3.311 \(\int \frac{x^5 \sqrt{a+b x^2+c x^4}}{d+e x^2} \, dx\)
Optimal. Leaf size=272 \[ -\frac{\left (-8 c^2 d e (b d-a e)-2 b c e^2 (b d-2 a e)-b^3 e^3+16 c^3 d^3\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{5/2} e^4}+\frac{\sqrt{a+b x^2+c x^4} \left ((2 c d-b e) (b e+4 c d)-2 c e x^2 (b e+2 c d)\right )}{16 c^2 e^3}+\frac{d^2 \sqrt{a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^4}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c e} \]
[Out]
(((2*c*d - b*e)*(4*c*d + b*e) - 2*c*e*(2*c*d + b*e)*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*c^2*e^3) + (a + b*x^2 +
c*x^4)^(3/2)/(6*c*e) - ((16*c^3*d^3 - b^3*e^3 - 2*b*c*e^2*(b*d - 2*a*e) - 8*c^2*d*e*(b*d - a*e))*ArcTanh[(b +
2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(5/2)*e^4) + (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*e^4)
________________________________________________________________________________________
Rubi [A] time = 0.573454, antiderivative size = 272, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used =
{1251, 1653, 814, 843, 621, 206, 724} \[ -\frac{\left (-8 c^2 d e (b d-a e)-2 b c e^2 (b d-2 a e)-b^3 e^3+16 c^3 d^3\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{5/2} e^4}+\frac{\sqrt{a+b x^2+c x^4} \left ((2 c d-b e) (b e+4 c d)-2 c e x^2 (b e+2 c d)\right )}{16 c^2 e^3}+\frac{d^2 \sqrt{a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^4}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c e} \]
Antiderivative was successfully verified.
[In]
Int[(x^5*Sqrt[a + b*x^2 + c*x^4])/(d + e*x^2),x]
[Out]
(((2*c*d - b*e)*(4*c*d + b*e) - 2*c*e*(2*c*d + b*e)*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*c^2*e^3) + (a + b*x^2 +
c*x^4)^(3/2)/(6*c*e) - ((16*c^3*d^3 - b^3*e^3 - 2*b*c*e^2*(b*d - 2*a*e) - 8*c^2*d*e*(b*d - a*e))*ArcTanh[(b +
2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(5/2)*e^4) + (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*e^4)
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 1653
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
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]
Rubi steps
\begin{align*} \int \frac{x^5 \sqrt{a+b x^2+c x^4}}{d+e x^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 \sqrt{a+b x+c x^2}}{d+e x} \, dx,x,x^2\right )\\ &=\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c e}+\frac{\operatorname{Subst}\left (\int \frac{\left (-\frac{3}{2} b d e-\frac{3}{2} e (2 c d+b e) x\right ) \sqrt{a+b x+c x^2}}{d+e x} \, dx,x,x^2\right )}{6 c e^2}\\ &=\frac{\left ((2 c d-b e) (4 c d+b e)-2 c e (2 c d+b e) x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c e}-\frac{\operatorname{Subst}\left (\int \frac{\frac{3}{4} d e (2 c d-b e) \left (4 b c d+b^2 e-4 a c e\right )+\frac{3}{4} e \left (16 c^3 d^3-b^3 e^3-2 b c e^2 (b d-2 a e)-8 c^2 d e (b d-a e)\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{24 c^2 e^4}\\ &=\frac{\left ((2 c d-b e) (4 c d+b e)-2 c e (2 c d+b e) x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c e}+\frac{\left (d^2 \left (c d^2-b d e+a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{2 e^4}-\frac{\left (16 c^3 d^3-b^3 e^3-2 b c e^2 (b d-2 a e)-8 c^2 d e (b d-a e)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{32 c^2 e^4}\\ &=\frac{\left ((2 c d-b e) (4 c d+b e)-2 c e (2 c d+b e) x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c e}-\frac{\left (d^2 \left (c d^2-b d e+a e^2\right )\right ) \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 )}{e^4}-\frac{\left (16 c^3 d^3-b^3 e^3-2 b c e^2 (b d-2 a e)-8 c^2 d e (b d-a 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^2 e^4}\\ &=\frac{\left ((2 c d-b e) (4 c d+b e)-2 c e (2 c d+b e) x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c e}-\frac{\left (16 c^3 d^3-b^3 e^3-2 b c e^2 (b d-2 a e)-8 c^2 d e (b d-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^{5/2} e^4}+\frac{d^2 \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 e^4}\\ \end{align*}
