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

Optimal. Leaf size=104 \[ \frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{5/2}}-\frac {3 b \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {a+b x^2+c x^4}}{4 c} \]

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Rubi [A]  time = 0.10, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1114, 742, 640, 621, 206} \begin {gather*} \frac {\left (3 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 )}{16 c^{5/2}}-\frac {3 b \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {a+b x^2+c x^4}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(-3*b*Sqrt[a + b*x^2 + c*x^4])/(8*c^2) + (x^2*Sqrt[a + b*x^2 + c*x^4])/(4*c) + ((3*b^2 - 4*a*c)*ArcTanh[(b + 2
*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*c^(5/2))

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 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 \frac {x^5}{\sqrt {a+b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac {x^2 \sqrt {a+b x^2+c x^4}}{4 c}+\frac {\operatorname {Subst}\left (\int \frac {-a-\frac {3 b x}{2}}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {3 b \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {a+b x^2+c x^4}}{4 c}+\frac {\left (3 b^2-4 a c\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 c^2}\\ &=-\frac {3 b \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {a+b x^2+c x^4}}{4 c}+\frac {\left (3 b^2-4 a c\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 )}{8 c^2}\\ &=-\frac {3 b \sqrt {a+b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {a+b x^2+c x^4}}{4 c}+\frac {\left (3 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 )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 88, normalized size = 0.85 \begin {gather*} \frac {\left (3 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 )+2 \sqrt {c} \left (2 c x^2-3 b\right ) \sqrt {a+b x^2+c x^4}}{16 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*(-3*b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4] + (3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a
+ b*x^2 + c*x^4])])/(16*c^(5/2))

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IntegrateAlgebraic [A]  time = 0.25, size = 91, normalized size = 0.88 \begin {gather*} \frac {\left (4 a c-3 b^2\right ) \log \left (-2 c^{5/2} \sqrt {a+b x^2+c x^4}+b c^2+2 c^3 x^2\right )}{16 c^{5/2}}+\frac {\left (2 c x^2-3 b\right ) \sqrt {a+b x^2+c x^4}}{8 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c^2) + ((-3*b^2 + 4*a*c)*Log[b*c^2 + 2*c^3*x^2 - 2*c^(5/2)*Sqrt[
a + b*x^2 + c*x^4]])/(16*c^(5/2))

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fricas [A]  time = 1.12, size = 203, normalized size = 1.95 \begin {gather*} \left [-\frac {{\left (3 \, b^{2} - 4 \, a c\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 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} - 3 \, b c\right )}}{32 \, c^{3}}, -\frac {{\left (3 \, b^{2} - 4 \, a c\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 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} - 3 \, b c\right )}}{16 \, c^{3}}\right ] \end {gather*}

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]

[-1/32*((3*b^2 - 4*a*c)*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)*sqr
t(c) - 4*a*c) - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c^2*x^2 - 3*b*c))/c^3, -1/16*((3*b^2 - 4*a*c)*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*sqrt(c*x^4 + b*x^2 + a)*(2*c^2*
x^2 - 3*b*c))/c^3]

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giac [A]  time = 0.24, size = 82, normalized size = 0.79 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{4} + b x^{2} + a} {\left (\frac {2 \, x^{2}}{c} - \frac {3 \, b}{c^{2}}\right )} - \frac {{\left (3 \, b^{2} - 4 \, a c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {5}{2}}} \end {gather*}

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]

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

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maple [A]  time = 0.01, size = 116, normalized size = 1.12 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{4 c}-\frac {a \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}+\frac {3 b^{2} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{8 c^{2}} \end {gather*}

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]

1/4*x^2*(c*x^4+b*x^2+a)^(1/2)/c-3/8*b*(c*x^4+b*x^2+a)^(1/2)/c^2+3/16*b^2/c^(5/2)*ln((c*x^2+1/2*b)/c^(1/2)+(c*x
^4+b*x^2+a)^(1/2))-1/4*a/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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \end {gather*}

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]

int(x^5/(a + b*x^2 + c*x^4)^(1/2), x)

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

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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