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

Optimal. Leaf size=562 \[ \frac{a^{3/2} e \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 d^2}+\frac{\sqrt{a+b x^2+c x^4} \left (-2 c e (5 b d-4 a e)+b^2 e^2-2 c e x^2 (2 c d-b e)+8 c^2 d^2\right )}{16 c d^2 e}-\frac{(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{3/2} d^2 e^2}+\frac{b e \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{3/2} d^2}-\frac{e \left (8 a c+b^2+2 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2}+\frac{3 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{16 \sqrt{c} d}+\frac{\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 d^2 e^2}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{2 d x^2}+\frac{3 \left (3 b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{8 d}-\frac{3 \sqrt{a} b \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{4 d} \]

[Out]

(3*(3*b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*d) - (e*(b^2 + 8*a*c + 2*b*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*
c*d^2) + ((8*c^2*d^2 + b^2*e^2 - 2*c*e*(5*b*d - 4*a*e) - 2*c*e*(2*c*d - b*e)*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16
*c*d^2*e) - (a + b*x^2 + c*x^4)^(3/2)/(2*d*x^2) - (3*Sqrt[a]*b*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2
+ c*x^4])])/(4*d) + (a^(3/2)*e*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(2*d^2) + (3*(b^2
+ 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*Sqrt[c]*d) + (b*(b^2 - 12*a*c)*e*ArcT
anh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(3/2)*d^2) - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2
- 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(3/2)*d^2*e^2) + (
(c*d^2 - b*d*e + a*e^2)^(3/2)*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a
+ b*x^2 + c*x^4])])/(2*d^2*e^2)

________________________________________________________________________________________

Rubi [A]  time = 0.923926, antiderivative size = 562, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {1251, 960, 732, 814, 843, 621, 206, 724, 734} \[ \frac{a^{3/2} e \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 d^2}+\frac{\sqrt{a+b x^2+c x^4} \left (-2 c e (5 b d-4 a e)+b^2 e^2-2 c e x^2 (2 c d-b e)+8 c^2 d^2\right )}{16 c d^2 e}-\frac{(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{3/2} d^2 e^2}+\frac{b e \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{3/2} d^2}-\frac{e \left (8 a c+b^2+2 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2}+\frac{3 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{16 \sqrt{c} d}+\frac{\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 d^2 e^2}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{2 d x^2}+\frac{3 \left (3 b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{8 d}-\frac{3 \sqrt{a} b \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(3*b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*d) - (e*(b^2 + 8*a*c + 2*b*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*
c*d^2) + ((8*c^2*d^2 + b^2*e^2 - 2*c*e*(5*b*d - 4*a*e) - 2*c*e*(2*c*d - b*e)*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16
*c*d^2*e) - (a + b*x^2 + c*x^4)^(3/2)/(2*d*x^2) - (3*Sqrt[a]*b*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2
+ c*x^4])])/(4*d) + (a^(3/2)*e*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(2*d^2) + (3*(b^2
+ 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*Sqrt[c]*d) + (b*(b^2 - 12*a*c)*e*ArcT
anh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(3/2)*d^2) - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2
- 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(3/2)*d^2*e^2) + (
(c*d^2 - b*d*e + a*e^2)^(3/2)*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a
+ b*x^2 + c*x^4])])/(2*d^2*e^2)

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 960

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], 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] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 732

