3.1068 \(\int \frac{1}{x^{3/2} (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=371 \[ -\frac{\sqrt [4]{c} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2^{3/4} a \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2^{3/4} a \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2^{3/4} a \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2^{3/4} a \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{2}{a \sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.567378, antiderivative size = 371, 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, 1368, 1510, 298, 205, 208} \[ -\frac{\sqrt [4]{c} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2^{3/4} a \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2^{3/4} a \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2^{3/4} a \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2^{3/4} a \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{2}{a \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(a + b*x^2 + c*x^4)),x]

[Out]

-2/(a*Sqrt[x]) - (c^(1/4)*(1 - 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)])/(2^(3/4)*a*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(1 + 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)])/(2^(3/4)*a*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(1 - 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)])/(2^(3/4)*a*(-b - Sqrt[b^2 - 4
*a*c])^(1/4)) + (c^(1/4)*(1 + 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)])/(2^(3/4)*a*(-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 1368

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

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

Mathematica [C]  time = 0.045767, size = 78, normalized size = 0.21 \[ -\frac{\text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{\text{$\#$1}^4 c \log \left (\sqrt{x}-\text{$\#$1}\right )+b \log \left (\sqrt{x}-\text{$\#$1}\right )}{2 \text{$\#$1}^5 c+\text{$\#$1} b}\& \right ]+\frac{4}{\sqrt{x}}}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(a + b*x^2 + c*x^4)),x]

[Out]

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

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Maple [C]  time = 0.26, size = 65, normalized size = 0.2 \begin{align*} -{\frac{1}{2\,a}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{{{\it \_R}}^{6}c+{{\it \_R}}^{2}b}{2\,{{\it \_R}}^{7}c+{{\it \_R}}^{3}b}\ln \left ( \sqrt{x}-{\it \_R} \right ) }}-2\,{\frac{1}{a\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(c*x^4+b*x^2+a),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 8.61433, size = 11429, normalized size = 30.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

