3.242 \(\int \frac{1}{x^2 (d+e x^2) (a+c x^4)} \, dx\)

Optimal. Leaf size=348 \[ -\frac{c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{4 \sqrt{2} a^{5/4} \left (a e^2+c d^2\right )}+\frac{c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{4 \sqrt{2} a^{5/4} \left (a e^2+c d^2\right )}+\frac{c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{5/4} \left (a e^2+c d^2\right )}-\frac{c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt{2} a^{5/4} \left (a e^2+c d^2\right )}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{d^{3/2} \left (a e^2+c d^2\right )}-\frac{1}{a d x} \]

[Out]

-(1/(a*d*x)) - (e^(5/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(d^(3/2)*(c*d^2 + a*e^2)) + (c^(3/4)*(Sqrt[c]*d + Sqrt[a]
*e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(5/4)*(c*d^2 + a*e^2)) - (c^(3/4)*(Sqrt[c]*d + Sqrt[
a]*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(5/4)*(c*d^2 + a*e^2)) - (c^(3/4)*(Sqrt[c]*d - Sqr
t[a]*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(5/4)*(c*d^2 + a*e^2)) + (c^(3/4)
*(Sqrt[c]*d - Sqrt[a]*e)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(5/4)*(c*d^2 + a
*e^2))

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Rubi [A]  time = 0.299538, antiderivative size = 348, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {1288, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{4 \sqrt{2} a^{5/4} \left (a e^2+c d^2\right )}+\frac{c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{4 \sqrt{2} a^{5/4} \left (a e^2+c d^2\right )}+\frac{c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{5/4} \left (a e^2+c d^2\right )}-\frac{c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt{2} a^{5/4} \left (a e^2+c d^2\right )}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{d^{3/2} \left (a e^2+c d^2\right )}-\frac{1}{a d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(a*d*x)) - (e^(5/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(d^(3/2)*(c*d^2 + a*e^2)) + (c^(3/4)*(Sqrt[c]*d + Sqrt[a]
*e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(5/4)*(c*d^2 + a*e^2)) - (c^(3/4)*(Sqrt[c]*d + Sqrt[
a]*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(5/4)*(c*d^2 + a*e^2)) - (c^(3/4)*(Sqrt[c]*d - Sqr
t[a]*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(5/4)*(c*d^2 + a*e^2)) + (c^(3/4)
*(Sqrt[c]*d - Sqrt[a]*e)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(5/4)*(c*d^2 + a
*e^2))

Rule 1288

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(
(f*x)^m*(d + e*x^2)^q)/(a + c*x^4), x], x] /; FreeQ[{a, c, d, e, f, m}, x] && IntegerQ[q] && IntegerQ[m]

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 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.25288, size = 389, normalized size = 1.12 \[ \frac{-\sqrt{d} \left (8 a^{5/4} e^2+\sqrt{2} c^{5/4} d^2 x \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )-\sqrt{2} c^{5/4} d^2 x \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )-\sqrt{2} \sqrt{a} c^{3/4} d e x \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )+\sqrt{2} \sqrt{a} c^{3/4} d e x \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )-2 \sqrt{2} c^{3/4} d x \left (\sqrt{a} e+\sqrt{c} d\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt{2} c^{3/4} d x \left (\sqrt{a} e+\sqrt{c} d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )+8 \sqrt [4]{a} c d^2\right )-8 a^{5/4} e^{5/2} x \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 a^{5/4} d^{3/2} x \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^(5/4)*e^(5/2)*x*ArcTan[(Sqrt[e]*x)/Sqrt[d]] - Sqrt[d]*(8*a^(1/4)*c*d^2 + 8*a^(5/4)*e^2 - 2*Sqrt[2]*c^(3/
4)*d*(Sqrt[c]*d + Sqrt[a]*e)*x*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 2*Sqrt[2]*c^(3/4)*d*(Sqrt[c]*d + Sqrt
[a]*e)*x*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + Sqrt[2]*c^(5/4)*d^2*x*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)
*x + Sqrt[c]*x^2] - Sqrt[2]*Sqrt[a]*c^(3/4)*d*e*x*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] - Sqr
t[2]*c^(5/4)*d^2*x*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] + Sqrt[2]*Sqrt[a]*c^(3/4)*d*e*x*Log[
Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2]))/(8*a^(5/4)*d^(3/2)*(c*d^2 + a*e^2)*x)

