3.393 \(\int \frac{1}{x^6 \sqrt{d+e x^2} (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=443 \[ -\frac{\left (b^2-a c\right ) \sqrt{d+e x^2}}{a^3 d x}-\frac{c \left (\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{a^3 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{c \left (-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{a^3 \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{2 b e \sqrt{d+e x^2}}{3 a^2 d^2 x}+\frac{b \sqrt{d+e x^2}}{3 a^2 d x^3}-\frac{8 e^2 \sqrt{d+e x^2}}{15 a d^3 x}+\frac{4 e \sqrt{d+e x^2}}{15 a d^2 x^3}-\frac{\sqrt{d+e x^2}}{5 a d x^5} \]

[Out]

-Sqrt[d + e*x^2]/(5*a*d*x^5) + (b*Sqrt[d + e*x^2])/(3*a^2*d*x^3) + (4*e*Sqrt[d + e*x^2])/(15*a*d^2*x^3) - ((b^
2 - a*c)*Sqrt[d + e*x^2])/(a^3*d*x) - (2*b*e*Sqrt[d + e*x^2])/(3*a^2*d^2*x) - (8*e^2*Sqrt[d + e*x^2])/(15*a*d^
3*x) - (c*(b^2 - a*c + (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)
/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(a^3*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 -
4*a*c])*e]) - (c*(b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c
])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(a^3*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sq
rt[b^2 - 4*a*c])*e])

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Rubi [A]  time = 1.43202, antiderivative size = 443, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {1303, 271, 264, 1692, 377, 205} \[ -\frac{\left (b^2-a c\right ) \sqrt{d+e x^2}}{a^3 d x}-\frac{c \left (\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{a^3 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{c \left (-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{a^3 \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{2 b e \sqrt{d+e x^2}}{3 a^2 d^2 x}+\frac{b \sqrt{d+e x^2}}{3 a^2 d x^3}-\frac{8 e^2 \sqrt{d+e x^2}}{15 a d^3 x}+\frac{4 e \sqrt{d+e x^2}}{15 a d^2 x^3}-\frac{\sqrt{d+e x^2}}{5 a d x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)),x]

[Out]

-Sqrt[d + e*x^2]/(5*a*d*x^5) + (b*Sqrt[d + e*x^2])/(3*a^2*d*x^3) + (4*e*Sqrt[d + e*x^2])/(15*a*d^2*x^3) - ((b^
2 - a*c)*Sqrt[d + e*x^2])/(a^3*d*x) - (2*b*e*Sqrt[d + e*x^2])/(3*a^2*d^2*x) - (8*e^2*Sqrt[d + e*x^2])/(15*a*d^
3*x) - (c*(b^2 - a*c + (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)
/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(a^3*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 -
4*a*c])*e]) - (c*(b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c
])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(a^3*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sq
rt[b^2 - 4*a*c])*e])

Rule 1303

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

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{x^6 \sqrt{d+e x^2} \left (a+b x^2+c x^4\right )} \, dx &=\int \left (\frac{1}{a x^6 \sqrt{d+e x^2}}-\frac{b}{a^2 x^4 \sqrt{d+e x^2}}+\frac{b^2-a c}{a^3 x^2 \sqrt{d+e x^2}}+\frac{-b \left (b^2-2 a c\right )-c \left (b^2-a c\right ) x^2}{a^3 \sqrt{d+e x^2} \left (a+b x^2+c x^4\right )}\right ) \, dx\\ &=\frac{\int \frac{-b \left (b^2-2 a c\right )-c \left (b^2-a c\right ) x^2}{\sqrt{d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{a^3}+\frac{\int \frac{1}{x^6 \sqrt{d+e x^2}} \, dx}{a}-\frac{b \int \frac{1}{x^4 \sqrt{d+e x^2}} \, dx}{a^2}+\frac{\left (b^2-a c\right ) \int \frac{1}{x^2 \sqrt{d+e x^2}} \, dx}{a^3}\\ &=-\frac{\sqrt{d+e x^2}}{5 a d x^5}+\frac{b \sqrt{d+e x^2}}{3 a^2 d x^3}-\frac{\left (b^2-a c\right ) \sqrt{d+e x^2}}{a^3 d x}+\frac{\int \left (\frac{-\frac{b c \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-c \left (b^2-a c\right )}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}}+\frac{\frac{b c \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-c \left (b^2-a c\right )}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}}\right ) \, dx}{a^3}-\frac{(4 e) \int \frac{1}{x^4 \sqrt{d+e x^2}} \, dx}{5 a d}+\frac{(2 b e) \int \frac{1}{x^2 \sqrt{d+e x^2}} \, dx}{3 a^2 d}\\ &=-\frac{\sqrt{d+e x^2}}{5 a d x^5}+\frac{b \sqrt{d+e x^2}}{3 a^2 d x^3}+\frac{4 e \sqrt{d+e x^2}}{15 a d^2 x^3}-\frac{\left (b^2-a c\right ) \sqrt{d+e x^2}}{a^3 d x}-\frac{2 b e \sqrt{d+e x^2}}{3 a^2 d^2 x}-\frac{\left (c \left (b^2-a c-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{a^3}-\frac{\left (c \left (b^2-a c+\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{a^3}+\frac{\left (8 e^2\right ) \int \frac{1}{x^2 \sqrt{d+e x^2}} \, dx}{15 a d^2}\\ &=-\frac{\sqrt{d+e x^2}}{5 a d x^5}+\frac{b \sqrt{d+e x^2}}{3 a^2 d x^3}+\frac{4 e \sqrt{d+e x^2}}{15 a d^2 x^3}-\frac{\left (b^2-a c\right ) \sqrt{d+e x^2}}{a^3 d x}-\frac{2 b e \sqrt{d+e x^2}}{3 a^2 d^2 x}-\frac{8 e^2 \sqrt{d+e x^2}}{15 a d^3 x}-\frac{\left (c \left (b^2-a c-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b+\sqrt{b^2-4 a c}-\left (-2 c d+\left (b+\sqrt{b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{a^3}-\frac{\left (c \left (b^2-a c+\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b-\sqrt{b^2-4 a c}-\left (-2 c d+\left (b-\sqrt{b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{a^3}\\ &=-\frac{\sqrt{d+e x^2}}{5 a d x^5}+\frac{b \sqrt{d+e x^2}}{3 a^2 d x^3}+\frac{4 e \sqrt{d+e x^2}}{15 a d^2 x^3}-\frac{\left (b^2-a c\right ) \sqrt{d+e x^2}}{a^3 d x}-\frac{2 b e \sqrt{d+e x^2}}{3 a^2 d^2 x}-\frac{8 e^2 \sqrt{d+e x^2}}{15 a d^3 x}-\frac{c \left (b^2-a c+\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} x}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{a^3 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}-\frac{c \left (b^2-a c-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} x}{\sqrt{b+\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{a^3 \sqrt{b+\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 = 1.79496, size = 383, normalized size = 0.86 \[ -\frac{\frac{a^2 \sqrt{d+e x^2} \left (3 d^2-4 d e x^2+8 e^2 x^4\right )}{d^3 x^5}+\frac{15 \left (b^2-a c\right ) \sqrt{d+e x^2}}{d x}+\frac{15 c \left (\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}+\frac{15 c \left (-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{5 a b \left (d-2 e x^2\right ) \sqrt{d+e x^2}}{d^2 x^3}}{15 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)),x]

