3.394 \(\int \frac{x^6}{(d+e x^2)^{3/2} (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=350 \[ \frac{2 \left (-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c \sqrt{b-\sqrt{b^2-4 a c}} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )^{3/2}}+\frac{2 \left (\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c \sqrt{\sqrt{b^2-4 a c}+b} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )^{3/2}}-\frac{d^2 x}{e \sqrt{d+e x^2} \left (a e^2-b d e+c d^2\right )}+\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c e^{3/2}} \]

[Out]

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

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Rubi [A]  time = 4.3278, antiderivative size = 507, normalized size of antiderivative = 1.45, number of steps used = 14, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1297, 288, 217, 206, 1692, 377, 205} \[ \frac{\left (-\frac{2 a^2 c e-a b^2 e-3 a b c d+b^3 d}{\sqrt{b^2-4 a c}}-a b e-a c d+b^2 d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}+\frac{\left (\frac{2 a^2 c e-a b^2 e-3 a b c d+b^3 d}{\sqrt{b^2-4 a c}}-a b e-a c d+b^2 d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}-\frac{d^2 x}{e \sqrt{d+e x^2} \left (a e^2-b d e+c d^2\right )}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac{(b d-a e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c \sqrt{e} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^6/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]

[Out]

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

Rule 1297

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Dist[(
d^2*f^4)/(c*d^2 - b*d*e + a*e^2), Int[(f*x)^(m - 4)*(d + e*x^2)^q, x], x] - Dist[f^4/(c*d^2 - b*d*e + a*e^2),
Int[((f*x)^(m - 4)*(d + e*x^2)^(q + 1)*Simp[a*d + (b*d - a*e)*x^2, x])/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[q] && LtQ[q, -1] && GtQ[m, 3]

Rule 288

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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{x^6}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx &=-\frac{\int \frac{x^2 \left (a d+(b d-a e) x^2\right )}{\sqrt{d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{c d^2-b d e+a e^2}+\frac{d^2 \int \frac{x^2}{\left (d+e x^2\right )^{3/2}} \, dx}{c d^2-b d e+a e^2}\\ &=-\frac{d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}-\frac{\int \left (\frac{b d-a e}{c \sqrt{d+e x^2}}-\frac{a (b d-a e)+\left (b^2 d-a c d-a b e\right ) x^2}{c \sqrt{d+e x^2} \left (a+b x^2+c x^4\right )}\right ) \, dx}{c d^2-b d e+a e^2}+\frac{d^2 \int \frac{1}{\sqrt{d+e x^2}} \, dx}{e \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}+\frac{\int \frac{a (b d-a e)+\left (b^2 d-a c d-a b e\right ) x^2}{\sqrt{d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{c \left (c d^2-b d e+a e^2\right )}+\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{e \left (c d^2-b d e+a e^2\right )}-\frac{(b d-a e) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{3/2} \left (c d^2-b d e+a e^2\right )}+\frac{\int \left (\frac{b^2 d-a c d-a b e+\frac{-b^3 d+3 a b c d+a b^2 e-2 a^2 c e}{\sqrt{b^2-4 a c}}}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}}+\frac{b^2 d-a c d-a b e-\frac{-b^3 d+3 a b c d+a b^2 e-2 a^2 c e}{\sqrt{b^2-4 a c}}}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}}\right ) \, dx}{c \left (c d^2-b d e+a e^2\right )}-\frac{(b d-a e) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{3/2} \left (c d^2-b d e+a e^2\right )}-\frac{(b d-a e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c \sqrt{e} \left (c d^2-b d e+a e^2\right )}+\frac{\left (b^2 d-a c d-a b e-\frac{b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{c \left (c d^2-b d e+a e^2\right )}+\frac{\left (b^2 d-a c d-a b e+\frac{b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{3/2} \left (c d^2-b d e+a e^2\right )}-\frac{(b d-a e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c \sqrt{e} \left (c d^2-b d e+a e^2\right )}+\frac{\left (b^2 d-a c d-a b e-\frac{b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{b-\sqrt{b^2-4 a c}-\left (-2 c d+\left (b-\sqrt{b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{c \left (c d^2-b d e+a e^2\right )}+\frac{\left (b^2 d-a c d-a b e+\frac{b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{b+\sqrt{b^2-4 a c}-\left (-2 c d+\left (b+\sqrt{b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}+\frac{\left (b^2 d-a c d-a b e-\frac{b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} x}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )}+\frac{\left (b^2 d-a c d-a b e+\frac{b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} x}{\sqrt{b+\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c \sqrt{b+\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{3/2} \left (c d^2-b d e+a e^2\right )}-\frac{(b d-a e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c \sqrt{e} \left (c d^2-b d e+a e^2\right )}\\ \end{align*}

Mathematica [B]  time = 11.2763, size = 10968, normalized size = 31.34 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^6/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]

[Out]

