3.355 \(\int \frac{x^5 \sqrt{d+e x^2}}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=324 \[ \frac{\left (-\frac{3 a b c e-2 a c^2 d+b^2 c d+b^3 (-e)}{\sqrt{b^2-4 a c}}+a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\left (\frac{3 a b c e-2 a c^2 d+b^2 c d+b^3 (-e)}{\sqrt{b^2-4 a c}}+a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{b \sqrt{d+e x^2}}{c^2}+\frac{\left (d+e x^2\right )^{3/2}}{3 c e} \]

[Out]

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

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Rubi [A]  time = 3.52866, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {1251, 897, 1287, 1166, 208} \[ \frac{\left (-\frac{3 a b c e-2 a c^2 d+b^2 c d+b^3 (-e)}{\sqrt{b^2-4 a c}}+a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\left (\frac{3 a b c e-2 a c^2 d+b^2 c d+b^3 (-e)}{\sqrt{b^2-4 a c}}+a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{b \sqrt{d+e x^2}}{c^2}+\frac{\left (d+e x^2\right )^{3/2}}{3 c e} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*Sqrt[d + e*x^2])/(a + b*x^2 + c*x^4),x]

[Out]

-((b*Sqrt[d + e*x^2])/c^2) + (d + e*x^2)^(3/2)/(3*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])*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^(5/2)*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])*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^(5/2)*Sqrt[2*c*d - (b + Sqrt[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 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1287

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

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^5 \sqrt{d+e x^2}}{a+b x^2+c x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 \sqrt{d+e x}}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (-\frac{d}{e}+\frac{x^2}{e}\right )^2}{\frac{c d^2-b d e+a e^2}{e^2}-\frac{(2 c d-b e) x^2}{e^2}+\frac{c x^4}{e^2}} \, dx,x,\sqrt{d+e x^2}\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{b e}{c^2}+\frac{x^2}{c}+\frac{b \left (c d^2-b d e+a e^2\right )-\left (b c d-b^2 e+a c e\right ) x^2}{c^2 e \left (\frac{c d^2-b d e+a e^2}{e^2}-\frac{(2 c d-b e) x^2}{e^2}+\frac{c x^4}{e^2}\right )}\right ) \, dx,x,\sqrt{d+e x^2}\right )}{e}\\ &=-\frac{b \sqrt{d+e x^2}}{c^2}+\frac{\left (d+e x^2\right )^{3/2}}{3 c e}+\frac{\operatorname{Subst}\left (\int \frac{b \left (c d^2-b d e+a e^2\right )+\left (-b c d+b^2 e-a c e\right ) x^2}{\frac{c d^2-b d e+a e^2}{e^2}-\frac{(2 c d-b e) x^2}{e^2}+\frac{c x^4}{e^2}} \, dx,x,\sqrt{d+e x^2}\right )}{c^2 e^2}\\ &=-\frac{b \sqrt{d+e x^2}}{c^2}+\frac{\left (d+e x^2\right )^{3/2}}{3 c 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}{-\frac{\sqrt{b^2-4 a c}}{2 e}-\frac{2 c d-b e}{2 e^2}+\frac{c x^2}{e^2}} \, dx,x,\sqrt{d+e x^2}\right )}{2 c^2 e^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}{\frac{\sqrt{b^2-4 a c}}{2 e}-\frac{2 c d-b e}{2 e^2}+\frac{c x^2}{e^2}} \, dx,x,\sqrt{d+e x^2}\right )}{2 c^2 e^2}\\ &=-\frac{b \sqrt{d+e x^2}}{c^2}+\frac{\left (d+e x^2\right )^{3/2}}{3 c 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 ) \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^{5/2} \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 ) \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^{5/2} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\\ \end{align*}