[Out]

-((15*(b^2 - a*c)*Sqrt[d + e*x^2])/(d*x) - (5*a*b*(d - 2*e*x^2)*Sqrt[d + e*x^2])/(d^2*x^3) + (a^2*Sqrt[d + e*x
^2]*(3*d^2 - 4*d*e*x^2 + 8*e^2*x^4))/(d^3*x^5) + (15*c*(b^2 - a*c + (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*ArcTa
n[(Sqrt[2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b - Sqrt[b
^2 - 4*a*c]]*Sqrt[2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e]) + (15*c*(b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c
])*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b
+ Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]))/(15*a^3)

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Maple [C]  time = 0.027, size = 350, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,{a}^{3}}\sqrt{e}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{4}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{3}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{2}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){\it \_Z}+c{d}^{4} \right ) }{\frac{c \left ( ac-{b}^{2} \right ){{\it \_R}}^{2}+2\, \left ( 4\,abce-a{c}^{2}d-2\,{b}^{3}e+{b}^{2}cd \right ){\it \_R}+a{c}^{2}{d}^{2}-{b}^{2}c{d}^{2}}{{{\it \_R}}^{3}c+3\,{{\it \_R}}^{2}be-3\,{{\it \_R}}^{2}cd+8\,{\it \_R}\,a{e}^{2}-4\,{\it \_R}\,bde+3\,{\it \_R}\,c{d}^{2}+b{d}^{2}e-c{d}^{3}}\ln \left ( \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{2}-{\it \_R} \right ) }}-{\frac{-ac+{b}^{2}}{{a}^{3}dx}\sqrt{e{x}^{2}+d}}-{\frac{1}{5\,ad{x}^{5}}\sqrt{e{x}^{2}+d}}+{\frac{4\,e}{15\,a{d}^{2}{x}^{3}}\sqrt{e{x}^{2}+d}}-{\frac{8\,{e}^{2}}{15\,a{d}^{3}x}\sqrt{e{x}^{2}+d}}+{\frac{b}{3\,{a}^{2}d{x}^{3}}\sqrt{e{x}^{2}+d}}-{\frac{2\,be}{3\,{a}^{2}{d}^{2}x}\sqrt{e{x}^{2}+d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x)

[Out]

-1/2/a^3*e^(1/2)*sum((c*(a*c-b^2)*_R^2+2*(4*a*b*c*e-a*c^2*d-2*b^3*e+b^2*c*d)*_R+a*c^2*d^2-b^2*c*d^2)/(_R^3*c+3
*_R^2*b*e-3*_R^2*c*d+8*_R*a*e^2-4*_R*b*d*e+3*_R*c*d^2+b*d^2*e-c*d^3)*ln(((e*x^2+d)^(1/2)-e^(1/2)*x)^2-_R),_R=R
ootOf(c*_Z^4+(4*b*e-4*c*d)*_Z^3+(16*a*e^2-8*b*d*e+6*c*d^2)*_Z^2+(4*b*d^2*e-4*c*d^3)*_Z+c*d^4))-(-a*c+b^2)*(e*x
^2+d)^(1/2)/a^3/d/x-1/5*(e*x^2+d)^(1/2)/a/d/x^5+4/15*e*(e*x^2+d)^(1/2)/a/d^2/x^3-8/15*e^2*(e*x^2+d)^(1/2)/a/d^
3/x+1/3*b*(e*x^2+d)^(1/2)/a^2/d/x^3-2/3*b*e*(e*x^2+d)^(1/2)/a^2/d^2/x

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2} + a\right )} \sqrt{e x^{2} + d} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^2 + a)*sqrt(e*x^2 + d)*x^6), 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(1/x^6/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),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(1/x**6/(c*x**4+b*x**2+a)/(e*x**2+d)**(1/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError