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

Optimal. Leaf size=417 \[ -\frac{\sqrt{c} \left (-2 a \left (e \left (d \sqrt{b^2-4 a c}-a e\right )+c d^2\right )+b d \left (d \sqrt{b^2-4 a c}-2 a e\right )+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{c} \left (-2 a \left (c d^2-e \left (d \sqrt{b^2-4 a c}+a e\right )\right )-b d \left (d \sqrt{b^2-4 a c}+2 a e\right )+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^2}-\frac{d \sqrt{d+e x^2}}{2 a x^2}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a} \]

[Out]

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

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Rubi [A]  time = 3.24265, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1251, 897, 1287, 199, 206, 1166, 208} \[ -\frac{\sqrt{c} \left (b d \left (d \sqrt{b^2-4 a c}-2 a e\right )-2 a e \left (d \sqrt{b^2-4 a c}-a e\right )-2 a c d^2+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{c} \left (-b d \left (d \sqrt{b^2-4 a c}+2 a e\right )+2 a e \left (d \sqrt{b^2-4 a c}+a e\right )-2 a c d^2+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^2}-\frac{d \sqrt{d+e x^2}}{2 a x^2}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1287

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

Rule 199

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

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 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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

Mathematica [A]  time = 1.62938, size = 380, normalized size = 0.91 \[ \frac{\frac{\sqrt{2} \sqrt{c} \left (\frac{\left (2 a \left (e \left (d \sqrt{b^2-4 a c}-a e\right )+c d^2\right )+b d \left (2 a e-d \sqrt{b^2-4 a c}\right )-b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\left (b d \left (d \sqrt{b^2-4 a c}+2 a e\right )-2 a e \left (d \sqrt{b^2-4 a c}+a e\right )+2 a c d^2-b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c}}+\sqrt{d} (2 b d-3 a e) \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )-\sqrt{d} \log (x) (2 b d-3 a e)-\frac{a d \sqrt{d+e x^2}}{x^2}}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.03, size = 555, normalized size = 1.3 \begin{align*} -{\frac{{x}^{3}b}{6\,{a}^{2}}{e}^{{\frac{3}{2}}}}+{\frac{{x}^{2}eb}{8\,{a}^{2}}\sqrt{e{x}^{2}+d}}-{\frac{3\,bdx}{4\,{a}^{2}}\sqrt{e}}-{\frac{7\,b}{24\,{a}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{x}{2\,a}{e}^{{\frac{3}{2}}}}+{\frac{e}{a}\sqrt{e{x}^{2}+d}}-{\frac{3\,bd}{8\,{a}^{2}}\sqrt{e{x}^{2}+d}}-{\frac{de}{2\,a} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}+{\frac{5\,b{d}^{2}}{8\,{a}^{2}} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}+{\frac{b{d}^{3}}{24\,{a}^{2}} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-3}}+{\frac{1}{4\,{a}^{2}}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{8}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{6}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{4}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){{\it \_Z}}^{2}+c{d}^{4} \right ) }{\frac{cd \left ( -2\,ae+bd \right ){{\it \_R}}^{6}+ \left ( 4\,{a}^{2}{e}^{3}-8\,abd{e}^{2}+2\,e{d}^{2}ca+4\,{b}^{2}{d}^{2}e-3\,bc{d}^{3} \right ){{\it \_R}}^{4}+d \left ( -4\,{a}^{2}{e}^{3}+8\,abd{e}^{2}-2\,e{d}^{2}ca-4\,{b}^{2}{d}^{2}e+3\,bc{d}^{3} \right ){{\it \_R}}^{2}+2\,ac{d}^{4}e-bc{d}^{5}}{{{\it \_R}}^{7}c+3\,{{\it \_R}}^{5}be-3\,{{\it \_R}}^{5}cd+8\,{{\it \_R}}^{3}a{e}^{2}-4\,{{\it \_R}}^{3}bde+3\,{{\it \_R}}^{3}c{d}^{2}+{\it \_R}\,b{d}^{2}e-{\it \_R}\,c{d}^{3}}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x-{\it \_R} \right ) }}+{\frac{b}{{a}^{2}}{d}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ) }-{\frac{1}{2\,ad{x}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{5}{2}}}}+{\frac{e}{2\,ad} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{3\,e}{2\,a}\sqrt{d}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/a^2*e^(3/2)*x^3*b+1/8/a^2*e*(e*x^2+d)^(1/2)*x^2*b-3/4/a^2*e^(1/2)*x*b*d-7/24/a^2*(e*x^2+d)^(3/2)*b+1/2/a*
e^(3/2)*x+1/a*(e*x^2+d)^(1/2)*e-3/8/a^2*(e*x^2+d)^(1/2)*b*d-1/2/a*d/((e*x^2+d)^(1/2)-e^(1/2)*x)*e+5/8/a^2*d^2/
((e*x^2+d)^(1/2)-e^(1/2)*x)*b+1/24/a^2*b*d^3/((e*x^2+d)^(1/2)-e^(1/2)*x)^3+1/4/a^2*sum((c*d*(-2*a*e+b*d)*_R^6+
(4*a^2*e^3-8*a*b*d*e^2+2*a*c*d^2*e+4*b^2*d^2*e-3*b*c*d^3)*_R^4+d*(-4*a^2*e^3+8*a*b*d*e^2-2*a*c*d^2*e-4*b^2*d^2
*e+3*b*c*d^3)*_R^2+2*a*c*d^4*e-b*c*d^5)/(_R^7*c+3*_R^5*b*e-3*_R^5*c*d+8*_R^3*a*e^2-4*_R^3*b*d*e+3*_R^3*c*d^2+_
R*b*d^2*e-_R*c*d^3)*ln((e*x^2+d)^(1/2)-e^(1/2)*x-_R),_R=RootOf(c*_Z^8+(4*b*e-4*c*d)*_Z^6+(16*a*e^2-8*b*d*e+6*c
*d^2)*_Z^4+(4*b*d^2*e-4*c*d^3)*_Z^2+c*d^4))+1/a^2*b*d^(3/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)-1/2/a/d/x^2*
(e*x^2+d)^(5/2)+1/2/a*e/d*(e*x^2+d)^(3/2)-3/2/a*e*d^(1/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out