{5 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \arctan \left (\frac{\sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + 2 \, \sqrt{x}}{\sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}}}\right )}{32 \, a c} + \frac{5 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \arctan \left (-\frac{\sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} - 2 \, \sqrt{x}}{\sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}}}\right )}{32 \, a c} - \frac{5 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \log \left (\sqrt{x} \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + x + \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}{64 \, a c} + \frac{5 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \log \left (-\sqrt{x} \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + x + \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}{64 \, a c} + \frac{5 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \log \left (\sqrt{x} \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + x + \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}{64 \, a c} - \frac{5 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \log \left (-\sqrt{x} \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + x + \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}{64 \, a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^4+a)^2,x, algorithm="giac")

[Out]

-1/4*x^(5/2)/((c*x^4 + a)*c) - 5/32*sqrt(-sqrt(2) + 2)*(a/c)^(5/8)*arctan((sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + 2*
sqrt(x))/(sqrt(sqrt(2) + 2)*(a/c)^(1/8)))/(a*c) - 5/32*sqrt(-sqrt(2) + 2)*(a/c)^(5/8)*arctan(-(sqrt(-sqrt(2) +
2)*(a/c)^(1/8) - 2*sqrt(x))/(sqrt(sqrt(2) + 2)*(a/c)^(1/8)))/(a*c) + 5/32*sqrt(sqrt(2) + 2)*(a/c)^(5/8)*arcta
n((sqrt(sqrt(2) + 2)*(a/c)^(1/8) + 2*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/(a*c) + 5/32*sqrt(sqrt(2) + 2)
*(a/c)^(5/8)*arctan(-(sqrt(sqrt(2) + 2)*(a/c)^(1/8) - 2*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/(a*c) - 5/6
4*sqrt(-sqrt(2) + 2)*(a/c)^(5/8)*log(sqrt(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a*c) + 5/64*sqr
t(-sqrt(2) + 2)*(a/c)^(5/8)*log(-sqrt(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a*c) + 5/64*sqrt(sq
rt(2) + 2)*(a/c)^(5/8)*log(sqrt(x)*sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a*c) - 5/64*sqrt(sqrt(2)
+ 2)*(a/c)^(5/8)*log(-sqrt(x)*sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a*c)