3.395 \(\int \frac{x^4}{(d+e x^2)^{3/2} (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=360 \[ -\frac{\left (-\frac{-a b e-2 a c d+b^2 d}{\sqrt{b^2-4 a c}}-a e+b d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac{\left (\frac{-a b e-2 a c d+b^2 d}{\sqrt{b^2-4 a c}}-a e+b d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}+\frac{d x}{\sqrt{d+e x^2} \left (a e^2-b d e+c d^2\right )} \]

[Out]

(d*x)/((c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x^2]) - ((b*d - a*e - (b^2*d - 2*a*c*d - a*b*e)/Sqrt[b^2 - 4*a*c])*A
rcTan[(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 - Sq
rt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)) - ((b*d - a*e + (b^2*d - 2*a
*c*d - a*b*e)/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^2 - b*d*e + a*
e^2))

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Rubi [A]  time = 1.26706, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {1297, 191, 1692, 377, 205} \[ -\frac{\left (-\frac{-a b e-2 a c d+b^2 d}{\sqrt{b^2-4 a c}}-a e+b d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac{\left (\frac{-a b e-2 a c d+b^2 d}{\sqrt{b^2-4 a c}}-a e+b d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}+\frac{d x}{\sqrt{d+e x^2} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x)/((c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x^2]) - ((b*d - a*e - (b^2*d - 2*a*c*d - a*b*e)/Sqrt[b^2 - 4*a*c])*A
rcTan[(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 - Sq
rt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)) - ((b*d - a*e + (b^2*d - 2*a
*c*d - a*b*e)/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^2 - b*d*e + a*
e^2))

Rule 1297

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Dist[(
d^2*f^4)/(c*d^2 - b*d*e + a*e^2), Int[(f*x)^(m - 4)*(d + e*x^2)^q, x], x] - Dist[f^4/(c*d^2 - b*d*e + a*e^2),
Int[((f*x)^(m - 4)*(d + e*x^2)^(q + 1)*Simp[a*d + (b*d - a*e)*x^2, x])/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[q] && LtQ[q, -1] && GtQ[m, 3]

Rule 191

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

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

Mathematica [B]  time = 11.195, size = 7792, normalized size = 21.64 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [C]  time = 0.03, size = 338, normalized size = 0.9 \begin{align*}{\frac{x}{cd}{\frac{1}{\sqrt{e{x}^{2}+d}}}}-8\,{\frac{{e}^{3/2}a}{c \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ) \left ( 2\,e{x}^{2}-2\,\sqrt{e}\sqrt{e{x}^{2}+d}x+2\,d \right ) }}+8\,{\frac{\sqrt{e}bd}{c \left ( 4\,a{e}^{2}-4\,deb+4\,c{d}^{2} \right ) \left ( 2\,e{x}^{2}-2\,\sqrt{e}\sqrt{e{x}^{2}+d}x+2\,d \right ) }}-2\,{\frac{\sqrt{e}}{4\,a{e}^{2}-4\,deb+4\,c{d}^{2}}\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 ( \left ( ae-bd \right ){{\it \_R}}^{2}+2\,d \left ( -3\,ae+bd \right ){\it \_R}+a{d}^{2}e-b{d}^{3} \right ) \ln \left ( \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{2}-{\it \_R} \right ) }{{{\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*x/d/(e*x^2+d)^(1/2)-8/c*e^(3/2)/(4*a*e^2-4*b*d*e+4*c*d^2)/(2*e*x^2-2*e^(1/2)*(e*x^2+d)^(1/2)*x+2*d)*a+8/c*
e^(1/2)/(4*a*e^2-4*b*d*e+4*c*d^2)/(2*e*x^2-2*e^(1/2)*(e*x^2+d)^(1/2)*x+2*d)*b*d-2*e^(1/2)/(4*a*e^2-4*b*d*e+4*c
*d^2)*sum(((a*e-b*d)*_R^2+2*d*(-3*a*e+b*d)*_R+a*d^2*e-b*d^3)/(_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^{4}}{{\left (c x^{4} + b x^{2} + a\right )}{\left (e x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

[Out]

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

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

[Out]

Exception raised: TypeError