\(\int \frac {\sqrt {x}}{(a+c x^4)^2} \, dx\) [752]

Optimal result

Integrand size = 15, antiderivative size = 308 \[ \int \frac {\sqrt {x}}{\left (a+c x^4\right )^2} \, dx=\frac {x^{3/2}}{4 a \left (a+c x^4\right )}-\frac {5 \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{c} \sqrt {x}}{\sqrt [8]{-a}}\right )}{16 \sqrt {2} (-a)^{13/8} c^{3/8}}+\frac {5 \arctan \left (1+\frac {\sqrt {2} \sqrt [8]{c} \sqrt {x}}{\sqrt [8]{-a}}\right )}{16 \sqrt {2} (-a)^{13/8} c^{3/8}}-\frac {5 \arctan \left (\frac {\sqrt [8]{c} \sqrt {x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{13/8} c^{3/8}}+\frac {5 \text {arctanh}\left (\frac {\sqrt [8]{c} \sqrt {x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{13/8} c^{3/8}}+\frac {5 \log \left (\sqrt [4]{-a}-\sqrt {2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt {x}+\sqrt [4]{c} x\right )}{32 \sqrt {2} (-a)^{13/8} c^{3/8}}-\frac {5 \log \left (\sqrt [4]{-a}+\sqrt {2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt {x}+\sqrt [4]{c} x\right )}{32 \sqrt {2} (-a)^{13/8} c^{3/8}} \]

[Out]

1/4*x^(3/2)/a/(c*x^4+a)-5/16*arctan(c^(1/8)*x^(1/2)/(-a)^(1/8))/(-a)^(13/8)/c^(3/8)+5/16*arctanh(c^(1/8)*x^(1/
2)/(-a)^(1/8))/(-a)^(13/8)/c^(3/8)+5/32*arctan(-1+c^(1/8)*2^(1/2)*x^(1/2)/(-a)^(1/8))/(-a)^(13/8)/c^(3/8)*2^(1
/2)+5/32*arctan(1+c^(1/8)*2^(1/2)*x^(1/2)/(-a)^(1/8))/(-a)^(13/8)/c^(3/8)*2^(1/2)+5/64*ln((-a)^(1/4)+c^(1/4)*x
-(-a)^(1/8)*c^(1/8)*2^(1/2)*x^(1/2))/(-a)^(13/8)/c^(3/8)*2^(1/2)-5/64*ln((-a)^(1/4)+c^(1/4)*x+(-a)^(1/8)*c^(1/
8)*2^(1/2)*x^(1/2))/(-a)^(13/8)/c^(3/8)*2^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {296, 335, 306, 303, 1176, 631, 210, 1179, 642, 304, 211, 214} \[ \int \frac {\sqrt {x}}{\left (a+c x^4\right )^2} \, dx=-\frac {5 \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{c} \sqrt {x}}{\sqrt [8]{-a}}\right )}{16 \sqrt {2} (-a)^{13/8} c^{3/8}}+\frac {5 \arctan \left (\frac {\sqrt {2} \sqrt [8]{c} \sqrt {x}}{\sqrt [8]{-a}}+1\right )}{16 \sqrt {2} (-a)^{13/8} c^{3/8}}-\frac {5 \arctan \left (\frac {\sqrt [8]{c} \sqrt {x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{13/8} c^{3/8}}+\frac {5 \text {arctanh}\left (\frac {\sqrt [8]{c} \sqrt {x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{13/8} c^{3/8}}+\frac {5 \log \left (-\sqrt {2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt {x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt {2} (-a)^{13/8} c^{3/8}}-\frac {5 \log \left (\sqrt {2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt {x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt {2} (-a)^{13/8} c^{3/8}}+\frac {x^{3/2}}{4 a \left (a+c x^4\right )} \]

[In]

Int[Sqrt[x]/(a + c*x^4)^2,x]

[Out]

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

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

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

Rule 214

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

Rule 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 303

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 304

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 306

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]] /; Free
Q[{a, b}, x] && IGtQ[n/4, 0] && IGtQ[m, 0] && LtQ[m, n/2] &&  !GtQ[a/b, 0]

Rule 335

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 631

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 642

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 1176

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 1179

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]

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

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {x}}{\left (a+c x^4\right )^2} \, dx=\frac {\frac {8 a^{5/8} x^{3/2}}{a+c x^4}+\frac {5 \sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {1-\frac {1}{\sqrt {2}}} \left (\sqrt [4]{a}-\sqrt [4]{c} x\right )}{\sqrt [8]{a} \sqrt [8]{c} \sqrt {x}}\right )}{c^{3/8}}-\frac {5 \sqrt {2+\sqrt {2}} \arctan \left (\frac {\sqrt {1+\frac {1}{\sqrt {2}}} \left (\sqrt [4]{a}-\sqrt [4]{c} x\right )}{\sqrt [8]{a} \sqrt [8]{c} \sqrt {x}}\right )}{c^{3/8}}+\frac {5 \sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{a} \sqrt [8]{c} \sqrt {x}}{\sqrt [4]{a}+\sqrt [4]{c} x}\right )}{c^{3/8}}-\frac {5 \sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt [8]{a} \sqrt [8]{c} \sqrt {-\left (\left (-2+\sqrt {2}\right ) x\right )}}{\sqrt [4]{a}+\sqrt [4]{c} x}\right )}{c^{3/8}}}{32 a^{13/8}} \]

