3.753 \(\int \frac{1}{\sqrt{x} (a+c x^4)^2} \, dx\)

Optimal. Leaf size=308 \[ \frac{\sqrt{x}}{4 a \left (a+c x^4\right )}-\frac{7 \log \left (-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \log \left (\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}-\frac{7 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}+1\right )}{16 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}} \]

[Out]

Sqrt[x]/(4*a*(a + c*x^4)) - (7*ArcTan[1 - (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(15/8)*c^(1/
8)) + (7*ArcTan[1 + (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(15/8)*c^(1/8)) + (7*ArcTan[(c^(1/
8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(15/8)*c^(1/8)) + (7*ArcTanh[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(15/8)*
c^(1/8)) - (7*Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(15/8)*c^(1/8
)) + (7*Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(15/8)*c^(1/8))

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Rubi [A]  time = 0.266109, 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 = {290, 329, 214, 212, 208, 205, 211, 1165, 628, 1162, 617, 204} \[ \frac{\sqrt{x}}{4 a \left (a+c x^4\right )}-\frac{7 \log \left (-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \log \left (\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}-\frac{7 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}+1\right )}{16 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(a + c*x^4)^2),x]

[Out]

Sqrt[x]/(4*a*(a + c*x^4)) - (7*ArcTan[1 - (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(15/8)*c^(1/
8)) + (7*ArcTan[1 + (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(15/8)*c^(1/8)) + (7*ArcTan[(c^(1/
8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(15/8)*c^(1/8)) + (7*ArcTanh[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(15/8)*
c^(1/8)) - (7*Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(15/8)*c^(1/8
)) + (7*Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(15/8)*c^(1/8))

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 214

Int[((a_) + (b_.)*(x_)^(n_))^(-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^(n/2)), x], x] + Dist[r/(2*a), Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a,
b}, x] && IGtQ[n/4, 1] &&  !GtQ[a/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]

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{1}{\sqrt{x} \left (a+c x^4\right )^2} \, dx &=\frac{\sqrt{x}}{4 a \left (a+c x^4\right )}+\frac{7 \int \frac{1}{\sqrt{x} \left (a+c x^4\right )} \, dx}{8 a}\\ &=\frac{\sqrt{x}}{4 a \left (a+c x^4\right )}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{a+c x^8} \, dx,x,\sqrt{x}\right )}{4 a}\\ &=\frac{\sqrt{x}}{4 a \left (a+c x^4\right )}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a}-\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{8 (-a)^{3/2}}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{8 (-a)^{3/2}}\\ &=\frac{\sqrt{x}}{4 a \left (a+c x^4\right )}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{-a}-\sqrt [4]{c} x^2} \, dx,x,\sqrt{x}\right )}{16 (-a)^{7/4}}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{-a}+\sqrt [4]{c} x^2} \, dx,x,\sqrt{x}\right )}{16 (-a)^{7/4}}+\frac{7 \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 (-a)^{7/4}}+\frac{7 \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 (-a)^{7/4}}\\ &=\frac{\sqrt{x}}{4 a \left (a+c x^4\right )}+\frac{7 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \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 (-a)^{7/4} \sqrt [4]{c}}+\frac{7 \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 (-a)^{7/4} \sqrt [4]{c}}-\frac{7 \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)^{15/8} \sqrt [8]{c}}-\frac{7 \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)^{15/8} \sqrt [8]{c}}\\ &=\frac{\sqrt{x}}{4 a \left (a+c x^4\right )}+\frac{7 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}}-\frac{7 \log \left (\sqrt [4]{-a}-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \log \left (\sqrt [4]{-a}+\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \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)^{15/8} \sqrt [8]{c}}-\frac{7 \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)^{15/8} \sqrt [8]{c}}\\ &=\frac{\sqrt{x}}{4 a \left (a+c x^4\right )}-\frac{7 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{15/8} \sqrt [8]{c}}-\frac{7 \log \left (\sqrt [4]{-a}-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}+\frac{7 \log \left (\sqrt [4]{-a}+\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{15/8} \sqrt [8]{c}}\\ \end{align*}

