3.322 \(\int \frac {\sqrt {x}}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=202 \[ -\frac {\sqrt [4]{c} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{c} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{5/4}}-\frac {\sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{5/4}}-\frac {2}{b \sqrt {x}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1584, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\sqrt [4]{c} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{c} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{5/4}}-\frac {\sqrt [4]{c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{5/4}}-\frac {2}{b \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(b*x^2 + c*x^4),x]

[Out]

-2/(b*Sqrt[x]) + (c^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^(5/4)) - (c^(1/4)*ArcTan[1
+ (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^(5/4)) - (c^(1/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt
[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(5/4)) + (c^(1/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(
2*Sqrt[2]*b^(5/4))

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 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -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 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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.13 \[ -\frac {2 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\frac {c x^2}{b}\right )}{b \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(b*x^2 + c*x^4),x]

[Out]

(-2*Hypergeometric2F1[-1/4, 1, 3/4, -((c*x^2)/b)])/(b*Sqrt[x])

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fricas [A]  time = 1.13, size = 142, normalized size = 0.70 \[ \frac {4 \, b x \left (-\frac {c}{b^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b c \sqrt {x} \left (-\frac {c}{b^{5}}\right )^{\frac {1}{4}} - \sqrt {-b^{3} c \sqrt {-\frac {c}{b^{5}}} + c^{2} x} b \left (-\frac {c}{b^{5}}\right )^{\frac {1}{4}}}{c}\right ) - b x \left (-\frac {c}{b^{5}}\right )^{\frac {1}{4}} \log \left (b^{4} \left (-\frac {c}{b^{5}}\right )^{\frac {3}{4}} + c \sqrt {x}\right ) + b x \left (-\frac {c}{b^{5}}\right )^{\frac {1}{4}} \log \left (-b^{4} \left (-\frac {c}{b^{5}}\right )^{\frac {3}{4}} + c \sqrt {x}\right ) - 4 \, \sqrt {x}}{2 \, b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 190, normalized size = 0.94 \[ -\frac {2}{b \sqrt {x}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{2} c^{2}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{2} c^{2}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{2} c^{2}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(b*sqrt(x)) - 1/2*sqrt(2)*(b*c^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(
b^2*c^2) - 1/2*sqrt(2)*(b*c^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b^2*c
^2) + 1/4*sqrt(2)*(b*c^3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^2) - 1/4*sqrt(2)*(b*c^
3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^2)

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maple [A]  time = 0.01, size = 140, normalized size = 0.69 \[ -\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} b}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} b}-\frac {\sqrt {2}\, \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} b}-\frac {2}{b \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/b/(b/c)^(1/4)*2^(1/2)*ln((x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)
^(1/2)))-1/2/b/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-1/2/b/(b/c)^(1/4)*2^(1/2)*arctan(2^(1
/2)/(b/c)^(1/4)*x^(1/2)-1)-2/b/x^(1/2)

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maxima [A]  time = 2.92, size = 186, normalized size = 0.92 \[ -\frac {c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{4 \, b} - \frac {2}{b \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqr
t(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt
(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x +
sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(
3/4)))/b - 2/(b*sqrt(x))

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mupad [B]  time = 4.53, size = 54, normalized size = 0.27 \[ \frac {{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{5/4}}-\frac {{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{5/4}}-\frac {2}{b\,\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(b*x^2 + c*x^4),x)

[Out]

((-c)^(1/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4)))/b^(5/4) - ((-c)^(1/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4)))/b^(
5/4) - 2/(b*x^(1/2))

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sympy [A]  time = 18.32, size = 170, normalized size = 0.84 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2}{5 c x^{\frac {5}{2}}} & \text {for}\: b = 0 \\- \frac {2}{b \sqrt {x}} & \text {for}\: c = 0 \\- \frac {2}{b \sqrt {x}} + \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} - \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} - \frac {\left (-1\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(5/2), Eq(b, 0) & Eq(c, 0)), (-2/(5*c*x**(5/2)), Eq(b, 0)), (-2/(b*sqrt(x)), Eq(c, 0)), (-2/
(b*sqrt(x)) + (-1)**(3/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(5/4)*(1/c)**(1/4)) - (-1)**
(3/4)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(5/4)*(1/c)**(1/4)) - (-1)**(3/4)*atan((-1)**(3/4
)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/(b**(5/4)*(1/c)**(1/4)), True))

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