3.1077 \(\int \frac{\sqrt{x}}{(a+b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=489 \[ \frac{x^{3/2} \left (-2 a c+b^2+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt [4]{c} \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}+b\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}+b\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{\sqrt{b^2-4 a c}-b}} \]

[Out]

(x^(3/2)*(b^2 - 2*a*c + b*c*x^2))/(2*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (c^(1/4)*(b - (b^2 - 20*a*c)/Sqrt[
b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(4*2^(3/4)*a*(b^2 - 4*a*c)*(-b
- Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(b + (b^2 - 20*a*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x]
)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(4*2^(3/4)*a*(b^2 - 4*a*c)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(b -
(b^2 - 20*a*c)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(4*2^(3/4
)*a*(b^2 - 4*a*c)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(b + (b^2 - 20*a*c)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2
^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(4*2^(3/4)*a*(b^2 - 4*a*c)*(-b + Sqrt[b^2 - 4*a*c])^(
1/4))

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Rubi [A]  time = 0.997934, antiderivative size = 489, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1115, 1366, 1510, 298, 205, 208} \[ \frac{x^{3/2} \left (-2 a c+b^2+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt [4]{c} \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}+b\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}+b\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{\sqrt{b^2-4 a c}-b}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a + b*x^2 + c*x^4)^2,x]

[Out]

(x^(3/2)*(b^2 - 2*a*c + b*c*x^2))/(2*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (c^(1/4)*(b - (b^2 - 20*a*c)/Sqrt[
b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(4*2^(3/4)*a*(b^2 - 4*a*c)*(-b
- Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(b + (b^2 - 20*a*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x]
)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(4*2^(3/4)*a*(b^2 - 4*a*c)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(b -
(b^2 - 20*a*c)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(4*2^(3/4
)*a*(b^2 - 4*a*c)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(b + (b^2 - 20*a*c)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2
^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(4*2^(3/4)*a*(b^2 - 4*a*c)*(-b + Sqrt[b^2 - 4*a*c])^(
1/4))

Rule 1115

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[
k/d, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(2*k))/d^2 + (c*x^(4*k))/d^4)^p, x], x, (d*x)^(1/k)], x]] /; FreeQ[
{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && FractionQ[m] && IntegerQ[p]

Rule 1366

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((d*x)^(m + 1)*(b
^2 - 2*a*c + b*c*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1))/(a*d*n*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(a*n*(p + 1)
*(b^2 - 4*a*c)), Int[(d*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[b^2*(m + n*(p + 1) + 1) - 2*a*c*(m + 2*n*(p
+ 1) + 1) + b*c*(m + n*(2*p + 3) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[n2, 2*n] && NeQ[b^2
- 4*a*c, 0] && IGtQ[n, 0] && ILtQ[p, -1]

Rule 1510

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Dist[e/
2 - (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[n2
, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 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{\sqrt{x}}{\left (a+b x^2+c x^4\right )^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^4+c x^8\right )^2} \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{3/2} \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (-b^2+10 a c-b c x^4\right )}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{2 a \left (b^2-4 a c\right )}\\ &=\frac{x^{3/2} \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\left (c \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^4} \, dx,x,\sqrt{x}\right )}{4 a \left (b^2-4 a c\right )}+\frac{\left (c \left (b+\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^4} \, dx,x,\sqrt{x}\right )}{4 a \left (b^2-4 a c\right )}\\ &=\frac{x^{3/2} \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (\sqrt{c} \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-b-\sqrt{b^2-4 a c}}-\sqrt{2} \sqrt{c} x^2} \, dx,x,\sqrt{x}\right )}{4 \sqrt{2} a \left (b^2-4 a c\right )}+\frac{\left (\sqrt{c} \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-b-\sqrt{b^2-4 a c}}+\sqrt{2} \sqrt{c} x^2} \, dx,x,\sqrt{x}\right )}{4 \sqrt{2} a \left (b^2-4 a c\right )}-\frac{\left (\sqrt{c} \left (b+\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-b+\sqrt{b^2-4 a c}}-\sqrt{2} \sqrt{c} x^2} \, dx,x,\sqrt{x}\right )}{4 \sqrt{2} a \left (b^2-4 a c\right )}+\frac{\left (\sqrt{c} \left (b+\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-b+\sqrt{b^2-4 a c}}+\sqrt{2} \sqrt{c} x^2} \, dx,x,\sqrt{x}\right )}{4 \sqrt{2} a \left (b^2-4 a c\right )}\\ &=\frac{x^{3/2} \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt [4]{c} \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b-\sqrt{b^2-4 a c}}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{-b-\sqrt{b^2-4 a c}}}+\frac{\sqrt [4]{c} \left (b+\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b+\sqrt{b^2-4 a c}}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{-b+\sqrt{b^2-4 a c}}}-\frac{\sqrt [4]{c} \left (b-\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b-\sqrt{b^2-4 a c}}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{-b-\sqrt{b^2-4 a c}}}-\frac{\sqrt [4]{c} \left (b+\frac{b^2-20 a c}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b+\sqrt{b^2-4 a c}}}\right )}{4\ 2^{3/4} a \left (b^2-4 a c\right ) \sqrt [4]{-b+\sqrt{b^2-4 a c}}}\\ \end{align*}