[In]

Integrate[Sqrt[x]/(a + c*x^4)^2,x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.92 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.16

method	result	size
derivativedivides	\(\frac {x^{\frac {3}{2}}}{4 a \left (x^{4} c +a \right )}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+a \right )}{\sum }\frac {\ln \left (\sqrt {x}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{32 a c}\)	\(50\)
default	\(\frac {x^{\frac {3}{2}}}{4 a \left (x^{4} c +a \right )}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+a \right )}{\sum }\frac {\ln \left (\sqrt {x}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{32 a c}\)	\(50\)

[In]

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

[Out]

1/4*x^(3/2)/a/(c*x^4+a)+5/32/a/c*sum(1/_R^5*ln(x^(1/2)-_R),_R=RootOf(_Z^8*c+a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {x}}{\left (a+c x^4\right )^2} \, dx=-\frac {5 \, \sqrt {2} {\left (\left (i - 1\right ) \, a c x^{4} + \left (i - 1\right ) \, a^{2}\right )} \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {1}{8}} \log \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} a^{5} c \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {3}{8}} + \sqrt {x}\right ) + 5 \, \sqrt {2} {\left (-\left (i + 1\right ) \, a c x^{4} - \left (i + 1\right ) \, a^{2}\right )} \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {1}{8}} \log \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} a^{5} c \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {3}{8}} + \sqrt {x}\right ) + 5 \, \sqrt {2} {\left (\left (i + 1\right ) \, a c x^{4} + \left (i + 1\right ) \, a^{2}\right )} \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {1}{8}} \log \left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} a^{5} c \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {3}{8}} + \sqrt {x}\right ) + 5 \, \sqrt {2} {\left (-\left (i - 1\right ) \, a c x^{4} - \left (i - 1\right ) \, a^{2}\right )} \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {1}{8}} \log \left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} a^{5} c \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {3}{8}} + \sqrt {x}\right ) + 10 \, {\left (a c x^{4} + a^{2}\right )} \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {1}{8}} \log \left (a^{5} c \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {3}{8}} + \sqrt {x}\right ) + 10 \, {\left (-i \, a c x^{4} - i \, a^{2}\right )} \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {1}{8}} \log \left (i \, a^{5} c \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {3}{8}} + \sqrt {x}\right ) + 10 \, {\left (i \, a c x^{4} + i \, a^{2}\right )} \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {1}{8}} \log \left (-i \, a^{5} c \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {3}{8}} + \sqrt {x}\right ) - 10 \, {\left (a c x^{4} + a^{2}\right )} \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {1}{8}} \log \left (-a^{5} c \left (-\frac {1}{a^{13} c^{3}}\right )^{\frac {3}{8}} + \sqrt {x}\right ) - 16 \, x^{\frac {3}{2}}}{64 \, {\left (a c x^{4} + a^{2}\right )}} \]

[In]

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

[Out]

-1/64*(5*sqrt(2)*((I - 1)*a*c*x^4 + (I - 1)*a^2)*(-1/(a^13*c^3))^(1/8)*log((1/2*I + 1/2)*sqrt(2)*a^5*c*(-1/(a^
13*c^3))^(3/8) + sqrt(x)) + 5*sqrt(2)*(-(I + 1)*a*c*x^4 - (I + 1)*a^2)*(-1/(a^13*c^3))^(1/8)*log(-(1/2*I - 1/2
)*sqrt(2)*a^5*c*(-1/(a^13*c^3))^(3/8) + sqrt(x)) + 5*sqrt(2)*((I + 1)*a*c*x^4 + (I + 1)*a^2)*(-1/(a^13*c^3))^(
1/8)*log((1/2*I - 1/2)*sqrt(2)*a^5*c*(-1/(a^13*c^3))^(3/8) + sqrt(x)) + 5*sqrt(2)*(-(I - 1)*a*c*x^4 - (I - 1)*
a^2)*(-1/(a^13*c^3))^(1/8)*log(-(1/2*I + 1/2)*sqrt(2)*a^5*c*(-1/(a^13*c^3))^(3/8) + sqrt(x)) + 10*(a*c*x^4 + a
^2)*(-1/(a^13*c^3))^(1/8)*log(a^5*c*(-1/(a^13*c^3))^(3/8) + sqrt(x)) + 10*(-I*a*c*x^4 - I*a^2)*(-1/(a^13*c^3))
^(1/8)*log(I*a^5*c*(-1/(a^13*c^3))^(3/8) + sqrt(x)) + 10*(I*a*c*x^4 + I*a^2)*(-1/(a^13*c^3))^(1/8)*log(-I*a^5*
c*(-1/(a^13*c^3))^(3/8) + sqrt(x)) - 10*(a*c*x^4 + a^2)*(-1/(a^13*c^3))^(1/8)*log(-a^5*c*(-1/(a^13*c^3))^(3/8)
+ sqrt(x)) - 16*x^(3/2))/(a*c*x^4 + a^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\left (a+c x^4\right )^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {x}}{\left (a+c x^4\right )^2} \, dx=\int { \frac {\sqrt {x}}{{\left (c x^{4} + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (207) = 414\).