Result too large to show

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Maple [C]  time = 0.034, size = 480, normalized size = 1.4 \begin{align*} -{\frac{x}{ce}{\frac{1}{\sqrt{e{x}^{2}+d}}}}+{\frac{1}{c}\ln \left ( \sqrt{e}x+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{3}{2}}}}-{\frac{bx}{{c}^{2}d}{\frac{1}{\sqrt{e{x}^{2}+d}}}}+8\,{\frac{{e}^{3/2}ab}{{c}^{2} \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ) \left ( 2\,e{x}^{2}-2\,\sqrt{e}\sqrt{e{x}^{2}+d}x+2\,d \right ) }}+8\,{\frac{\sqrt{e}ad}{c \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ) \left ( 2\,e{x}^{2}-2\,\sqrt{e}\sqrt{e{x}^{2}+d}x+2\,d \right ) }}-8\,{\frac{\sqrt{e}{b}^{2}d}{{c}^{2} \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ) \left ( 2\,e{x}^{2}-2\,\sqrt{e}\sqrt{e{x}^{2}+d}x+2\,d \right ) }}+2\,{\frac{\sqrt{e}}{c \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ) }\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{4}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{3}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{2}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){\it \_Z}+c{d}^{4} \right ) }{\frac{ \left ( \left ( abe+acd-{b}^{2}d \right ){{\it \_R}}^{2}+2\, \left ( 2\,{a}^{2}{e}^{2}-3\,abde-ac{d}^{2}+{b}^{2}{d}^{2} \right ){\it \_R}+ab{d}^{2}e+ac{d}^{3}-{b}^{2}{d}^{3} \right ) \ln \left ( \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{2}-{\it \_R} \right ) }{{{\it \_R}}^{3}c+3\,{{\it \_R}}^{2}be-3\,{{\it \_R}}^{2}cd+8\,{\it \_R}\,a{e}^{2}-4\,{\it \_R}\,bde+3\,{\it \_R}\,c{d}^{2}+b{d}^{2}e-c{d}^{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x)

[Out]

-1/c*x/e/(e*x^2+d)^(1/2)+1/c/e^(3/2)*ln(e^(1/2)*x+(e*x^2+d)^(1/2))-1/c^2*b*x/d/(e*x^2+d)^(1/2)+8/c^2*e^(3/2)/(
4*a*e^2-4*b*d*e+4*c*d^2)/(2*e*x^2-2*e^(1/2)*(e*x^2+d)^(1/2)*x+2*d)*a*b+8/c*e^(1/2)/(4*a*e^2-4*b*d*e+4*c*d^2)/(
2*e*x^2-2*e^(1/2)*(e*x^2+d)^(1/2)*x+2*d)*a*d-8/c^2*e^(1/2)/(4*a*e^2-4*b*d*e+4*c*d^2)/(2*e*x^2-2*e^(1/2)*(e*x^2
+d)^(1/2)*x+2*d)*b^2*d+2/c*e^(1/2)/(4*a*e^2-4*b*d*e+4*c*d^2)*sum(((a*b*e+a*c*d-b^2*d)*_R^2+2*(2*a^2*e^2-3*a*b*
d*e-a*c*d^2+b^2*d^2)*_R+a*b*d^2*e+a*c*d^3-b^2*d^3)/(_R^3*c+3*_R^2*b*e-3*_R^2*c*d+8*_R*a*e^2-4*_R*b*d*e+3*_R*c*
d^2+b*d^2*e-c*d^3)*ln(((e*x^2+d)^(1/2)-e^(1/2)*x)^2-_R),_R=RootOf(c*_Z^4+(4*b*e-4*c*d)*_Z^3+(16*a*e^2-8*b*d*e+
6*c*d^2)*_Z^2+(4*b*d^2*e-4*c*d^3)*_Z+c*d^4))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{{\left (c x^{4} + b x^{2} + a\right )}{\left (e x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate(x^6/((c*x^4 + b*x^2 + a)*(e*x^2 + d)^(3/2)), 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^6/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\left (d + e x^{2}\right )^{\frac{3}{2}} \left (a + b x^{2} + c x^{4}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(e*x**2+d)**(3/2)/(c*x**4+b*x**2+a),x)

[Out]

Integral(x**6/((d + e*x**2)**(3/2)*(a + b*x**2 + c*x**4)), x)

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Giac [A]  time = 1.19012, size = 101, normalized size = 0.29 \begin{align*} -\frac{c^{2} d^{2} x}{{\left (c^{3} d^{2} e - b c^{2} d e^{2} + a c^{2} e^{3}\right )} \sqrt{x^{2} e + d}} - \frac{e^{\left (-\frac{3}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

-c^2*d^2*x/((c^3*d^2*e - b*c^2*d*e^2 + a*c^2*e^3)*sqrt(x^2*e + d)) - 1/2*e^(-3/2)*log((x*e^(1/2) - sqrt(x^2*e
+ d))^2)/c