\(\int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx\) [968]

Optimal result

Integrand size = 21, antiderivative size = 124 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\left (15 b^2+16 a c+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}-\frac {b \left (5 b^2+12 a c\right ) \arctan \left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{32 c^{7/2}} \]

[Out]

-1/32*b*(12*a*c+5*b^2)*arctan(1/2*(-2*c*x^2+b)/c^(1/2)/(-c*x^4+b*x^2+a)^(1/2))/c^(7/2)-1/6*x^4*(-c*x^4+b*x^2+a
)^(1/2)/c-1/48*(10*b*c*x^2+16*a*c+15*b^2)*(-c*x^4+b*x^2+a)^(1/2)/c^3

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1128, 756, 793, 635, 210} \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {b \left (12 a c+5 b^2\right ) \arctan \left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{32 c^{7/2}}-\frac {\left (16 a c+15 b^2+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c} \]

[In]

Int[x^7/Sqrt[a + b*x^2 - c*x^4],x]

[Out]

-1/6*(x^4*Sqrt[a + b*x^2 - c*x^4])/c - ((15*b^2 + 16*a*c + 10*b*c*x^2)*Sqrt[a + b*x^2 - c*x^4])/(48*c^3) - (b*
(5*b^2 + 12*a*c)*ArcTan[(b - 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4])])/(32*c^(7/2))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 756

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 793

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p +
3))), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(
a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

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^3}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right ) \\ & = -\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\text {Subst}\left (\int \frac {x \left (-2 a-\frac {5 b x}{2}\right )}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )}{6 c} \\ & = -\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\left (15 b^2+16 a c+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}+\frac {\left (b \left (5 b^2+12 a c\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )}{32 c^3} \\ & = -\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\left (15 b^2+16 a c+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}+\frac {\left (b \left (5 b^2+12 a c\right )\right ) \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 )}{16 c^3} \\ & = -\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\left (15 b^2+16 a c+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}-\frac {b \left (5 b^2+12 a c\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{32 c^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\frac {\sqrt {a+b x^2-c x^4} \left (-15 b^2-16 a c-10 b c x^2-8 c^2 x^4\right )}{48 c^3}+\frac {\left (5 b^3+12 a b c\right ) \arctan \left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2-c x^4}}\right )}{16 c^{7/2}} \]

[In]

Integrate[x^7/Sqrt[a + b*x^2 - c*x^4],x]

[Out]

(Sqrt[a + b*x^2 - c*x^4]*(-15*b^2 - 16*a*c - 10*b*c*x^2 - 8*c^2*x^4))/(48*c^3) + ((5*b^3 + 12*a*b*c)*ArcTan[(S
qrt[c]*x^2)/(-Sqrt[a] + Sqrt[a + b*x^2 - c*x^4])])/(16*c^(7/2))

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {\left (8 c^{2} x^{4}+10 b c \,x^{2}+16 a c +15 b^{2}\right ) \sqrt {-c \,x^{4}+b \,x^{2}+a}}{48 c^{3}}+\frac {b \left (12 a c +5 b^{2}\right ) \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{32 c^{\frac {7}{2}}}\)	\(93\)
pseudoelliptic	\(\frac {-16 c^{\frac {5}{2}} x^{4} \sqrt {-c \,x^{4}+b \,x^{2}+a}-20 b \,c^{\frac {3}{2}} x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}-32 a \,c^{\frac {3}{2}} \sqrt {-c \,x^{4}+b \,x^{2}+a}-30 b^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}-36 \arctan \left (\frac {-2 c \,x^{2}+b}{2 \sqrt {c}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}}\right ) a b c -15 \arctan \left (\frac {-2 c \,x^{2}+b}{2 \sqrt {c}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}}\right ) b^{3}}{96 c^{\frac {7}{2}}}\)	\(166\)
default	\(-\frac {x^{4} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{6 c}-\frac {5 b \,x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{24 c^{2}}-\frac {5 b^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{16 c^{3}}+\frac {5 b^{3} \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{32 c^{\frac {7}{2}}}+\frac {3 b a \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{8 c^{\frac {5}{2}}}-\frac {a \sqrt {-c \,x^{4}+b \,x^{2}+a}}{3 c^{2}}\)	\(168\)
elliptic	\(-\frac {x^{4} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{6 c}-\frac {5 b \,x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{24 c^{2}}-\frac {5 b^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{16 c^{3}}+\frac {5 b^{3} \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{32 c^{\frac {7}{2}}}+\frac {3 b a \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{8 c^{\frac {5}{2}}}-\frac {a \sqrt {-c \,x^{4}+b \,x^{2}+a}}{3 c^{2}}\)	\(168\)