Mathematica [A] time = 0.426532, size = 267, normalized size = 0.98 \[ \frac{2 \sqrt{c} \left (e \sqrt{a+b x^2+c x^4} \left (2 c e \left (4 a e-3 b d+b e x^2\right )-3 b^2 e^2+4 c^2 \left (6 d^2-3 d e x^2+2 e^2 x^4\right )\right )+24 c^2 d^2 \sqrt{e (a e-b d)+c d^2} \tanh ^{-1}\left (\frac{-2 a e+b \left (d-e x^2\right )+2 c d x^2}{2 \sqrt{a+b x^2+c x^4} \sqrt{e (a e-b d)+c d^2}}\right )\right )-3 \left (8 c^2 d e (a e-b d)-2 b c e^2 (b d-2 a e)-b^3 e^3+16 c^3 d^3\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{96 c^{5/2} e^4} \]
Antiderivative was successfully verified.
[In]
Integrate[(x^5*Sqrt[a + b*x^2 + c*x^4])/(d + e*x^2),x]
[Out]
(-3*(16*c^3*d^3 - b^3*e^3 - 2*b*c*e^2*(b*d - 2*a*e) + 8*c^2*d*e*(-(b*d) + a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[
c]*Sqrt[a + b*x^2 + c*x^4])] + 2*Sqrt[c]*(e*Sqrt[a + b*x^2 + c*x^4]*(-3*b^2*e^2 + 2*c*e*(-3*b*d + 4*a*e + b*e*
x^2) + 4*c^2*(6*d^2 - 3*d*e*x^2 + 2*e^2*x^4)) + 24*c^2*d^2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*ArcTanh[(-2*a*e + 2*
c*d*x^2 + b*(d - e*x^2))/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + b*x^2 + c*x^4])]))/(96*c^(5/2)*e^4)
________________________________________________________________________________________
Maple [B] time = 0.043, size = 1049, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^5*(c*x^4+b*x^2+a)^(1/2)/(e*x^2+d),x)
[Out]
1/6*(c*x^4+b*x^2+a)^(3/2)/c/e-1/8/e*b/c*x^2*(c*x^4+b*x^2+a)^(1/2)-1/16/e*b^2/c^2*(c*x^4+b*x^2+a)^(1/2)-1/8/e*b
/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*a+1/32/e*b^3/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+
b*x^2+a)^(1/2))-1/4/e^2*d*(c*x^4+b*x^2+a)^(1/2)*x^2-1/8/e^2*d/c*(c*x^4+b*x^2+a)^(1/2)*b-1/4/e^2*d/c^(1/2)*ln((
1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*a+1/16/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
))*b^2+1/2*d^2/e^3*(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*d^2/e^3*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*d^3/e^4*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)-1/2*d^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))*a+1/2*d^3/e^4/((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-1/2*d^4/e^5/((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
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^5*(c*x^4+b*x^2+a)^(1/2)/(e*x^2+d),x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
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^5*(c*x^4+b*x^2+a)^(1/2)/(e*x^2+d),x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5} \sqrt{a + b x^{2} + c x^{4}}}{d + e x^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**5*(c*x**4+b*x**2+a)**(1/2)/(e*x**2+d),x)
[Out]
Integral(x**5*sqrt(a + b*x**2 + c*x**4)/(d + e*x**2), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{4} + b x^{2} + a} x^{5}}{e x^{2} + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^5*(c*x^4+b*x^2+a)^(1/2)/(e*x^2+d),x, algorithm="giac")
[Out]
integrate(sqrt(c*x^4 + b*x^2 + a)*x^5/(e*x^2 + d), x)