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Maple [A]  time = 0.01, size = 390, normalized size = 1.1 \begin{align*} -{\frac{ec\sqrt{2}}{ \left ( 4\,a{e}^{2}+4\,c{d}^{2} \right ) a}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }-{\frac{ec\sqrt{2}}{ \left ( 4\,a{e}^{2}+4\,c{d}^{2} \right ) a}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }-{\frac{ec\sqrt{2}}{ \left ( 8\,a{e}^{2}+8\,c{d}^{2} \right ) a}\sqrt [4]{{\frac{a}{c}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ) }-{\frac{cd\sqrt{2}}{ \left ( 8\,a{e}^{2}+8\,c{d}^{2} \right ) a}\ln \left ({ \left ({x}^{2}-\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ({x}^{2}+\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-{\frac{cd\sqrt{2}}{ \left ( 4\,a{e}^{2}+4\,c{d}^{2} \right ) a}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-{\frac{cd\sqrt{2}}{ \left ( 4\,a{e}^{2}+4\,c{d}^{2} \right ) a}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-{\frac{1}{adx}}-{\frac{{e}^{3}}{d \left ( a{e}^{2}+c{d}^{2} \right ) }\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c/(a*e^2+c*d^2)/a*e*(1/c*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c*a)^(1/4)*x+1)-1/4*c/(a*e^2+c*d^2)/a*e*(1/c*
a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c*a)^(1/4)*x-1)-1/8*c/(a*e^2+c*d^2)/a*e*(1/c*a)^(1/4)*2^(1/2)*ln((x^2+(1/c*
a)^(1/4)*x*2^(1/2)+(1/c*a)^(1/2))/(x^2-(1/c*a)^(1/4)*x*2^(1/2)+(1/c*a)^(1/2)))-1/8*c/(a*e^2+c*d^2)/a*d/(1/c*a)
^(1/4)*2^(1/2)*ln((x^2-(1/c*a)^(1/4)*x*2^(1/2)+(1/c*a)^(1/2))/(x^2+(1/c*a)^(1/4)*x*2^(1/2)+(1/c*a)^(1/2)))-1/4
*c/(a*e^2+c*d^2)/a*d/(1/c*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c*a)^(1/4)*x+1)-1/4*c/(a*e^2+c*d^2)/a*d/(1/c*a)^(
1/4)*2^(1/2)*arctan(2^(1/2)/(1/c*a)^(1/4)*x-1)-1/a/d/x-1/d*e^3/(a*e^2+c*d^2)/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2
))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 13.5046, size = 8235, normalized size = 23.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*a*e^2*x*sqrt(-e/d)*log((e*x^2 - 2*d*x*sqrt(-e/d) - d)/(e*x^2 + d)) + (a*c*d^3 + a^2*d*e^2)*x*sqrt(-(2*
c^2*d*e + (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d
^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4
*e^4))*log(-(c^3*d^2 - a*c^2*e^2)*x + (a^2*c^2*d^2*e - a^3*c*e^3 - (a^4*c^2*d^5 + 2*a^5*c*d^3*e^2 + a^6*d*e^4)
*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*
c*d^2*e^6 + a^9*e^8)))*sqrt(-(2*c^2*d*e + (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d
^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2
*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))) - (a*c*d^3 + a^2*d*e^2)*x*sqrt(-(2*c^2*d*e + (a^2*c^2*d^4 + 2*a^3*c*d^
2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*
d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))*log(-(c^3*d^2 - a*c^2*e^2)*x
- (a^2*c^2*d^2*e - a^3*c*e^3 - (a^4*c^2*d^5 + 2*a^5*c*d^3*e^2 + a^6*d*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 +
a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))*sqrt(-(2*c^2
*d*e + (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8
+ 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^
4))) + (a*c*d^3 + a^2*d*e^2)*x*sqrt(-(2*c^2*d*e - (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2
*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8
)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))*log(-(c^3*d^2 - a*c^2*e^2)*x + (a^2*c^2*d^2*e - a^3*c*e^3 + (a^
4*c^2*d^5 + 2*a^5*c*d^3*e^2 + a^6*d*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*
c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))*sqrt(-(2*c^2*d*e - (a^2*c^2*d^4 + 2*a^3*c*d^2*e
^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4
*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))) - (a*c*d^3 + a^2*d*e^2)*x*sqrt
(-(2*c^2*d*e - (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*
c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2
+ a^4*e^4))*log(-(c^3*d^2 - a*c^2*e^2)*x - (a^2*c^2*d^2*e - a^3*c*e^3 + (a^4*c^2*d^5 + 2*a^5*c*d^3*e^2 + a^6*d
*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4
*a^8*c*d^2*e^6 + a^9*e^8)))*sqrt(-(2*c^2*d*e - (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*
c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))
/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))) - 4*c*d^2 - 4*a*e^2)/((a*c*d^3 + a^2*d*e^2)*x), -1/4*(4*a*e^2*x*s
qrt(e/d)*arctan(x*sqrt(e/d)) - (a*c*d^3 + a^2*d*e^2)*x*sqrt(-(2*c^2*d*e + (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4
*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4
*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))*log(-(c^3*d^2 - a*c^2*e^2)*x + (a^2*c^2
*d^2*e - a^3*c*e^3 - (a^4*c^2*d^5 + 2*a^5*c*d^3*e^2 + a^6*d*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^
4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))*sqrt(-(2*c^2*d*e + (a^2
*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3
*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))) + (a*c
*d^3 + a^2*d*e^2)*x*sqrt(-(2*c^2*d*e + (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*
e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^
2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))*log(-(c^3*d^2 - a*c^2*e^2)*x - (a^2*c^2*d^2*e - a^3*c*e^3 - (a^4*c^2*d^5 +
2*a^5*c*d^3*e^2 + a^6*d*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2
+ 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))*sqrt(-(2*c^2*d*e + (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^
4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^
8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))) - (a*c*d^3 + a^2*d*e^2)*x*sqrt(-(2*c^2*d*
e - (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4
*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))
*log(-(c^3*d^2 - a*c^2*e^2)*x + (a^2*c^2*d^2*e - a^3*c*e^3 + (a^4*c^2*d^5 + 2*a^5*c*d^3*e^2 + a^6*d*e^4)*sqrt(
-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*
e^6 + a^9*e^8)))*sqrt(-(2*c^2*d*e - (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2
+ a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d
^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))) + (a*c*d^3 + a^2*d*e^2)*x*sqrt(-(2*c^2*d*e - (a^2*c^2*d^4 + 2*a^3*c*d^2*e^2
+ a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^
4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))*log(-(c^3*d^2 - a*c^2*e^2)*x - (a^
2*c^2*d^2*e - a^3*c*e^3 + (a^4*c^2*d^5 + 2*a^5*c*d^3*e^2 + a^6*d*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c
^3*e^4)/(a^5*c^4*d^8 + 4*a^6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))*sqrt(-(2*c^2*d*e -
(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-(c^5*d^4 - 2*a*c^4*d^2*e^2 + a^2*c^3*e^4)/(a^5*c^4*d^8 + 4*a^
6*c^3*d^6*e^2 + 6*a^7*c^2*d^4*e^4 + 4*a^8*c*d^2*e^6 + a^9*e^8)))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4))) +
4*c*d^2 + 4*a*e^2)/((a*c*d^3 + a^2*d*e^2)*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**2/(e*x**2+d)/(c*x**4+a),x)

