3.1078 \(\int \frac{1}{\sqrt{x} (a+b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=503 \[ \frac{c^{3/4} \left (-3 b \sqrt{b^2-4 a c}-28 a c+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{c^{3/4} \left (3 b \sqrt{b^2-4 a c}-28 a c+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{c^{3/4} \left (-3 b \sqrt{b^2-4 a c}-28 a c+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{c^{3/4} \left (3 b \sqrt{b^2-4 a c}-28 a c+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{\sqrt{x} \left (-2 a c+b^2+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

(Sqrt[x]*(b^2 - 2*a*c + b*c*x^2))/(2*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (c^(3/4)*(3*b^2 - 28*a*c - 3*b*Sqr
t[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^(1/4)*a*(b^2 - 4*a*c)^(
3/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (c^(3/4)*(3*b^2 - 28*a*c + 3*b*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^(1/4)*a*(b^2 - 4*a*c)^(3/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))
+ (c^(3/4)*(3*b^2 - 28*a*c - 3*b*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^(1/4)*a*(b^2 - 4*a*c)^(3/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (c^(3/4)*(3*b^2 - 28*a*c + 3*b*Sqr
t[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^(1/4)*a*(b^2 - 4*a*c)^
(3/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))

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Rubi [A]  time = 1.2756, antiderivative size = 503, 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, 1345, 1422, 212, 208, 205} \[ \frac{c^{3/4} \left (-3 b \sqrt{b^2-4 a c}-28 a c+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{c^{3/4} \left (3 b \sqrt{b^2-4 a c}-28 a c+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{c^{3/4} \left (-3 b \sqrt{b^2-4 a c}-28 a c+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{4 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{c^{3/4} \left (3 b \sqrt{b^2-4 a c}-28 a c+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{4 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{\sqrt{x} \left (-2 a c+b^2+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(b^2 - 2*a*c + b*c*x^2))/(2*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (c^(3/4)*(3*b^2 - 28*a*c - 3*b*Sqr
t[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^(1/4)*a*(b^2 - 4*a*c)^(
3/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (c^(3/4)*(3*b^2 - 28*a*c + 3*b*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^(1/4)*a*(b^2 - 4*a*c)^(3/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))
+ (c^(3/4)*(3*b^2 - 28*a*c - 3*b*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^(1/4)*a*(b^2 - 4*a*c)^(3/2)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - (c^(3/4)*(3*b^2 - 28*a*c + 3*b*Sqr
t[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^(1/4)*a*(b^2 - 4*a*c)^
(3/2)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))