Mathematica [C]  time = 0.0113292, size = 51, normalized size = 0.17 \[ \frac{7 \sqrt{x} \, _2F_1\left (\frac{1}{8},1;\frac{9}{8};-\frac{c x^4}{a}\right )}{4 a^2}+\frac{\sqrt{x}}{4 a \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(a + c*x^4)^2),x]

[Out]

Sqrt[x]/(4*a*(a + c*x^4)) + (7*Sqrt[x]*Hypergeometric2F1[1/8, 1, 9/8, -((c*x^4)/a)])/(4*a^2)

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Maple [C]  time = 0.012, size = 50, normalized size = 0.2 \begin{align*}{\frac{1}{4\,a \left ( c{x}^{4}+a \right ) }\sqrt{x}}+{\frac{7}{32\,ac}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+a \right ) }{\frac{1}{{{\it \_R}}^{7}}\ln \left ( \sqrt{x}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^4+a)^2/x^(1/2),x)

[Out]

1/4*x^(1/2)/a/(c*x^4+a)+7/32/a/c*sum(1/_R^7*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*} -7 \, c \int \frac{x^{\frac{7}{2}}}{8 \,{\left (a^{2} c x^{4} + a^{3}\right )}}\,{d x} + \frac{7 \, c x^{\frac{9}{2}} + 8 \, a \sqrt{x}}{4 \,{\left (a^{2} c x^{4} + a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+a)^2/x^(1/2),x, algorithm="maxima")

[Out]

-7*c*integrate(1/8*x^(7/2)/(a^2*c*x^4 + a^3), x) + 1/4*(7*c*x^(9/2) + 8*a*sqrt(x))/(a^2*c*x^4 + a^3)

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Fricas [B]  time = 1.68283, size = 1381, normalized size = 4.48 \begin{align*} \frac{28 \, \sqrt{2}{\left (a c x^{4} + a^{2}\right )} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} \arctan \left (\sqrt{2} \sqrt{a^{4} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{4}} + \sqrt{2} a^{2} \sqrt{x} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} + x} a^{13} c \left (-\frac{1}{a^{15} c}\right )^{\frac{7}{8}} - \sqrt{2} a^{13} c \sqrt{x} \left (-\frac{1}{a^{15} c}\right )^{\frac{7}{8}} + 1\right ) + 28 \, \sqrt{2}{\left (a c x^{4} + a^{2}\right )} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} \arctan \left (\sqrt{2} \sqrt{a^{4} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{4}} - \sqrt{2} a^{2} \sqrt{x} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} + x} a^{13} c \left (-\frac{1}{a^{15} c}\right )^{\frac{7}{8}} - \sqrt{2} a^{13} c \sqrt{x} \left (-\frac{1}{a^{15} c}\right )^{\frac{7}{8}} - 1\right ) + 7 \, \sqrt{2}{\left (a c x^{4} + a^{2}\right )} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} \log \left (a^{4} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{4}} + \sqrt{2} a^{2} \sqrt{x} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} + x\right ) - 7 \, \sqrt{2}{\left (a c x^{4} + a^{2}\right )} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} \log \left (a^{4} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{4}} - \sqrt{2} a^{2} \sqrt{x} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} + x\right ) + 56 \,{\left (a c x^{4} + a^{2}\right )} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} \arctan \left (\sqrt{a^{4} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{4}} + x} a^{13} c \left (-\frac{1}{a^{15} c}\right )^{\frac{7}{8}} - a^{13} c \sqrt{x} \left (-\frac{1}{a^{15} c}\right )^{\frac{7}{8}}\right ) + 14 \,{\left (a c x^{4} + a^{2}\right )} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} \log \left (a^{2} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} + \sqrt{x}\right ) - 14 \,{\left (a c x^{4} + a^{2}\right )} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} \log \left (-a^{2} \left (-\frac{1}{a^{15} c}\right )^{\frac{1}{8}} + \sqrt{x}\right ) + 16 \, \sqrt{x}}{64 \,{\left (a c x^{4} + a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+a)^2/x^(1/2),x, algorithm="fricas")

[Out]