Mathematica [C]  time = 0.228606, size = 149, normalized size = 0.3 \[ -\frac{\left (a+b x^2+c x^4\right ) \text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{\text{$\#$1}^4 b c \log \left (\sqrt{x}-\text{$\#$1}\right )-10 a c \log \left (\sqrt{x}-\text{$\#$1}\right )+b^2 \log \left (\sqrt{x}-\text{$\#$1}\right )}{2 \text{$\#$1}^5 c+\text{$\#$1} b}\& \right ]+4 x^{3/2} \left (-2 a c+b^2+b c x^2\right )}{8 a \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4*x^(3/2)*(b^2 - 2*a*c + b*c*x^2) + (a + b*x^2 + c*x^4)*RootSum[a + b*#1^4 + c*#1^8 & , (b^2*Log[Sqrt[x] - #
1] - 10*a*c*Log[Sqrt[x] - #1] + b*c*Log[Sqrt[x] - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ])/(8*a*(-b^2 + 4*a*c)*(a + b*
x^2 + c*x^4))

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Maple [C]  time = 0.27, size = 146, normalized size = 0.3 \begin{align*} 2\,{\frac{1}{c{x}^{4}+b{x}^{2}+a} \left ( -1/4\,{\frac{bc{x}^{7/2}}{a \left ( 4\,ac-{b}^{2} \right ) }}+1/4\,{\frac{ \left ( 2\,ac-{b}^{2} \right ){x}^{3/2}}{a \left ( 4\,ac-{b}^{2} \right ) }} \right ) }-{\frac{1}{8\,a \left ( 4\,ac-{b}^{2} \right ) }\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{bc{{\it \_R}}^{6}+ \left ( -10\,ac+{b}^{2} \right ){{\it \_R}}^{2}}{2\,{{\it \_R}}^{7}c+{{\it \_R}}^{3}b}\ln \left ( \sqrt{x}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-1/4*b/a/(4*a*c-b^2)*c*x^(7/2)+1/4*(2*a*c-b^2)/a/(4*a*c-b^2)*x^(3/2))/(c*x^4+b*x^2+a)-1/8/a/(4*a*c-b^2)*sum
((b*c*_R^6+(-10*a*c+b^2)*_R^2)/(2*_R^7*c+_R^3*b)*ln(x^(1/2)-_R),_R=RootOf(_Z^8*c+_Z^4*b+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{b c x^{\frac{7}{2}} +{\left (b^{2} - 2 \, a c\right )} x^{\frac{3}{2}}}{2 \,{\left ({\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )}} - \int -\frac{b c x^{\frac{5}{2}} +{\left (b^{2} - 10 \, a c\right )} \sqrt{x}}{4 \,{\left ({\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*c*x^(7/2) + (b^2 - 2*a*c)*x^(3/2))/((a*b^2*c - 4*a^2*c^2)*x^4 + a^2*b^2 - 4*a^3*c + (a*b^3 - 4*a^2*b*c)
*x^2) - integrate(-1/4*(b*c*x^(5/2) + (b^2 - 10*a*c)*sqrt(x))/((a*b^2*c - 4*a^2*c^2)*x^4 + a^2*b^2 - 4*a^3*c +
(a*b^3 - 4*a^2*b*c)*x^2), x)

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

[Out]

Timed out

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

[Out]

Timed out

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

[Out]

Timed out