\(\int \frac {x^5}{(a+b x^2+c x^4)^{3/2}} \, dx\) [983]

Optimal result

Integrand size = 20, antiderivative size = 115 \[ \int \frac {x^5}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^2 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1128, 752, 654, 635, 212} \[ \int \frac {x^5}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 c^{3/2}}+\frac {x^2 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 654

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 752

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 1128

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*} \text {integral}& = \frac {1}{2} \text {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 {\text {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 {\text {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 {\text {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*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.82 \[ \int \frac {x^5}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {b^2 x^2+a \left (b-2 c x^2\right )}{c \left (-b^2+4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 c^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.92

method	result	size
pseudoelliptic	\(-\frac {x^{2}}{2 c \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {b \left (b \,x^{2}+2 a \right )}{8 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (a c -\frac {b^{2}}{4}\right ) c}+\frac {-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )}{2 c^{\frac {3}{2}}}\)	\(106\)
default	\(-\frac {x^{2}}{2 c \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (2 c \,x^{2}+b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}\right )}{4 c}+\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}\)	\(124\)
elliptic	\(-\frac {x^{2}}{2 c \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (2 c \,x^{2}+b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}\right )}{4 c}+\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}\)	\(124\)

[In]

int(x^5/(c*x^4+b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*x^2/c/(c*x^4+b*x^2+a)^(1/2)+1/8*b*(b*x^2+2*a)/(c*x^4+b*x^2+a)^(1/2)/(a*c-1/4*b^2)/c+1/2/c^(3/2)*(-ln(2)+l
n((2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*c^(1/2)+b)/c^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 387, normalized size of antiderivative = 3.37 \[ \int \frac {x^5}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} x^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (a b c + {\left (b^{2} c - 2 \, a c^{2}\right )} x^{2}\right )}}{4 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{4} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2}\right )}}, -\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (a b c + {\left (b^{2} c - 2 \, a c^{2}\right )} x^{2}\right )}}{2 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{4} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2}\right )}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {x^5}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[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 deta

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int \frac {x^5}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\frac {{\left (b^{2} - 2 \, a c\right )} x^{2}}{b^{2} c - 4 \, a c^{2}} + \frac {a b}{b^{2} c - 4 \, a c^{2}}}{\sqrt {c x^{4} + b x^{2} + a}} - \frac {\log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {3}{2}}} \]

[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)

Mupad [B] (verification not implemented)

Time = 14.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.73 \[ \int \frac {x^5}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )}{2\,c^{3/2}}+\frac {\frac {a\,b}{2}-x^2\,\left (a\,c-\frac {b^2}{2}\right )}{2\,c\,\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^4+b\,x^2+a}} \]

[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))