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

Optimal. Leaf size=491 \[ \frac{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right )}{2 c^2}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right )}{2 c^2}+\frac{e x \sqrt{d+e x^2}}{2 c}+\frac{d \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 1.80219, antiderivative size = 491, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {1293, 195, 217, 206, 1692, 402, 377, 205} \[ \frac{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right )}{2 c^2}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right )}{2 c^2}+\frac{e x \sqrt{d+e x^2}}{2 c}+\frac{d \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1293

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Dist[(
e*f^2)/c, Int[(f*x)^(m - 2)*(d + e*x^2)^(q - 1), x], x] - Dist[f^2/c, Int[((f*x)^(m - 2)*(d + e*x^2)^(q - 1)*S
imp[a*e - (c*d - b*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] && GtQ[q, 0] && GtQ[m, 1] && LeQ[m, 3]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

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

Mathematica [B]  time = 6.27491, size = 14032, normalized size = 28.58 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^2*(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 = 0.029, size = 382, normalized size = 0.8 \begin{align*} -{\frac{{x}^{2}}{4\,c}{e}^{{\frac{3}{2}}}}+{\frac{ex}{4\,c}\sqrt{e{x}^{2}+d}}-{\frac{d}{8\,c}\sqrt{e}}+{\frac{{d}^{2}}{8\,c}\sqrt{e} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-2}}+{\frac{b}{{c}^{2}}{e}^{{\frac{3}{2}}}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) }-{\frac{3\,d}{2\,c}\sqrt{e}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) }+{\frac{1}{2\,{c}^{2}}\sqrt{e}\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 ( ac{e}^{2}-{b}^{2}{e}^{2}+2\,bcde-{c}^{2}{d}^{2} \right ){{\it \_R}}^{2}+2\, \left ( -2\,ab{e}^{3}+3\,{e}^{2}dac+{b}^{2}d{e}^{2}-2\,bc{d}^{2}e+{c}^{2}{d}^{3} \right ){\it \_R}+ac{d}^{2}{e}^{2}-{b}^{2}{d}^{2}{e}^{2}+2\,bc{d}^{3}e-{c}^{2}{d}^{4}}{{{\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}}\ln \left ( \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{2}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*e^(3/2)/c*x^2+1/4*e*x*(e*x^2+d)^(1/2)/c-1/8*e^(1/2)/c*d+1/8*e^(1/2)/c*d^2/((e*x^2+d)^(1/2)-e^(1/2)*x)^2+e
^(3/2)/c^2*ln((e*x^2+d)^(1/2)-e^(1/2)*x)*b-3/2*e^(1/2)/c*ln((e*x^2+d)^(1/2)-e^(1/2)*x)*d+1/2*e^(1/2)/c^2*sum((
(a*c*e^2-b^2*e^2+2*b*c*d*e-c^2*d^2)*_R^2+2*(-2*a*b*e^3+3*a*c*d*e^2+b^2*d*e^2-2*b*c*d^2*e+c^2*d^3)*_R+a*c*d^2*e
^2-b^2*d^2*e^2+2*b*c*d^3*e-c^2*d^4)/(_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{{\left (e x^{2} + d\right )}^{\frac{3}{2}} x^{2}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)^(3/2)*x^2/(c*x^4 + b*x^2 + a), 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^2*(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^{2} \left (d + e x^{2}\right )^{\frac{3}{2}}}{a + b x^{2} + c x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.43818, size = 78, normalized size = 0.16 \begin{align*} \frac{\sqrt{x^{2} e + d} x e}{2 \, c} - \frac{{\left (3 \, c d e - 2 \, b e^{2}\right )} e^{\left (-\frac{1}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(x^2*e + d)*x*e/c - 1/4*(3*c*d*e - 2*b*e^2)*e^(-1/2)*log((x*e^(1/2) - sqrt(x^2*e + d))^2)/c^2