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

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

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.07, size = 136, normalized size = 0.89 \[ \frac {2 \sqrt {c} \sqrt {a+b x^2+c x^4} \left (b \left (8 c^2 x^4-52 a c\right )+24 c^2 x^2 \left (a+2 c x^4\right )+15 b^3-10 b^2 c x^2\right )-3 \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{768 c^{7/2}} \]

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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fricas [A]  time = 0.97, size = 303, normalized size = 1.98 \[ \left [\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{1536 \, c^{4}}, \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{768 \, c^{4}}\right ] \]

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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giac [A]  time = 0.23, size = 134, normalized size = 0.88 \[ \frac {1}{384} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {{\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {7}{2}}} \]

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 - 12*a*c^2)/c^3)*x^2 + (15*b^3 - 52*a*b*c)/c^
3) + 1/256*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/
c^(7/2)

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maple [A]  time = 0.02, size = 247, normalized size = 1.61 \[ -\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,x^{2}}{16 c}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} x^{2}}{64 c^{2}}-\frac {a^{2} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 a \,b^{2} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {5}{2}}}-\frac {5 b^{4} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {7}{2}}}+\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} x^{2}}{8 c}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a b}{32 c^{2}}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3}}{128 c^{3}}-\frac {5 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b}{48 c^{2}} \]

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((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*a-5/256*b^4/c^(
7/2)*ln((c*x^2+1/2*b)/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((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 4.64, size = 193, normalized size = 1.26 \[ \frac {x^2\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{8\,c}-\frac {a\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}+\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{8\,c}-\frac {5\,b\,\left (\frac {\left (8\,c\,\left (c\,x^4+a\right )-3\,b^2+2\,b\,c\,x^2\right )\,\sqrt {c\,x^4+b\,x^2+a}}{24\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^4+b\,x^2+a}+\frac {2\,c\,x^2+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}\right )}{16\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^2 + c*x^4)^(1/2),x)

[Out]

(x^2*(a + b*x^2 + c*x^4)^(3/2))/(8*c) - (a*((b/(4*c) + x^2/2)*(a + b*x^2 + c*x^4)^(1/2) + (log((a + b*x^2 + c*
x^4)^(1/2) + (b/2 + c*x^2)/c^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(8*c) - (5*b*(((8*c*(a + c*x^4) - 3*b^2 + 2*b
*c*x^2)*(a + b*x^2 + c*x^4)^(1/2))/(24*c^2) + (log(2*(a + b*x^2 + c*x^4)^(1/2) + (b + 2*c*x^2)/c^(1/2))*(b^3 -
4*a*b*c))/(16*c^(5/2))))/(16*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{5} \sqrt {a + b x^{2} + c x^{4}}\, dx \]

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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