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 1345

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

Rule 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 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{1}{\sqrt{x} \left (a+b x^2+c x^4\right )^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^4+c x^8\right )^2} \, dx,x,\sqrt{x}\right )\\ &=\frac{\sqrt{x} \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{b^2-2 a c-4 \left (b^2-4 a c\right )-3 b c x^4}{a+b x^4+c x^8} \, dx,x,\sqrt{x}\right )}{2 a \left (b^2-4 a c\right )}\\ &=\frac{\sqrt{x} \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 (3 b^2-28 a c-3 b \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\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 )^{3/2}}+\frac{\left (c \left (3 b^2-28 a c+3 b \sqrt{b^2-4 a c}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\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 )^{3/2}}\\ &=\frac{\sqrt{x} \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 (3 b^2-28 a c-3 b \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 a \left (b^2-4 a c\right )^{3/2} \sqrt{-b-\sqrt{b^2-4 a c}}}+\frac{\left (c \left (3 b^2-28 a c-3 b \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 a \left (b^2-4 a c\right )^{3/2} \sqrt{-b-\sqrt{b^2-4 a c}}}-\frac{\left (c \left (3 b^2-28 a c+3 b \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 a \left (b^2-4 a c\right )^{3/2} \sqrt{-b+\sqrt{b^2-4 a c}}}-\frac{\left (c \left (3 b^2-28 a c+3 b \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 a \left (b^2-4 a c\right )^{3/2} \sqrt{-b+\sqrt{b^2-4 a c}}}\\ &=\frac{\sqrt{x} \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{c^{3/4} \left (3 b^2-28 a c-3 b \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 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{c^{3/4} \left (3 b^2-28 a c+3 b \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 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}+\frac{c^{3/4} \left (3 b^2-28 a c-3 b \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 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{c^{3/4} \left (3 b^2-28 a c+3 b \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 \sqrt [4]{2} a \left (b^2-4 a c\right )^{3/2} \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.234815, size = 153, normalized size = 0.3 \[ -\frac{\left (a+b x^2+c x^4\right ) \text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{3 \text{$\#$1}^4 b c \log \left (\sqrt{x}-\text{$\#$1}\right )-14 a c \log \left (\sqrt{x}-\text{$\#$1}\right )+3 b^2 \log \left (\sqrt{x}-\text{$\#$1}\right )}{\text{$\#$1}^3 b+2 \text{$\#$1}^7 c}\& \right ]+4 \sqrt{x} \left (-2 a c+b^2+b c x^2\right )}{8 a \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4*Sqrt[x]*(b^2 - 2*a*c + b*c*x^2) + (a + b*x^2 + c*x^4)*RootSum[a + b*#1^4 + c*#1^8 & , (3*b^2*Log[Sqrt[x] -
#1] - 14*a*c*Log[Sqrt[x] - #1] + 3*b*c*Log[Sqrt[x] - #1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ])/(8*a*(-b^2 + 4*a*c)*(
a + b*x^2 + c*x^4))

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Maple [C]  time = 0.264, size = 144, normalized size = 0.3 \begin{align*} 2\,{\frac{1}{c{x}^{4}+b{x}^{2}+a} \left ( -1/4\,{\frac{bc{x}^{5/2}}{a \left ( 4\,ac-{b}^{2} \right ) }}+1/4\,{\frac{ \left ( 2\,ac-{b}^{2} \right ) \sqrt{x}}{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{-3\,{{\it \_R}}^{4}bc+14\,ac-3\,{b}^{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(1/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^(5/2)+1/4*(2*a*c-b^2)/a/(4*a*c-b^2)*x^(1/2))/(c*x^4+b*x^2+a)+1/8/a/(4*a*c-b^2)*sum
((-3*_R^4*b*c+14*a*c-3*b^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{{\left (3 \, b^{2} c - 14 \, a c^{2}\right )} x^{\frac{9}{2}} +{\left (3 \, b^{3} - 13 \, a b c\right )} x^{\frac{5}{2}} + 4 \,{\left (a b^{2} - 4 \, a^{2} c\right )} \sqrt{x}}{2 \,{\left (a^{3} b^{2} - 4 \, a^{4} c +{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{4} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )}} - \int \frac{{\left (3 \, b^{2} c - 14 \, a c^{2}\right )} x^{\frac{7}{2}} +{\left (3 \, b^{3} - 17 \, a b c\right )} x^{\frac{3}{2}}}{4 \,{\left (a^{3} b^{2} - 4 \, a^{4} c +{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{4} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((3*b^2*c - 14*a*c^2)*x^(9/2) + (3*b^3 - 13*a*b*c)*x^(5/2) + 4*(a*b^2 - 4*a^2*c)*sqrt(x))/(a^3*b^2 - 4*a^4
*c + (a^2*b^2*c - 4*a^3*c^2)*x^4 + (a^2*b^3 - 4*a^3*b*c)*x^2) - integrate(1/4*((3*b^2*c - 14*a*c^2)*x^(7/2) +
(3*b^3 - 17*a*b*c)*x^(3/2))/(a^3*b^2 - 4*a^4*c + (a^2*b^2*c - 4*a^3*c^2)*x^4 + (a^2*b^3 - 4*a^3*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(1/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(1/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(1/x^(1/2)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

Timed out