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

Optimal. Leaf size=390 \[ -\frac{\left (-\frac{3 a b c e-2 a c^2 d+b^2 c d+b^3 (-e)}{\sqrt{b^2-4 a c}}+a c e+b^2 (-e)+b c d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (\frac{3 a b c e-2 a c^2 d+b^2 c d+b^3 (-e)}{\sqrt{b^2-4 a c}}+a c e+b^2 (-e)+b c d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{(c d-2 b e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{e}}+\frac{x \sqrt{d+e x^2}}{2 c} \]

[Out]

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

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Rubi [A]  time = 2.91949, antiderivative size = 390, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1291, 388, 217, 206, 1692, 377, 205} \[ -\frac{\left (-\frac{3 a b c e-2 a c^2 d+b^2 c d+b^3 (-e)}{\sqrt{b^2-4 a c}}+a c e+b^2 (-e)+b c d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (\frac{3 a b c e-2 a c^2 d+b^2 c d+b^3 (-e)}{\sqrt{b^2-4 a c}}+a c e+b^2 (-e)+b c d\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{(c d-2 b e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{e}}+\frac{x \sqrt{d+e x^2}}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1291

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

Rule 388

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

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

Mathematica [B]  time = 6.41774, size = 10915, normalized size = 27.99 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [C]  time = 0.033, size = 290, normalized size = 0.7 \begin{align*}{\frac{x}{2\,c}\sqrt{e{x}^{2}+d}}+{\frac{d}{2\,c}\ln \left ( \sqrt{e}x+\sqrt{e{x}^{2}+d} \right ){\frac{1}{\sqrt{e}}}}+{\frac{b}{{c}^{2}}\sqrt{e}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) }+{\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 ( ace-{b}^{2}e+bcd \right ){{\it \_R}}^{2}+2\, \left ( -2\,ab{e}^{2}+acde+{b}^{2}de-bc{d}^{2} \right ){\it \_R}+e{d}^{2}ca-{b}^{2}{d}^{2}e+bc{d}^{3}}{{{\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^4*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a),x)

[Out]

1/2*x*(e*x^2+d)^(1/2)/c+1/2/c*d/e^(1/2)*ln(e^(1/2)*x+(e*x^2+d)^(1/2))+1/c^2*e^(1/2)*b*ln((e*x^2+d)^(1/2)-e^(1/
2)*x)+1/2/c^2*e^(1/2)*sum(((a*c*e-b^2*e+b*c*d)*_R^2+2*(-2*a*b*e^2+a*c*d*e+b^2*d*e-b*c*d^2)*_R+e*d^2*c*a-b^2*d^
2*e+b*c*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{\sqrt{e x^{2} + d} x^{4}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(e*x^2 + d)*x^4/(c*x^4 + b*x^2 + a), 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)^(1/2)/(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} \sqrt{d + e x^{2}}}{a + b x^{2} + c x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28065, size = 72, normalized size = 0.18 \begin{align*} -\frac{{\left (c d - 2 \, b e\right )} e^{\left (-\frac{1}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{4 \, c^{2}} + \frac{\sqrt{x^{2} e + d} x}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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