-1/2*(4*a*x*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 - (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b
^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^
13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)))*arctan(1/2*((b^6 - 7*a*b^4*c + 13*a^2*b^2*c^2 - 4*a^3*c^3 + (
a^5*b^5 - 8*a^6*b^3*c + 16*a^7*b*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*
b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))*sqrt((b^8*c^8 - 6*a*b^6*c^9 + 11*a^2*b^4*c^10 - 6*a^3*b
^2*c^11 + a^4*c^12)*x - 1/2*sqrt(1/2)*(b^13*c^5 - 13*a*b^11*c^6 + 65*a^2*b^9*c^7 - 155*a^3*b^7*c^8 + 175*a^4*b
^5*c^9 - 79*a^5*b^3*c^10 + 12*a^6*b*c^11 + (a^5*b^12*c^5 - 16*a^6*b^10*c^6 + 100*a^7*b^8*c^7 - 305*a^8*b^6*c^8
+ 460*a^9*b^4*c^9 - 304*a^10*b^2*c^10 + 64*a^11*c^11)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3
+ a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 -
(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*
b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2))) - (b^10*c^4 - 10
*a*b^8*c^5 + 35*a^2*b^6*c^6 - 50*a^3*b^4*c^7 + 25*a^4*b^2*c^8 - 4*a^5*c^9 + (a^5*b^9*c^4 - 11*a^6*b^7*c^5 + 41
*a^7*b^5*c^6 - 56*a^8*b^3*c^7 + 16*a^9*b*c^8)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4
)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))*sqrt(x))*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c
+ 5*a^2*b*c^2 - (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 +
a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)))/(
b^8*c^5 - 6*a*b^6*c^6 + 11*a^2*b^4*c^7 - 6*a^3*b^2*c^8 + a^4*c^9)) - 4*a*x*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3
*c + 5*a^2*b*c^2 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3
+ a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2))
)*arctan(-1/2*((b^6 - 7*a*b^4*c + 13*a^2*b^2*c^2 - 4*a^3*c^3 - (a^5*b^5 - 8*a^6*b^3*c + 16*a^7*b*c^2)*sqrt((b^
8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^1
3*c^3)))*sqrt((b^8*c^8 - 6*a*b^6*c^9 + 11*a^2*b^4*c^10 - 6*a^3*b^2*c^11 + a^4*c^12)*x - 1/2*sqrt(1/2)*(b^13*c^
5 - 13*a*b^11*c^6 + 65*a^2*b^9*c^7 - 155*a^3*b^7*c^8 + 175*a^4*b^5*c^9 - 79*a^5*b^3*c^10 + 12*a^6*b*c^11 - (a^
5*b^12*c^5 - 16*a^6*b^10*c^6 + 100*a^7*b^8*c^7 - 305*a^8*b^6*c^8 + 460*a^9*b^4*c^9 - 304*a^10*b^2*c^10 + 64*a^
11*c^11)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12
*b^2*c^2 - 64*a^13*c^3)))*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^
8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^1
3*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)))*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 + (a^5*b^4
- 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*
a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2))) - (b^10*c^4 - 10*a*b^8*c^
5 + 35*a^2*b^6*c^6 - 50*a^3*b^4*c^7 + 25*a^4*b^2*c^8 - 4*a^5*c^9 - (a^5*b^9*c^4 - 11*a^6*b^7*c^5 + 41*a^7*b^5*
c^6 - 56*a^8*b^3*c^7 + 16*a^9*b*c^8)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b
^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))*sqrt(x)*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*
c^2 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(
a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2))))/(b^8*c^5 -
6*a*b^6*c^6 + 11*a^2*b^4*c^7 - 6*a^3*b^2*c^8 + a^4*c^9)) - a*x*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*
b*c^2 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)
/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)))*log(1/2*s
qrt(1/2)*(b^11 - 13*a*b^9*c + 63*a^2*b^7*c^2 - 138*a^3*b^5*c^3 + 128*a^4*b^3*c^4 - 32*a^5*b*c^5 - (a^5*b^10 -
16*a^6*b^8*c + 98*a^7*b^6*c^2 - 280*a^8*b^4*c^3 + 352*a^9*b^2*c^4 - 128*a^10*c^5)*sqrt((b^8 - 6*a*b^6*c + 11*a
^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))*sqrt(sqrt(1
/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2
*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*
a^6*b^2*c + 16*a^7*c^2)))*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^
8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^1
3*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)) + (b^4*c^4 - 3*a*b^2*c^5 + a^2*c^6)*sqrt(x)) + a*x*sqrt(sqrt(1/
2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*
b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a
^6*b^2*c + 16*a^7*c^2)))*log(-1/2*sqrt(1/2)*(b^11 - 13*a*b^9*c + 63*a^2*b^7*c^2 - 138*a^3*b^5*c^3 + 128*a^4*b^
3*c^4 - 32*a^5*b*c^5 - (a^5*b^10 - 16*a^6*b^8*c + 98*a^7*b^6*c^2 - 280*a^8*b^4*c^3 + 352*a^9*b^2*c^4 - 128*a^1
0*c^5)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b
^2*c^2 - 64*a^13*c^3)))*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*
c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2
*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)))*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 + (a^5*b^4 -
8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^
11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)) + (b^4*c^4 - 3*a*b^2*c^5 + a
^2*c^6)*sqrt(x)) - a*x*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 - (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c
^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*
c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)))*log(1/2*sqrt(1/2)*(b^11 - 13*a*b^9*c + 63*a^2*b^7*
c^2 - 138*a^3*b^5*c^3 + 128*a^4*b^3*c^4 - 32*a^5*b*c^5 + (a^5*b^10 - 16*a^6*b^8*c + 98*a^7*b^6*c^2 - 280*a^8*b
^4*c^3 + 352*a^9*b^2*c^4 - 128*a^10*c^5)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^
10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2
- (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10
*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)))*sqrt(-(b^5 - 5*
a*b^3*c + 5*a^2*b*c^2 - (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^
2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*
c^2)) + (b^4*c^4 - 3*a*b^2*c^5 + a^2*c^6)*sqrt(x)) + a*x*sqrt(sqrt(1/2)*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 -
(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*
b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)))*log(-1/2*sqrt(1/
2)*(b^11 - 13*a*b^9*c + 63*a^2*b^7*c^2 - 138*a^3*b^5*c^3 + 128*a^4*b^3*c^4 - 32*a^5*b*c^5 + (a^5*b^10 - 16*a^6
*b^8*c + 98*a^7*b^6*c^2 - 280*a^8*b^4*c^3 + 352*a^9*b^2*c^4 - 128*a^10*c^5)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4
*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))*sqrt(sqrt(1/2)*sq
rt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 - (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c
^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)))/(a^5*b^4 - 8*a^6*b^
2*c + 16*a^7*c^2)))*sqrt(-(b^5 - 5*a*b^3*c + 5*a^2*b*c^2 - (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*sqrt((b^8 - 6*
a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/(a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)
))/(a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)) + (b^4*c^4 - 3*a*b^2*c^5 + a^2*c^6)*sqrt(x)) + 4*sqrt(x))/(a*x)

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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**(3/2)/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2} + a\right )} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

integrate(1/((c*x^4 + b*x^2 + a)*x^(3/2)), x)