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

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 724

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

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2+c x^4\right )^{3/2}}{x^3 \left (d+e x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^{3/2}}{x^2 (d+e x)} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{\left (a+b x+c x^2\right )^{3/2}}{d x^2}-\frac{e \left (a+b x+c x^2\right )^{3/2}}{d^2 x}+\frac{e^2 \left (a+b x+c x^2\right )^{3/2}}{d^2 (d+e x)}\right ) \, dx,x,x^2\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^{3/2}}{x^2} \, dx,x,x^2\right )}{2 d}-\frac{e \operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^{3/2}}{x} \, dx,x,x^2\right )}{2 d^2}+\frac{e^2 \operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx,x,x^2\right )}{2 d^2}\\ &=-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{2 d x^2}+\frac{3 \operatorname{Subst}\left (\int \frac{(b+2 c x) \sqrt{a+b x+c x^2}}{x} \, dx,x,x^2\right )}{4 d}+\frac{e \operatorname{Subst}\left (\int \frac{(-2 a-b x) \sqrt{a+b x+c x^2}}{x} \, dx,x,x^2\right )}{4 d^2}-\frac{e \operatorname{Subst}\left (\int \frac{(b d-2 a e+(2 c d-b e) x) \sqrt{a+b x+c x^2}}{d+e x} \, dx,x,x^2\right )}{4 d^2}\\ &=\frac{3 \left (3 b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{8 d}-\frac{e \left (b^2+8 a c+2 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2}+\frac{\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2 e}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{2 d x^2}-\frac{3 \operatorname{Subst}\left (\int \frac{-4 a b c-c \left (b^2+4 a c\right ) x}{x \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{16 c d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (4 c e (b d-2 a e)^2-d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )-\frac{1}{2} (2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{16 c d^2 e}-\frac{e \operatorname{Subst}\left (\int \frac{8 a^2 c-\frac{1}{2} b \left (b^2-12 a c\right ) x}{x \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{16 c d^2}\\ &=\frac{3 \left (3 b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{8 d}-\frac{e \left (b^2+8 a c+2 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2}+\frac{\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2 e}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{2 d x^2}+\frac{(3 a b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{4 d}+\frac{\left (3 \left (b^2+4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{16 d}-\frac{\left (a^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{2 d^2}+\frac{\left (b \left (b^2-12 a c\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{32 c d^2}+\frac{\left (c d^2-b d e+a e^2\right )^2 \operatorname{Subst}\left (\int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{2 d^2 e^2}-\frac{\left ((2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{32 c d^2 e^2}\\ &=\frac{3 \left (3 b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{8 d}-\frac{e \left (b^2+8 a c+2 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2}+\frac{\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2 e}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{2 d x^2}-\frac{(3 a b) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x^2}{\sqrt{a+b x^2+c x^4}}\right )}{2 d}+\frac{\left (3 \left (b^2+4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{8 d}+\frac{\left (a^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x^2}{\sqrt{a+b x^2+c x^4}}\right )}{d^2}+\frac{\left (b \left (b^2-12 a c\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{16 c d^2}-\frac{\left (c d^2-b d e+a e^2\right )^2 \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac{-b d+2 a e-(2 c d-b e) x^2}{\sqrt{a+b x^2+c x^4}}\right )}{d^2 e^2}-\frac{\left ((2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{16 c d^2 e^2}\\ &=\frac{3 \left (3 b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{8 d}-\frac{e \left (b^2+8 a c+2 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2}+\frac{\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c d^2 e}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{2 d x^2}-\frac{3 \sqrt{a} b \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{4 d}+\frac{a^{3/2} e \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 d^2}+\frac{3 \left (b^2+4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{16 \sqrt{c} d}+\frac{b \left (b^2-12 a c\right ) e \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{3/2} d^2}-\frac{(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{3/2} d^2 e^2}+\frac{\left (c d^2-b d e+a e^2\right )^{3/2} \tanh ^{-1}\left (\frac{b d-2 a e+(2 c d-b e) x^2}{2 \sqrt{c d^2-b d e+a e^2} \sqrt{a+b x^2+c x^4}}\right )}{2 d^2 e^2}\\ \end{align*}

Mathematica [A]  time = 0.513455, size = 240, normalized size = 0.43 \[ \frac{1}{4} \left (-\frac{2 \left (x^2 \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x^2-2 c d x^2}{2 \sqrt{a+b x^2+c x^4} \sqrt{e (a e-b d)+c d^2}}\right )+d e \sqrt{a+b x^2+c x^4} \left (a e-c d x^2\right )\right )}{d^2 e^2 x^2}+\frac{\sqrt{a} (2 a e-3 b d) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{d^2}-\frac{\sqrt{c} (2 c d-3 b e) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{e^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.023, size = 1207, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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