3.8.62 \(\int \frac {x^5}{(a+b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=115 \[ \frac {x^2 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 c^{3/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 115, 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, 738, 640, 621, 206} \begin {gather*} \frac {x^2 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && 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}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac {x^2 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {\operatorname {Subst}\left (\int \frac {2 a+b x}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{b^2-4 a c}\\ &=\frac {x^2 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{2 c}\\ &=\frac {x^2 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\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 )}{c}\\ &=\frac {x^2 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 107, normalized size = 0.93 \begin {gather*} \frac {\frac {2 \sqrt {c} \left (a \left (b-2 c x^2\right )+b^2 x^2\right )}{\sqrt {a+b x^2+c x^4}}-\left (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 c^{3/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.46, size = 96, normalized size = 0.83 \begin {gather*} \frac {a b-2 a c x^2+b^2 x^2}{c \left (4 a c-b^2\right ) \sqrt {a+b x^2+c x^4}}-\frac {\log \left (-2 c^{3/2} \sqrt {a+b x^2+c x^4}+b c+2 c^2 x^2\right )}{2 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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fricas [A]  time = 1.62, size = 387, normalized size = 3.37 \begin {gather*} \left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} x^{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 \, \sqrt {c x^{4} + b x^{2} + a} {\left (a b c + {\left (b^{2} c - 2 \, a c^{2}\right )} x^{2}\right )}}{4 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{4} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2}\right )}}, -\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} x^{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 \, \sqrt {c x^{4} + b x^{2} + a} {\left (a b c + {\left (b^{2} c - 2 \, a c^{2}\right )} x^{2}\right )}}{2 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{4} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(((b^2*c - 4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*x^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*sqrt(c*x^4 + b*x^2 + a)*(a*b*c + (b^2*c - 2*a*
c^2)*x^2))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^4 + (b^3*c^2 - 4*a*b*c^3)*x^2), -1/2*(((b^2*c - 4*a*
c^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*x^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sq
rt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*sqrt(c*x^4 + b*x^2 + a)*(a*b*c + (b^2*c - 2*a*c^2)*x^2))/(a*b^2*c^2 - 4*
a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^4 + (b^3*c^2 - 4*a*b*c^3)*x^2)]

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giac [A]  time = 0.26, size = 101, normalized size = 0.88 \begin {gather*} -\frac {\frac {{\left (b^{2} - 2 \, a c\right )} x^{2}}{b^{2} c - 4 \, a c^{2}} + \frac {a b}{b^{2} c - 4 \, a c^{2}}}{\sqrt {c x^{4} + b x^{2} + a}} - \frac {\log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 149, normalized size = 1.30 \begin {gather*} \frac {b^{2} x^{2}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c}+\frac {b^{3}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}-\frac {x^{2}}{2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c}+\frac {\ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}+\frac {b}{4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, 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)^(3/2),x)

[Out]

-1/2*x^2/c/(c*x^4+b*x^2+a)^(1/2)+1/4*b/c^2/(c*x^4+b*x^2+a)^(1/2)+1/2*b^2/c/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)*x
^2+1/4*b^3/c^2/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)+1/2/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 \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)^(3/2),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 [B]  time = 4.76, size = 84, normalized size = 0.73 \begin {gather*} \frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )}{2\,c^{3/2}}+\frac {\frac {a\,b}{2}-x^2\,\left (a\,c-\frac {b^2}{2}\right )}{2\,c\,\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^4+b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^2 + c*x^4)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(x**5/(a + b*x**2 + c*x**4)**(3/2), x)

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