3.4.15 \(\int \frac {\sqrt {a+b x^2+c x^4}}{x^3 (d+e x^2)} \, dx\) [315]

Optimal. Leaf size=361 \[ -\frac {\sqrt {a+b x^2+c x^4}}{2 d x^2}-\frac {b \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} d}+\frac {\sqrt {a} e \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 d^2}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 d}-\frac {b e \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} d^2}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} d^2}+\frac {\sqrt {c d^2-b d e+a e^2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x^2}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x^2+c x^4}}\right )}{2 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.35, antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {1265, 974, 746, 857, 635, 212, 738, 748} \begin {gather*} \frac {\sqrt {a e^2-b d e+c d^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}+\frac {\sqrt {a} 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 {b e \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} d^2}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} d^2}-\frac {\sqrt {a+b x^2+c x^4}}{2 d x^2}-\frac {b \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} d}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Sqrt[a + b*x^2 + c*x^4]/(d*x^2) - (b*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(4*Sqrt[
a]*d) + (Sqrt[a]*e*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(2*d^2) + (Sqrt[c]*ArcTanh[(b +
2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(2*d) - (b*e*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 +
c*x^4])])/(4*Sqrt[c]*d^2) - ((2*c*d - b*e)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4*Sqr
t[c]*d^2) + (Sqrt[c*d^2 - b*d*e + a*e^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)

Rule 212

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

Rule 635

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 738

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 746

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 748

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]

Rule 857

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 974

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 1265

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.67, size = 158, normalized size = 0.44 \begin {gather*} \frac {-\frac {d \sqrt {a+b x^2+c x^4}}{x^2}+2 \sqrt {-c d^2+b d e-a e^2} \tan ^{-1}\left (\frac {\sqrt {c} \left (d+e x^2\right )-e \sqrt {a+b x^2+c x^4}}{\sqrt {-c d^2+b d e-a e^2}}\right )+\frac {(b d-2 a e) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1003\) vs. \(2(297)=594\).
time = 0.16, size = 1004, normalized size = 2.78 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(1/2)/x^3/(e*x^2+d),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^3/(e*x^2+d),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.63, size = 1130, normalized size = 3.13 \begin {gather*} \left [\frac {2 \, \sqrt {c d^{2} - b d e + a e^{2}} a x^{2} \log \left (-\frac {8 \, c^{2} d^{2} x^{4} + 8 \, b c d^{2} x^{2} + {\left (b^{2} + 4 \, a c\right )} d^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c d x^{2} + b d - {\left (b x^{2} + 2 \, a\right )} e\right )} \sqrt {c d^{2} - b d e + a e^{2}} + {\left ({\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 8 \, a^{2}\right )} e^{2} - 2 \, {\left (4 \, b c d x^{4} + {\left (3 \, b^{2} + 4 \, a c\right )} d x^{2} + 4 \, a b d\right )} e}{x^{4} e^{2} + 2 \, d x^{2} e + d^{2}}\right ) - 4 \, \sqrt {c x^{4} + b x^{2} + a} a d - {\left (b d x^{2} - 2 \, a x^{2} e\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right )}{8 \, a d^{2} x^{2}}, \frac {4 \, \sqrt {-c d^{2} + b d e - a e^{2}} a x^{2} \arctan \left (-\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c d x^{2} + b d - {\left (b x^{2} + 2 \, a\right )} e\right )} \sqrt {-c d^{2} + b d e - a e^{2}}}{2 \, {\left (c^{2} d^{2} x^{4} + b c d^{2} x^{2} + a c d^{2} + {\left (a c x^{4} + a b x^{2} + a^{2}\right )} e^{2} - {\left (b c d x^{4} + b^{2} d x^{2} + a b d\right )} e\right )}}\right ) - 4 \, \sqrt {c x^{4} + b x^{2} + a} a d - {\left (b d x^{2} - 2 \, a x^{2} e\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right )}{8 \, a d^{2} x^{2}}, \frac {\sqrt {c d^{2} - b d e + a e^{2}} a x^{2} \log \left (-\frac {8 \, c^{2} d^{2} x^{4} + 8 \, b c d^{2} x^{2} + {\left (b^{2} + 4 \, a c\right )} d^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c d x^{2} + b d - {\left (b x^{2} + 2 \, a\right )} e\right )} \sqrt {c d^{2} - b d e + a e^{2}} + {\left ({\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 8 \, a^{2}\right )} e^{2} - 2 \, {\left (4 \, b c d x^{4} + {\left (3 \, b^{2} + 4 \, a c\right )} d x^{2} + 4 \, a b d\right )} e}{x^{4} e^{2} + 2 \, d x^{2} e + d^{2}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2} + a} a d + {\left (b d x^{2} - 2 \, a x^{2} e\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right )}{4 \, a d^{2} x^{2}}, \frac {2 \, \sqrt {-c d^{2} + b d e - a e^{2}} a x^{2} \arctan \left (-\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c d x^{2} + b d - {\left (b x^{2} + 2 \, a\right )} e\right )} \sqrt {-c d^{2} + b d e - a e^{2}}}{2 \, {\left (c^{2} d^{2} x^{4} + b c d^{2} x^{2} + a c d^{2} + {\left (a c x^{4} + a b x^{2} + a^{2}\right )} e^{2} - {\left (b c d x^{4} + b^{2} d x^{2} + a b d\right )} e\right )}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2} + a} a d + {\left (b d x^{2} - 2 \, a x^{2} e\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right )}{4 \, a d^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^3/(e*x^2+d),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x^{3} \left (d + e x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(1/2)/x**3/(e*x**2+d),x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 4.63, size = 216, normalized size = 0.60 \begin {gather*} \frac {{\left (c d^{2} - b d e + a e^{2}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e - a e^{2}}}\right )}{\sqrt {-c d^{2} + b d e - a e^{2}} d^{2}} + \frac {{\left (b d - 2 \, a e\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} d^{2}} + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} b + 2 \, a \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^3/(e*x^2+d),x, algorithm="giac")

[Out]

(c*d^2 - b*d*e + a*e^2)*arctan(-((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e -
a*e^2))/(sqrt(-c*d^2 + b*d*e - a*e^2)*d^2) + 1/2*(b*d - 2*a*e)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))
/sqrt(-a))/(sqrt(-a)*d^2) + 1/2*((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*b + 2*a*sqrt(c))/(((sqrt(c)*x^2 - sqr
t(c*x^4 + b*x^2 + a))^2 - a)*d)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c\,x^4+b\,x^2+a}}{x^3\,\left (e\,x^2+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________