[Out]

Timed out

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Giac [A]  time = 1.12155, size = 470, normalized size = 1.35 \begin{align*} -\frac{{\left (\left (a c^{3}\right )^{\frac{1}{4}} a c e + \left (a c^{3}\right )^{\frac{3}{4}} d\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x + \sqrt{2} \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{c}\right )^{\frac{1}{4}}}\right )}{2 \,{\left (\sqrt{2} a^{2} c^{2} d^{2} + \sqrt{2} a^{3} c e^{2}\right )}} - \frac{{\left (\left (a c^{3}\right )^{\frac{1}{4}} a c e + \left (a c^{3}\right )^{\frac{3}{4}} d\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x - \sqrt{2} \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{c}\right )^{\frac{1}{4}}}\right )}{2 \,{\left (\sqrt{2} a^{2} c^{2} d^{2} + \sqrt{2} a^{3} c e^{2}\right )}} - \frac{{\left (\left (a c^{3}\right )^{\frac{1}{4}} a c e - \left (a c^{3}\right )^{\frac{3}{4}} d\right )} \log \left (x^{2} + \sqrt{2} x \left (\frac{a}{c}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{c}}\right )}{4 \,{\left (\sqrt{2} a^{2} c^{2} d^{2} + \sqrt{2} a^{3} c e^{2}\right )}} + \frac{{\left (\left (a c^{3}\right )^{\frac{1}{4}} a c e - \left (a c^{3}\right )^{\frac{3}{4}} d\right )} \log \left (x^{2} - \sqrt{2} x \left (\frac{a}{c}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{c}}\right )}{4 \,{\left (\sqrt{2} a^{2} c^{2} d^{2} + \sqrt{2} a^{3} c e^{2}\right )}} - \frac{\arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\frac{5}{2}}}{{\left (c d^{3} + a d e^{2}\right )} \sqrt{d}} - \frac{1}{a d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a*c^3)^(1/4)*a*c*e + (a*c^3)^(3/4)*d)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(sqrt
(2)*a^2*c^2*d^2 + sqrt(2)*a^3*c*e^2) - 1/2*((a*c^3)^(1/4)*a*c*e + (a*c^3)^(3/4)*d)*arctan(1/2*sqrt(2)*(2*x - s
qrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(sqrt(2)*a^2*c^2*d^2 + sqrt(2)*a^3*c*e^2) - 1/4*((a*c^3)^(1/4)*a*c*e - (a*c^3
)^(3/4)*d)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a^2*c^2*d^2 + sqrt(2)*a^3*c*e^2) + 1/4*((a*c^
3)^(1/4)*a*c*e - (a*c^3)^(3/4)*d)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a^2*c^2*d^2 + sqrt(2)*
a^3*c*e^2) - arctan(x*e^(1/2)/sqrt(d))*e^(5/2)/((c*d^3 + a*d*e^2)*sqrt(d)) - 1/(a*d*x)