3.921 \(\int x^5 \sqrt{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=153 \[ \frac{\left (5 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^3}-\frac{\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{256 c^{7/2}}-\frac{5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{3/2}}{8 c} \]

[Out]

((5*b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(128*c^3) - (5*b*(a + b*x^2 + c*x^4)^(3/2))/(48*c^2) +
(x^2*(a + b*x^2 + c*x^4)^(3/2))/(8*c) - ((b^2 - 4*a*c)*(5*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[
a + b*x^2 + c*x^4])])/(256*c^(7/2))

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Rubi [A]  time = 0.129359, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1114, 742, 640, 612, 621, 206} \[ \frac{\left (5 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^3}-\frac{\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{256 c^{7/2}}-\frac{5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{3/2}}{8 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

((5*b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(128*c^3) - (5*b*(a + b*x^2 + c*x^4)^(3/2))/(48*c^2) +
(x^2*(a + b*x^2 + c*x^4)^(3/2))/(8*c) - ((b^2 - 4*a*c)*(5*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[
a + b*x^2 + c*x^4])])/(256*c^(7/2))

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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])

Rubi steps

\begin{align*} \int x^5 \sqrt{a+b x^2+c x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \sqrt{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{x^2 \left (a+b x^2+c x^4\right )^{3/2}}{8 c}+\frac{\operatorname{Subst}\left (\int \left (-a-\frac{5 b x}{2}\right ) \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{8 c}\\ &=-\frac{5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{3/2}}{8 c}+\frac{\left (5 b^2-4 a c\right ) \operatorname{Subst}\left (\int \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{32 c^2}\\ &=\frac{\left (5 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^3}-\frac{5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac{\left (\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{256 c^3}\\ &=\frac{\left (5 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^3}-\frac{5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac{\left (\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{128 c^3}\\ &=\frac{\left (5 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^3}-\frac{5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac{\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{256 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0698704, size = 136, normalized size = 0.89 \[ \frac{2 \sqrt{c} \sqrt{a+b x^2+c x^4} \left (b \left (8 c^2 x^4-52 a c\right )+24 c^2 x^2 \left (a+2 c x^4\right )-10 b^2 c x^2+15 b^3\right )-3 \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{768 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]*(15*b^3 - 10*b^2*c*x^2 + 24*c^2*x^2*(a + 2*c*x^4) + b*(-52*a*c + 8*c^2*x^4)
) - 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(768*c^(7/
2))

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Maple [A]  time = 0.167, size = 247, normalized size = 1.6 \begin{align*}{\frac{{x}^{2}}{8\,c} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,b}{48\,{c}^{2}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}{x}^{2}}{64\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,{b}^{3}}{128\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{b}^{2}a}{32}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{5\,{b}^{4}}{256}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{7}{2}}}}-{\frac{a{x}^{2}}{16\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{ab}{32\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{{a}^{2}}{16}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(c*x^4+b*x^2+a)^(1/2),x)

[Out]

1/8*x^2*(c*x^4+b*x^2+a)^(3/2)/c-5/48*b*(c*x^4+b*x^2+a)^(3/2)/c^2+5/64*b^2/c^2*(c*x^4+b*x^2+a)^(1/2)*x^2+5/128*
b^3/c^3*(c*x^4+b*x^2+a)^(1/2)+3/32*b^2/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*a-5/256*b^4/c^(
7/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-1/16*a/c*(c*x^4+b*x^2+a)^(1/2)*x^2-1/32*a/c^2*(c*x^4+b*x^
2+a)^(1/2)*b-1/16*a^2/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66946, size = 698, normalized size = 4.56 \begin{align*} \left [\frac{3 \,{\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{c} - 4 \, a c\right ) + 4 \,{\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \,{\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{1536 \, c^{4}}, \frac{3 \,{\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \,{\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \,{\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{768 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1536*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 + 4*sqrt(c*x^4 + b*x^2 +
a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + 4*(48*c^4*x^6 + 8*b*c^3*x^4 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a
*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4, 1/768*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*arctan(1/2*sqrt(c
*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*(48*c^4*x^6 + 8*b*c^3*x^4 + 15*b^3*c -
52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \sqrt{a + b x^{2} + c x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral(x**5*sqrt(a + b*x**2 + c*x**4), x)

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Giac [A]  time = 1.43727, size = 197, normalized size = 1.29 \begin{align*} \frac{1}{384} \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \,{\left (6 \, x^{2} + \frac{b}{c}\right )} x^{2} - \frac{5 \, b^{2} c^{3} - 12 \, a c^{4}}{c^{5}}\right )} x^{2} + \frac{15 \, b^{3} c^{2} - 52 \, a b c^{3}}{c^{5}}\right )} + \frac{{\left (5 \, b^{4} c^{2} - 24 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*x^2 + b/c)*x^2 - (5*b^2*c^3 - 12*a*c^4)/c^5)*x^2 + (15*b^3*c^2 - 52*a*b
*c^3)/c^5) + 1/256*(5*b^4*c^2 - 24*a*b^2*c^3 + 16*a^2*c^4)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*
sqrt(c) - b))/c^(11/2)