3.332 \(\int \frac{x^5}{(d+e x^2) \sqrt{a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=173 \[ -\frac{(b e+2 c d) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4 c^{3/2} e^2}+\frac{d^2 \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^2 \sqrt{a e^2-b d e+c d^2}}+\frac{\sqrt{a+b x^2+c x^4}}{2 c e} \]

[Out]

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

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Rubi [A]  time = 0.31374, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {1251, 1653, 843, 621, 206, 724} \[ -\frac{(b e+2 c d) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4 c^{3/2} e^2}+\frac{d^2 \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^2 \sqrt{a e^2-b d e+c d^2}}+\frac{\sqrt{a+b x^2+c x^4}}{2 c e} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

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

Mathematica [A]  time = 0.327012, size = 167, normalized size = 0.97 \[ \frac{-\frac{(b e+2 c d) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{c^{3/2}}+\frac{2 d^2 \tanh ^{-1}\left (\frac{-2 a e+b d-b e x^2+2 c d x^2}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{\sqrt{a e^2-b d e+c d^2}}+\frac{2 e \sqrt{a+b x^2+c x^4}}{c}}{4 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

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

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Maple [A]  time = 0.02, size = 267, normalized size = 1.5 \begin{align*}{\frac{1}{2\,ce}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{b}{4\,e}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{d}{2\,{e}^{2}}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{d}^{2}}{2\,{e}^{3}}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}-deb+c{d}^{2}}{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ({x}^{2}+{\frac{d}{e}} \right ) }+2\,\sqrt{{\frac{a{e}^{2}-deb+c{d}^{2}}{{e}^{2}}}}\sqrt{c \left ({x}^{2}+{\frac{d}{e}} \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({x}^{2}+{\frac{d}{e}} \right ) }+{\frac{a{e}^{2}-deb+c{d}^{2}}{{e}^{2}}}} \right ) \left ({x}^{2}+{\frac{d}{e}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}-deb+c{d}^{2}}{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\sqrt{c x^{4} + b x^{2} + a}{\left (e x^{2} + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^5/(sqrt(c*x^4 + b*x^2 + a)*(e*x^2 + d)), x)

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Fricas [B]  time = 130.366, size = 2921, normalized size = 16.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\left (d + e x^{2}\right ) \sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(e*x**2+d)/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral(x**5/((d + e*x**2)*sqrt(a + b*x**2 + c*x**4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\sqrt{c x^{4} + b x^{2} + a}{\left (e x^{2} + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^5/(sqrt(c*x^4 + b*x^2 + a)*(e*x^2 + d)), x)