1/64*(28*sqrt(2)*(a*c*x^4 + a^2)*(-1/(a^15*c))^(1/8)*arctan(sqrt(2)*sqrt(a^4*(-1/(a^15*c))^(1/4) + sqrt(2)*a^2
*sqrt(x)*(-1/(a^15*c))^(1/8) + x)*a^13*c*(-1/(a^15*c))^(7/8) - sqrt(2)*a^13*c*sqrt(x)*(-1/(a^15*c))^(7/8) + 1)
+ 28*sqrt(2)*(a*c*x^4 + a^2)*(-1/(a^15*c))^(1/8)*arctan(sqrt(2)*sqrt(a^4*(-1/(a^15*c))^(1/4) - sqrt(2)*a^2*sq
rt(x)*(-1/(a^15*c))^(1/8) + x)*a^13*c*(-1/(a^15*c))^(7/8) - sqrt(2)*a^13*c*sqrt(x)*(-1/(a^15*c))^(7/8) - 1) +
7*sqrt(2)*(a*c*x^4 + a^2)*(-1/(a^15*c))^(1/8)*log(a^4*(-1/(a^15*c))^(1/4) + sqrt(2)*a^2*sqrt(x)*(-1/(a^15*c))^
(1/8) + x) - 7*sqrt(2)*(a*c*x^4 + a^2)*(-1/(a^15*c))^(1/8)*log(a^4*(-1/(a^15*c))^(1/4) - sqrt(2)*a^2*sqrt(x)*(
-1/(a^15*c))^(1/8) + x) + 56*(a*c*x^4 + a^2)*(-1/(a^15*c))^(1/8)*arctan(sqrt(a^4*(-1/(a^15*c))^(1/4) + x)*a^13
*c*(-1/(a^15*c))^(7/8) - a^13*c*sqrt(x)*(-1/(a^15*c))^(7/8)) + 14*(a*c*x^4 + a^2)*(-1/(a^15*c))^(1/8)*log(a^2*
(-1/(a^15*c))^(1/8) + sqrt(x)) - 14*(a*c*x^4 + a^2)*(-1/(a^15*c))^(1/8)*log(-a^2*(-1/(a^15*c))^(1/8) + sqrt(x)
) + 16*sqrt(x))/(a*c*x^4 + a^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(1/(c*x**4+a)**2/x**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.36933, size = 613, normalized size = 1.99 \begin{align*} \frac{7 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{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^{2}} + \frac{7 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{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^{2}} + \frac{7 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{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^{2}} + \frac{7 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{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^{2}} + \frac{7 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{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^{2}} - \frac{7 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{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^{2}} + \frac{7 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{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^{2}} - \frac{7 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{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^{2}} + \frac{\sqrt{x}}{4 \,{\left (c x^{4} + a\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+a)^2/x^(1/2),x, algorithm="giac")

[Out]

7/32*sqrt(sqrt(2) + 2)*(a/c)^(1/8)*arctan((sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + 2*sqrt(x))/(sqrt(sqrt(2) + 2)*(a/c
)^(1/8)))/a^2 + 7/32*sqrt(sqrt(2) + 2)*(a/c)^(1/8)*arctan(-(sqrt(-sqrt(2) + 2)*(a/c)^(1/8) - 2*sqrt(x))/(sqrt(
sqrt(2) + 2)*(a/c)^(1/8)))/a^2 + 7/32*sqrt(-sqrt(2) + 2)*(a/c)^(1/8)*arctan((sqrt(sqrt(2) + 2)*(a/c)^(1/8) + 2
*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/a^2 + 7/32*sqrt(-sqrt(2) + 2)*(a/c)^(1/8)*arctan(-(sqrt(sqrt(2) +
2)*(a/c)^(1/8) - 2*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/a^2 + 7/64*sqrt(sqrt(2) + 2)*(a/c)^(1/8)*log(sqr
t(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/a^2 - 7/64*sqrt(sqrt(2) + 2)*(a/c)^(1/8)*log(-sqrt(x)*sq
rt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/a^2 + 7/64*sqrt(-sqrt(2) + 2)*(a/c)^(1/8)*log(sqrt(x)*sqrt(-sqr
t(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/a^2 - 7/64*sqrt(-sqrt(2) + 2)*(a/c)^(1/8)*log(-sqrt(x)*sqrt(-sqrt(2)
+ 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/a^2 + 1/4*sqrt(x)/((c*x^4 + a)*a)