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

Optimal. Leaf size=292 \[ \frac{\left (-\sqrt{b^2-4 a c} (c d-b e)+2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (\sqrt{b^2-4 a c} (c d-b e)+2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d+e x^2}}{c} \]

[Out]

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

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Rubi [A]  time = 3.59965, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {1251, 824, 826, 1166, 208} \[ \frac{\left (-\sqrt{b^2-4 a c} (c d-b e)+2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (\sqrt{b^2-4 a c} (c d-b e)+2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d+e x^2}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 824

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

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{x^3 \sqrt{d+e x^2}}{a+b x^2+c x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x \sqrt{d+e x}}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{\sqrt{d+e x^2}}{c}+\frac{\operatorname{Subst}\left (\int \frac{-a e+(c d-b e) x}{\sqrt{d+e x} \left (a+b x+c x^2\right )} \, dx,x,x^2\right )}{2 c}\\ &=\frac{\sqrt{d+e x^2}}{c}+\frac{\operatorname{Subst}\left (\int \frac{-a e^2-d (c d-b e)+(c d-b e) x^2}{c d^2-b d e+a e^2+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x^2}\right )}{c}\\ &=\frac{\sqrt{d+e x^2}}{c}-\frac{\left (b c d-b^2 e+2 a c e-\sqrt{b^2-4 a c} (c d-b e)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x^2}\right )}{2 c \sqrt{b^2-4 a c}}+\frac{\left (b c d-b^2 e+2 a c e+\sqrt{b^2-4 a c} (c d-b e)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x^2}\right )}{2 c \sqrt{b^2-4 a c}}\\ &=\frac{\sqrt{d+e x^2}}{c}+\frac{\left (b c d-b^2 e+2 a c e-\sqrt{b^2-4 a c} (c d-b e)\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} c^{3/2} \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+2 a c e+\sqrt{b^2-4 a c} (c d-b e)\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} c^{3/2} \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 = 0.571325, size = 308, normalized size = 1.05 \[ \frac{\frac{\left (-c d \sqrt{b^2-4 a c}+b e \sqrt{b^2-4 a c}+2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{2} \sqrt{c} \sqrt{b^2-4 a c} \sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}+\frac{\left (-c d \sqrt{b^2-4 a c}+b e \sqrt{b^2-4 a c}-2 a c e+b^2 e-b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \sqrt{c} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\sqrt{d+e x^2}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.023, size = 275, normalized size = 0.9 \begin{align*} -{\frac{x}{2\,c}\sqrt{e}}+{\frac{1}{2\,c}\sqrt{e{x}^{2}+d}}+{\frac{d}{2\,c} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}+{\frac{1}{4\,c}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{8}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{6}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{4}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){{\it \_Z}}^{2}+c{d}^{4} \right ) }{\frac{ \left ( -be+cd \right ){{\it \_R}}^{6}+ \left ( -4\,a{e}^{2}+3\,deb-3\,c{d}^{2} \right ){{\it \_R}}^{4}+d \left ( 4\,a{e}^{2}-3\,deb+3\,c{d}^{2} \right ){{\it \_R}}^{2}+b{d}^{3}e-c{d}^{4}}{{{\it \_R}}^{7}c+3\,{{\it \_R}}^{5}be-3\,{{\it \_R}}^{5}cd+8\,{{\it \_R}}^{3}a{e}^{2}-4\,{{\it \_R}}^{3}bde+3\,{{\it \_R}}^{3}c{d}^{2}+{\it \_R}\,b{d}^{2}e-{\it \_R}\,c{d}^{3}}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/c*e^(1/2)*x+1/2*(e*x^2+d)^(1/2)/c+1/2/c*d/((e*x^2+d)^(1/2)-e^(1/2)*x)+1/4/c*sum(((-b*e+c*d)*_R^6+(-4*a*e^
2+3*b*d*e-3*c*d^2)*_R^4+d*(4*a*e^2-3*b*d*e+3*c*d^2)*_R^2+b*d^3*e-c*d^4)/(_R^7*c+3*_R^5*b*e-3*_R^5*c*d+8*_R^3*a
*e^2-4*_R^3*b*d*e+3*_R^3*c*d^2+_R*b*d^2*e-_R*c*d^3)*ln((e*x^2+d)^(1/2)-e^(1/2)*x-_R),_R=RootOf(c*_Z^8+(4*b*e-4
*c*d)*_Z^6+(16*a*e^2-8*b*d*e+6*c*d^2)*_Z^4+(4*b*d^2*e-4*c*d^3)*_Z^2+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^{3}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate(sqrt(e*x^2 + d)*x^3/(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^3*(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^{3} \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**3*(e*x**2+d)**(1/2)/(c*x**4+b*x**2+a),x)

[Out]

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

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Giac [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^3*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Timed out