Time = 0.45 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.50 \[ \int \frac {\sqrt {x}}{\left (a+c x^4\right )^2} \, dx=\frac {x^{\frac {3}{2}}}{4 \, {\left (c x^{4} + a\right )} a} - \frac {5 \, \left (\frac {a}{c}\right )^{\frac {3}{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 )}{16 \, a^{2} \sqrt {2 \, \sqrt {2} + 4}} - \frac {5 \, \left (\frac {a}{c}\right )^{\frac {3}{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 )}{16 \, a^{2} \sqrt {2 \, \sqrt {2} + 4}} + \frac {5 \, \left (\frac {a}{c}\right )^{\frac {3}{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 )}{16 \, a^{2} \sqrt {-2 \, \sqrt {2} + 4}} + \frac {5 \, \left (\frac {a}{c}\right )^{\frac {3}{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 )}{16 \, a^{2} \sqrt {-2 \, \sqrt {2} + 4}} + \frac {5 \, \left (\frac {a}{c}\right )^{\frac {3}{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 )}{32 \, a^{2} \sqrt {2 \, \sqrt {2} + 4}} - \frac {5 \, \left (\frac {a}{c}\right )^{\frac {3}{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 )}{32 \, a^{2} \sqrt {2 \, \sqrt {2} + 4}} - \frac {5 \, \left (\frac {a}{c}\right )^{\frac {3}{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 )}{32 \, a^{2} \sqrt {-2 \, \sqrt {2} + 4}} + \frac {5 \, \left (\frac {a}{c}\right )^{\frac {3}{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 )}{32 \, a^{2} \sqrt {-2 \, \sqrt {2} + 4}} \]

[In]

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

[Out]

1/4*x^(3/2)/((c*x^4 + a)*a) - 5/16*(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*sqrt(2*sqrt(2) + 4)) - 5/16*(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*sqrt(2*sqrt(2) + 4)) + 5/16*(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*sqrt(-2*sqrt(2) + 4)) + 5/16*(a/c)^(3/8)*a
rctan(-(sqrt(sqrt(2) + 2)*(a/c)^(1/8) - 2*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/(a^2*sqrt(-2*sqrt(2) + 4)
) + 5/32*(a/c)^(3/8)*log(sqrt(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a^2*sqrt(2*sqrt(2) + 4)) -
5/32*(a/c)^(3/8)*log(-sqrt(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a^2*sqrt(2*sqrt(2) + 4)) - 5/3
2*(a/c)^(3/8)*log(sqrt(x)*sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a^2*sqrt(-2*sqrt(2) + 4)) + 5/32*
(a/c)^(3/8)*log(-sqrt(x)*sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/(a^2*sqrt(-2*sqrt(2) + 4))

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {x}}{\left (a+c x^4\right )^2} \, dx=\frac {x^{3/2}}{4\,a\,\left (c\,x^4+a\right )}-\frac {5\,\mathrm {atan}\left (\frac {c^{1/8}\,\sqrt {x}}{{\left (-a\right )}^{1/8}}\right )}{16\,{\left (-a\right )}^{13/8}\,c^{3/8}}-\frac {\mathrm {atan}\left (\frac {c^{1/8}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/8}}\right )\,5{}\mathrm {i}}{16\,{\left (-a\right )}^{13/8}\,c^{3/8}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,c^{1/8}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )}{{\left (-a\right )}^{1/8}}\right )\,\left (\frac {5}{32}-\frac {5}{32}{}\mathrm {i}\right )}{{\left (-a\right )}^{13/8}\,c^{3/8}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,c^{1/8}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )}{{\left (-a\right )}^{1/8}}\right )\,\left (\frac {5}{32}+\frac {5}{32}{}\mathrm {i}\right )}{{\left (-a\right )}^{13/8}\,c^{3/8}} \]

[In]

int(x^(1/2)/(a + c*x^4)^2,x)

[Out]

x^(3/2)/(4*a*(a + c*x^4)) - (5*atan((c^(1/8)*x^(1/2))/(-a)^(1/8)))/(16*(-a)^(13/8)*c^(3/8)) - (atan((c^(1/8)*x
^(1/2)*1i)/(-a)^(1/8))*5i)/(16*(-a)^(13/8)*c^(3/8)) + (2^(1/2)*atan((2^(1/2)*c^(1/8)*x^(1/2)*(1/2 - 1i/2))/(-a
)^(1/8))*(5/32 - 5i/32))/((-a)^(13/8)*c^(3/8)) + (2^(1/2)*atan((2^(1/2)*c^(1/8)*x^(1/2)*(1/2 + 1i/2))/(-a)^(1/
8))*(5/32 + 5i/32))/((-a)^(13/8)*c^(3/8))