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

Optimal. Leaf size=366 \[ \frac{\left (-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\left (\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c^2 \sqrt{e}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c e^{3/2}}+\frac{x \sqrt{d+e x^2}}{2 c e} \]

[Out]

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

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Rubi [A]  time = 1.1727, antiderivative size = 366, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1303, 217, 206, 321, 1692, 377, 205} \[ \frac{\left (-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\left (\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c^2 \sqrt{e}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c e^{3/2}}+\frac{x \sqrt{d+e x^2}}{2 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[d + e*x^2])/(2*c*e) + ((b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b - Sqr
t[b^2 - 4*a*c])*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(c^2*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[2*
c*d - (b - Sqrt[b^2 - 4*a*c])*e]) + ((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])])/(c^2*Sqrt[b + Sqrt[b^2 - 4*a*c]]*S
qrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]) - (d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c*e^(3/2)) - (b*ArcTanh[
(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(c^2*Sqrt[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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 321

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

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{x^6}{\sqrt{d+e x^2} \left (a+b x^2+c x^4\right )} \, dx &=\int \left (-\frac{b}{c^2 \sqrt{d+e x^2}}+\frac{x^2}{c \sqrt{d+e x^2}}+\frac{a b+\left (b^2-a c\right ) x^2}{c^2 \sqrt{d+e x^2} \left (a+b x^2+c x^4\right )}\right ) \, dx\\ &=\frac{\int \frac{a b+\left (b^2-a c\right ) x^2}{\sqrt{d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{c^2}-\frac{b \int \frac{1}{\sqrt{d+e x^2}} \, dx}{c^2}+\frac{\int \frac{x^2}{\sqrt{d+e x^2}} \, dx}{c}\\ &=\frac{x \sqrt{d+e x^2}}{2 c e}+\frac{\int \left (\frac{b^2-a c+\frac{b \left (-b^2+3 a c\right )}{\sqrt{b^2-4 a c}}}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}}+\frac{b^2-a c-\frac{b \left (-b^2+3 a c\right )}{\sqrt{b^2-4 a c}}}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}}\right ) \, dx}{c^2}-\frac{b \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{c^2}-\frac{d \int \frac{1}{\sqrt{d+e x^2}} \, dx}{2 c e}\\ &=\frac{x \sqrt{d+e x^2}}{2 c e}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c^2 \sqrt{e}}+\frac{\left (b^2-a c-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{c^2}+\frac{\left (b^2-a c+\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{c^2}-\frac{d \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{2 c e}\\ &=\frac{x \sqrt{d+e x^2}}{2 c e}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c e^{3/2}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c^2 \sqrt{e}}+\frac{\left (b^2-a c-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\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 )}{c^2}+\frac{\left (b^2-a c+\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}\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 )}{c^2}\\ &=\frac{x \sqrt{d+e x^2}}{2 c e}+\frac{\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 )}{c^2 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}+\frac{\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 )}{c^2 \sqrt{b+\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c e^{3/2}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c^2 \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 1.14884, size = 355, normalized size = 0.97 \[ \frac{\frac{2 \left (-\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}+\frac{2 \left (\frac{b \left (b^2-3 a c\right )}{\sqrt{b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{2 b \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{\sqrt{e}}-\frac{c d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{3/2}}+\frac{c x \sqrt{d+e x^2}}{e}}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.023, size = 269, normalized size = 0.7 \begin{align*}{\frac{x}{2\,ce}\sqrt{e{x}^{2}+d}}-{\frac{d}{2\,c}\ln \left ( \sqrt{e}x+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{3}{2}}}}-{\frac{b}{{c}^{2}}\ln \left ( \sqrt{e}x+\sqrt{e{x}^{2}+d} \right ){\frac{1}{\sqrt{e}}}}+{\frac{1}{2\,{c}^{2}}\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{ \left ( ac-{b}^{2} \right ){{\it \_R}}^{2}+2\, \left ( -2\,abe-acd+{b}^{2}d \right ){\it \_R}+ac{d}^{2}-{b}^{2}{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 ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*(e*x^2+d)^(1/2)/c/e-1/2/c*d/e^(3/2)*ln(e^(1/2)*x+(e*x^2+d)^(1/2))-1/c^2*b*ln(e^(1/2)*x+(e*x^2+d)^(1/2))/
e^(1/2)+1/2/c^2*e^(1/2)*sum(((a*c-b^2)*_R^2+2*(-2*a*b*e-a*c*d+b^2*d)*_R+a*c*d^2-b^2*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=RootOf(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22523, size = 74, normalized size = 0.2 \begin{align*} \frac{\sqrt{x^{2} e + d} x e^{\left (-1\right )}}{2 \, c} + \frac{{\left (c d + 2 \, b e\right )} e^{\left (-\frac{3}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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