Mathematica [A]  time = 7.81822, size = 591, normalized size = 1.82 \[ \frac{c \left (d+e x^2\right )^{7/2} \left (\frac{e^2 \left (\sqrt{2} \sqrt{\frac{c \left (d+e x^2\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}} \left (5 b e \left (3 e \sqrt{b^2-4 a c}+c \left (d+e x^2\right )\right )+c \left (d+e x^2\right ) \left (-5 e \sqrt{b^2-4 a c}+4 c d-6 c e x^2\right )+30 a c e^2-15 b^2 e^2\right )-15 e^2 \left (b \sqrt{b^2-4 a c}+2 a c-b^2\right ) \tanh ^{-1}\left (\sqrt{2} \sqrt{\frac{c \left (d+e x^2\right )}{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )\right )}{\left (e \left (b-\sqrt{b^2-4 a c}\right )-2 c d\right ) \left (e \left (\sqrt{b^2-4 a c}-b\right )+2 c d\right )^2 \left (\frac{c \left (d+e x^2\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}\right )^{7/2}}-\frac{e^2 \left (\sqrt{2} \sqrt{\frac{c \left (d+e x^2\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \left (-5 b e \left (3 e \sqrt{b^2-4 a c}-c \left (d+e x^2\right )\right )+c \left (d+e x^2\right ) \left (5 e \sqrt{b^2-4 a c}+4 c d-6 c e x^2\right )+30 a c e^2-15 b^2 e^2\right )+15 e^2 \left (b \sqrt{b^2-4 a c}-2 a c+b^2\right ) \tanh ^{-1}\left (\sqrt{2} \sqrt{\frac{c \left (d+e x^2\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )\right )}{\left (e \left (\sqrt{b^2-4 a c}+b\right )-2 c d\right )^3 \left (\frac{c \left (d+e x^2\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )^{7/2}}\right )}{30 \sqrt{2} e^4 \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*Sqrt[d + e*x^2])/(a + b*x^2 + c*x^4),x]

[Out]

(c*(d + e*x^2)^(7/2)*((e^2*(Sqrt[2]*Sqrt[(c*(d + e*x^2))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)]*(-15*b^2*e^2 +
30*a*c*e^2 + c*(d + e*x^2)*(4*c*d - 5*Sqrt[b^2 - 4*a*c]*e - 6*c*e*x^2) + 5*b*e*(3*Sqrt[b^2 - 4*a*c]*e + c*(d +
e*x^2))) - 15*(-b^2 + 2*a*c + b*Sqrt[b^2 - 4*a*c])*e^2*ArcTanh[Sqrt[2]*Sqrt[(c*(d + e*x^2))/(2*c*d - b*e + Sq
rt[b^2 - 4*a*c]*e)]]))/((-2*c*d + (b - Sqrt[b^2 - 4*a*c])*e)*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)^2*((c*(d + e
*x^2))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e))^(7/2)) - (e^2*(Sqrt[2]*Sqrt[(c*(d + e*x^2))/(2*c*d - (b + Sqrt[b^
2 - 4*a*c])*e)]*(-15*b^2*e^2 + 30*a*c*e^2 + c*(d + e*x^2)*(4*c*d + 5*Sqrt[b^2 - 4*a*c]*e - 6*c*e*x^2) - 5*b*e*
(3*Sqrt[b^2 - 4*a*c]*e - c*(d + e*x^2))) + 15*(b^2 - 2*a*c + b*Sqrt[b^2 - 4*a*c])*e^2*ArcTanh[Sqrt[2]*Sqrt[(c*
(d + e*x^2))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]]))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)^3*((c*(d + e*x^2))/
(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^(7/2))))/(30*Sqrt[2]*Sqrt[b^2 - 4*a*c]*e^4)

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Maple [C]  time = 0.026, size = 332, normalized size = 1. \begin{align*}{\frac{1}{3\,ce} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{bx}{2\,{c}^{2}}\sqrt{e}}-{\frac{b}{2\,{c}^{2}}\sqrt{e{x}^{2}+d}}-{\frac{bd}{2\,{c}^{2}} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}-{\frac{1}{4\,{c}^{2}}\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 ( ace-{b}^{2}e+bcd \right ){{\it \_R}}^{6}+ \left ( -4\,ab{e}^{2}+acde+3\,{b}^{2}de-3\,bc{d}^{2} \right ){{\it \_R}}^{4}+d \left ( 4\,ab{e}^{2}-acde-3\,{b}^{2}de+3\,bc{d}^{2} \right ){{\it \_R}}^{2}-ac{d}^{3}e+{b}^{2}{d}^{3}e-c{d}^{4}b}{{{\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^5*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a),x)

[Out]

1/3*(e*x^2+d)^(3/2)/c/e+1/2/c^2*b*e^(1/2)*x-1/2*b*(e*x^2+d)^(1/2)/c^2-1/2/c^2*b*d/((e*x^2+d)^(1/2)-e^(1/2)*x)-
1/4/c^2*sum(((a*c*e-b^2*e+b*c*d)*_R^6+(-4*a*b*e^2+a*c*d*e+3*b^2*d*e-3*b*c*d^2)*_R^4+d*(4*a*b*e^2-a*c*d*e-3*b^2
*d*e+3*b*c*d^2)*_R^2-a*c*d^3*e+b^2*d^3*e-c*d^4*b)/(_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^{5}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate(sqrt(e*x^2 + d)*x^5/(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^5*(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^{5} \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**5*(e*x**2+d)**(1/2)/(c*x**4+b*x**2+a),x)

[Out]

Integral(x**5*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^5*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Timed out