[In]

int(x^7/(-c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/48*(8*c^2*x^4+10*b*c*x^2+16*a*c+15*b^2)/c^3*(-c*x^4+b*x^2+a)^(1/2)+1/32*b*(12*a*c+5*b^2)/c^(7/2)*arctan(c^(
1/2)*(x^2-1/2*b/c)/(-c*x^4+b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.01 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\left [-\frac {3 \, {\left (5 \, b^{3} + 12 \, a b c\right )} \sqrt {-c} \log \left (8 \, c^{2} x^{4} - 8 \, b c x^{2} + b^{2} - 4 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {-c} - 4 \, a c\right ) + 4 \, {\left (8 \, c^{3} x^{4} + 10 \, b c^{2} x^{2} + 15 \, b^{2} c + 16 \, a c^{2}\right )} \sqrt {-c x^{4} + b x^{2} + a}}{192 \, c^{4}}, -\frac {3 \, {\left (5 \, b^{3} + 12 \, a b c\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {c}}{2 \, {\left (c^{2} x^{4} - b c x^{2} - a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 10 \, b c^{2} x^{2} + 15 \, b^{2} c + 16 \, a c^{2}\right )} \sqrt {-c x^{4} + b x^{2} + a}}{96 \, c^{4}}\right ] \]

[In]

integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/192*(3*(5*b^3 + 12*a*b*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)*sqrt(-c) - 4*a*c) + 4*(8*c^3*x^4 + 10*b*c^2*x^2 + 15*b^2*c + 16*a*c^2)*sqrt(-c*x^4 + b*x^2 + a))/c^4, -1/9
6*(3*(5*b^3 + 12*a*b*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*(8*c^3*x^4 + 10*b*c^2*x^2 + 15*b^2*c + 16*a*c^2)*sqrt(-c*x^4 + b*x^2 + a))/c^4]

Sympy [F]

\[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\int \frac {x^{7}}{\sqrt {a + b x^{2} - c x^{4}}}\, dx \]

[In]

integrate(x**7/(-c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral(x**7/sqrt(a + b*x**2 - c*x**4), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.23 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {\sqrt {-c x^{4} + b x^{2} + a} x^{4}}{6 \, c} - \frac {5 \, \sqrt {-c x^{4} + b x^{2} + a} b x^{2}}{24 \, c^{2}} - \frac {5 \, b^{3} \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{32 \, c^{\frac {7}{2}}} - \frac {3 \, a b \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{8 \, c^{\frac {5}{2}}} - \frac {5 \, \sqrt {-c x^{4} + b x^{2} + a} b^{2}}{16 \, c^{3}} - \frac {\sqrt {-c x^{4} + b x^{2} + a} a}{3 \, c^{2}} \]

[In]

integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

-1/6*sqrt(-c*x^4 + b*x^2 + a)*x^4/c - 5/24*sqrt(-c*x^4 + b*x^2 + a)*b*x^2/c^2 - 5/32*b^3*arcsin(-(2*c*x^2 - b)
/sqrt(b^2 + 4*a*c))/c^(7/2) - 3/8*a*b*arcsin(-(2*c*x^2 - b)/sqrt(b^2 + 4*a*c))/c^(5/2) - 5/16*sqrt(-c*x^4 + b*
x^2 + a)*b^2/c^3 - 1/3*sqrt(-c*x^4 + b*x^2 + a)*a/c^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90 \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {1}{48} \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{c} + \frac {5 \, b}{c^{2}}\right )} + \frac {15 \, b^{2} + 16 \, a c}{c^{3}}\right )} - \frac {{\left (5 \, b^{3} + 12 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {-c} x^{2} - \sqrt {-c x^{4} + b x^{2} + a}\right )} \sqrt {-c} + b \right |}\right )}{32 \, \sqrt {-c} c^{3}} \]

[In]

integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/48*sqrt(-c*x^4 + b*x^2 + a)*(2*x^2*(4*x^2/c + 5*b/c^2) + (15*b^2 + 16*a*c)/c^3) - 1/32*(5*b^3 + 12*a*b*c)*l
og(abs(2*(sqrt(-c)*x^2 - sqrt(-c*x^4 + b*x^2 + a))*sqrt(-c) + b))/(sqrt(-c)*c^3)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx=\int \frac {x^7}{\sqrt {-c\,x^4+b\,x^2+a}} \,d x \]

[In]

int(x^7/(a + b*x^2 - c*x^4)^(1/2),x)

[Out]

int(x^7/(a + b*x^2 - c*x^4)^(1/2), x)